10. A frog leaps from its resting position at the

Two rocks, a and b, are thrown horizontally from the top of a cliff. rock a has an initial speed of 10 meters per second and roc

Arte-miy333 [17]

Both rocks hit the ground at the same time and at the same distance from the base of the cliff. Option A is correct.

<h3>What is speed ?</h3>

Speed is defined as the rate of change of the distance or the height attained. it is a time-based quantity. it is denoted by u for the initial speed while u for the final speed. its SI unit is m/sec.

If the resistance is ignored, things falling near the Earth's surface have the same estimated acceleration due to gravity of 9.8 meters per second.

As a result, the objects' acceleration is the same, and their velocity is growing at a consistent pace.

So that both rocks hit the ground at the same time and at the same distance from the base of the cliff.

Hence, option A is correct.

To learn more about the sped, refer to the link;

brainly.com/question/7359669

#SPJ1

5 0

2 years ago

In a typical badminton swing the racket is in contact with the birdy for about 0.0010 seconds. If the 0.045kg birdy acquires a s

zepelin [54]

Answer:

3015 N

Explanation:

From Newton's second law, we know that;

F.t = mv

F = force on the ball= ?

m= mass of the ball

v= velocity

F= mv/t

F= 0.045 × 67/0.0010

F= 3.015/0.0010

F= 3015 N

3 0

2 years ago

The density of a sample can be obtained by dividing its __________ by its _______________.

mixer [17]

Density can be calculated and found by dividing the sample's mass by its volume. D=m/v

3 0

3 years ago

Two stones are launched from the top of a tall building. One stoneis thrown in a direction 30.0^\circ above the horizontal with

Butoxors [25]

Answer:

Part A)

t(1) > t(2), the stone thrown 30 above the horizontal spends more time in the air.

Part B)

x(f1) > x(f2), the first stone will land farther away from the building.

Explanation:

Part A)

Let's use the parabolic motion equation to solve it. Let's define the variables:

y(i) is the initial height, it is a constant.
y(f) is the final height, in our case is 0
v(i) is the initial velocity (v(i)=16 m/s)
θ1 is the first angle, 30°
θ2 is the first angle, -30°

For the first stone

$y_{f1}=y_{i1}+v*sin(\theta_{1})t_{1}-0.5gt_{1}^{2}$

$0=y_{i1}+16*sin(30)t_{1}-0.5*9.81*t_{1}^{2}$

$0=y_{i1}+8t_{1}-4.905*t_{1}^{2}$ (1)

For the second stone

$0=y_{i2}+16*sin(-30)t_{2}-4.905t_{2}^{2}$

$0=y_{i2}-8t_{2}-4.905t_{2}^{2}$ (2)

If we solve the equation (1) we will have:

$t_{1}=\frac{-8\pm \sqrt{64+19.62*y_{i}}}{-9.81}$

We can do the same procedure for the equation (2)

$t_{1}=\frac{8\pm \sqrt{64+19.62*y_{i}}}{-9.81}$

We can analyze each solution to see which one spends more time in the air.

It is easy to see that the value inside the square root of each equation is always greater than 8, assuming that the height of the building is > 0. Now, to get positive values of t(1) and t(2) we need to take the negative option of the square root.

Therefore, t(1) > t(2), it means that the stone thrown 30 above the horizontal spends more time in the air.

Part B)

We can use the equation of the horizontal position here.

First stone

$x_{f1}=x_{i1}+vcos(30)t_{1}$