Answer:
Since velocity is considered a vector quantity and vectors show you in what direction it goes the object wont be moving forward but backwards because its a negative and negatives go to the left and that would be considered backwards.
Explanation:
Hope this helps
Calculate the magnitude of the internal normal force, NB, shear force, VB, and moment, MB, at point B.
Answer:
The answer is "[tex]\bold{MB= \frac{L^2 \ WB}{12}}[/tex]"
Explanation:
please find the complete question in the attached file:
From A to B
[tex]reqAB = \int^{\frac{1}{2}}_{0} \frac{WB}{\frac{L}{2}} x \ dx = \frac{L\ WB}{4}[/tex]
From symmetry [tex]Ay = Cy = FeqAB[/tex]
The FeqAB position is found on a table of common centres:
[tex]\frac{2}{3}( \frac{L}{2})[/tex]
[tex]\to \Sigma_{NB=0} FX = NB=0\\\\\to \Sigma_{VB=0} Fy = \frac{LWB}{4} - \frac{LWB}{4} -VB =0 \\\\\to \Sigma M_{.CCW \ A} = \frac{L}{3} - \frac{LWB}{4} - \frac{L}{2}VB+MB =0 \\\\MB= \frac{L^2 \ WB}{12}[/tex]
A proton in a cyclotron has a velocity of 2x108 m/s, which is nearly the speed of light. The uncertainty in the velocity is 1%. What is the minimum uncertainty in the position of the proton in meters
Answer:
The minimum uncertainty in the position of the proton is 1.578 x 10⁻¹⁴ m
Explanation:
Given;
velocity of the proton, v = 2 x 10⁸ m/s
uncertainty in the velocity, = 1 % = 0.01 v = 0.01 x 2 x 10⁸ m/s = 2 x 10⁶ m/s
The momentum of the proton is given by;
ΔP = mv = (1.67 x 10⁻²⁷)(2 x 10⁶ ) = 3.34 x 10⁻²¹ kgm/s
The minimum uncertainty in the position of the proton is given by;
[tex]\delta x = \frac{h}{4\pi \delta P}\\\\ \delta x =\frac{6.626*10^{-34}}{4\pi *3.34*10^{-21}}\\\\ \delta x = 1.578*10^{-14} \ m[/tex]
Therefore, the minimum uncertainty in the position of the proton is 1.578 x 10⁻¹⁴ m
During the 3.7 min a 7.4 A current is set up in a wire, how many (a) coulombs and (b) electrons pass through any cross section of the wire's width?
Answer:
1,642.8 coulombs and 1.39*10²¹ electrons pass through any cross section the width of the wire.
Explanation:
Electric current is the circulation of electric charges in an electric circuit, while electric current intensity (I) is the amount of electricity or electric charge (Q) that circulates through a circuit in unit time (t).
Then, the intensity of electric current is expressed as:
[tex]I=\frac{Q}{t}[/tex]
Where:
I is the intensity expressed in Amps (A) Q is the electric charge expressed in Coulombs (C) t is the time expressed in seconds (s)In this case:
I= 7.4 AQ= ?t= 3.7 min= 222 s (being 1 min= 60 s)Replacing:
[tex]7.4 A=\frac{Q}{222 s}[/tex]
Solving:
Q= 7.4 A* 222 s
Q= 1,642.8 C
A Coulomb represents about 6.24*10¹⁸ electrons, so you can apply the following rule of three: if 1 C represents 6.24*10¹⁸ electrons, 222 C how many electrons does it represent?
[tex]electrons=\frac{222 C*6.24*10^{18} }{1C}[/tex]
electrons= 1.39*10²¹
1,642.8 coulombs and 1.39*10²¹ electrons pass through any cross section the width of the wire.
Which muscles pump blood through your heart and body?
a
Cardiac
b
Skeletal
c
Smooth
d
Strong
A heating element is made by maintaining a potential difference of 75.0 V across the length of a Nichrome wire that has a 3.73 × 10–6 m2 cross section. Nichrome has a resistivity of 5.00 × 10–7 Ω·m. (a) If the element dissipates 7880 W, what is its length? (b) If 107 V is used to obtain the same dissipation rate, what should the length be?
Answer:
[tex]P=V^2/R[/tex]
[tex]R=V^2/P[/tex]
Upon substituting the values
a) R= 75*75/7800 = 0.72ohm
Also
R=ρ*L/A
L=R*A/ρ
L=[tex]\frac{0.72*3.73*10^-6 }{5*10^-7}[/tex] = 5.44m
b) when the voltage 107 is used
R=107*107/7800 = 1.468OHM
L=[tex]\frac{1.468*3.73*10^-6 }{5*10^-7}[/tex] =11.1m
What is the potential energy of a 20-kg safe sitting on a shelf 0.5 meters
above the ground?
A-20j
B-10j
C-98j
D-196j
Answer:
C - 98j
Explanation:
Answer:
c- 98j
hope it helped
The John Hancock Center in Chicago is the tallest building in the United States in which there are residential apartments. The Hancock Center is 343 m tall. Suppose a resident accidentally causes a chunk of ice to fall from the roof. What would be the velocity of the ice as it hits the ground
Answer:
Explanation:
The chunk of ice will fall with acceleration of 9.8 m /s²
initial velocity u = 0 ,
v² = u² + 2 gH
H = 343 m
g = 9.8 m /s²
v is final velocity
v² = 0 + 2 x 9.8 x 343
v² = 6723
v = 82 m /s approx .
The velocity of the the chunk of ice as it hits the ground is 82m/s.
Given the data in the question;
Height of John Hancock Center; [tex]h = 343m[/tex]Before the chunk of ice to fell off the roof, its was initially at rest, so
Initial velocity; [tex]u = 0[/tex]Final velocity as the ice hits the ground; [tex]v = \ ?[/tex]
To determine the final velocity, we use the third equation of motion:
[tex]v^2 = u^2 + 2as[/tex]
Where v is final velocity, u is the initial velocity, s is distance or height of the building and a is acceleration due to gravity{ since the ice will be under gravity as it falls, ([tex]a = g = 9.8m/s^2[/tex]) }
We substitute our values into the equation
[tex]v^2 = [0^ 2] + [ 2\ *\ 9.8m/s^2\ *\ 343m]\\\\v^2 = 2\ *\ 9.8m/s^2\ *\ 343m\\\\v^2 = 6722.8m^2/s^2\\\\v = \sqrt{6722.8m^2/s^2}\\\\v = 81.99m/s\\\\v = 82m/s[/tex]
Therefore, the velocity of the the chunk of ice as it hits the ground is 82m/s.
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What change to the atomic model helped solve the problem seen in Rutherford’s model?
A 55kg skateboarder wants to just make it to the upper edge of a quarter pipe, a pipe that is one-quarter of a circle with a radius of 3.10m. What speed does he need at the bottom?
Answer:
Explanation:
A 52.0 kg skateboarder wants to just make it to the upper edge of a "quarter pipe," a track that is one-quarter of a circle with a radius of 2.60 m .
What speed does he need at the bottom
Let the lowest point of the circle = (0,0)
Center of circle = (0, 2.6)
The height of the circle = 2 * radius = 2 * 2.60 = 5.20 m
Highest point = (0, 5.2)
As the skateboarder moves around ¼ of a vertical circle, the skateboarder moves from the lowest position to a position that is ½ the way up to the highest position. This is the point that is 2.6 meters directly to the right of left of the center = (2.6, 2.6)
As the skateboarder has moved 2.6 m upward, the potential energy increase = m * g * ∆h = 52.0 * 9.8 * 2.6
During this same time, the kinetic energy decreased from the maximum to 0. The decrease of KE = the increase of PE
½ * m * ∆v^2 = m * g * ∆h
½ * 52 * ∆v^2 = 52.0 * 9.8 * 2.6
½ * ∆v^2 = 9.8 * 2.6
∆v = (2 * 9.8 * 2.6)^0.5 = 7.14 m/s
The velocity at highest point, (2.6,2.6) is 0 m/s
So the velocity at the lowest point must be 7.14 m/s
Let’s see if the skateboarder has enough KE to move upward 2.6 m.
KE at bottom = ½ * 52 * 7.14^2 = 1325.5
PE = 52 * 9.8 * h
52 * 9.8 * h = 1325.5
h = 2.6 m
The answer is OK
The speed needed by the skateboarder at the bottom is 7.8 m/s.
The given parameters;
mass of the skateboarder, m = 55 kgradius of the quarter pipe, r = 3.1 mApply the principle of conservation of mechanical energy to determine the sped of the skateboarder at the bottom;
[tex]P.E_{top} = K.E_{bottom}\\\\mgh = \frac{1}{2}mv^2\\\\gh = \frac{1}{2} v^2\\\\v^2 = 2gh\\\\v = \sqrt{2gh} \\\\v = \sqrt{2\times 9.8 \times 3.1} \\\\v = 7.8 \ m/s[/tex]
Thus, the speed needed by the skateboarder at the bottom is 7.8 m/s.
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a force of 193 pounds makes an angle of 86.27 with a second force. The resultant of the two forces makes an angle of 31.4 to the first force. Find the magnitudes of the second force and of the resultant.
Answer:
Explanation:
Let force P = 193 pounds
second force = Q
Angle that resultant makes with force P is θ
Tanθ = Q sin86.27 / (P + Q cos86.27 )
Tan 31.4 = Q sin86.27 / (193 + Q cos86.27 )
.61 = .99 Q / (193 + .065Q )
117.73 + .04 Q = .99Q
117.73 = .95 Q
Q = 123.93 pounds .
The log left the cliff with a horizontal velocity of 7.5 m/s. When it reached the water, it had a vertical velocity of -8.9 m/s. What was the log's speed (two dimensional) before it entered the water
Answer:
v = 11.6 m/s
Explanation:
Assuming no other forces acting upon the log than gravity (which acts only in the vertical direction) in the horizontal direction, velocity must be constant at any point.So, before it entered the water, its velocity has two components, as follows:vₓ = 7.5 m/svy = -8.9 m/sWe can find the magnitude of the velocity vector, just applying Pythagorean Theorem to both components, as follows:[tex]v = \sqrt{v_{x}^{2} + v_{y}^{2}} = \sqrt{((7.5m/s)^{2} + (-8.9m/s)^{2}[/tex]
⇒ v = 11.6 m/s
The tendon from Lissa’s knee extensor muscles attaches to the tibia bone 1.5 in. (4 cm) below the center of her knee joint, and her foot is 15 in. (38 cm) away from her knee joint. What arc length does Lissa’s foot move through when her knee extensor muscles contract and their point of insertion on the tibia moves through an arc length of 2 in. (5 cm)?
Answer:
s₂ = 0.475 m = 47.5 cm
Explanation:
The arc length and the angle of rotation are related through the formula:
s = rθ
where,
s = arc length
r = radius of curvature
θ = angle of rotation
First, we consider the arc length covered by the point of insertion of extensor muscles.
s₁ = r₁θ
where,
s₁ = arc length covered by insertion of extensor muscle = 5 cm
r₁ = length of insertion from knee = 4 cm
θ = Angle of Rotation = ?
Therefore,
5 cm = (4 cm)(θ)
θ = (5 cm)/(4 cm)
θ = 1.25 rad
Now, we consider the arc length covered by the foot.
s₂ = r₂θ
where,
s₂ = arc length covered by the foot = ?
r₂ = distance from knee to foot = 38 cm = 0.38 m
The angle of rotation will be the same for the foot as the insertion.
Therefore,
s₂ = (0.38 m)(1.25 rad)
s₂ = 0.475 m = 47.5 cm
How does the temperature affect the amount of weathering
A segment of wire of total length 3 m carries a 10 A current and is formed into a semicircle. Determine the magnitude of the magnetic field (in micro-tesla) at the center of the circle along which the wire is placed.
Answer:
[tex]B=3.29\ \mu T[/tex]
Explanation:
Given that,
Length of a wire is 3 m
Current in the wore, I = 10 A
The wire formed a semicircle.
We need to find the magnitude of the magnetic field (in micro-tesla) at the center of the circle along which the wire is placed.
The magnetic field at the center of semicircle is given by :
[tex]B=\dfrac{\mu_o I}{4R}[/tex]
R is radius of semicircle,
[tex]R=\dfrac{3}{\pi}=0.954\ m[/tex]
So,
[tex]B=\dfrac{4\pi \times 10^{-7}\times 10I}{4(0.954)}\\\\B=3.29\times 10^{-6}\ T\\\\B=3.29\ \mu T[/tex]
So, the magnitude of the magnetic field is [tex]3.29\ \mu T[/tex].
When trying to stop the train, Spiderman applies a 4000 N force over 200 seconds. What is his impulse on the train?
Answer:
Initial speed of train = 35.76 m/s (80 mph)
I figured out the speed of the train (which equals 80mph) by viewing the movie. In one of the scenes, it is shown that the train reaches 80mph after Doc Ock desroys the break on the train.
1mph = 0.44704mps
80(0.44704) = 35.7632
= 35.76m/s
vf = 0m/s
vi = 35.76m/s
Mass of train = approximately 137,100 kg (302,254 lbs) (found in research)
Time taken to stop train = approximately 46 seconds
a = vf - vi
t
= 0 - 35.76
46
= -.7773913043
= -.78 m/s^2
(The number is negative since the accelerarion of the train is slowing down)
According to Newton's Second Law, a = f/m. When rearranged, the formula reads f = m(a). As shown in the free body diagram, the weight of the train equals the force due to gravity.
(force = tension)
F = m(a)
= 137,100(-.78)
= -106938 N
106938 N is the calculated tension of Spider-Man's webs
I also calculated the distance it took for Spider-Man to stop the train:
Approximate distance traveled by train:
d = 1/2(a)(t)^2
= 1/2(-.78)(46)^2
= 825.24 m
In order for Spider-Man to have been able to completely stop the train before falling off the edge of the tracks, I found that the tension of his webs must have equaled 106938N.
In order to determine the tension of Spider-Man's webs, I needed to know the mass of the train. Since the movie was unclear on what type of trian was used, I researched online to find the mass of the train. Knowing the mass of the train, i was then able to calculate the tension of Spider-Man's webs and the distance the train took to stop.
Explanation:
The most distant galaxy we have observed is more than
13.2 billion light-years away.
What does this evidence indicate?
e
A. That the universe is less than 13.2 billion years old
B. That the universe will exist for at least 13.2 billion more years
C. That the universe is at least 13.2 billion years old
D. That the universe is infinitely large and has always existed
Answer:
That the universe is at least 13.2 billion years old
Explanation:
The most distant galaxy we have observed is more than 13.2 billion light-years away. this evidence indicates. That the universe is at least 13.2 billion years old
What is the universe?
The universe is the sum total of all existent matter and space. The universe is estimated to be at least 10 billion light-years across and contains an enormous number of galaxies;
It has been expanding since the Big Bang some 13 billion years ago.
The most distant galaxy we have observed is more than 13.2 billion light-years away. this evidence indicates. That the universe is at least 13.2 billion years old
Hence option c is correct.
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Gray used a pulley to lift a 300 N object a distance of 3 m. It took Gray 30 seconds to lift the object. How much WORK did Gray exert?
Answer:
900J
Explanation:
The Work done can be calculated using
Work done = force × distance
According to this question, force = 300N, distance = 3m
Work done = 300 × 3
Work done = 900J
Gray exerted 900J of work
Calculate the peak voltage (in Volts) of a generator that rotates its 200-turn, 0.100 m diameter coil at 3.5 x 103 rpm in a 0.7 T field. You should round your answer to the nearest integer, do not include unit.
Angular velocity of the coil :
[tex]\omega=\dfrac{2\pi n}{60}[/tex]
[tex]\omega=\dfrac{2\times \pi\times 3500}{60}\\\\\omega =366.52\ rad/s[/tex]
Now,
Peak voltage is given by :
[tex]V=NAB\omega\\\\V= 200\times \dfrac{\pi (0.1)^2}{4}\times 0.7\times 366.52\\\\V=403.01\ V[/tex]
Hence, this is the required solution.
If you pour liquid into a tall, narrow glass, you may hear sound with a steadily rising pitch. What is the source of the sound
Answer:
the vibration of the glass tube creating sound waves within itself. reason is the rising pitch is that the liquid rise shortens the length of the vibrating area within the tube.
Explanation:
In a class experiment to determine information about free-fall acceleration, a watermelon and a pumpkin are each set to fall from the back of the stands at your football stadium.
a. If the watermelon and the pumpkin are both dropped at the same time, which one will hit the ground first?
b. If the watermelon is thrown downward with an initial speed of 10 m/s and the pumpkin is dropped, which one will hit the ground first? Show calculation to support your answer.
c. If it takes the watermelon 1 second to reach the ground when it is thrown downward at 10 m/s, how tall are the stands?
d. How long does it take the pumpkin to reach the ground if it is dropped from this height you calculated in part c?
Answer:
a) They both hit the ground at the same time
b) Watermelon will hit first, since its speed is faster than the pumpkin speed
c) h = 29,6 m
d)t = 3,02 sec
Explanation:
Equations for fall free movement are:
vf = v₀ + g*t when v₀ = 0 (dropped case) vf = g*t
h = v₀*t + 1/2*g*t²
a) For both ( watermelon and pumpkin) the equation of speed is the same:
vf = g*t² Both will have the same speed second through second
They both hit the ground at the same time
b) Now is watermelon is thrown with v₀ = 10 m/s
Watermelon will hit first since its speed is faster than the pumpkin speed
vf(watermelon) = 10 + g*t
vf₂ (pumpkin) = g*t
c) h = v₀*t + (1/2)*g*t²
h = (10)*1 + (1/2)*9,8*1
h = 10 + 19,6
h = 29,6 m
d) h = g*t
t = 29,6/9,8
t = 3,02 sec
Based on these photographs, how do Mars's surface and atmosphere differ from Earth's?
Answer:
Explanation:
Mars does not appear to have any water, while Earth is mostly water. Mars also has yellow-brown sky (pink-brown at sunrise and sunset), which suggests that it has a different atmospheric makeup.
Answer:
Mars does not appear to have any water, while Earth is mostly water. Mars also has yellow-brown sky (pink-brown at sunrise and sunset), which suggests that it has a different atmospheric makeup.
Explanation:
A uniform string of length 2.5 m and mass 0.01 kg is placed under a tension of 10 N. (5 pts) A) What is the frequency of the fundamental mode
Answer:
The fundamental frequency is 10 Hz
Explanation:
The frequency of the fundamental mode of the string is given by;
[tex]f_o =\frac{1}{2l}\sqrt{\frac{T}{\mu }[/tex]
where;
L is length of the string = 2.5 m
T is tension of the string, = 10 N
μ is mass per unit length = m / L = 0.01 / 2.5 = 0.004 kg/m
Substitute the given values and solve for the frequency;
[tex]f_o =\frac{1}{2l}\sqrt{\frac{T}{\mu }}\\\\ f_o =\frac{1}{2*2.5}\sqrt{\frac{10}{0.004}}\\\\ f_o = 0.2\sqrt{2500}\\\\ f_o = 0.2 *50\\\\f_o = 10 \ Hz[/tex]
Therefore, the fundamental frequency is 10 Hz
A 75.0 kg skier slides down a 75.0 m high slope without friction. The velocity of the skier at the bottom of the slope is
Answer:
v = 38.34 m
Explanation:
Given that,
Mass of a skier, m = 75 kg
It falls down a 75.0 m high slope without friction, h = 75 m
We need to find the velocity of the skier at the bottom of the slope. It is based on the concept of conservation of energy. So, v is the required velocity. So,
[tex]mgh=\dfrac{1}{2}mv^2\\\\v=\sqrt{2gh} \\\\v=\sqrt{2\times 9.8\times 75} \\\\v=38.34\ m/s[/tex]
So, the velocity of the skier at the bottom of the slope is 38.34 m.
A 40.0 kg box is being pushed across the floor to the right with 200 N of applied force. There is 120 N of frictional force opposing the motion.
What is the net force acting on the box?
What is the acceleration of the box?
A mass that has a force of the 10N to the right, 7N up, 3N down, and 4N to the left. What is the net external force on that mass? If the mass is 2kg find the acceleration.
Two points are on a disk that rotates about an axis perpendicular to the plane of the disk at its center. Point B is 3 times as far from the axis as point A. If the linear speed of point B is V, then the linear speed of point A is:_______
a. 9V.
b. 3V.
c. V.
d. V/3.
e. V/9.
Answer:
D. V/3.
Explanation:
V = rw
At point a
Linear velocity
= Va = rw
At point b
We have,
V = 3rw
So linear velocity of a
= V/Va = 3rw/re
I cross multiples and after which I got Va = v/3
Please view attachment
Answer:
The linear speed of point A is [tex]\frac{V}{3}[/tex] (Option D)Explanation:
The relation between linear and angular velocity,
[tex]V = rw[/tex]
Now, for point A the linear velocity is
[tex]V_A = rw[/tex]
And for point B
[tex]V = 3rw[/tex]
Therefore, the linear velocity of the point A
[tex]\frac{V}{V_A} = \frac{3rw}{rw}\\\\V_A = \frac{V}{3}[/tex]
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What would be the luminosity of the Sun if its surface temperature were 4000 K and its radius were 2.0 AU ?
Answer:
0.04558
Explanation:
Luminosity is defined as the total amount of energy a star can produce in a second. Luminosity is given by the formula
σ * A * T⁴
where;
σ = Stefan-Boltzmann constant
= 5.670367 * 10⁻⁸
A= Surface Area
=4πr²
T= Temperature in Kelvin
In the question,
Temperature= 4000K
Radius= 2.0 A.U
Substituting the variables into the equation,
0.000000056704 * 4(3.14)(2.0²) * 4000⁴
= 0.000000056704 * 50.24 * 16000
= 0.04558
3. A 0.14-kg ball traveling with a speed of 43 m/s is brought to rest in a catcher's mitt.
Determine the magnitude of the impulse exerted by the mitt on the ball.
Answer:
change of momentum
= mass * change in velocity
= 0.14 * (36 - 0)
= 5.04 kg m/s ---answer
Explanation:
Which is one way that ultrasound technology may be used?
O repairing fractured bones
O locating misplaced electrical equipment
O cleaning dirty surgical tools
O canceling out unwanted noise
Answer:
Locating misplaced electrical equipment.
Explanation:
Ultrasound is primarily used by doctors to locate fractured bones or damaged organs, so I assume it can be used to locate other things too.
A scout hikes 2.35KM east of a campsite he takes a break for lunch and then hikes another 1.25 KM north of the location where he ate lunch. what distance is the scout from the campsite
Answer:
mk7000000.0000000000