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stich3 [128]
3 years ago
14

A rocket ship is moving through space at 1200 m/s. The rocket ship accelerates at a rate of 4 m/s in the same direction. What is

the rocket ship's speed after 200 seconds of acceleration?
Physics
1 answer:
svetoff [14.1K]3 years ago
3 0

Answer:

Vf = 2000 [m/s]

Explanation:

To solve this problem we must use the following equation of kinematics.

v_{f} =v_{o}+a*t\\

where:

Vf = final velocity [m/s]

Vo = initial velocity = 1200 [m/s]

a = acceleration = 4 [m/s]

t = 200 [s]

Vf = 1200 + (4*200)

Vf = 1200 + 800

Vf = 2000 [m/s]

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A powerful searchlight shines on a man. The man's cross-sectional area is 0.500m2 perpendicular to the light beam, and the inten
babymother [125]

Answer:

The magnitude of the force the light beam exerts on the man is 5.9 x 10⁻⁵N

(b) the force the light beam exerts is much too small to be felt by the man.

Explanation:

Given;

cross-sectional area of the man, A = 0.500m²

intensity of light, I = 35.5kW/m²

If all the incident light were absorbed, the pressure of the incident light on the man can be calculated as follows;

P = I/c

where;

P is the pressure of the incident light

I is the intensity of the incident light

c is the speed of light

P = \frac{I}{c} =\frac{35500}{3*10^8} = 1.18*10^{-4} \ N/m^2

F = PA

where;

F is the force of the incident light on the man

P is the pressure of the incident light on the man

A is the cross-sectional area of the man

F = 1.18 x 10⁻⁴ x 0.5 = 5.9 x 10⁻⁵ N

The magnitude of the force the light beam exerts on the man is 5.9 x 10⁻⁵ N

Therefore, the force the light beam exerts is much too small to be felt by the man.

8 0
4 years ago
A 80 kg bungee jumper is on a bridge that is 100 meters above a river. Attached to the jumper is a bungee cord that is 50 meters
Usimov [2.4K]

Answer:

a) 70,560 J

b) 88.2 N/m

Explanation:

The spring potential will equal the change in gravity potential

PS = PE = mgh = 80(9.8)(100 - 10) = 70,560 J

PS = ½kx²

k = 2PS/x² = 2(70560)/(100 - 50 - 10)² = 88.2 N/m

7 0
3 years ago
The 60-mm-diameter shaft is made of 6061-T6 aluminum having an allowable shear stress of τallow = 80 MPa. Determine the maximum
grin007 [14]
The maximum allowable torque must correspond to the allowable shear stress for maximization. To solve this, we use the torsion formula:

Max. Allowable Shear Stress = Maximum Torque ÷ Cross-Sectional Area
8 x 10^6 Pa = Maximum Torque ÷ pi*(d/2)²
Maximum Torque = 8 x 10^6 Pa * pi*(0.06/2)² m²
Maximum Torque = 22,619.47 J or
Maximum Torque = 22.62 kJ

As for the second question, I have no reference figure so I am unable to answer it. I hope I was still able to help you, though.
5 0
4 years ago
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