is it true that the composition of the atmosphere changes every few kilometers as you move away from earth

An object is thrown directly up (positive direction) with a velocity (vo) of 20.0 m/s and do= 0. How high does it rise (v = 0 cm

Anit [1.1K]

Answer:

The value of d is 20.4 m.

(C) is correct option.

Explanation:

Given that,

Initial velocity = 20 m/s

Final velocity = 0

We need to calculate the time

Using equation of motion

$v = u+gt$

Where, u = Initial velocity

v = Final velocity

Put the value into the formula

$t = \dfrac{20}{9.8}$

$t= 2.04\ sec$

We need to calculate the distance

Using equation of motion

$s = ut+\dfrac{1}{2}gt^2$

$s=0+\dfrac{1}{2}\times9.8\times(2.04)^2$

$s=20.4\ m$

Hence, The value of d is 20.4 m.

4 0

4 years ago

The kinetic energy of an object increases as its ____ its potential energy

tankabanditka [31]

The kinetic energy of an object increases as its decreases <span>its potential energy as the sum of energy will remain constant.

In short, Your Answer would be "Decreases"

Hope this helps!</span>

8 0

3 years ago

5) [Honors]A seagull, ascending straight upward at 5.2 m/s, drops a shell when it is 12.5m above the ground. (A)

jolli1 [7]

Answer:

(B) 13.9 m

(C) 1.06 s

Explanation:

Given:

v₀ = 5.2 m/s

y₀ = 12.5 m

(A) The acceleration in free fall is -9.8 m/s².

(B) At maximum height, v = 0 m/s.

v² = v₀² + 2aΔy

(0 m/s)² = (5.2 m/s)² + 2 (-9.8 m/s²) (y − 12.5 m)

y = 13.9 m

(C) When the shell returns to a height of 12.5 m, the final velocity v is -5.2 m/s.

v = at + v₀

-5.2 m/s = (-9.8 m/s²) t + 5.2 m/s

t = 1.06 s

3 0

3 years ago

In February 1955, a paratrooper fell 370 m from an airplane without being able to open his chute but happened to land in snow, s

nevsk [136]

a) 0.94 m

The work done by the snow to decelerate the paratrooper is equal to the change in kinetic energy of the man:

$W=\Delta K\\-F d = \frac{1}{2}mv^2 - \frac{1}{2}mu^2$

where:

$F=1.1 \cdot 10^5 N$ is the force applied by the snow

d is the displacement of the man in the snow, so it is the depth of the snow that stopped him

m = 68 kg is the man's mass

v = 0 is the final speed of the man

u = 55 m/s is the initial speed of the man (when it touches the ground)

and where the negative sign in the work is due to the fact that the force exerted by the snow on the man (upward) is opposite to the displacement of the man (downward)

Solving the equation for d, we find:

$d=\frac{1}{2F}mu^2 = \frac{(68 kg)(55 m/s)^2}{2(1.1\cdot 10^5 N)}=0.94 m$

b) -3740 kg m/s

The magnitude of the impulse exerted by the snow on the man is equal to the variation of momentum of the man:

$I=\Delta p = m \Delta v$

where

m = 68 kg is the mass of the man

$\Delta v = 0-55 m/s = -55 m/s$ is the change in velocity of the man

Substituting,

$I=(68 kg)(-55 m/s)=-3740 kg m/s$

7 0

3 years ago

If you throw a raw egg against a wall, you'll break it, but if you throw it with the same speed into a sagging sheet it won't br

SashulF [63]

<span>The egg doesn't break when it hits the sheet because the impact time is longer. Momentum means the egg is slowed rather than coming to an abrubt halt. The softer the object that the egg hits, the longer the time it takes to break. A sheet is so soft that the force is never high enough for the egg to break.</span>

4 0

3 years ago