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IgorC [24]
3 years ago
15

John traveled East at 10 m/s for ten

Physics
1 answer:
NNADVOKAT [17]3 years ago
5 0

Answer:

Option D

130 m

Explanation:

From the concept of speed, distance=speed*time

When traveling East, displacement=10*10=100 m (since he takes 10 seconds while traveling at a speed of 10 m/s)

When traveling West, displacement=5*6=30 m (since John takes 6 seconds to travel at a speed of 5 m/s)

Total displacement=Displacement East+ Displacement West=100+30=130 m

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AlexFokin [52]

Answer:

K=0.023J

Explanation:

From the question we are told that:

Mass m=0.15

Velocity v=0.5m/s

Angular Velocity \omega=8.4rad/s

Generally the equation for Kinetic Energy is mathematically given by

K=\frac{1}{2}M(v^2+\frac{1}{2}R^2\omega^2)

K=\frac{1}{2}0.15(0.5^2+\frac{1}{2}(0.038)^2.(8.4rad/s^2))

K=0.023J

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2 years ago
I have an astronomy question... Spinning up the solar nebula. The orbital speed of the material in the solar nebula at Pluto's a
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<span>The angular momentum of a particle in orbit is 

l = m v r 

Assuming that no torques act and that angular momentum is conserved then if we compare two epochs "1" and "2" 

m_1 v_1 r_1 = m_2 v_2 r_2 

Assuming that the mass did not change, conservation of angular momentum demands that 

v_1 r_1 = v_2 r_2 

or 

v1 = v_2 (r_2/r_1) 

Setting r_1 = 40,000 AU and v_2 = 5 km/s and r_2 = 39 AU (appropriate for Pluto's orbit) we have 

v_2 = 5 km/s (39 AU /40,000 AU) = 4.875E-3 km/s

Therefore, </span> the orbital speed of this material when it was 40,000 AU from the sun is <span>4.875E-3 km/s.

I hope my answer has come to your help. Thank you for posting your question here in Brainly.
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3 years ago
If a 2kg ball rolls down a ramp that is 15 meters long in 25 seconds, what is the
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2 years ago
a body of a mass 1kg suspended from a spring is found to stretch the spring by 10cm. (A) what is the spring constant? what is pe
Nikitich [7]

Answer:

(A) The spring constant is <u>98 N/m.</u>

(B) The period of oscillation is <u>0.635 s.</u>

Explanation:

Given:

Mass of the body is, m=1\ kg

Extension length of the spring is, x=10\ cm=0.1\ m

Now, let 'k' be the spring constant.

The force acting on the body is due to gravity only and is equal to its weight.

So, weight of the body is given as:

F_g=mg\\F_g=1\times 9.8\\F_g=9.8\ N

Now, we know that for a spring-mass system, the net force acting on the body is equal to the product of the spring constant and extension length. Therefore,

F_g=kx\\9.8=k(0.1)\\k=\frac{9.8}{0.1}=98\ N/m

Hence, the spring constant is 98 N/m.

(B)

Period of oscillation of a body of spring-mass system in SHM is given as:

T=2\pi\sqrt{\frac{m}{k}}

Plug in the given values and solve for period 'T'. This gives,

T=2\pi\sqrt{\frac{1}{98}}\\T=2\pi\times 0.101\\T=0.635\ s

Therefore, the period of oscillation is 0.635 s.

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3 years ago
Which advantage of genetic technologies would these people most benefit from?
Westkost [7]

Answer:

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2 years ago
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