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LiRa [457]
1 year ago
9

A 50 kg woman, riding on a 10 kg cart, is moving east at 5.0 m/s. the woman jumps off the cart and hits the ground running at 7.

0 m/s, eastward, relative to the ground. calculate the velocity of the cart after she jumps off.
Physics
1 answer:
pav-90 [236]1 year ago
4 0
To answer this question, we will use the law of conservation of momentum which states that:
(m1+m2)Vi = m1V1 + m2V2 where:
m1 is the mass of the woman = 50 kg
m2 is the mass of the cart = 10 kg
Vi is the initial velocity (of woman and cart combined) = 5 m/sec
V1 is the final velocity of the woman = 7 m/sec
V2 is the final velocity of the cart that we need to calculate

Substitute with the givens in the above equation to get the final velocity of the cart as follows:
(50+10)(5) = (50)(7) + (10)V2
10V2 = -50
V2 = -5 m/sec
Note that the negative sign indicates that the cart is moving in an opposite direction to the others.
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Which statement about energy conservation BEST explains why a bouncing basketball will not remain in motion forever?
bearhunter [10]

Answer: d

Explanation:

7 0
1 year ago
1)After catching the ball, Sarah throws it back to Julie. However, Sarah throws it too hard so it is over Julie's head when it r
DENIUS [597]

Answer:

1)

v_{oy}=11.29\ m/s

2)

y=7.39\ m

Explanation:

<u>Projectile Motion</u>

When an object is launched near the Earth's surface forming an angle \theta with the horizontal plane, it describes a well-known path called a parabola. The only force acting (neglecting the effects of the wind) is the gravity, which acts on the vertical axis.

The heigh of an object can be computed as

\displaystyle y=y_o+V_{oy}t-\frac{gt^2}{2}

Where y_o is the initial height above the ground level, v_{oy} is the vertical component of the initial velocity and t is the time

The y-component of the speed is

v_y=v_{oy}-gt

1) We'll find the vertical component of the initial speed since we have not enough data to compute the magnitude of v_o

The object will reach the maximum height when v_y=0. It allows us to compute the time to reach that point

v_{oy}-gt_m=0

Solving for t_m

\displaystyle t_m=\frac{v_{oy}}{g}

Thus, the maximum heigh is

\displaystyle y_m=y_o+\frac{v_{oy}^2}{2g}

We know this value is 8 meters

\displaystyle y_o+\frac{v_{oy}^2}{2g}=8

Solving for v_{oy}

\displaystyle v_{oy}=\sqrt{2g(8-y_o)}

Replacing the known values

\displaystyle v_{oy}=\sqrt{2(9.8)(8-1.5)}

\displaystyle v_{oy}=11.29\ m/s

2) We know at t=1.505 sec the ball is above Julie's head, we can compute

\displaystyle y=y_o+V_{oy}t-\frac{gt^2}{2}

\displaystyle y=1.5+(11.29)(1.505)-\frac{9.8(1.505)^2}{2}

\displaystyle y=1.5\ m+16,991\ m-11.098\ m

y=7.39\ m

5 0
2 years ago
A 70 kg student jumps down to form a 1 m high platform. She forgets to bend her knees and her downward motion stops in 0.02 seco
34kurt

Answer:

15,505 N

Explanation:

Using the principle of conservation of energy, the potential energy loss of the student equals the kinetic energy gain of the student

-ΔU = ΔK

-(U₂ - U₁) = K₂ - K₁ where U₁ = initial potential energy = mgh , U₂ = final potential energy = 0, K₁ = initial kinetic energy = 0 and K₂ = final kinetic energy = 1/2mv²

-(0 - mgh) = 1/2mv² - 0

mgh = 1/2mv² where m = mass of student = 70kg, h = height of platform  = 1 m, g = acceleration due to gravity = 9.8 m/s² and v = final velocity of student as he hits the ground.

mgh = 1/2mv²

gh = 1/2v²

v² = 2gh

v = √(2gh)

v = √(2 × 9.8 m/s² × 1 m)

v = √(19.6 m²/s²)

v = 4.43 m/s

Upon impact on the ground and stopping, impulse I = Ft = m(v' - v) where F = force, t = time = 0.02 s, m =mass of student = 70 kg, v = initial velocity on impact = 4.43 m/s and v'= final velocity at stopping = 0 m/s

So Ft = m(v' - v)

F = m(v' - v)/t

substituting the values of the variables, we have

F = 70 kg(0 m/s - 4.43 m/s)/0.02 s

= 70 kg(- 4.43 m/s)/0.02 s

= -310.1 kgm/s ÷ 0.02 s

= -15,505 N

So, the force transmitted to her bones is 15,505 N

3 0
1 year ago
The human ear canal is, on average, 2.5cm long and aids in hearing by acting like a resonant cavity that is closed on one end an
Troyanec [42]

Answer:

3400 Hz

Explanation:

We know that

1 cm = 0.01 m

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V = Speed of sound = 340 ms⁻¹

f = First resonant frequency

The human ear canal behaves as a closed pipe and for a closed pipe, nth resonant frequency is given as

f = \frac{(2n - 1)V}{4L}

for first resonant frequency, we have n = 1

Inserting the values

f = \frac{(2(1) - 1) 340}{4(0.025)}

f = \frac{340}{4(0.025)}

f = 3400 Hz

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2 years ago
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Kobotan [32]
Any two-dimensional vector in cartesian (x,y) coordinates can be broken down into individual horizontal and vertical components using trigonometry. If a train goes up a hill with 15 degree incline at a speed of 22 m/s, the horizontal component is 22cos(15)=21.3 m/s and the vertical component is 22sin(15)=5.5 m/s. 
8 0
1 year ago
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