Which one of the following is an example of an indicator?

The time for a sound wave to travel between two people is 0.80 s,

sasho [114]

Answer:

Explanation:

Using the below formula

Speed of sound = ( distance between observers) *2/(total time taken)

Now putt the given values ,

time taken = 0.80 sec

distance = 256 m

hence

V of sound= 256*2/0.80

V of sound = 640 m/sec

4 0

3 years ago

SI UNIT OF GRAVITY is the

pickupchik [31]

Answer:

G

Explanation:

4 0

2 years ago

50 points !! I need help asap.......Consider a 2-kg bowling ball sits on top of a building that is 40 meters tall. It falls to t

r-ruslan [8.4K]

1) At the top of the building, the ball has more potential energy

2) When the ball is halfway through the fall, the potential energy and the kinetic energy are equal

3) Before hitting the ground, the ball has more kinetic energy

4) The potential energy at the top of the building is 784 J

5) The potential energy halfway through the fall is 392 J

6) The kinetic energy halfway through the fall is 392 J

7) The kinetic energy just before hitting the ground is 784 J

Explanation:

1)

The potential energy of an object is given by

$PE=mgh$

where

m is the mass

g is the acceleration of gravity

h is the height relative to the ground

While the kinetic energy is given by

$KE=\frac{1}{2}mv^2$

where v is the speed of the object

When the ball is sitting on the top of the building, we have

$h=40 m$ , therefore the potential energy is not zero
$v=0$ , since the ball is at rest, therefore the kinetic energy is zero

This means that the ball has more potential energy than kinetic energy.

2)

When the ball is halfway through the fall, the height is

$h=20 m$

So, half of its initial height. This also means that the potential energy is now half of the potential energy at the top (because potential energy is directly proportional to the height).

The total mechanical energy of the ball, which is conserved, is the sum of potential and kinetic energy:

$E=PE+KE=const.$

At the top of the building,

$E=PE_{top}$

While halfway through the fall,

$PE_{half}=\frac{PE_{top}}{2}=\frac{E}{2}$

And the mechanical energy is

$E=PE_{half} + KE_{half} = \frac{PE_{top}}{2}+KE_{half}=\frac{E}{2}+KE_{half}$

which means

$KE_{half}=\frac{E}{2}$

So, when the ball is halfway through the fall, the potential energy and the kinetic energy are equal, and they are both half of the total energy.

3)

Just before the ball hits the ground, the situation is the following:

The height of the ball relative to the ground is now zero: $h=0$ . This means that the potential energy of the ball is zero: $PE=0$

The kinetic energy, instead, is not zero: in fact, the ball has gained speed during the fall, so $v\neq 0$ , and therefore the kinetic energy is not zero

Therefore, just before the ball hits the ground, it has more kinetic energy than potential energy.

4)

The potential energy of the ball as it sits on top of the building is given by

$PE=mgh$

where:

m = 2 kg is the mass of the ball

$g=9.8 m/s^2$ is the acceleration of gravity

h = 40 m is the height of the building, where the ball is located

Substituting the values, we find the potential energy of the ball at the top of the building:

$PE=(2)(9.8)(40)=784 J$

5)

The potential energy of the ball as it is halfway through the fall is given by

$PE=mgh$

where:

m = 2 kg is the mass of the ball

$g=9.8 m/s^2$ is the acceleration of gravity

h = 20 m is the height of the ball relative to the ground

Substituting the values, we find the potential energy of the ball halfway through the fall:

$PE=(2)(9.8)(20)=392 J$

6)

The kinetic energy of the ball halfway through the fall is given by

$KE=\frac{1}{2}mv^2$

where

m = 2 kg is the mass of the ball

v = 19.8 m/s is the speed of the ball when it is halfway through the fall

Substituting the values into the equation, we find the kinetic energy of the ball when it is halfway through the fall:

$KE=\frac{1}{2}(2)(19.8)^2=392 J$

We notice that halfway through the fall, half of the initial potential energy has converted into kinetic energy.

7)

The kinetic energy of the ball just before hitting the ground is given by

$KE=\frac{1}{2}mv^2$

where:

m = 2 kg is the mass of the ball

v = 28 m/s is the speed of the ball just before hitting the ground

Substituting the values into the equation, we find the kinetic energy of the ball just before hitting the ground:

$KE=\frac{1}{2}(2)(28)^2=784 J$

We notice that when the ball is about to hit the ground, all the potential energy has converted into kinetic energy.

Learn more about kinetic and potential energy:

brainly.com/question/6536722

brainly.com/question/1198647

brainly.com/question/10770261

#LearnwithBrainly

4 0

3 years ago

g A change in the initial _____ of a projectile changes the range and maximum height of the projectile.

docker41 [41]

Answer:

Velocity.

Explanation:

Projectile motion is characterized as the motion that an object undergoes when it is thrown into the air and it is only exposed to acceleration due to gravity.

As per the question, 'any change in the initial velocity of the projectile(object having gravity as the only force) would lead to a change in the range as well as the maximum height of the projectile.' To illustrate numerically:

Horizontal range: As per expression:

R= ( $u^{2}$ *sin2θ)/g

the range depending on the square of the initial velocity.

Maximum height: As per expression:

H= ( $u^{2}$ * $sin^{2}$ θ )/2g

the maximum distance also depends upon square of the initial velocity.

7 0

3 years ago

E14. A ball rolls off a table with a horizontal velocity of 5 m/s. If

Shkiper50 [21]

a) Vertical velocity: 5.9 m/s

b) Horizontal velocity: 5 m/s

Explanation:

a)

The motion of the ball is the motion of a projectile, which consists of two independent motions:

- A uniform motion (constant velocity) along the horizontal direction)

- A uniformly accelerated motion (constant acceleration) along the vertical direction

Here we want to find the vertical component of the ball's velocity. This can be done by using the suvat equation for the vertical motion:

$v_y = u_y +gt$

where:

$v_y$ is the vertical velocity at time t

$u_y=0$ is the initial vertical velocity (zero because the ball has been thrown horizontally)

$g=10 m/s^2$ is the acceleration of gravity (here we take downward as positive direction)

Substituting t = 0.6 s, which is the total time of flight, we find the vertical velocity of the ball just before it hits the ground:

$v_y=0+(9.8)(0.6)=5.9 m/s$

b)

The motion along the vertical direction is an accelerated motion, because there is a force (the force of gravity) acting on the ball and that it causes an acceleration in the ball.

However, there are no forces acting in the horizontal direction on the ball (if we neglect the air resistance): this means that the acceleration of the ball in the horizontal direction is zero.

As a consequence, this also means that the horizontal component of the ball's velocity is constant during the motion.

Since the ball was thrown from the table with an initial horizontal velocity of 5 m/s, this means that the horizontal velocity of the ball just before it hits the floor is still

$v_x = 5 m/s$

8 0

3 years ago