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stiks02 [169]
3 years ago
5

If it had been thrown with twice the speed in the same direction, it would have hit the ground in If it had been thrown with twi

ce the speed in the same direction, it would have hit the ground in
3.20 s.
1.60 s.
0.800 s.
0.400 s.
0.200 s.
Physics
1 answer:
Semmy [17]3 years ago
7 0

Answer:

The ball will reach the ground in 0.8s

Option C

Explanation:

Given:

- Takes t = 0.8 s for ball to reach ground when thrown horizontal from top of a building.

Find:

If it had been thrown with twice the speed in the same direction, it would have hit the ground in how many second.

Solution:

- We know that the amount of time taken to hit the ground is determined by the vertical distance i.e height at which it is thrown. The displacement of ball from top is given by:

                                    S_y = S_o + V_i,y*t + 0.5*g*t^2

- We know that the S_o = height of the building.

 We also know that the ball os thrown horizontally; hence, y-component of initial velocity is zero. V_y,i = 0

                                    0 = h + 0 + 0.5*g*t^2

- Hence, the time taken t is:

                                    t = sqrt ( 2h / g)

- The time taken to reach the ground is independent of the initial speed. Hence, the ball will reach the ground in 0.8s .

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A 27-g steel-jacketed bullet is fired with a velocity of 640 m/s toward a steel plate and ricochets along path CD with a velocit
Dmitry [639]

Answer:

F = - 3.56*10⁵ N

Explanation:

To attempt this question, we use the formula for the relationship between momentum and the amount of movement.

I = F t = Δp

Next, we try to find the time that the average speed in the contact is constant (v = 600m / s), so we say

v = d / t

t = d / v

Given that

m = 26 g = 26 10⁻³ kg

d = 50 mm = 50 10⁻³ m

t = d/v

t = 50 10⁻³ / 600

t = 8.33 10⁻⁵ s

F t = m v - m v₀

This is so, because the bullet bounces the speed sign after the crash is negative

F = m (v-vo) / t

F = 26*10⁻³ (-500 - 640) / 8.33*10⁻⁵

F = - 3.56*10⁵ N

The negative sign is as a result of the force exerted against the bullet

6 0
3 years ago
In a lab experiment, a student is trying to apply the conservation of momentum. Two identical balls, each with a mass of 1.0 kg,
Studentka2010 [4]

Answer:

Second Trial satisfy principle of conservation of momentum

Explanation:

Given mass of ball A and ball B =\ 1.0\ Kg.

Let mass of ball A and B\ is\ m  

Final velocity of ball A\ is\ v_1

Final velocity of ball B\ is\ v_2

initial velocity of ball A\ is\ u_1

Initial velocity of ball B\ is\ u_2

Momentum after collision =mv_1+mv_2

Momentum before collision = mu_1+mu_2

Conservation of momentum in a closed system states that, moment before collision should be equal to moment after collision.

Now, mu_1+mu_2=mv_1+mv_2

Plugging each trial in this equation we get,

First Trial

mu_1+mu_2=mv_1+mv_2\\1(1)+1(-2)=1(-2)+1(-1)\\1-2=-2-1\\-1=-3

momentum before collision \neq moment after collision

Second Trial

mu_1+mu_2=mv_1+mv_2\\1(.5)+1(-1.5)=1(-.5)+1(-.5)\\.5-1.5=-.5-.5\\-1=-1

moment before collision = moment after collision

Third Trial

mu_1+mu_2=mv_1+mv_2\\1(2)+1(1)=1(1)+1(-2)\\2+1=1-2\\3=-1

momentum before collision \neq moment after collision

Fourth Trial

mu_1+mu_2=mv_1+mv_2\\1(.5)+1(-1)=1(1.5)+1(-1.5)\\.5-1=1.5-1.5\\-.5=0

momentum before collision \neq moment after collision

We can see only Trial- 2 shows the conservation of momentum in a closed system.

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A batter hits two baseballs with the same force. One hits the ground near third base. The other is a home run out of the park. W
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