Imagine a 10kg block moving with a speed of 20m/s. calculate the kinetic energy of this block

A person runs 400 meters in a straight line. What is their distance and displacement.

suter [353]

Distance = 400 m, Displacement = 400 m in the direction of the straight line.

7 0

3 years ago

What is the wavelength of the radio waves from an FM station operating at a frequency of 99.5 MHz

ser-zykov [4K]

Answer:

Wavelength, $\lambda=3.01\ m$

Explanation:

It is given that,

Frequency, f = 99.5 MHz = 99.5 × 10⁶ Hz

We need to find the wavelength of the radio waves from an FM station operating at above frequency. The relationship between the frequency and the wavelength is given by :

$c=f\lambda$

$\lambda=\dfrac{c}{f}{$

c = speed of light

$\lambda=\dfrac{3\times 10^8\ m/s}{99.5\times 10^6\ Hz}$

$\lambda=3.01\ m$

So, the wavelength of the radio waves from an FM station is 3.01 m. Hence, this is the required solution.

3 0

3 years ago

A person's prescription for bifocals is -0.25 diopter for distant vision and +2.50 diopters for near vision, the near vision bei

Darina [25.2K]

Answer:

net power is + 2.25 D

Explanation:

Given data

distance vision = -0.25 D

near vision = + 2.50 D

to find out

net power

solution

we have given a person lens power for near is - 0.25 diopter and lens power for near power is +2.50 diopter so

net power is sum of both the power vision

so

net power = distance + near power

put both value we get net power

net power = ( -0.25 D) + ( + 2.50 D)

net power = + 2.25 D

so net power is + 2.25 D

6 0

3 years ago

Suppose the entire solar nebula had cooled to 50 K before the solar wind cleared the early solar system of its gases. How would

Butoxors [25]

Suppose the entire solar nebula had cooled to 50 K before the solar wind cleared the early solar system of its gases. How would the composition and sizes of the planets of the inner solar system be different from what we see today is given below

Explanation:

1.In astronomy or planetary science, the frost line, also known as the snow line or ice line, is the particular distance in the solar nebula from the central protostar where it is cold enough for volatile compounds such as water, ammonia, methane, carbon dioxide, carbon monoxide to condense into solid ice grains.

2.The frost line in the solar nebula lies between Mars and Jupiter. It is the distance where it was cold enough for hydrogen compounds to condense into ices. Frost line: Explain how temperature differences led to the formation of two distinct types of planets.

3. The frost line is the point moving away from the Sun where it is cool enough for hydrogen compounds to freeze. Since the solar nebula was hotter near the center of the disk, hydrogen compounds such as water stayed gaseous in the inner solar system. Outside of the frost line, they froze.

4.The solar nebula flattened into a rotating disk. As gas became dense and hot, then it spins faster and pulled towards the center whereby the sun is formed. Solar nebula they collapse where the protostellar disk rotates. In the center of of nebula, there is a fusion begins and then sun is being formed

5.When it comes to the formation of our Solar System, the most widely accepted view is known as the Nebular Hypothesis. In essence, this theory states that the Sun, the planets, and all other objects in the Solar System formed from nebulous material billions of years ago.

4 0

2 years ago

Bill throws a tennis ball to his dog. He throws the ball at a speed of 15 m/s at an angle of 30° to the horizontal. Assume he th

Sidana [21]

1a) Bill and the dog must have a speed of 13.0 m/s

1b) The speed of the dog must be 22.5 m/s

2a) The ball passes over the outfielder's head at 3.33 s

2b) The ball passes 1.2 m above the glove

2c) The player can jump after 2.10 s or 3.13 s after the ball has been hit

2d) One solution is when the player is jumping up, the other solution is when the player is falling down

Explanation:

1a)

The motion of the ball in this problem is a projectile motion, so it follows a parabolic path which consists of two independent motions:

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

- An accelerated motion with constant acceleration (acceleration of gravity) in the vertical direction

In part a), we want to know at what speed Bill and the dog have to run in order to intercept the ball as it lands on the ground: this means that Bill and the dog must have the same velocity as the horizontal velocity of the ball.

The ball's initial speed is

u = 15 m/s

And the angle of projection is

$\theta=30^{\circ}$

So, the ball's horizontal velocity is

$v_x = u cos \theta = (15)(cos 30)=13.0 m/s$

And therefore, Bill and the dog must have this speed.

1b)

For this part, we have to consider the vertical motion of the ball first.

The vertical position of the ball at time t is given by

$y=u_yt+\frac{1}{2}at^2$

where

$u_y = u sin \theta = (15)(sin 30) = 7.5 m/s$ is the initial vertical velocity

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

The ball is at a position of y = 2 m above the ground when:

$2=7.5t + \frac{1}{2}(-9.8)t^2\\4.9t^2-7.5t+2=0$

Which has two solutions: t=0.34 s and t=1.19 s. We are told that the ball is falling to the ground, so we have to consider the second solution, t = 1.19 s.

The horizontal distance covered by the ball during this time is

$d=v_x t =(13.0)(1.19)=15.5 m$

The dog must be there 0.5 s before, so at a time

t' = t - 0.5 = 0.69 s

So, the speed of the dog must be

$v_x' = \frac{d}{t'}=\frac{15.5}{0.69}=22.5 m/s$

2a)

Here we just need to consider the horizontal motion of the ball.

The horizontal distance covered is

$d=98 m$

while the horizontal velocity of the ball is

$v_x = u cos \theta = (34)(cos 30)=29.4 m/s$

where u = 34 m/s is the initial speed.

So, the time taken for the ball to cover this distance is

$t=\frac{d}{v_x}=\frac{98}{29.4}=3.33 s$

2b)

Here we need to calculate the vertical position of the ball at t = 3.33 s.

The vertical position is given by

$y= h + u_y t + \frac{1}{2}at^2$

where

h = 1.2 m is the initial height

$u_y = u sin \theta = (34)(sin 30)=17.0 m/s$ is the initial vertical velocity

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

Substituting t = 3.33 s,

$y=1.2+(17)(3.33)+\frac{1}{2}(-9.8)(3.33)^2=3.5 m$

And sinc the glove is at a height of y' = 2.3 m, the difference in height is

y - y' = 3.5 - 2.3 = 1.2 m

2c)

In order to intercept the ball, he jumps upward at a vertical speed of

$u_y' = 7 m/s$

So its position of the glove at time t' is

$y'= h' + u_y' t' + \frac{1}{2}at'^2$

where h' = 2.3 m is the initial height of the glove, and t' is the time from the moment when he jumps. To catch the ball, the height must be

y' = y = 3.5 m (the height of the ball)

Substituting and solving for t', we find

$3.5 = 2.3 + 7t' -4.9t'^2\\4.9t'^2-7t'+12 = 0$

Which has two solutions: t' = 0.20 s, t' = 1.23 s. But this is the time t' that the player takes to reach the same height of the ball: so the corresponding time after the ball has been hit is

$t'' = t -t'$

So we have two solutions:

$t'' = 3.33 s - 0.20 s = 3.13 s\\t'' = 3.33 s - 1.23 s = 2.10 s$

So, the player can jump after 2.10 s or after 3.13 s.

2d)

The reason for the two solutions is the following: the motion of the player is a free fall motion, so initially he jump upwards, then because of gravity he is accelerated downward, and therefore eventually he reaches a maximum height and then he falls down.

Therefore, the two solutions corresponds to the two different part of the motion.

The first solution, t'' = 2.10 s, is the time at which the player catches the ball while he is in motion upward.

On the other hand, the second solution t'' = 3.13 s, is the time at which the player catches the ball while falling down.

Learn more about projectile motion:

brainly.com/question/8751410

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7 0

3 years ago