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2 years ago

9

Suppose the Pathfinder moves at a rate of 0.2m/s for 20 seconds and then turns around and travels at the same speed for 3 second

s, what is the Pathfinder's current position?

1 answer:

kotykmax [81]2 years ago

7 0

Answer:

PLEASE BE MORE SPECIFIC

Explanation:

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A comet is in an elliptical orbit around the Sun. Its closest approach to the Sun is a distance of 4.7 1010 m (inside the orbit

Lubov Fominskaja [6]

Answer:

58515.9 m/s

Explanation:

We are given that

$d_1=4.7\times 10^{10} m$

$v_i=9.5\times 10^4 m/s$

$d_2=6\times 10^{12} m$

We have to find the speed (vf).

Work done by surrounding particles=W=0 Therefore, initial energy is equal to final energy.

$K_i+U_i=K_f+U_f$

$\frac{1}{2}mv^2_i-\frac{GmM}{d_1}=\frac{1}{2}mv^2_f-\frac{GmM}{d_2}$

$\frac{1}{2}v^2_i-\frac{GM}{d_1}+\frac{GM}{d_2}=\frac{1}{2}v^2_f$

$v^2_f=2(\frac{1}{2}v^2_i-\frac{GM}{d_1}+\frac{GM}{d_2})$

$v_f=\sqrt{2(\frac{1}{2}v^2_i-\frac{GM}{d_1}+\frac{GM}{d_2})}$

Using the formula

$v_f=\sqrt{v^2_i+2GM(\frac{1}{d_2}-\frac{1}{d_1})}$

$v_f=\sqrt{(9.5\times 10^4)^2+2\times 6.7\times 10^{-11}\times 1.98\times 10^{30}(\frac{1}{6\times 10^{12}}-\frac{1}{4.7\times 10^{10})}$

Where mass of sun= $M=1.98\times 10^{30} kg$

$G=6.7\times 10^{-11}$

$v_f=58515.9 m/s$

4 0

3 years ago

A playground merry-go-round has a mass of 50 kg and a diameter of 4.0 m. There are 4 children who want to ride on it. They have

mixer [17]

Answer:

B) I1 = 1680 kg.m^2 I2 = 1120 kg.m^2

C) V = 0.84m/s T = 29.92s

D) ω2 = 0.315 rad/s

Explanation:

The moment of inertia when they are standing on the edge:

$I1 = 1/2*M*R^2 + (m1+m2+m3+m4)*R^2$ where M is the mass of the merry-go-round.

I1 = 1680 kg.m^2

The moment of inertia when they are standing half way to the center:

$I2 = 1/2*M*R^2 + (m1+m2+m3+m4)*(R/2)^2$

I2 = 1120 kg.m^2

The tangencial velocity is given by:

V = ω1*R = 0.84m/s

Period of rotation:

T = 2π / ω1 = 29.92s

Assuming that there is no friction and their parents are not pushing anymore, we can use conservation of the angular momentum to calculate the new angular velocity:

I1*ω1 = I2*ω2 Solving for ω2:

ω2 = I1*ω1 / I2 = 0.315 rad/s

5 0

3 years ago

A sample of aluminum has a mass of 4.86 grams and a volume of 1.8 cubic centimeters. What is the density of aluminum rounded to

TEA [102]

Answer:

3g/cm³

Explanation:

2.7

The seven adds 1 to the 2 making it 3.0

6 0

3 years ago

A ball rolling down a hill for 9 seconds accelerates from 3 m/s to 34.5 m/s. What is the ball's rate of acceleration?

Delicious77 [7]

Answer:njjikkmokkmmm

Explanation:

4 0

3 years ago

2 kilogram bowling ball sits on top of a building that is 40 m the object has k e or g p e or both be sure to include the correc

Licemer1 [7]

GPE=mass*GFS*height
2kg*9.8N/Kg*40m -I've used the Grabitational field strength of the earth's
surface

=784J

7 0

3 years ago