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4 years ago

A tennis ball is dropped from 1.65 m above

Physics

1 answer:

dexar [7]4 years ago

7 0

Answer:

The velocity with which the ball strikes the ground = -5.7 m/s

Explanation:

To find the velocity with which the tennis ball hits the ground, we only need to worry about what happens up to that point. We can ignore the rebound for this part. Given:

d = -1.65

a = -9.8

vi = 0

vf = ?

$v_f^2 = v_i^2 + 2ad\\v_f^2=0+2(-9.8)(-1.65)\\v_f^2=32.34\\v_f=-5.7$

*Keep in mind that the square root gives us two answers, a positve and a negative one. We use the negative one here because the final speed is downwards and the question says down is negative.

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023 (part 1 of 2) 10.0 points

Annette [7]

Answer:

Part 1

The angular speed is approximately 1.31947 rad/s

Part 2

The change in kinetic energy due to the movement is approximately 675.65 J

Explanation:

The given parameters are;

The rotation rate of the merry-go-round, n = 0.21 rev/s

The mass of the man on the merry-go-round = 99 kg

The distance of the point the man stands from the axis of rotation = 2.8 m

Part 1

The angular speed, ω = 2·π·n = 2·π × 0.21 rev/s ≈ 1.31947 rad/s

The angular speed is constant through out the axis of rotation

Therefore, when the man walks to a point 0 m from the center, the angular speed ≈ 1.31947 rad/s

Part 2

Given that the kinetic energy of the merry-go-round is constant, the change in kinetic energy, for a change from a radius of of the man from 2.8 m to 0 m, is given as follows;

$\Delta KE_{rotational} = \dfrac{1}{2} \cdot I \cdot \omega ^2 = \dfrac{1}{2} \cdot m \cdot v ^2$

I = m·r²

Where;

m = The mass of the man alone = 99 kg

r = The distance of the point the man stands from the axis, r = 2.8 m

v = The tangential velocity = ω/r

ω ≈ 1.31947 rad/s

Therefore, we have;

I = 99 × 2.8² = 776.16 kg·m²

$\Delta KE_{rotational}$ = 1/2 × 776.16 kg·m² × (1.31947)² ≈ 675.65 J

3 0

3 years ago

Jonah rode his skateboard 350 meters down a straight road at a constant velocity. He skated

Marysya12 [62]

Answer:

250 m/min down the road

Explanation:

Velocity is equivalent to speed but it considers the direction of the object. Velocity is also calculated by dividing the distance travelled by time. Therefore, $v=\frac {d}{t}$ where d and t are distance and time respectively. Given that d is given as 350 m and t is 1.4 s then by substitution $v=\frac {350}{1.4}=250 m/min$ and the direction is down the road.

Velocity is 250 m/min down the road

8 1

3 years ago

A dentist’s drill starts from rest. After 3.20 s of constant angu-lar acceleration, it turns at a rate of 2.51 3 104 re v/m i n.

Gekata [30.6K]

Answer:

$ΔTita = 4205.6 rad$

Explanation:

$w_{i}$ means initial angular velocity, which is 0 rev/min

$w_{f}$ means final angular velocity, which is $2.513*10^{4}rev/min$

t means time t= 3.20 s

one revolution is equivalent to 2πrad so the final angular velocity is:

$w_{f}$ = (2π/60) *2.513*10^{4} rad/s

$w_{f}$ = 2628.5 rad/s

so the angular acceleration, α will be:

α = 2628.5 rad/s / 3.20 s

$a = 821.40 rad/s^{2}$

so the rotational motion about a fixed axis is:

$w^{2} _{f} =w^{2} _{i}$ + 2αΔTita where ΔTita is the angle in radians

so now find the ΔTita the subject of the formula

ΔTita = $\frac{w^{2} _{f}-w^{2} _{i} }{2a}$

$ΔTita = ((2628.5)^{2} - (0 rev/min)^{2}) / 2* 821.40$

$ΔTita = 4205.6 rad$

7 0

4 years ago

Read 2 more answers

A charged particle enters into a uniform magnetic field such that its velocity vector is perpendicular to the magnetic field vec

Anit [1.1K]

Answer:

The charged particle will follow a circular path.

Explanation:

The magnetic force exerted on the charged particle due to the magnetic field is given by:

$F=qvB sin \theta$

where

q is the charge

v is the velocity of the particle

B is the magnetic field

$\theta$ is the angle between v and B

In this problem, the velocity is perpendicular to the magnetic field, so $\theta = 90^{\circ}, sin \theta=1$ and the force is simply

$F=qvB$

Moreover, the force is perpendicular to both B and v, according to the right-hand rule. Therefore, we have:

- a force that is always perpendicular to the velocity, v

- a force which is constant in magnitude (because the magnitude of v or B does not change)

--> this means that the force acts as a centripetal force, so it will keep the charged particle in a uniform circular motion. So, the correct answer is

The charged particle will follow a circular path.

8 0

4 years ago

An object initially at rest accelerates at 5 meters per second2 until it attains a speed of 30 meters per second. What distance

Tcecarenko [31]

Answer: 90m

Explanation:

Use Equation for distance:

S=a*t²/2

use eqation for acceleraton a=(Vf-Vs)/t

Vs- starting speed

Vf - final speed

a-accelaration

---------------------------------

a=5m/s²

Vs=0m/s

Vf=30m/s

use

a=(Vf-Vs)/t

to find time

t=(Vf-Vs)/a

t=30m/s/5m/s²

t=6s

Now calculate distance that object travel using:

S=(a*t²)72

s=(5m/s²*(6s)²)/2

S=90m

4 0

3 years ago

Read 2 more answers