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3 years ago

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Two ships of equal mass are 110 m apart. What is the acceleration of either ship due to the gravitational attraction of the othe

r? Treat the ships as particles and assume each has a mass of 39,000 metric tons. (Give the magnitude of your answer in m/s2.)

1 answer:

dusya [7]3 years ago

7 0

Answer:

Acceleration of the ship, $a=2.14\times 10^{-7}\ m/s^2$

Explanation:

It is given that,

Mass of both ships, $m=39000\ metric\ tons=39\times 10^6\ kg$

Distance between two ships, d = 110 m

The gravitational force between two ships is given by :

$F=G\dfrac{m^2}{d^2}$

$F=6.67\times 10^{-11}\ Nm^2/kg^2\times \dfrac{(39\times 10^6\ kg)^2}{(110\ m)^2}$

F = 8.38 N

Let a is the acceleration. Now, using second law of motion as :

$a=\dfrac{F}{m}$

$a=\dfrac{8.38\ N}{39\times 10^6\ kg}$

$a=2.14\times 10^{-7}\ m/s^2$

So, the acceleration of either ship due to the gravitational attraction of the other is $2.14\times 10^{-7}\ m/s^2$ . Hence, this is the required solution.

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Krista is playing tennis at the park. When the tennis ball flies toward her, Krista hits the ball with her racket, which causes

xenn [34]

Answer:

B. When the racket hits the tennis ball with a force, the tennis ball applies an equal but opposite force to the racket.

Explanation:

According to the Newton's third law of motion every action has equal and opposite reaction. So, when the force is applied by the racket on the ball then the ball also applies an equal intensity of force in the opposite direction on the racket. It is just that the the force on the racket is absorbed by the player holding it.

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3 years ago

Physics help please

zhuklara [117]

Answer: 37.981 m/s

Explanation:

This situation is related to projectile motion or parabolic motion, in which the travel of the ball has two components: <u>x-component</u> and <u>y-component.</u> Being their main equations as follows:

<u>x-component: </u>

$x=V_{o}cos\theta t$ (1)

Where:

$x=52 m$ is the point where the ball strikes ground horizontally

$V_{o}$ is the ball's initial speed

$\theta=0$ because we are told the ball is thrown horizontally

$t$ is the time since the ball is thrown until it hits the ground

<u>y-component: </u>

$y=y_{o}+V_{o}sin\theta t+\frac{gt^{2}}{2}$ (2)

Where:

$y_{o}=120m$ is the initial height of the ball

$y=0$ is the final height of the ball (when it finally hits the ground)

$g=-9.8m/s^{2}$ is the acceleration due gravity

Knowing this, let's start by finding $t$ from (2):

<u></u>

$0=y_{o}+V_{o}sin(0\°) t+\frac{gt^{2}}{2}$ (3)

$0=y_{o}+\frac{gt^{2}}{2}$

$t=\sqrt{\frac{-2 y_{o}}{g}}$ (4)

$t=\sqrt{\frac{-2 (120 m)}{-9.8m/s^{2}}}$ (5)

$t=4.948 s$ (6)

Then, we have to substitute (6) in (1):

$x=V_{o}cos(0\°) t$ (7)

And find $V_{o}$ :

$V_{o}=\frac{x}{t}$ (8)

$V_{o}=\frac{52 m}{4.948 s}$ (9)

$V_{o}=10.509 m/s$ (10)

On the other hand, since we are dealing with constant acceleration (due gravity) we can use the following equation to find the value of the ball's final velocity $V$ :

$V=V_{o} + gt$ (11)

$V=10.509 m/s + (-9.8 m/s^{2})(4.948 s)$ (12)

$V=-37.981 m/s$ (13) This is the ball's final velocity, and the negative sign indicates its direction is downwards.

However, we were asked to find the <u>ball's final speed</u>, which is the module of the ball's final vleocity vector. This module is always positive, hence the speed of the ball just before it strikes the ground is 37.981 m/s (positive).

5 0

2 years ago

If an all electric house uses 2400 kWh in a month, what will the amount of the bill for electricity be at a rate of 9.5 cents pe

alexira [117]

Answer:

<h2>$228</h2>

Explanation:

Given data

Total electricity consumption = 2400 kWh

Rate of consumption = 9.5 cent per kWh

converting cent to dollar we have. 9.5/100= $0.095

hence the bill is 2400*0.095= $228

6 0

3 years ago

Light travels approximately 982,080,000 ft/s, and one year has approximately 32,000,000 seconds. A light year is the distance li

Anastasy [175]

Answer: 3.142656 × 10^16 feet

Explanation: Given that the

Speed = 982,080,000 ft/s, and

Time = 32,000,000 seconds

The formula for speed is:

Speed = distance/ time

Make distance the subject of formula

Since the time is second in one year and speed is ft/s, substitute both into the formula

Distance = speed × time

distance = 982,080,000 × 32,000,000

Distance = 3.142656 × 10^16 feet.

The distance of one light year in feet is 3.142656 × 10^16

8 0

3 years ago

Un engrane que gira con una velocidad de 20 rad/s, es acelerado durante 5 segundos hasta alcanzar una velocidad de 35 rad/s

larisa [96]

Answer:

a) La aceleración angular es: $\alpha=2\: rad/s^{2}$

b) El engranaje gira 125 radianes.

c) El engranaje hara aproximadamente 20 revoluciones.

Explanation:

a)

La aceleración angular se define como:

$\alpha=\frac{\Delta \omega}{\Delta t}$

Donde:

Δω es la diferencia de velocidad angular (en otras palabras ω(final)-ω(inicial))
Δt es el tiempo en el que occure el cambio de velocidad angular

$\alpha=\frac{35-25}{5}$

$\alpha=2\: rad/s^{2}$

b)

El desplazamiento angular puede ser calculado usando la siguiente ecuación:

$\theta=\theta_{i}+\omega_{i}t+\frac{1}{2}\alpha t^{2}$

Aqui el angulo inicial es 0, por lo tanto.

$\theta=20(5)+\frac{1}{2}(2)(5)^{2}$

$\theta=125\: rad$

El engranaje gira 125 radianes.

c)

Lo que debemos hacer aquí es convertir radianes a revoluciones.

Recordemos que 2π rad = 1 rev

Entonces:

$\theta=125\: rad \times \frac{1\: rev}{2\pi\: rad}=19.89\: rev$

Por lo tanto el engranaje hara aproximadamente 20 revoluciones.

Espero te haya sido de ayuda!

6 0

3 years ago