Answers
2: A
3: C
4: D
These are the answers
Related Questions
Tsunamis are fast-moving waves often generated by underwater earthquakes. In the deep ocean their amplitude is barely noticable, but upon reaching shore, they can rise up to the astonishing height of a six-story building. One tsunami, generated off the Aleutian islands in Alaska, had a wavelength of 612 km and traveled a distance of 3920 km in 4.09 h. (a) What was the speed (in m/s) of the wave
Answers
Answer:
V = 266.23 m/s
Explanation:
The speed of the wave can easily be given by the following formula:
V = S/t
where,
V = Speed of the Wave = ?
S = Distance Covered by Wave = 3920 km
S = Distance Covered by Wave = (3920 km)(1000 m/1 km)
S = Distance Covered by Wave = 3.92 x 10⁶ m
t = Time taken by the wave to cover the distance = 4.09 h
t = Time taken by the wave to cover the distance = (4.09 h)(3600 s/1 h)
t = Time taken by the wave to cover the distance = 14724 s
Therefore,
V = (3.92 x 10⁶ m)/(14724 s)
V = 266.23 m/s
Two wires carry current I1 = 47 A and I2 = 29 A in the opposite directions parallel to the x-axis at y1 = 9 cm and y2 = 15 cm. Where on the y-axis (in cm) is the magnetic field zero?
Answers
Answer:
Approximately [tex]25\; \rm cm[/tex], assuming that the permeability between the two wires is constant, and that the two wires are of infinite lengths.
Explanation:
Note that [tex]y_1 < y_2[/tex], which means that the first wire (with current [tex]I_1[/tex]) is underneath the second wire (with current [tex]I_2[/tex].) Let [tex]y[/tex] denote the [tex]y[/tex]-coordinate (in [tex]\rm cm[/tex]) of a point of interest. The two wires partition this region into three parts:
The region under the first wire, where [tex]y < y_1[/tex];The region between the first and the second wire, where [tex]y_1 < y < y_2[/tex]; andThe region above the second wire, where [tex]y > y_2[/tex].Apply the right-hand rule to find the direction of the magnetic field in each of the three regions. Assume that [tex]I_1[/tex] points to the left while [tex]I_2[/tex] points to the right. Let [tex]B_1[/tex] and [tex]B_2[/tex] denote the magnetic field due to [tex]I_1[/tex] and [tex]I_2[/tex], respectively.
In the region below the first wire, [tex]B_1[/tex] points out of the [tex]x y[/tex]-plane while [tex]B_2[/tex] points into the [tex]x y[/tex]-plane.In the region between the two wires, both [tex]B_1[/tex] and [tex]B_2[/tex] point into the [tex]x y[/tex]-plane.In the region above the second wire, [tex]B_1[/tex] points into the [tex]x y[/tex]-plane while [tex]B_2[/tex] points out of the [tex]x y[/tex]-plane.The (net) magnetic field on this plane would be zero only in regions where [tex]B_1[/tex] and[tex]B_2[/tex] points in opposite directions. That rules out the region between the two wires.
At a distance of [tex]R[/tex] away from a wire with current [tex]I[/tex] and infinite length, the formula for the magnitude of the magnetic field [tex]B[/tex] due to that wire is:
[tex]\displaystyle B = \frac{\mu\, I}{2\pi\, R}[/tex],
where [tex]\mu[/tex] is the permeability of the space between the wire and the point of interest (should be constant.) The exact value of [tex]\mu[/tex] does not affect the answer to this question, as long as it is constant throughout this region.
Note that the value of [tex]R[/tex] in this formula is supposed to be positive. Let [tex]R_1[/tex] and [tex]R_2[/tex] denote the distance between the point of interest and the two wires, respectively.
In the region under the first wire, [tex]R_1 = y_1 - y = 9 - y[/tex], while [tex]R_2 = y_2 - y = 15 - y[/tex]. In the region above the second wire, [tex]R_1 = y - 9[/tex], while [tex]R_2 = y - 15[/tex].Make sure that given the corresponding range of [tex]y[/tex], these distances are all positive.
Strength of the magnetic field due to the first wire: [tex]\displaystyle B_1 = \frac{\mu\, I_1}{4\pi\, R_1}[/tex].Strength of the magnetic field due to the second wire: [tex]\displaystyle B_1 = \frac{\mu\, I_2}{4\pi\, R_2}[/tex].For the net magnetic field to be zero at a certain [tex]y[/tex]-value, the strength of the two magnetic fields at that point should match. That is:
[tex]B_1 = B_2[/tex].
[tex]\displaystyle \frac{\mu\, I_1}{4\pi\, R_1} = \frac{\mu\, I_2}{4\pi\, R_2}[/tex].
Simplify this equation:
[tex]\displaystyle \frac{I_1}{R_1} = \frac{I_2}{R_2}[/tex].
In the region under the first wire (where [tex]y < 9[/tex],) this equation becomes [tex]\displaystyle \frac{47}{9 - y} = \frac{29}{15 - y}[/tex].In the region above the second wire (where [tex]y > 15[/tex],) this equation becomes [tex]\displaystyle \frac{47}{y - 9} = \frac{29}{y - 15}[/tex].These two equations give the same result: [tex]y = 25[/tex]. However, based on the respective assumptions on the value of [tex]y[/tex], this value corresponds to the region above the second wire.
The distance covered by a car at a time, t is given by x = 20t + 6t4, calculate
(i) the instantaneous velocity when t = 1
(ii) the instantaneous acceleration when t =1
Answers
Answer:
(i) v = 44 m/s
(ii) a = 72 m/s^2
Explanation:
You have the following equation for the potion of a car:
[tex]x=20t+6t^4[/tex]
(i) The instantaneous velocity is the derivative of x in time:
[tex]\frac{dx}{dt}=20+(6)(4)t^3=20+24t^3[/tex]
for t = 1 is:
[tex]v(t=1)=\frac{dx}{dt}=20+24(1)^3=44m/s[/tex]
(ii) The instantaneous acceleration is the derivative of the velocity:
[tex]\frac{dv}{dt}=24(3)t^2=72t^2[/tex]
for t = 1
[tex]a(t=1)=\frac{dv}{dt}=72(1)^2=72m/s^2[/tex]
At a track meet, a runner runs the 100–m dash in 16 s. What was the runner's average speed?
Answers
Answer:
6.25 m/s
Explanation:
Remember that speed(m/s) = distance(m) / time(s)
Which of the following statements is true for an elastic collision? A. Both momentum and kinetic energy are conserved. B. Momentum is conserved, but kinetic energy is not conserved. C. Kinetic energy is conserved, but momentum is not conserved. D. The amount of momentum lost by one object is the same as the amount gained by the other object. E. The amount of kinetic energy lost by one object is the same as the amount gained by the other object.
Answers
Answer:
A, D and E
Explanation:
The idea behind elastic is take an elastic band for instance if you stretch it, it will assume it's original shape and form when it returns to its original position hence nothing is lost.
Similarly for elastic collision momentum and kinetic energy are conserved hence; the momentum lost by one is gained by the other and the kinetic energy lost by one is gained by the other.
We have that for the Question ,it can be said that the the appropriate option that identifies elastic collision are
Both momentum and kinetic energy are conservedThe amount of momentum lost by one object is the same as the amount gained by the other objectThe amount of kinetic energy lost by one object is the same as the amount gained by the other object.Therefore Option A,D,C
From the question we are told
Which of the following statements is true for an elastic collision?
A. Both momentum and kinetic energy are conserved.
B. Momentum is conserved, but kinetic energy is not conserved.
C. Kinetic energy is conserved, but momentum is not conserved.
D. The amount of momentum lost by one object is the same as the amount gained by the other object.
E. The amount of kinetic energy lost by one object is the same as the amount gained by the other object.
Generally
Elastic Collision
Elastic collision sees an encounter between two bodies have a net kinetic energy between two bodies remains the same.
Therefore
The the appropriate option that identifies elastic collision are
Both momentum and kinetic energy are conservedThe amount of momentum lost by one object is the same as the amount gained by the other objectThe amount of kinetic energy lost by one object is the same as the amount gained by the other object.Therefore Option A,D,C
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The latent heat of fusion of water at 0 °C is 6.025 kJ mol'' and the molar heat
capacities (C ) of water and ice are 75.3 and 37.7JK'mol', respectively. Calculate
AH for the freezing of 1 mol of supercooled water at -10.0°C.
Answers
Answer:
[tex]\Delta H_{tot} = 2258.025\,kJ[/tex]
Explanation:
The amount of heat released from water is equal to the sum of latent and sensible heats. Let suppose that water is initially at a temperature of [tex]25^{\circ}C[/tex]. Then:
[tex]\Delta H_{tot} = \Delta H_{s, w} + \Delta H_{f,w} + \Delta H_{s,i}[/tex]
[tex]\Delta H_{tot} = n\cdot (c_{w}\cdot \Delta T_{w} + L_{f} + c_{i}\cdot \Delta T_{i})[/tex]
Finally, the amount of heat released from water is now computed by replacing variables:
[tex]\Delta H_{tot} = (1\,mol)\cdot \left[\left(75.3\,\frac{kJ}{mol\cdot K} \right)\cdot (25^{\circ}C-0^{\circ}C)+ 6.025\,\frac{kJ}{mol} + \left(37.7\,\frac{kJ}{mol\cdot K} \right)\cdot (0 + 10^{\circ}C)\right][/tex][tex]\Delta H_{tot} = 2258.025\,kJ[/tex]
Two parallel wires are separated by 5.60 cm, each carrying 2.65 A of current in the same direction. (a) What is the magnitude of the force per unit length between the wires? N/m (b) Is the force attractive or repulsive? attractive repulsive
Answers
Explanation:
It is given that,
The separation between two parallel wires, r = 5.6 cm = 0.056 m
Current in both the wires is 2.65 A
(a) We need to find the magnitude of the force per unit length between the wires. It can be given by :
[tex]\dfrac{F}{l}=\dfrac{\mu_o I_1I_2}{2\pi r}\\\\\dfrac{F}{l}=\dfrac{4\pi \times 10^{-7}\times 2.65\times 2.65}{2\pi \times 0.056}\\\\\dfrac{F}{l}=2.5\times 10^{-5}\ N/m[/tex]
(b) As the current is in same direction, the wires will attract each other.
A cylindrical capacitor is made of two concentric cylinders. The inner cylinder has radius r1 = 4 mm, and the outer one a radius r2= 8 mm. The common length of the cylinders is L = 150 m. What is the potential energy stored in this capacitor when a potential difference V = 4 V is applied between the inner and outer cylinder?
Answers
Answer:
E = 9.62*10^-8 J
Explanation:
The energy stored in a capacitor is given by the following formula:
[tex]E=\frac{1}{2}CV^2[/tex] (1)
E: energy stored
C: capacitance
V: potential difference of the capacitor = 4 V
The capacitance for a concentric cylindrical capacitor is:
[tex]C=\frac{2\pi \epsilon_o L}{ln(\frac{r_2}{r_1})}[/tex] (2)
L: length of the capacitor = 150m
r2: radius of the outer cylinder = 8mm = 8*10^-3m
r1: radius of the inner cylinder = 4mm = 4*10^-3m
εo: dielectric permittivity of vacuum = 8.85*10^-12C^2/Nm^2
You replace the expression (2) into the equation (1) and replace the values of all parameters:
[tex]E=\frac{1}{2}(\frac{2\pi \epsilon_o L}{ln(\frac{r_2}{r_1})})V^2\\\\E=\frac{\pi \epsilon_o L}{ln(\frac{r_2}{r_1})}V^2\\\\E=\frac{\pi (8.85*10^{-12}C^2/Nm^2)(150m)}{ln(\frac{8*10^{-3}m}{4*10^{-3}m})}(4V)^2\\\\E=9.62*10^{-8}J[/tex]
The energy stored in the cylindrical capacitor is 9.62*10-8 J
In a football game a kicker attempts a field goal. The ball remains in contact with the kicker's foot for 0.0580 s, during which time it experiences an acceleration of 376 m/s2. The ball is launched at an angle of 59.9° above the ground. Determine the (a) horizontal and (b) vertical components of the launch velocity.
Answers
Answer:
V₀ₓ = 10.94 m/s
V₀y = 18.87 m/s
Explanation:
To find the launch velocity, we use 1st equation of motion.
Vf = Vi + at
where,
Vf = Final Velocity of Ball = Launch Speed = V₀ = ?
Vi = Initial Velocity = 0 m/s (Since ball was initially at rest)
a = acceleration = 376 m/s²
t = time = 0.058 s
Therefore,
V₀ = 0 m/s + (376 m/s²)(0.058 s)
V₀ = 21.81 m/s
Now, for x-component:
V₀ₓ = V₀ Cos θ
where,
V₀ₓ = x-component of launch velocity = ?
θ = Angle with horizontal = 59.9⁰
V₀ₓ = (21.81 m/s)(Cos 59.9°)
V₀ₓ = 10.94 m/s
for y-component:
V₀ₓ = V₀ Sin θ
where,
V₀y = y-component of launch velocity = ?
θ = Angle with horizontal = 59.9⁰
V₀y = (21.81 m/s)(Sin 59.9°)
V₀y = 18.87 m/s
If the coefficient of kinetic friction between tires and dry pavement is 0.84, what is the shortest distance in which you can stop an automobile by locking the brakes when traveling at 29.0 m/s
Answers
Answer:
The shortest distance in which you can stop the automobile by locking the brakes is 53.64 m
Explanation:
Given;
coefficient of kinetic friction, μ = 0.84
speed of the automobile, u = 29.0 m/s
To determine the the shortest distance in which you can stop an automobile by locking the brakes, we apply the following equation;
v² = u² + 2ax
where;
v is the final velocity
u is the initial velocity
a is the acceleration
x is the shortest distance
First we determine a;
From Newton's second law of motion
∑F = ma
F is the kinetic friction that opposes the motion of the car
-Fk = ma
but, -Fk = -μN
-μN = ma
-μmg = ma
-μg = a
- 0.8 x 9.8 = a
-7.84 m/s² = a
Now, substitute in the value of a in the equation above
v² = u² + 2ax
when the automobile stops, the final velocity, v = 0
0 = 29² + 2(-7.84)x
0 = 841 - 15.68x
15.68x = 841
x = 841 / 15.68
x = 53.64 m
Thus, the shortest distance in which you can stop the automobile by locking the brakes is 53.64 m
A basketball player is 4.22 m from
the goal when she shoots the ball
at 9.20 m/s at a 69.0° angle.
What is the height of the ball
(above the release point)
when it reaches the goal?
(Unit = m)
Enter
Answers
Answer: The height above the release point is 2.96 meters.
Explanation:
The acceleration of the ball is the gravitational acceleration in the y axis.
A = (0, -9.8m/s^)
For the velocity we can integrate over time and get:
V(t) = (9.20m/s*cos(69°), -9.8m/s^2*t + 9.20m/s^2*sin(69°))
for the position we can integrate it again over time, but this time we do not have any integration constant because the initial position of the ball will be (0,0)
P(t) = (9.20*cos(69°)*t, -4.9m/s^2*t^2 + 9.20m/s^2*sin(69°)*t)
now, the time at wich the horizontal displacement is 4.22 m will be:
4.22m = 9.20*cos(69°)*t
t = (4.22/ 9.20*cos(69°)) = 1.28s
Now we evaluate the y-position in this time:
h = -4.9m/s^2*(1.28s)^2 + 9.20m/s^2*sin(69°)*1.28s = 2.96m
The height above the release point is 2.96 meters.
Answer:
3.10 m
Explanation:
Given;
Initial speed of ball u = 9.20 m/s
Angle θ = 69°
Horizontal distance from goal d = 4.22m
Resolving the initial velocity into horizontal and vertical components;
The horizontal component of the initial velocity;
uh = ucosθ
Substituting the given values;
uh = 9.2cos69°
uh = 3.30 m/s
The time taken for it to cover the horizontal distance of 4.22 m (to reach the goal);
Time = distance/speed = d/uh
time = 4.22/3.30 =. 1.279s
The time taken to reach the goal is 1.279 seconds.
To determine the height of the ball, we will resolve the vertical component of the initial velocity;
Vertical component of the ball velocity is;
Uv = usinθ
Uv = 9.20sin69°
Uv = 8.59 m/s
Applying the equation of motion;
Height h = ut - 0.5gt^2
Velocity u = Uv = 8.59 m/s
Time t = 1.279s
Acceleration due to gravity g = 9.8 m/s^2
Substituting the values;
h = 8.69(1.279) - 0.5(9.8×1.279^2)
Height h = 3.0988891 = 3.10 m
The height of ball above the release point is 3.10m
an object with momentum must also have
a. acceleration
b. kinetic energy
c. impulse
d. force
Answers
An object with momentum must also have impulse.
An object with momentum must also have: c. impulse.
Momentum can be defined as the product of the mass possessed by an object and its velocity. Also, it is a vector quantity and as such has both magnitude and direction.
Mathematically, momentum is giving by the formula;
[tex]Momentum = Mass \times Velocity[/tex]
In Physics, the impulse experienced by an object is always equal to the change in momentum of the object. This is due to the force acting on an object.
[tex]Impulse = Change\;in\;momentum\\\\Force \times time = m \Delta V[/tex]
In conclusion, an object with momentum must also have impulse.
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A girl walks South at 2.7 m/s. What is the y component of her velocity?
Answers
Answer: horizontal components
Explanation:use the Cardinals point
What is the meaning of powee
Answers
Hope it helps
Good luck on your assignment
a. When the electric field between the plates is 75% of the dielectric strength and energy density of the stored energy is 2800 J/m3, what is the value of the dielectric constant k?
Answers
Answer: The value of the dielectric constant k = 1.8
Explanation:
If C= ε A/d and
Electrostatic energy W = 1/2CV^2
Substitutes C in the first formula into the energy formula.
W = 1/2 ε A/d × V^2
Let us remember that electric field strength E is the ratio of potential V to the distance d. Where V = Ed
Substitute V = Ed into the energy W.
W = 1/2 × ε A/d ×( Ed )^2
W = 1/2 × ε A/d × E^2 × d^2
d will cancel one of the ds
W = 1/2 × ε Ad × E^2
W/Ad = 1/2 × ε × E^2
W/V = 1/2 × ε E^2
Where Ad = volume V
E = dielectric strength
εo = permittivity of free space = 8.84 x 10^-12 F/m
W/V = 2800 J/m^3
Let first calculate the dielectric strength
2800 = 1/2 × 8.84×10^-12 × E^2
5600 = 8.84×10^-12E^2
E^2 = 5600/8.84×10^-12
E = sqrt( 6.3 × 10^14)
E = 25 × 10^7
75% of E = 18.9 × 10^6Jm
The permittivity of the material will be achieved by using the same formula
2800 = 1/2 × ε E^2
2800 = 0.5 × ε × (18.9×10^6)^2
2800 = ε × 1.78 × 10^14
ε = 2800/1.78×10^14
ε = 1.57 × 10^-11
Dielectric constant k = relative permittivity
Relative permittivity is the ratio of the permittivity of the material to the permittivity of the vacuum in a free space. That is
k = 1.57×10^-11/8.84×10^-12
k = 1.776
k = 1.8 approximately
Therefore, the value of the dielectric constant k is 1.8
A catapult launches a test rocket vertically upward from a well, giving the rocket an initial speed of 80.6 m/s at ground level. The engines then fire, and the rocket accelerates upward at 3.90 m/s2 until it reaches an altitude of 1150 m. At that point its engines fail, and the rocket goes into free fall, with an acceleration of ?9.80 m/s2. (You will need to consider the motion while the engine is operating and the free-fall motion separately.)
(a) For what time interval is the rocket in motion above the ground?
(b) What is its maximum altitude?
(c) What is its velocity just before it hits the ground?
Answers
Before the engines fail, the rocket's altitude at time t is given by
[tex]y_1(t)=\left(80.6\dfrac{\rm m}{\rm s}\right)t+\dfrac12\left(3.90\dfrac{\rm m}{\mathrm s^2}\right)t^2[/tex]
and its velocity is
[tex]v_1(t)=80.6\dfrac{\rm m}{\rm s}+\left(3.90\dfrac{\rm m}{\mathrm s^2}\right)t[/tex]
The rocket then reaches an altitude of 1150 m at time t such that
[tex]1150\,\mathrm m=\left(80.6\dfrac{\rm m}{\rm s}\right)t+\dfrac12\left(3.90\dfrac{\rm m}{\mathrm s^2}\right)t^2[/tex]
Solve for t to find this time to be
[tex]t=11.2\,\mathrm s[/tex]
At this time, the rocket attains a velocity of
[tex]v_1(11.2\,\mathrm s)=124\dfrac{\rm m}{\rm s}[/tex]
When it's in freefall, the rocket's altitude is given by
[tex]y_2(t)=1150\,\mathrm m+\left(124\dfrac{\rm m}{\rm s}\right)t-\dfrac g2t^2[/tex]
where [tex]g=9.80\frac{\rm m}{\mathrm s^2}[/tex] is the acceleration due to gravity, and its velocity is
[tex]v_2(t)=124\dfrac{\rm m}{\rm s}-gt[/tex]
(a) After the first 11.2 s of flight, the rocket is in the air for as long as it takes for [tex]y_2(t)[/tex] to reach 0:
[tex]1150\,\mathrm m+\left(124\dfrac{\rm m}{\rm s}\right)t-\dfrac g2t^2=0\implies t=32.6\,\mathrm s[/tex]
So the rocket is in motion for a total of 11.2 s + 32.6 s = 43.4 s.
(b) Recall that
[tex]{v_f}^2-{v_i}^2=2a\Delta y[/tex]
where [tex]v_f[/tex] and [tex]v_i[/tex] denote final and initial velocities, respecitively, [tex]a[/tex] denotes acceleration, and [tex]\Delta y[/tex] the difference in altitudes over some time interval. At its maximum height, the rocket has zero velocity. After the engines fail, the rocket will keep moving upward for a little while before it starts to fall to the ground, which means [tex]y_2[/tex] will contain the information we need to find the maximum height.
[tex]-\left(124\dfrac{\rm m}{\rm s}\right)^2=-2g(y_{\rm max}-1150\,\mathrm m)[/tex]
Solve for [tex]y_{\rm max}[/tex] and we find that the rocket reaches a maximum altitude of about 1930 m.
(c) In part (a), we found the time it takes for the rocket to hit the ground (relative to [tex]y_2(t)[/tex]) to be about 32.6 s. Plug this into [tex]v_2(t)[/tex] to find the velocity before it crashes:
[tex]v_2(32.6\,\mathrm s)=-196\frac{\rm m}{\rm s}[/tex]
That is, the rocket has a velocity of 196 m/s in the downward direction as it hits the ground.
Calculate the average speed of a car between 0 And 10 second give the unit
Answers
Answer:
I think the distance should be given in the question because the formula for getting the
average speed= total distance/total time taken
A cave rescue team lifts an injured spelunker directly upward and out of a sinkhole by means of a motor-driven cable. The lift is performed in three stages, each requiring a vertical distance of 5.90 m: (a) the initially stationary spelunker is accelerated to a speed of 2.30 m/s; (b) he is then lifted at the constant speed of 2.30 m/s; (c) finally he is decelerated to zero speed. How much work is done on the 69.0 kg rescue by the force lifting him during each stage
Answers
Answer:
W₃ = 3310.49 J , W3 = 3310.49 J
Explanation:
We can solve this exercise in parts, the first with acceleration, the second with constant speed and the third with deceleration. Therefore it is work we calculate it in these three sections
We start with the part with acceleration, the distance traveled is y = 5.90 m and the final speed is v = 2.30 m / s. Let's calculate the acceleration with kinematics
v2 = v₀² + 2 a₁ y
as they rest part of the rest the ricial speed is zero
v² = 2 a₁ y
a₁ = v² / 2y
a₁ = 2.3² / (2 5.90)
a₁ = 0.448 m / s²
with this acceleration we can calculate the applied force, using Newton's second law
F -W = m a₁
F = m a₁ + m g
F = m (a₁ + g)
F = 69 (0.448 + 9.8)
F = 707.1 N
Work is defined by
W₁ = F.y = F and cos tea
As the force lifts the man, this and the displacement are parallel, therefore the angle is zero
W₁ = 707.1 5.9
W₁ = 4171.89 J W3 = 3310.49 J
Let's calculate for the second part
the speed is constant, therefore they relate it to zero
F - W = 0
F = W
F = m g
F = 60 9.8
F = 588 A
the job is
W² = 588 5.9
W2 = 3469.2 J
finally the third part
in this case the initial speed is 2.3 m / s and the final speed is zero
v² = v₀² + 2 a₂ y
0 = vo2₀² + 2 a₂ y
a₂ = -v₀² / 2 y
a₂ = - 2.3²/2 5.9
a2 = - 0.448 m / s²
we calculate the force
F - W = m a₂
F = m (g + a₂)
F = 60 (9.8 - 0.448)
F = 561.1 N
we calculate the work
W3 = F and
W3 = 561.1 5.9
W3 = 3310.49 J
total work
W_total = W1 + W2 + W3
W_total = 4171.89 +3469.2 + 3310.49
w_total = 10951.58 J
what is the study of physilogy
Answers
Answer:
physiology is the study of mechanisms and functions in a living system
Explanation:
it's basically a part of biology that works with how a living organism or body part functions
How do you solve this?
Answers
Answer:
The circuit is in series connection,
the same current will flow through i.e
I1 = I2 = I3 = 2A.
Explanation:
We'll begin by calculating the total resistance in the circuit. This is shown below:
R1 = 4Ω
R2 = 3Ω
RT =..?
Total resistance in series can be obtained as follow:
RT = R1 + R2
RT = 4 + 3
RT = 7Ω
Next we shall determine the total current flowing in the circuit:
Voltage (V) = 14V
Resistance (R) = 7Ω
Current (I) =..?
V = IR
14 = I x 7
Divide both side by 7
I = 14/7
I = 2A.
Since the circuit is in series connection,
the same current will flow through i.e
I1 = I2 = I3 = 2A.
Proposed Kinematic Exercise I
Consider a body moving on a straight path with a speed given by v=(3t^2-6t) m/s, where t is in seconds. If the body is at x=4m when t=0s, determine (a) its position at t=4s and (b) its acceleration at t=2s. (c) What is the direction of body movement between t=0s and t=4s? Justify your answer.
Answers
Answer:
(a) 20 m
(b) 6 m/s²
(c) Between t=0 and t=2, the body moves to the left.
Between t=2 and t=4, the body moves to the right.
Explanation:
v = 3t² − 6t
x(0) = 4
(a) Position is the integral of velocity.
x = ∫ v dt
x = ∫ (3t² − 6t) dt
x = t³ − 3t² + C
Use initial condition to find value of C.
4 = 0³ − 3(0)² + C
4 = C
x = t³ − 3t² + 4
Find position at t = 4.
x = 4³ − 3(4)² + 4
x = 20
(b) Acceleration is the derivative of velocity.
a = dv/dt
a = 6t − 6
Find acceleration at t = 2.
a = 6(2) − 6
a = 6
(c) v = 3t² − 6t
v = 3t (t − 2)
The velocity is 0 at t = 0 and t = 2. Evaluate the intervals.
When 0 < t < 2, v < 0.
When t > 2, v > 0.
A soccer ball is released from rest at the top of a grassy incline. After 8.6 seconds, the ball travels 87 meters and 1.0 s after this, the ball reaches the bottom of the incline. (a) What was the magnitude of the ball's acceleration, assume it to be constant
Answers
Answer:
a) a = 2.35 m/s^2
Explanation:
(a) In order to calculate the magnitude of the acceleration of the ball, you use the following formula, for the position of the ball:
[tex]x=v_ot+\frac{1}{2}at^2[/tex] (1)
x: position of the ball after t seconds = 87 m
t: time = 8.6 s
a: acceleration of the ball = ?
vo: initial velocity of the ball = 0 m/s
You solve the equation (1) for a:
[tex]x=0+\frac{1}{2}at^2\\\\a=\frac{2x}{t^2}[/tex]
You replace the values of the parameters in the previous equation:
[tex]a=\frac{2(87m)}{(8.6s)^2}=2.35\frac{m}{s^2}[/tex]
The acceleration of the ball is 2.35 m/s^2
. How many times higher could an astronaut jump on the Moon than on Earth if her takeoff speed is the same in both locations (gravitational acceleration on the Moon is about on-sixth of that on Earth)?
1) On the Moon, the jump height is about 3 times higher than on Earth.
2) On the Moon, the jump height is about 6 times higher than on Earth.
3) On the Moon, the jump height is about 36 times higher than on Earth.
4) On the Moon, the jump height is more than on Earth by about a times.
Answers
Answer:
On the Moon, the jump height is about 6 times higher than on Earth.
Explanation:
The maximum height reached by an object is given by the formula as follows :
[tex]H=\dfrac{v^2}{2g}[/tex]
v is the speed of object
g is acceleration due to gravity on Earth's surface
If g' is acceleration on surface of Moon, g' = g/6
Let H' is the height reached by object on Moon, then,
[tex]H'=\dfrac{v^2}{2g'}\\\\H'=\dfrac{v^2}{2\times \dfrac{g}{6}}\\\\H'=6\times \dfrac{v^2}{2g}\\\\H'=6H[/tex]
On the Moon, the jump height is about 6 times higher than on Earth. Hence, the correct option is (2).
Make predictions for the following five situations based on what you observed in part A. Check your answers only after you have made all five predictions. 1. The magnet is turned so that the south pole is near the wire while the switch is closed. Prediction Observation: 1. 2. The leads to the power supply are reversed (consider both orientations of the magnet). Prediction: Observation: 3. The north pole of the magnet is held near the wire but the switch remains open. Prediction: Observation 4. The north pole of the magnet is held: (a) closer to the wire and (b) farther from the wire. Prediction: Observation 5. The magnet is turned so that it is parallel to the wire while the switch is closed. Prediction Observation Resolve any discrepancies between your predictions and your observations. (Hint: Recall the right- hand rule used to determine the direction of the force exerted by a magnetic field on moving charges.) II. The magnetie field of a current-carrying wire A. Suppose you place a small magnet in a magnetic field and allow it to rotate freely. How will the magnet orient relative to the external magnetic field lines?
Answers
Answer:
the answer is D.
Explanation: Mark me brainlest
A sample of a compound weighs 2.54 grams. how many ounces (oz.) is this equivalent to?
a. 71.9 oz
b. 0.0139 oz
c. 0.0896 oz
d. 30.8 oz
Answers
The correct answer is C. 0.0896 oz
Explanation:
Both grams and ounces are units use to measure the mass of objects. Additionally, 1 gram represents a smaller amount of mass than the one represented by 1 ounce. Indeed, 1 ounce is equivalent to 28.349 grams and therefore, 1 gram is equivalent to 0.035 ounces ( 1 ounces x 1 gram / 28.349 grams = 0.03527 ounces).
According to this, you can convert grams to ounces by multiplying the amount of grams given by 0.035, which is the equivalent of 1 gram to ounces. This means, to determine how many ounces there are in 2.54 you multiply it by 0.035 which is equivalent to 0.0896 ounces (2.54 x 0.03527 = 0.0896 oz).
A hydraulic lift has two pistons: a small one of cross-sectional area 5.00 cm2 and a large one of cross-sectional area 300 cm2. What minimum force must be applied to the small piston in order for the large piston to lift a truck of mass 4500 kg
Answers
Answer:
735.75 N
Explanation:
We are given;
Area of small lift; A1 = 5 cm²
Area of large lift; A2 = 300 cm²
Mass required to lift large lift;m = 4500 kg
Since Force = mg, then, Force required to lift large lift;F2 = 4500 × 9.81 = 44145 N
By Pascal's law,
F1/A1 = F2/A2
We want to find minimum force that must be applied to the small piston;F1, so let's make F1 the subject.
Thus;
F1 = (F2 × A1)/A2
Plugging in the relevant values, we have;
F1 = (44145 × 5)/300
F1 = 735.75 N
The cornea behaves as a thin lens of focal length approximately 1.80 {\rm cm}, although this varies a bit. The material of whichit is made has an index of refraction of 1.38, and its front surface is convex,with a radius of curvature of 5.00 {\rm mm}.(Note: The results obtained here are not strictlyaccurate, because, on one side, the cornea has a fluid with arefractive index different from that of air.)a) If this focal length is in air, what is the radius ofcurvature of the back side of the cornea? (in mm)b) The closest distance at which a typical person can focus onan object (called the near point) is about 25.0 {\rm cm}, although this varies considerably with age. Wherewould the cornea focus the image of an 10.0 {\rm mm}-tall object at the near point? (in mm)c) What is the height of the image in part B? (mm)d) Is this image real or virtual? Is it erect orinverted?
Answers
Answer:
The height of the image will be "1.16 mm".
Explanation:
The given values are:
Object distance, u = 25 cm
Focal distance, f = 1.8 cm
On applying the lens formula, we get
⇒ [tex]\frac{1}{v} -\frac{1}{u} =\frac{1}{f}[/tex]
On putting estimate values, we get
⇒ [tex]\frac{1}{v} -\frac{1}{(-25)} =\frac{1}{1.8}[/tex]
⇒ [tex]\frac{1}{v} =\frac{1}{1.8} -\frac{1}{25}[/tex]
⇒ [tex]v=1.94 \ cm[/tex]
As a result, the image would be established mostly on right side and would be true even though v is positive.
By magnification,
[tex]m=\frac{v}{u}[/tex] and [tex]m=\frac{h_{1}}{h_{0}}[/tex]
⇒ [tex]\frac{v}{u} =\frac{h_{1}}{h_{0}}[/tex]
⇒ [tex]\frac{1.94}{25}=\frac{{h_{1}}}{15}[/tex]
⇒ [tex]{h_{1}}=1.16 \ mm[/tex]
How much work will it take to lift a 2-kg pair of hiking boots 2 meters off the
ground and onto a shelf in your closet?
O A. 2.45 J
B. 4J
C. 39.2 J
D. 20 J
Answers
Answer:39.2 (B)
Explanation:
Smooth and continuous data is a characteristic of a(n)
analog signal
binary signal
digital signal
pulse signal
Answers
Answer:
D
Explanation:
Continuous data is a characteristic of analog signals though been smooth is a subject of question. The smooth signal is after it's been processed. However a wave can still stay continuous defined by what we call duty circle.
Smooth and continuous data is one of the major features of an analog signal.
Analog signal isn't as accurate as the digital signal and it contains minimum
and maximum values which could be positive or negative.
Analog signal comprises of smooth and continuous data. This is explained
by one time-varying quantity representing another time-based variable.In
this type of signal , a variable is an analog of the other.
Analog signals examples include the following : Human voice, Thermometers, analog clocks etc
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A charge of 6.65 mC is placed at each corner of a square 0.500 m on a side.
Required:
a. Determine the direction of the force on a charge.
b. Determine the direction of the force on a charge.
1. along the line between the charge and the center of the square outward of the center
2. along the line between the charge and the center of the square toward the center
3. along the side of the square toward the other charge that lies on the side
4. along the side of the square outward of the other charge that lies on the side
Answers
Answer:
Explanation:
Given that:
length of side , a = 0.5 m
charge , q = 6.65 mC
length of diagonal , d = 0.5 * sqrt(2)
d = 0.707 m
F is the force due to adjacent particle ,
F1 is the force due to diagonal particle
Now , for the net charge on a particle
Fnet = 2 * F * cos(45) + F1
Fnet = 2*cos(45) * k * q^2/a^2 + k * q^2/d^2
Fnet = 9*10^9 * 0.00665^2 * (2* cos(45)/.5^2 + 1/.707^2)
Fnet = 3.05 *10^6 N
the magnitude of net force acting on each particle is 3.05 *10^6 N
part B)
for the direction of particle
d) along the line between the charge and the center of the square outward of the center