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krek1111 [17]
2 years ago
10

A rocket lifts a payload upward from the surface of Earth. The radius of Earth is R, and the weight of the payload on the surfac

e of Earth is W. The force of Earth's gravity on the payload is W/2 when the rocket's distance from the center of the Earth is
A. R
B. sqrt(2)R
C. 2R
D. 2sqrt(2)R
E. 4R
Physics
1 answer:
topjm [15]2 years ago
4 0

Answer:

the correct answer is B      r =√2   R

Explanation:

To solve this exercise we will use the force of gravity, at two points when the rocket is on the platform and when it is at the desired height.

Rocket with cargo on the platform

              F_{g} = G \frac{ m M }{R^2} = W

where m is the mass of the rocket and the charge, M the mass of the earth and R the distance from the center of the earth

Rocket loaded at desired height

              F_{g} = G \frac{m M}{r^2 } = \frac{W}{2}

we write our two equations

        W = g m M / R²

        \frac{W}{2} = G m M / r²

we match to solve them

         Gm M / R² = 2 Gm M / r²

         1 / R² = 2 / r²

          r =√2   R

therefore the correct answer is B

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Two swimmers begin a race and Swimmer A completes each length of the pool in 30 seconds, while Swimmer B completes each length i
Anni [7]

Answer:1.33 times

Explanation:

Given

A takes 30 sec to complete each length

While B takes 40 sec to complete each length

Let L be the Length of Swimming pool Length

and V_a and V_b be the length of A & B

Thus after 5 minutes A traveled

x_a=\frac{L}{30}\times 5\times 60=10L

x_b=\frac{L}{40}\times 5\times 60=7.5L

so A is 2.5L units ahead of B

A has traveled 1.33 times than B in 5 mins

6 0
2 years ago
Find the angle (above the horizontal) at which a projectile achieves its maximum range, if y=y0.
KatRina [158]
The answer is 45 degrees. 
According to the Kinematics of projectile motion, if the purpose is to maximize range, optimum angle of landing is always 45 degrees.If the purpose is to maximize range & projection height is zero, the optimum angle of projection (and landing) is 45 degrees.
5 0
2 years ago
1. For each of the following scenarios, describe the force providing the centripetal force for the motion: a. a car making a tur
GenaCL600 [577]

Complete Question

For each of the following scenarios, describe the force providing the centripetal force for the motion:

a. a car making a turn

b. a child swinging around a pole

c. a person sitting on a bench facing the center of a carousel

d. a rock swinging on a string

e. the Earth orbiting the Sun.

Answer:

Considering a

    The force providing the centripetal force is the frictional force on the tires \

          i.e  \mu mg  =  \frac{mv^2}{r}

    where \mu is the coefficient of static friction

Considering b

   The force providing the centripetal force is the force experienced by the boys  hand on the pole

Considering c

     The force providing the centripetal force is the normal from the bench due to the boys weight

Considering d

     The force providing the centripetal force is the tension on the string

Considering e

      The force providing the centripetal force is the force of gravity between the earth and the sun

Explanation:

6 0
2 years ago
Capillary waves travel what than long waves
7nadin3 [17]
Faster than. Hope this helps!!!
6 0
2 years ago
Read 2 more answers
The drawing shows an adiabatically isolated cylinder that is divided initially into two identical parts by an adiabatic partitio
Sveta_85 [38]

Answer:

temperature on left side is 1.48 times the temperature on right

Explanation:

GIVEN DATA:

\gamma = 5/3

T1 = 525 K

T2 = 275 K

We know that

P_1 = \frac{nRT_1}{v}

P_2 = \frac{nrT_2}{v}

n and v remain same at both side. so we have

\frac{P_1}{P_2} = \frac{T_1}{T_2} = \frac{525}{275} = \frac{21}{11}

P_1 = \frac{21}{11} P_2 ..............1

let final pressure is P and temp  T_1 {f} and T_2 {f}

P_1^{1-\gamma} T_1^{\gamma} = P^{1 - \gamma}T_1 {f}^{\gamma}

P_1^{-2/3} T_1^{5/3} = P^{-2/3} T_1 {f}^{5/3} ..................2

similarly

P_2^{-2/3} T_2^{5/3} = P^{-2/3} T_2 {f}^{5/3} .............3

divide 2 equation by 3rd equation

\frac{21}{11}^{-2/3} \frac{21}{11}^{5/3} = [\frac{T_1 {f}}{T_2 {f}}]^{5/3}

T_1 {f} = 1.48 T_2 {f}

thus, temperature on left side is 1.48 times the temperature on right

6 0
2 years ago
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