a force is something that causes change in motion. what provides the force that made the cart speed up?

A soft tennis ball is dropped onto a hard floor from a height of 1.50 m and rebounds to a height of 1.10 m. (a) Calculate its ve

Gemiola [76]

Answer:

(a) v = 5.42m/s

(b) vo = 4.64m/s

(c) a = 2874.28m/s^2

(d) Δy = 5.11*10^-3m

Explanation:

(a) The velocity of the ball before it hits the floor is given by:

$v=\sqrt{2gh}$ (1)

g: gravitational acceleration = 9.8m/s^2

h: height where the ball falls down = 1.50m

$v=\sqrt{2(9.8m/s^2)(1.50m)}=5.42\frac{m}{s}$

The speed of the ball is 5.42m/s

(b) To calculate the velocity of the ball, after it leaves the floor, you use the information of the maximum height reached by the ball after it leaves the floor.

You use the following formula:

$h_{max}=\frac{v_o^2}{2g}$ (2)

vo: velocity of the ball where it starts its motion upward

You solve for vo and replace the values of the parameters:

$v_o=\sqrt{2gh_{max}}=\sqrt{2(9.8m/s^2)(1.10m)}=4.64\frac{m}{s}$

The velocity of the ball is 4.64m/s

(c) The acceleration is given by:

$a=\frac{\Delta v}{\Delta t}=\frac{v_o-v}{3.50*10^{-3}s}=\frac{4.64m/s-(-5.42m)/s}{3.50*10^{-3}s}=2874.285\frac{m}{s^2}$

$a=\frac{\Delta v}{\Delta t}=\frac{v_o-v}{3.50*10^{-3}s}=\frac{4.64m/s-5.42m/s}{3.50*10^{-3}s}=-222.85\frac{m}{s^2}$

The acceleration of the ball is 2874.28/s^2

(d) The compression of the ball is:

$\Deta y=\frac{v^2}{2(a)}=\frac{(5.42m/s)^2}{2(2874.28m/s^2)}=5.11*10^{-3}m$

THe compression of the ball when it strikes the floor is 5.11*10^-3m

4 0

3 years ago

Unlike a roller coaster, the seats in a Ferris wheel swivel so that the rider is always seated upright. An 80-ft-diameter Ferris

telo118 [61]

Answer:

a

$F_A =425.42 \ N$

b

$F_A_H = 358.58 \ N$

Explanation:

From the question we are told that

The diameter of the Ferris wheel is $d = 80 \ ft = \frac{80}{3.281} = 24.383$

The period of the Ferris wheel is $T = 24 \ s$

The mass of the passenger is $m_g = 40 \ kg$

The apparent weight of the passenger at the lowest point is mathematically represented as

$F_A_L = F_c + W$

Where $F_c$ is the centripetal force on the passenger, which is mathematically represented as

$F_c =m * r * w^2$

Where $w$ is the angular velocity which is mathematically represented as

$w = \frac{2* \pi }{T}$

substituting values

$w = \frac{2* 3.142 }{24}$

$w = 0.2618 \ rad/s$

and r is the radius which is evaluated as $r = \frac{d}{2}$

substituting values

$r = \frac{24.383}{2}$

$r = 12.19 \ ft$

So

$F_c = 40 * 12.19* (0.2618)^2$

$F_c = 33.42 \ N$

W is the weight which is mathematically represented as

$W = 40 * 9.8$

$W = 392 \ N$

So

$F_A = 33.42 + 392$

$F_A =425.42 \ N$

The apparent weight of the passenger at the highest point is mathematically represented as

$F_A_H = W- F_c$

substituting values

$F_A_H = 392 - 33.42$

$F_A_H = 358.58 \ N$

4 0

3 years ago

What does the principal quantum number determine?

morpeh [17]

principal quantum number (n) represents the relative overall energy of each orbital

Hope this helps!

4 0

2 years ago

At what frequency should a 200-turn, flat coil of cross sectional area of 300 cm2 be rotated in a uniform 30-mT magnetic field t

stellarik [79]

Answer:

The frequency of the coil is 7.07 Hz

Explanation:

Given;

number of turn of the coil, N = 200 turn

area of the coil, A = 300 cm² = 0.03 m²

magnitude of magnetic field, B = 30 mT = 0.03 T

maximum value of induced emf, E = 8 V

The maximum induced emf in the coil is given by;

E = NBAω

E = NBA(2πf)

$f = \frac{E_{max}}{2\pi*NBA}$

where;

f is the frequency of the coil

$f = \frac{E_{max}}{2\pi*NBA}\\\\f = \frac{8}{2\pi(200)(0.03)(0.03)} \\\\f = 7.07 \ Hz$

Therefore, the frequency of the coil is 7.07 Hz

5 0

2 years ago

Using the data provided in the graph and the table, interpolate to find the potential energy (U) at a height (H) of 2.5 meters a

Assoli18 [71]

Science questions require you to calculate values without ... Calculation questions require you to find a specific value based on the figures provided. ... would happen past the edges of the graph or between values on a table. ... at 75 m, this gap in data is what makes this aninterpolation question!

3 0

3 years ago

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