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Ber [7]
3 years ago
12

For a projectile launched from the vertically from the surface the earth with initial velocity v0, the velocity of the projectil

e when at distance r from the center of the earth satisifes the equation:
v^2 = v0^2 + k^2(1/r - 1/R)

Here, k is a positive constant.
Write the differential equation governing the distance r(t)
Physics
1 answer:
UNO [17]3 years ago
3 0

Answer:

(dr/dt) = √(v₀² + k²/r - k²/R)

Explanation:

v² = v₀² + k²[(1/r) - (1/R)]

v = velocity of Body launched from the surface of the earth

v = initial velocity of body

k = a constant

r = distance from the centre of the earth

R = radius of the earth

v² = v₀² + k²/r - k²/R

v = √(v₀² + k²/r - k²/R)

But v = dr/dt

(dr/dt) = √(v₀² + k²/r - k²/R)

(dr/√(v₀² + k²/r - k²/R)) = dt

To go one step beyond and integrate the differential equation

∫ (dr/√(v₀² + k²/r - k²/R)) = ∫ dt

Integrating the left hand side from 0 to r and the right hand side from 0 to t

Note, v₀, k and R are all constants

(- 4r²/k²)[√(v₀² + k²/r - k²/R)] = t

√(v₀² + k²/r - k²/R) = (- k² t/4r²)

(v₀² + k²/r - k²/R) = (- k² t/4r²)²

(v₀² + k²/r - k²/R) = (k⁴t²/16r⁴)

(k⁴t²/16r⁴) + (k²/r) = [v₀² - (k²/R)]

k² [(k²t²/16r⁴) + (1/r)] = [v₀² - (k²/R)]

[(k²t²/16r⁴) + (1/r)] = [(v₀²/k²) - (1/R)]

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a car with a mass of 2,000 kilograms is moving around a circular curve at a uniform velocity of 25 meters per second. the curve
Ber [7]
Fc=mv^2/r so we get 

2000kg*(25m/s)^2/(80m)= 15625N of force 

hope this helps! Thank You!!

4 0
2 years ago
Suppose a baseball pitcher throws the ball to his catcher.
amm1812

a) Same

b) Same

c) Same

d) Throw the ball takes longer

e) F is larger when the ball is catched

Explanation:

a)

The change in speed of an object is given by:

\Delta v = |v-u|

where

u is the initial velocity of the object

v is the final velocity of the object

The change in speed is basically the magnitude of the change in velocity (because velocity is a vector, while speed is a scalar, so it has no direction).

In this problem:

- In situation 1 (pitcher throwing the ball), the initial velocity is

u = 0 (because the ball starts from rest)

while the final velocity is v, so the change in speed is

\Delta v=|v-0|=|v|

- In situation 2 (catcher receiving the ball), the initial velocity is now

u = v

while the final velocity is now zero (ball coming to rest), so the change in speed is

\Delta v =|0-v|=|-v|

Which means that the two situations have same change in speed.

b)

The change in momentum of an object is given by

\Delta p = m \Delta v

where

m is the mass of the object

\Delta v is the change in velocity

If we want to compare only the magnitude of the change in momentum of the object, then it is given by

|\Delta p|=m|\Delta v|

- In situation 1 (pitcher throwing the ball), the change in momentum is

\Delta p = m|\Delta v|=m|v|=mv

- In situation 2 (catcher receiving the ball), the change in momentum is

\Delta p = m\Delta v = m|-v|=mv

So, the magnitude of the change in momentum is the same (but the direction is opposite)

c)

The impulse exerted on an object is equal to the change in momentum of the object:

I=\Delta p

where

I is the impulse

\Delta p is the change in momentum

As we saw in part b), the change in momentum of the ball in the two situations is the same, therefore the impulse exerted on the ball will also be the same, in magnitude.

However, the direction will be opposite, as the change in momentum has opposite direction in the two situations.

d)

To compare the time of impact in the two situations, we have to look closer into them.

- When the ball is thrown, the hand "moves together" with the ball, from back to ahead in order to give it the necessary push. We can verify therefore that the time is longer in this case.

- When the ball is cacthed, the hand remains more or less "at rest", it  doesn't move much, so the collision lasts much less than the previous situation.

Therefore, we can say that the time of impact is longer when the ball is thrown, compared to when it is catched.

e)

The impulse exerted on an object can also be rewritten as the product between the force applied on the object and the time of impact:

I=F\Delta t

where

I is the impulse

F is the force applied

\Delta t is the time of impact

This can be rewritten as

F=\frac{I}{\Delta t}

In this problem, in the two situations,

- I (the impulse) is the same in both situations

- \Delta t when the ball is thrown is larger than when it is catched

Therefore, since F is inversely proportional to \Delta t, this means that the force is larger when the ball is catched.

6 0
3 years ago
A student uses 60 Newtons of force to climb 18 meters in 6 seconds. Calculate her Power.
puteri [66]

Answer:

15

Explanation:

P=W/T

T=6sec

W=?

F=60N

S=18m

W=F X S. .s indicate displacement

W=60x18

W=108

So p=108 j/6sec

P=15watt

5 0
2 years ago
A student is trying to determine the acceleration of a feather as she drops it to the ground. if the student is looking to achie
Anna [14]

The coordinate system should have the origin at the point where the feather is dropped and the downward direction is to be taken as positive.

All falling bodies experience acceleration towards the center of the Earth due to the force of gravitational attraction exerted on the object by the Earth. A feather, when dropped experiences an acceleration in the downward direction. Since the acceleration of the feather is in the downward direction, a feather, when dropped with zero initial velocity, has its velocity vector directed in the direction of its acceleration.

If the downward direction is taken as positive, the falling feather can be said to have a positive velocity and a positive acceleration.

5 0
3 years ago
In a drill during basketball practice, a player runs the length of the 30-meter court and back. The player does this three times
Sergeeva-Olga [200]

Answer:

0 m/s

Explanation:

Average velocity of an object is given by the net displacement divided by time taken. Displacement is equal to the shortest path covered by the object.

In this problem, a player runs the length of the 30-meter court and back. The player does this three times in 60 seconds.

As the player runs the court and returns to the original point. It would mean that the shortest path covered is 0.

Average velocity = displacement/time

v=0/30

v = 0 m/s

Hence, the correct option is (1).

6 0
3 years ago
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