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tino4ka555 [31]

2 years ago

15

Deimos, a moon of Mars, is about 12 km in diameter with mass 1.5×1015 kg. Suppose you are stranded alone on Deimos and want to p

lay a one-person game of baseball. You would be the pitcher, and you would be the batter!
Part A

With what speed would you have to throw a baseball so that it would go into a circular orbit just above the surface and return to you so you could hit it?

Express your answer with the appropriate units.

Part C

How long (in hours) after throwing the ball should you be ready to hit it?

I am stuck on how to solve this problem. Thanks!

1 answer:

yulyashka [42]2 years ago

3 0

Answer:

a) v = 4.08 m / s , b) T = 2.58 h

Explanation:

For this exercise, let's start by using Newton's second law where force is the attraction of gravity

F = m a

Force is

F = G m $M_{D}$ / r²

The acceleration is centripetal

a = v² / r

Let's replace

G m $M_{D}$ / r² = m v² / r

G $M_{D}$ / r = v²

v = √ G $M_{D}$ / r

The radius is half the diameter

r = d / 2

r = 12 10³/2 = 6.0 10³ m

v = √ (6.67 10⁻¹¹ 1.5 10¹⁵/6 10³

v = √ 16,675

v = 4.08 m / s

If you throw the ball with this speed, take a full turn

c) The constant speed module only changes the direction, so we can use the relationship

v = d / t

The time is

t = d / v

The distance along the circle and this time is called Period

d = 2π r

d = 2π 6 10³

d = 37,70 10³ m

Let's calculate

T = 37.70 10³ / 4.08

T = 9.24 10³ s

Let's reduce to hours

T = 9.24 10³ s (1h / 3600 s)

T = 2.58 h

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Case 1: A DJ starts up her phonograph player. The turntable accelerates uniformly from rest, and takes t₁ = 11.9 seconds to get

olga_2 [115]

Answer:

Part a)

$\omega = 8.17 rad/s$

Part b)

$N = 7.74 rev$

Part c)

$\alpha = 0.69 rad/s^2$

Part d)

$\alpha = 0.48 rad/s^2$

Part e)

$t = 9.14 s$

Explanation:

Part a)

Angular speed is given as

$\omega = 2\pi f$

$\omega = 2\pi(\frac{78}{60})$

$\omega = 8.17 rad/s$

Part b)

Since turn table is accelerating uniformly

so we will have

$\theta = \frac{\omega_f + \omega_i}{2} t$

$\theta = \frac{8.17 + 0}{2}(11.9)$

$2N\pi = 48.6$

$N = 7.74 rev$

Part c)

angular acceleration is given as

$\alpha = \frac{\omega_f - \omega_i}{t}$

$\alpha = \frac{8.17 - 0}{11.9}$

$\alpha = 0.69 rad/s^2$

Part d)

When its angular speed changes to 120 rpm

then we will have

$\omega_2 = 2\pi (\frac{120}{60})$

$\omega_2 = 12.56 rad/s$

number of turns revolved is 15 times

so we have

$\omega_f^2 - \omega_i^2 = 2 \alpha \theta$

$12.56^2 - 8.17^2 = 2\alpha (2\pi\times 15)$

$\alpha = 0.48 rad/s^2$

Part e)

now for uniform acceleration we have

$\omega_f - \omega_i = \alpha t$

$12.56 - 8.17 = 0.48 t$

$t = 9.14 s$

7 0

2 years ago

A solenoid of length 18 cm consists of closely spaced coils of wire wrapped tightly around a wooden core. The magnetic field str

Kisachek [45]

Answer:

$B_2 = 1.71 mT$

Explanation:

As we know that the magnetic field near the center of solenoid is given as

$B = \frac{\mu_0 N i}{L}$

now we know that initially the length of the solenoid is L = 18 cm and N number of turns are wounded on it

So the magnetic field at the center of the solenoid is 2 mT

now we pulled the coils apart and the length of solenoid is increased as L = 21 cm

so we have

$\frac{B_1}{B_2} = \frac{L_2}{L_1}$

now plug in all values in it

$\frac{2.0 mT}{B_2} = \frac{21}{18}$

$B_2 = 1.71 mT$

3 0

2 years ago

A tennis player serves a tennis ball such that it is moving horizontally when it leaves the racquet. When the ball travels a hor

nalin [4]

Answer:

$u_x=38.13\ m/s$

Explanation:

Given that initially ball moves in the horizontal direction ,it means that the velocity in the vertical direction is zero.

Horizontal distance = 13 m

Vertical distance = 57 cm

Lets take time to cover 57 cm distance in vertical direction is t.

We know that g is the constant acceleration in the vertical direction so we can apply the equation of motion in the vertical direction.

$S=u_yt+\dfrac{1}{2}gt^2$

Here $u_y=0$

S= 57 cm

$0.57=0\times t+\dfrac{1}{2}\times 9.81\times t^2$

t=0.34 s

Now in the horizontal direction

$x=u_xt$

Here x=13 m

t= 0.34 s

So

$13=u_x\times 0.34$

$u_x=38.13\ m/s$

So the initial speed of ball is 38.13 m/s.

7 0

2 years ago

A wave has a frequency of 15,500 Hz and a wavelength of 0.20 m. What is the

Hoochie [10]

Answer:

3100 m/s

Explanation:

The relationship between frequency and wavelength of a wave is given by the wave equation:

$v=f\lambda$

where

v is the speed of the wave

f is its frequency

$\lambda$ is the wavelength

For the wave in this problem,

f = 15,500 Hz

$\lambda=0.20 m$

Therefore, the wave speed is

$v=(15500)(0.20)=3100 m/s$

4 0

2 years ago

Suggest reasons why poaching for subsistence is likely to be less damaging to the biodiversity of an area than poaching for prof

dlinn [17]

An example for ruining a biodiversity is fishing. The two factors that have contributed to increased fishing in deep ocean waters in recent years are the human population growth and decreased fishing opportunities inshore. Increase population growth increases the demand for food which also leads to increase in fish demand. Because the fish demand is high, inshore fishing opportunities decrease that is why deep ocean waters is the new venue for fishing. This may sound absurd but poaching for subsistence is likely to be less damaging to he biodiversity <span>of an area than poaching for profit. Because the people do not care anymore to the biodiversity that they interrupted just to get back more profit. They do not care what must be taken from it like getting bigger fishes and leaving the smaller ones behind to maintain productivity.</span>

5 0

2 years ago