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devlian [24]
3 years ago
12

Over a time interval of 1.99 years, the velocity of a planet orbiting a distant star reverses direction, changing from +20.7 km/

s to -22.0 km/s. Find (a) the total change in the planet's velocity (in m/s) and (b) its average acceleration (in m/s2) during this interval. Include the correct algebraic sign with your answers to convey the directions of the velocity and the acceleration.
Physics
1 answer:
madam [21]3 years ago
8 0

Answer:

(a) - 42700 m/s

(b) - 6.8 x 10^-4 m/s^2

Explanation:

initial velocity of star, u = 20.7 km/s

Final velocity of star, v = - 22 km/s

time, t = 1.99 years

Convert velocities into m/s and time into second

So, u = 20700 m / s

v = - 22000 m/s

t = 1.99 x 365.25 x 24 x 3600 = 62799624 second

(a) Change in planet's velocity = final velocity - initial velocity

  = - 22000 - 20700 = - 42700 m/s

(b) Accelerate is defined as the rate of change of velocity.

Acceleration = change in velocity / time

                     = ( - 42700 ) / (62799624) = - 6.8 x 10^-4 m/s^2

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goldenfox [79]

Answer:

660V

Explanation:

V=IR

V=?, I=11A,R=60w

V=60 ×11

=660V

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a rocket, initially at rest, steadily gains speed at a rate of 13.0m/s^2 for 6.40 during take-off. How far did the rocket travel
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Two speedboats are traveling at the same speed relative to the water in opposite directions in a moving river. An observer on th
DENIUS [597]

Answer:

a) vboat = 5.95 m/s  b) vriver= 1.05 m/s

Explanation:

a) As observed from the shore, the speed of the boats can be expressed as the vector sum, of the boat speed relative to the water and the river speed relative to the shore, as follows:

vb₁s = vb₁w + vrs

In one case, the boat is moving in the same direction as the water:

vb₁s = vb₁w + vrs = 7.0 m/s (1)

For the other boat, it is clear that is moving in an opposite direction:

vb₂s = vb₂w - vrs = 4.9 m/s (2)

As  we know that vb₁w = vb₂w, adding both sides, we can remove the river speed from the equation, as follows:

vb₁w = vb₂w =  \frac{7.0 m/s + 4.9 m/s}{2} =5.95 m/s

b) Replacing this value in (1) and solving for vriver, we have:

vriver = 7.0 m/s - 5.95 m/s = 1.05 m/s

(we could have arrived to the same result subtracting both sides in (1), and (2))

3 0
3 years ago
A rope of negligible mass supports a block that weighs 30.0 N. as shown below. The maximum tension the rope can support without
Mars2501 [29]

Answer:

a = 6.7 m/s²

Explanation:

The formula to apply here is :

Mass * acceleration = Tension -weight

m*a = Ft -mg  -----where m is mass of block, a is acceleration , Ft  is force due to tension.

Given in the question that ;

Mass of block = 3 kg

Ft = 50 N

g= 10

m*a = Ft -mg

3*a = 50-30

3a= 20

a= 20/3

a= 6.7 m/s²

8 0
3 years ago
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