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bulgar [2K]
3 years ago
5

Calculate the speed of an object that moves 200 m in 4 s?

Physics
1 answer:
Rasek [7]3 years ago
7 0
50m/s

I divided the 200m by 4s (so 200/4) to get 50m per second
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A rock is thrown vertically upward from some height above the ground. It rises to some maximum height and falls back to the grou
True [87]

Answer:

At the highest point the velocity is zero, the acceleration is directed downward.

Explanation:

This is a free-fall problem, in the case of something being thrown or dropped, the acceleration is equal to -gravity, so -9.80m/s^2. So, the acceleration is never 0 here.

I attached an image from my lecture today, I find it to be helpful. You can see that because of gravity the acceleration is pulled downwards.

At the highest point the velocity is 0, but it's changing direction and that's why there's still an acceleration there.

6 0
3 years ago
When a 0.622 kg basketball hits the floor, its velocity changes from 4.23 m/s down to 3.85 m/s up. If the ball was in contact wi
SVEN [57.7K]

when the ball hits the floor and bounces back the momentum of the ball changes.

the rate of change of momentum is the force exerted by the floor on it.

the equation for the force exerted is

f = rate of change of momentum

f = \frac{mv - mu}{t}

v is the final velocity which is - 3.85 m/s

u is initial velocity - 4.23 m/s

m = 0.622 kg

time is the impact time of the ball in contact with the floor - 0.0266 s

substituting the values

f = \frac{0.622 kg (3.85 m/s - (-)4.23 m/s)}{0.0266}

since the ball is going down, we take that as negative and ball going upwards as positive.

f = 189 N

the force exerted from the floor is 189 N

4 0
3 years ago
Using equations, determine the temperature, pressure and density of the air for a aircraft flying at 19.5 km. Is this aircraft s
Viefleur [7K]

Answer:

a) - 72.5°c

b) pressure = 3625.13 Pa

c) density =  0.063 kg/m^3

d) it is a subsonic aircraft

Explanation:

a) Determine Temperature

Temperature at 19.5 km ( 19500 m )

T = -131 + ( 0.003 * altitude in meters )

  =  -131 + ( 0.003 * 19500 ) = - 72.5°c

b) Determine pressure and density at 19.5 km altitude

Given :

Po (atmospheric pressure at sea level )  = 101kpa

R ( gas constant of air ) = 0.287 KJ/Kgk

T = -72.5°c ≈ 200.5 k

pressure = 3625.13 Pa

hence density = 0.063 kg/m^3

attached below is the remaining part of the solution

C) determine if the aircraft is subsonic or super sonic

Velocity ( v ) = \sqrt{CRT}  =  \sqrt{1.4*287*200.5 } = 283.8 m/s

hence it is a subsonic aircraft

4 0
3 years ago
The width of the central maxima, formed from light of wavelength 575 nm behind a single slit that has a width of 115 μm, is 1.15
Lady bird [3.3K]

Answer:

 L  = 1.15 m

Explanation:

The diffraction phenomenon is described by the equation

        a sin θ = m λ

Where a is the width of the slit, λ  the wavelength and m is an integer, the order of diffraction is left.

The diffraction measurements are made on a screen that is far from the slit, and the angles in the experiment are very small, let's use trigonometry

          tan θ = y / L

          tan θ = sint θ / cos θ≈ sin θ

We substitute in the first equation

           a (y / L) = m λ

The first maximum occurs for m = 1

The distance is measured from the center point of maximum, which coincides with the center of the slit, in this case the distance is the total width of the central maximum, so the distance (y) measured from the center is

         y = 1.15 / 2 = 0.575 cm

         y = 0.575 10⁻² m

Let's clear the distance to the screen (L)

       L = a y / λ  

Let's calculate

     L = 115 10⁻⁶  0.575 10⁻² / 575 10⁻⁹

     L  = 1.15 m

3 0
3 years ago
What type of planets does the asteroid belt orbit?​
mihalych1998 [28]

Answer:

Main Asteroid Belt: The majority of known asteroids orbit within the asteroid belt between Mars and Jupiter, generally with not very elongated orbits. The belt is estimated to contain between 1.1 and 1.9 million asteroids larger than 1 kilometer (0.6 mile) in diameter, and millions of smaller ones.

Explanation:

6 0
3 years ago
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