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3 years ago

A ball is thrown straight up. For which situation are both the instantaneous velocity and the acceleration zero?

Physics

1 answer:

leva [86]3 years ago

6 0

The correct option is b i.e none is correct.

The instantaneous velocity would be zero at the top but the acceleration would not be zero at that point. This is because the acceleration is the rate of change of velocity. since, the velocity is not constant, hence the acceleration would not be zero. The correct answer would be option b: None is correct.

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A train locomotive is pulling two cars of the same mass behind it. Determine the ratio of the tension in the coupling (think of

Anna007 [38]

Answer:

The ratio is $\frac{F_{T1}}{F_{T2}} = 2$

Explanation:

The diagram for this question is shown on the first uploaded image

Here we are assume the acceleration of the train is a

which makes the acceleration of each car a

From the question we are told that

Considering the second car

The force causing it s movement is mathematically represented as

$F_{T2} = ma$

Considering the first car

The force causing it s movement is mathematically represented as

$F = F_{T1} -F_{T2} = ma$

=> $F_{T1} -ma = ma$

=> $F_{T1} = 2 ma$

=> $\frac{F_{T1}}{ma} = 2$

=> $\frac{F_{T1}}{F_{T2}} = 2$

7 0

3 years ago

Before colliding, the momentum of block A is -100 kg*m/, and block B is -150 kg*m/s. After, block A has a momentum -200 kg*m/s.

rjkz [21]

Answer:

Momentum of block B after collision = $-50\ kg\ ms^{-1}$

Explanation:

Given

Before collision:

Momentum of block A = $p_{A1}$ = $-100\ kg\ ms^{-1}$

Momentum of block B = $p_{B1}$ = $-150\ kg\ ms^{-1}$

After collision:

Momentum of block A = $p_{A2}$ = $-200\ kg\ ms^{-1}$

Applying law of conservation of momentum to find momentum of block B after collision $p_{B2}$ .

$p_{A1}+p_{B1}=p_{A2}+p_{B2}$

Plugging in the given values and simplifying.

$-100-150=-200+p_{B2}$

$-250=-200+p_{B2}$

Adding 200 to both sides.

$200-250=-200+p_{B2}+200$

$-50=p_{B2}$

∴ $p_{B2}=-50\ kg\ ms^{-1}$

Momentum of block B after collision = $-50\ kg\ ms^{-1}$

6 0

2 years ago

A system dissipates 12 J of heat into the surroundings; meanwhile, 28 J of work is done on the system. What is the change of the

Stels [109]

Answer:

+16 J

Explanation:

We can solve the problem by using the 1st law of thermodynamics:

$\Delta U = Q-W$

where

$\Delta U$ is the change of the internal energy of the system

Q is the heat (positive if supplied to the system, negative if dissipated by the system)

W is the work done (positive if done by the system, negative if done by the surroundings on the system)

In this case we have:

Q = -12 J is the heat dissipated by the system

W = -28 J is the work done ON the system

Substituting into the equation, we find the change in internal energy of the system:

$\Delta U=-12 J-(-28 J)=+16 J$

3 0

3 years ago

Which tool would you use to measure the amount of rainfall

mariarad [96]

A rain gauge! Hope this helps!

3 0

3 years ago

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A book is thrown downward from the library window with a speed of 2.0 s m and lands on the ground 5.0 m. We can ignore air resis

OlgaM077 [116]

Answer:

Explanation:

Using the equation of motion v² = u²+2as where;

v is the final velocity

u is the initial velocity

a is the acceleration due to gravity

s is the height of fall

Given

u = 2.0m/s

s = 5.0m

g = 9.81m/s²

Required

final velocity of the book in m/s

Substitute the given parameters into the formula;

v² = u²+2as

v² = 2²+2(9.81)(5)

v² = 4+98.1

v² = 102.1

v = √102.1

v = 10.1m/s

<em>Hence the final velocity of the book in m/s is 10.1m/s</em>

6 0

3 years ago

Read 2 more answers