Answer:
FALSE
Explanation:
A 100-g bullet is fired from a rifle having a barrel 0.600 m long. Choose the origin to be at the location where the bullet begins to move. Then the force (in newtons) exerted by the expanding gas on the bullet is 15 000 1 10 000x 2 25 000x2, where x is in meters. (a) Determine the work done by the gas on the bullet as the bullet travels the length of the barrel. (b) What If
Answer:
9000 J
Explanation:
weight of bullet = 100 g
length of gun barrel = 0.6m
A) determine the work done by the gas on the bullet as the bullet travels the length of the barrel
The work done is an integral of force over distance in this scenario
lets represent the force using some constants ( arbitrary )
Fx = a + bx + cx^2
where : a = 15000, b = 10000, c = - 25000
hence work done ( W ) = [tex]ax_{f } + \frac{bx^{2} _{f} }{2} + \frac{cx^{3} _{f} }{3}[/tex]
where ; xf = 0.6
hence work done(W)
= 15000(0.6) + [(10000*0.6^2)/2] + [(-25000*0.6^3)/3]
= 9000 J
119) You throw a rock horizontally off a cliff with a speed of 20 m/s and no significant air resistance. After 2.0 s, the magnitude of the velocity of the rock is closest to
Answer: 44.54 m/s
Explanation:
Let's break the problem into horizontal and vertical.
Horizontal:
Here we do not have any force, only a constant horizontal velocity, so we can write:
Vx(t) = 20m/s.
Vertical:
Here we have the gravitational acceleration acting on the rock, then we can write:
Ay(t) = -9.8m/s^2
For the vertical velocity, we need to integrate over time and get:
Vy(t) = (-9.8m/s^2)*t + V0
Where V0 is the initial vertical velocity, but the rock was thrown horizontally, so we do not have an initial velocity in the vertical axis.
Vy(t) = (-9.8m/s^2)*t.
Ok, now we want to know the magnitude of the velocity when t = 2.0s.
The magnitude will be equal to:
V = √( Vx(2s)^2 + Vy(2s)^2)
= √( (20m/s)^2 + (-2s*9.8m/s^2)^2) = 44.54 m/s
The magnitude of the velocity after 2 seconds is around 44.54 m/s
A particle with charge q = +5.00 C initially moves at v = (1.00 î + 7.00 ĵ ) m/s. If it encounters a magnetic field B = 10 k, find the magnetic force vector on the particle.
Given :
Force on object moving with constant velocity due to magnetic field is given by :
[tex]\vec{F}=q\vec{v}\times \vec{B}\\\\\vec{F}=-5[ ( \hat{i} + 7\hat{j}) \times ( 10\hat{k})]\\\\\vec{F}=-5[ 10(-\hat{j})+70(\hat{i})]\\\\\vec{F}=50\hat{j}-350\hat{i}\ N[/tex]
Now,
[tex]F = \sqrt{50^2+350^2}\\\\F=353.55\ N[/tex]
Hence, this is the required solution.
A particle with charge q = +5.00 C that is moving with some velocity, then the magnetic force vector on the particle will be 353.55 N.
What is a vector?The phrase "vector" in physics and mathematics is used popularly to refer to some values that cannot be stated by a simple integer (a scalar) or to certain elements of high - dimensional space.
For things like displacements, forces, and velocity can have both a magnitude and direction, vectors were initially introduced in geometry and physics. Similar to how lengths, masses, and time are represented by real numbers, these quantities were represented by geometric vectors.
In some contexts, tuples—finite sequences of numbers with a definite length—are sometimes referred to as vectors.
From the data given in the question,
F = qv × B
F = -5 [(î +7 ĵ) ×(10 k)]
F = 50 ĵ - 350 î N
Now, calculate the force,
F = √50² + 350²
F = 353.55 N.
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The microwave background radiation is observed at a wavelength of 1.9 millimeters. The red shift of these photons is observed to be raoughly 1100. What would be the emitted wavelength?
Answer:
1730nm
Explanation:
Observed wavelength λ= 1.9 millimeters
Red shift of photons = 1100
Emitted wavelength = λo
Red shift = λ - λo/ λo = 1100
= [(1.9x10^-3 - λo)/ λo] = 1100
= 1.9x10^-3 - λo = 1100 λo
= 1.9x10^-3 = 1100 λo + λo
= 1.9x10^-3 = 1101 λo
We divide through by 1101 to get the value of lambda
1.9x10^-3/1101 = λo
λo = 0.000001726
λo = 1726nm
This is approximately
λo = 1730nm
Thank you!
Which of The following statements best describes the objects motion between zero and four seconds?
Answer:
A OR C
Nearly A.
Explanation:
A long thin rod of length 2L rotates with a constant angular acceleration of 8.0 rad/s2 about an axis that is perpendicular to the rod and passes through its center. What is the ratio of the tangential acceleration of a point on the end of the rod to that of a point a distance L/2 from the end of the rod
Answer:
Explanation:
angular acceleration ω = 8 rad /s²
tangential acceleration a = angular acceleration x radius of circle of rotation
For a point on the end of the rod , radius of circle = L .
tangential acceleration = 8 L .
For a point a distance L/2 from the end of the rod, radius of circle = L /2 .
tangential acceleration = 8 L/2 = 4 L .
It is so because angular acceleration will be same for this point , only radius changes .
required ratio = 8L / 4L
= 2.
Two trains travel from New Orleans to Memphis in 5 h. The first train travels at a constant velocity of 80 mph, but the velocity of the second train varies. What was the second train's average velocity during the trip
Answer:
Explanation:
The first train travels at a constant velocity of 80 mph and it takes 5 h to cover the distance so distance between two station = 80 x 5 = 400 m
second train too takes 5 h to cover the same distance so its average velocity
= total distance / total time
= 400 / 5 = 80 m /h
average velocity = 80 m /h
a decompression chamber used by deep sea divers has a volume of 10.1m^3 and operates at an internal pressure of 4.25 atm. What volume, in cubic meters would the air in the chamber occupy if it were at 1.00 atm pressure, assuming no tempreature change.
Answer:
V2=42.93cm3
Explanation:
According to Gay-Lussac's Law: "The Pressure Temperature Law. This law states that the pressure of a given amount of gas held at constant volume is directly proportional to the Kelvin temperature. With an increase in temperature, the pressure will go up."
P1V1=P2V2
Given data
P1=4.25atm
V1=10.1cm³
P2=1atm
V2=?
V2=P1V1/P2
V2=(4.25*10.1)/1
V2=42.93cm3
Calculate the linear acceleration of a car, the 0.300-m radius tires of which have an angular acceleration of 17.0 rad/s2. Assume no slippage and give your answer in m/s2.
Answer:
5.1m/s^2Explanation:
In this problem we want to solve for the linear acceleration
Given data
radius r= 0.3m
angular acceleration α=17rad/s^2
we know that the expression relating angular acceleration and linear acceleration is given as
a=rα
Substituting we have
a=0.3*17
a=5.1m/s^2
A small, 300 gg cart is moving at 1.30 m/sm/s on a frictionless track when it collides with a larger, 4.00 kgkg cart at rest. After the collision, the small cart recoils at 0.860 m/sm/s . Part A What is the speed of the large cart after the collision
Answer:
v₂f = 0.16 m/s (to the right)
Explanation:
Assuming no external forces acting during the collision, as stated by Newton's 2nd law, total momentum must be conserved.The initial momentum is due to the small cart only, because the big one is at rest.Assuming that the small cart is going to the right and taking this direction as the positive one, we can write the following expression for the initial momentum p₀ :[tex]p_{o} = m_{1} * v_{o1} = 0.3 kg * 1.3 m/s (1)[/tex]
The final momentum can be written as follows:[tex]p_{f} = m_{1} * v_{1f} + m_{2} * v_{2f} = 0.3 kg* (-0.86m/s) + 4.00 kg* v_{2f} (2)[/tex]
As p₀ = pf, from (1) and (2), we can solve for v2f, as follows:
[tex]v_{2f} = \frac{0.3kg* (1.3 + 0.86) m/s}{4.00kg} = \frac{0.3kg*2.16m/s}{4.00kg} = 0.16 m/s[/tex]
As the sign of v2f is positive, we conclude that it starts to move in the same direction that m1 was originally going (to the right), whilst m1 recoils, which means that after the collision, it bounces back from the larger mass.If the ball is released from rest at a height of 0.63 m above the bottom of the track on the no-slip side, what is its angular speed when it is on the frictionless side of the track
Answer:
When the ball is on the frictionless side of the track , the angular speed is 89.7 rad/s.
Explanation:
Consider the ball is a solid sphere of radius 3.8 cm and mass 0.14 kg .
Given , mass, m=0.14 kg
Ball is released from rest at a height of, h= 0.83 m
Solid sphere of radius, R = 3.8 cm
=0.038 m
From the conservation of energy
ΔK = ΔU
[tex]\frac{1}{2}mv^2 +\frac{1}{2} I\omega^2=mgh[/tex]
Here , [tex]I=\frac{2}{5} MR^2 , v= R \omega[/tex]
[tex]\frac{1}{2}mv^2\frac{1}{2}(\frac{2}{5}MR^2)(\frac{v}{R^2} )=mgh[/tex]
[tex]\frac{1}{2} [v^2+\frac{2}{5}v^2]= gh[/tex]
[tex]\frac{7}{10} v^2=gh[/tex]
[tex]0.7v^2=gh[/tex]
v=[tex]\sqrt{[gh/(0.7)][/tex]
=[tex]\sqrt{ [(9.8 m/s^2)(0.83 m) / (0.7) ][/tex]
= 3.408 m/s
Hence, angular speed when it is on the frictionless side of the track,
[tex]\omega=\frac{v}{R}[/tex]
= (3.408 m/s)/(0.038 m)
[tex]\omega[/tex] = 89.7 rad/s
Hence , the angular speed is 89.7 rad/s
ufl.edu A wire of length 27.7 cm carrying a current of 4.11 mA is to be formed into a circular coil and placed in a uniform magnetic field of magnitude 5.85 mT. The torque on the coil from the field is maximized. (a) What is the angle between and the coil's magnetic dipole moment? (b) What is the number of turns in the coil? (c) What is the magnitude of that maximum torque?
Answer:
A)90°
Explanation:
ufl.edu A wire of length 27.7 cm carrying a current of 4.11 mA is to be formed into a circular coil and placed in a uniform magnetic field of magnitude 5.85 mT. The torque on the coil from the field is maximized. (a) What is the angle between and the coil's magnetic dipole moment? (b) What is the number of turns in the coil? (c) What is the magnitude of that maximum torque?
(a) What is the angle between and the coil's magnetic dipole moment.
The angle between and the coil's magnetic dipole moment is 90°, because it is perpendicular to the field since there is maximum torgue.
(b) What is the number of turns in the coil?
A tuna jumps out of the water with an initial velocity of 44 feet per second
Answer:
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Explanation:
plz do so it is true
will give brainliest pls help
Answer:
a and c
Explanation:
1.A large beach ball weighs 4.0 N. One person pushes it with a force of 7.0 N due South while another person pushes it 5.0 N due East. Find the acceleration on the beach ball.
2. What is the weight of a 70 kg astronaut on the earth, on the moon, (g=1.6 m/s2), on Venus (g = 18.7 m/s2) and in outer space traveling at a constant velocity
Answer:
1. force applied southward = -4 j
force applied eastward = 5 i
total force applied = 5i - 4j
magnitude of total force applied = √(5)²+(-4)²
magnitude of total force applied = √25 + 16 = √41
magnitude of total force applied = 6.4N
But the beach ball also weighs 4 N,
which means that a force of 4N is required to overcome the inertia of the ball
Net force applied on the ball = total force applied - force applied by inertia
Net force applied on the ball = 6.4 - 4
Net force applied on the ball = 2.4 N
Mass of the ball:
Mass of the ball = weight of the ball / gravitational constant
Mass of the ball = 4 / 9.8 = 0.4 kg
Acceleration of Ball:
from newton's second law of motion:
F = ma
replacing the variables
2.4 = 0.4 * a (where a is the acceleration of the ball)
a = 2.4/0.4
a = 6 m/s²
2. Mass of astronaut = 70 kg
Weight of Earth:
Weight = Mass * acceleration due to gravity
Weight = 70 * 9.8
Weight = 686 N
Weight on Moon:
Weight = Mass * acceleration due to gravity
Weight = 70 * 1.6 (we are given that g = 1.6 on moon)
Weight = 112 N
Weight on Venus:
Weight = Mass * acceleration due to gravity
Weight = 70 * 18.7 (we are given that g = 18.7 on Venus)
Weight = 1309 N
s this statement true or false?
Hurricanes are classified by three stages in which updrafts of air billow up, updrafts and downdrafts swirl the wind up and down, then downdrafts cause clouds to come apart.
true
false
Answer:
True
Explanation:
If the rock is kicked at initial velocity of 12 m/s from a height of 106 m above the ground what horizontal distance does the rock travel before striking the ground?
Answer:
74 m.
Explanation:
From the question given above, the following data were obtained:
Height (h) = 186 m
Initial velocity (u) = 12 m/s
Horizontal distance (d) =?
Next, we shall determine the time taken for the rock to get to the ground. This can be obtained as follow:
Height (h) = 186 m
Acceleration due to gravity (g) = 9.8 m/s²
Time (t) =?
H = ½gt²
186 = ½ × 9.8 × t²
186 = 4.9 × t²
Divide both side by 4.9
t² = 186/4.9
Take the square root of both side
t = √(186/4.9)
t = 6.16 s
Thus, the time taken for the rock to get to the ground is 6.16 s
Finally, we shall determine the horizontal distance travelled by the rock as follow:
Initial velocity (u) = 12 m/s
Time (t) = 6.16 s
Horizontal distance (d) =?
Horizontal distance (d) = Initial velocity (u) × Time (t)
d = u × t
d = 12 × 6.16
d = 73.92 ≈ 74 m
Therefore, the horizontal distance travelled by the rock is 74 m
A Ferris wheel rotates at an angular velocity of 0.25 rad/s. Starting from rest, it reaches its operating speed with an average angular acceleration of 0.027 rad/s2. How long does it take the wheel to come up to operating speed?
Answer:
t = 9.25 s
Explanation:
Given that,
Initial angular velocity, [tex]\omega_o=0[/tex] (at rest)
Final angular velocity, [tex]\omega_o=0.25\ rad/s[/tex]
Angular acceleration, [tex]\alpha =0.027\ rad/s^2[/tex]
We need to find the time it take the wheel to come up to operating speed. We know that the angular acceleration in terms of angular speed is given by :
[tex]\alpha =\dfrac{\omega_f-\omega_o}{t}\\\\t=\dfrac{\omega_f-\omega_o}{\alpha }\\\\t=\dfrac{0.25-0}{0.027}\\\\t=9.25\ s[/tex]
So, it will reach up to the operating speed in 9.25 s.
A fully loaded, slow-moving freight elevator has a cab with a total mass of 1200 kg, which is required to travel upward 35 m in 3.5 min, starting and ending at rest. The elevator's counterweight has a mass of only 940 kg, so the elevator motor must help pull the cab upward. What average power is required of the force the motor exerts on the cab via the cable
Answer:
425.1 W
Explanation:
We are given;
Counter mass of elevator; m_c = 940 kg
Cab mass of elevator; m_d = 1200 kg
Distance from rest upwards; d = 35 m
Time to cover distance; t = 3.5 min
Now, this elevator will have 3 forces acting on it namely;
Force due to the counter weight of the elevator; F_c
Force due to the cab weight on the elevator; F_d
Force exerted by the motor; F_m
Now, from Newton's 2nd law of motion,
The force exerted by the motor on the elevator can be given by the relationship;
F_m = F_d - F_c
Now,
F_d = m_d × g
F_d = 1200 × 9.81
F_d = 11772 N
F_c = m_c × g
F_c = 940 × 9.81
F_c = 9221.4 N
Thus;
F_m = 11772 - 9221.4
F_m = 2550.6 N
Now, the average power required of the force the motor exerts on the cab via the cable is given by;
P_m = F_m × v
Where v is the velocity of the elevator.
The velocity is calculated from;
v = distance/time
v = 35/3.5
v = 10 m/min
Converting to m/s gives;
v = 10/60 m/s = 1/6 m/s
Thus;
P_m = 2550.6 × 1/6
P_m = 425.1 W
What is the most logical explanation for why kilo is a more important prefix to remember than tera
When the momentum of an object increases with respect to time, what is true of the net force acting on it?
a. It is zero, because the net force is equal to the rate of change of the momentum.
b. It is zero, because the net force is equal to the product of the momentum and the time interval.
c. It is nonzero, because the net force is equal to the rate of change of the momentum.
d. It is nonzero, because the net force is equal to the product of the momentum and the time interval.
Answer:
C. It is nonzero, because the net force is equal to the rate of change of the momentum
Explanation:
Momentum of an object is given as;
P = mv = ft
f = mv/t
or f = P/t
where;
P is the momentum of the object
m is mass of the object
v is velocity of the object
f is the applied force on the object
t is time
From the given equation above; the net force acting on the object is equal to the rate of change of the momentum, thus it is non-zero.
Therefore, the correct option is "C"
C. It is nonzero, because the net force is equal to the rate of change of the momentum.
The net force acting on the object is non-zero, because the net force is equal to the rate of change of the momentum. Hence, option (c) is correct.
The problem is based on the impulse-momentum concept. As per the impulse -momentum concept, the magnitude of impulse produced by an object is equal to the change in momentum. Then the expression is,
[tex]I = \Delta P[/tex]
here,
I is the magnitude of impulse. And its value is,
[tex]I = F \times t[/tex]
t is the impact time or time interval.
F is the magnitude of net force.
[tex]\Delta P[/tex] is the change in momentum.
Solving as,
[tex]I= \Delta P\\\\F \times t= \Delta P\\\\F = \dfrac{\Delta P}{t}[/tex]
Clearly, the net force is dependent of the change in momentum with respect to time. So, on increasing the momentum change with respect to time, the net force will also increase with non-zero value.
Thus, we can conclude that the net force acting on the object is non-zero, because the net force is equal to the rate of change of the momentum. Hence, option (c) is correct.
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What height does a frictionless playground slide need so that a 35 kg child reaches the bottom at a speed of 6.5 m/s?
Answer:
h = 2.15 m
Explanation:
Given that,
Mass of the child, m = 35 kg
The child reaches the bottom at a speed of 6.5 m/s
We need to find the height slide by a frictionless playground slide. Let it is equal to h. Using the conservation of energy to find it as follows :
[tex]\dfrac{1}{2}mv^2=mgh\\\\h=\dfrac{v^2}{2g}\\\\h=\dfrac{(6.5)^2}{2\times 9.8}\\\\h=2.15\ m[/tex]
So, it will slide to a height of 2.15 m
The height slid by the frictionless playground can be calculated by equating the kinetic and potential energy equations which is 2.16 meters
Mass of child = 35 kg Velocity, v = 6.5 m/s g = 9.8 m/s²K.E = P.E
K. E = 0.5mv² ; P.E = mgh
0.5mv² = mgh
gh = 0.5v²
h = [(0.5v²) ÷ g]
h = [(0.5 × 6.5²) ÷ 9.8]
h = (21.125 ÷ 9.8)
h = 2.155 m
Therefore, the height slid by a frictionless playground is 2.16 meters
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What happens if atoms lose energy during a change of state?
Answer: they become slower.
Explanation: because they get pulled together.
Answer:
The atoms are pulled together by attractive forces and become more organized.
Explanation:
Atoms lose energy as a gas changes to a solid. ... Atoms gain energy as a solid changes to a liquid. If atoms energy during a change of state, they are pulled together by attractive forces and become more organized.
Reference by Quizlet.
You are walking in Paris alongside the Eiffel Tower and suddenly a croissant smacks you on the
head and knocks you to the ground. From your handy dandy tourist guidebook, you find that the
height of the Eiffel Tower is 300.5 m. If you neglect air resistance, calculate how many seconds
the croissant dropped before it tagged you on the head.
What is the answer
If we put a flammable material in a point where many light rays converge we can set it on fire. With what type of spherical mirror would it be possible to do this
Answer:
Concave mirror
Explanation:
A concave mirror is a type that has the property of converging rays of light that are parallel to the principal axis to a point of focus. Thus, for the convergence of many rays of light (i.e a beam of light) as required in the given question, a concave mirror would serve the purpose. Curved mirrors have various applications in optics.
Convert 200km/h to m/s
Answer:
200 kilometers per hour =
55.556 meters per second
Explanation:
have a great day
Answer:
200km × 1000m/1km = 200000 m
1h × 60 mins/h × 60sec/min = 3600s
now you divide 200000m/3600s
= 55.55m/s
A disk rotates about its central axis starting from rest and accelerates with constant angular acceleration. At one time it is rotating at 4.40 rev/s; 60.0 revolutions later, its angular speed is 15.0 rev/s. Calculate (a) the angular acceleration (rev/s2), (b) the time required to complete the 60.0 revolutions, (c) the time required to reach the 4.40 rev/s angular speed, and (d) the number of revolutions from rest until the time the disk reaches the 4.40 rev/s angular speed.
Answer:
The answer is below
Explanation:
a) Using the formula:
[tex]\omega^2=\omega_o^2+2\alpha \theta\\\\\omega=final\ velocity=15\ rev/s,w_o=initial\ velocity=4.4\ rev/s, \\\theta=distance=60\ rev\\\\Substituting:\\\\15^2=4.4^2+2(60)\alpha\\\\2(60)\alpha=15^2-4.4^2\\\\2(60)\alpha=205.64\\\\\alpha=1.71\ rev/s^2[/tex]
b) The disk is initially at rest. Using the formula:
[tex]\theta=\omega_ot+\frac{1}{2}\alpha t^2 \\\\but\ \omega_o=0(rest), \theta=60\ rev,\alpha=1.71\ rev/s^2\\\\Subsituting:\\\\60=0+\frac{1}{2}(1.71) t^2\\\\t=8.4\ s[/tex]
c)
[tex]\omega=\omega_o+\alpha t \\\\but\ \omega_o=0(rest), \omega=4.4 \ rev/s\ rev,\alpha=1.71\ rev/s^2\\\\Subsituting:\\\\4.4=1.71t\\\\t=2.6\ s[/tex]
d)
[tex]\omega^2=\omega_o^2+2\alpha \theta\\\\\omega=final\ velocity=4.4\ rev/s,w_o=initial\ velocity=0\ rev/s, \\\alpha=1.71\ rev/s^2\\\\Substituting:\\\\4.4^2=0+2(1.71)\theta\\\\19.36=3.42\theta\\\\\theta=5.66\ rev[/tex]
State two function of the mass M placed on the metre rule
correct answer is mass media is correct answer OK
Jack drops a stone from rest off of the top of a bridge that is 23.4 m above the ground. After the stone falls 7.2 m, Jill throws a second stone straight down. Both rocks hit the water at the exact same time. What was the initial velocity of Jill's rock
Explanation:dropped stone hits water:
24.4-4.9t^2=0
t=2.23
stone has fallen 6.6m:
4.9t^2 = 6.6
t = 1.16
So, Jill's thrown stone only has 2.23-1.16=1.07 seconds to hit the water:
24+1.07v-4.9*1.07^2 = 0
v = -17.19 m/s
Both the rocks hit the water at the exact same time, and the initial velocity of Jill's rock will be -19.28 m/s.
What is velocity?A vector measurement of the rate and direction of motion is what is meant by the term "velocity." Simply said, velocity is the rate of movement in a single direction. Velocity may be used to gauge both the speed of a rocket launching into space and the speed of a car travelling north on a busy freeway.
According to the question, initial velocity will be 0.
By using equation of motion :
Let Jill is x and Jack is y.
s=ut+1/2at² where,
a is acceleration,
u is initial velocity and,
t is the time period.
23.4=(-1/2)(9.8)×t²
t(y)=[tex]\sqrt{(23.4)/(4.9)}[/tex]
=2.185 seconds.
Time taken by Jack's stone that it travel the first 7.2 meters.
t=[tex]\sqrt{(2*7.2)/9.8}[/tex] =1.21 seconds.
So, time taken by stone thrown by jack,
t(Jill)=t(jack)-t
=23.4=-ut(x)-1/2(gt²)(y)
-23.4=(-u×0.975)-(1/2×9.8×(0.975)²)
u= -(23.4-4.6/0.975)
u=-19.28 m/s
Hence, the initial velocity of the Jill's stone will be -19.28 m/s.
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a 214kg boat is sinking in the ocean.its acceleration towards the sea floor is 6.12 m/s^2. the force of gravity that draws the boat down is partially offset by the buoyant force of the water, so what is the force pulling on the boat as it sinks?