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Anvisha [2.4K]
2 years ago
13

A marble is dropped straight down from a distance h above the floor.

Physics
1 answer:
Mrac [35]2 years ago
7 0
Fm=Fe and am>ae
Hopefully this helps
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80 foot-pounds of work is needed to move the sofa in Tyler's apartment. Which of the following statements is true?
erastova [34]
D is the correct answer
hop it helped.
3 0
2 years ago
pitot tube on an airplane flying at a standard sea level reads 1.07 x 105 N/m2. What is the velocity of the airplane?
Allushta [10]

Answer:

V_infinty=98.772 m/s

Explanation:

complete question is:

The following problem assume an inviscid, incompressible flow. Also, standard sea level density and pressure are 1.23kg/m3(0.002377slug/ft3) and 1.01imes105N/m2(2116lb/ft2), respectively. A Pitot tube on an airplane flying at standard sea level reads 1.07imes105N/m2. What is the velocity of the airplane?

<u>solution:</u>

<u>given:</u>

<em>p_o=1.07*10^5 N/m^2</em>

<em>ρ_infinity=1.23 kg/m^2</em>

<em>p_infinity=1.01*10^5 N/m^2</em>

p_o=p_infinity+(1/2)*(ρ_infinity)*V_infinty^2

V_infinty^2=9756.097

V_infinty=98.772 m/s

8 0
2 years ago
A 1.0-kg ball has a velocity of 12 m/s downward just before it strikes the ground and bounces up with a velocity of 12 m/s upwar
Nezavi [6.7K]

Answer:

The change in momentum of the ball is 24 kg-m/s  

Explanation:

It is given that,

Mass of the ball, m = 1 kg

Initial velocity of the ball, u = -12 m/s (in downwards)

Final velocity of the ball, v = +12 m/s (in upward)

We need to find the change in momentum of the ball.

Initial momentum of the ball, p_i=mu=1\ kg\times (-12\ m/s)=-12\ kg-m/s

Final momentum of the ball, p_f=mv=1\ kg\times (12\ m/s)=12\ kg-m/s

Change in momentum of the ball, \Delta p=p_f-p_i

\Delta p=12-(-12)=24\ kg-m/s

So, the change in momentum of the ball is 24 kg-m/s. Hence, this is the required solution.

3 0
2 years ago
Which is found farthest from the center of an atom?
xxTIMURxx [149]
Electron

Hope this helps
5 0
1 year ago
A 100-kg box is sitting on a 10 degree incline with a coefficient of friction of 0.5. At what angle must the incline be raised t
Marta_Voda [28]

Answer:

26.56°

Explanation:

μ= 0.5

F = ?

N =?

Fs = μN

Fs = force acting due to friction (on an inclined plane)

N = normal force.

μ= coefficient of friction

μ= tanΘ

μ= tan 0.5

Θ = tan⁻¹ 0.5

Θ = 26.56°

The angle in which is required for the box to start moving across the inclined plane is 26.56°

5 0
1 year ago
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