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chubhunter [2.5K]
2 years ago
7

A ball is released from the top of a tower of height h meters. it takes T seconds to reach the ground . what is the position of

the ball in T/3 seconds ?​
Physics
1 answer:
alekssr [168]2 years ago
3 0

Answer:the

8/9 h

Explanation:

Height  =   1/2 a T^2     now change to T/3

 now height = 1/2 a (T/3)^2   =<u> 1/9</u>  1/2 a T^2     <===== it is 1/9 of the way down   or   8/9 h

 

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A positively charged metal sphere, A, is held close to but not touching and identical uncharged sphere, sphere B. Sphere A is no
Yuri [45]

Answer:

The sphere C carries no net charge.

Explanation:

  • When brougth close to the charged sphere A, as charges can move freely in  a conductor, a charge equal and opposite to the one on the sphere A, appears on the sphere B surface facing to the sphere A.
  • As sphere B must remain neutral (due to the principle of conservation of charge) an equal charge, but of opposite sign, goes to the surface also, on the opposite part of the sphere.
  • If sphere A is removed, a charge movement happens in the sphere B, in such a way, that no net charge remains on the surface.
  • If in such state, if  the sphere B (assumed again uncharged completely, without any local charges on the surface), is touched by an initially uncharged sphere C, due to the conservation of  charge principle, no net  charge can be built on sphere C.
3 0
4 years ago
A ball with a mass of 5.0 g is moving at a speed of 2.0 m/s. Would doubling the mass or doubling the speed have a greater effect
Sloan [31]

Answer:

Explanation:

doubling the speed will have a greater impact on kinetic energy as KE is a product of mass and the square of velocity.

KE = ½mv²

Base KE = ½(0.005)2.0² = 0.01 J

doubling the mass

        KE = ½(0.010)2.0² = 0.02 J

doubling the velocity

        KE = ½(0.005)4.0² = 0.04 J

8 0
3 years ago
In which situation will screening be used to separate a mixture?
EleoNora [17]

Hello!


I believe the answer is

A) if the mixture is a solid suspended in a gas

8 0
4 years ago
Read 2 more answers
A(n) 0.212 kg baseball is thrown with a speed of 18.8 m/s. It is hit straight back at the pitcher with a final speed of 21.2 m/s
Alona [7]

Answer:

Impulse will be 8.4788 kgm/sec

Explanation:

We have given mass of the ball m = 0.212 kg

Initial velocity u = 18.8 m/sec

We know that momentum P = mass×velocity

Taking initial velocity as negative

So initial momentum = 0.212×-18.8 = -3.9856 kgm/sec

Final velocity is given as v = 21.2 m/sec

So final momentum = 0.212×21.2 = 4.4944 kgm/sec

We know that impulse is given by change in momentum

So impulse = final momentum-initial momentum = 4.4944-(-3.9844 ) = 8.4788 kgm/sec

7 0
3 years ago
A particle's velocity is described by the function vx = t 2 - 7t + 10 m/s, where t is in s.
mart [117]

velocity is given as

v = t^2 - 7t +10

when the object reached to its turning its velocity will be zero

0 = t^2 - 7t +10

(t - 5)(t-2) = 0

so it will reach its turning at two time instant

t_1 = 2 s and t_2 = 5 s

Part b)

For acceleration we know that

a = \frac{dv}{dt}

a = \frac{d}{dt}(t^2 - 7t + 10)

a = 2t - 7

now at t = 2

a_1 = 2*2 - 7 = -3 m/s^2

also at t=5

a_2 = 2*5 - 7 = 3 m/s^2


4 0
3 years ago
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