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MrMuchimi
4 years ago
5

PLS HELP ANSWER QUICK

Physics
1 answer:
Ipatiy [6.2K]4 years ago
8 0
Weight changes due to the gravitational pull in mars,basically gravity
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____ crust forms the ocean floors
Fittoniya [83]

Answer:

lithosphere

Explanation:

6 0
3 years ago
Read 2 more answers
Three ropes A, B and C are tied together in one single knot K. (See figure.)
masha68 [24]

The tension in the rope B is determined as 10.9 N.

<h3>Vertical angle of cable B</h3>

tanθ = (6 - 4)/(5 - 0)

tan θ = (2)/(5)

tan θ = 0.4

θ = arc tan(0.4) = 21.8 ⁰

<h3>Angle between B and C</h3>

θ = 21.8 ⁰ + 21.8 ⁰ = 43.6⁰

Apply cosine rule to determine the tension in rope B;

A² = B² + C² - 2BC(cos A)

B = C

A² = B² + B² - (2B²)(cos A)

A² = 2B² - 2B²(cos 43.6)

A² = 0.55B²

B² = A²/0.55

B² = 65.3/0.55

B² = 118.73

B = √(118.73)

B = 10.9 N

Thus, the tension in the rope B is determined as 10.9 N.

Learn more about tension here: brainly.com/question/24994188

#SPJ1

6 0
2 years ago
an athlete sprints from 150 m south of the finish line to 65 m south of the finish line in 5.0s what is his average velocity
Ilya [14]
  • Total distance=150m-65m=85m
  • Time=5s

\\ \rm\longrightarrow Avg\:Velocity=\dfrac{Total\:Distance}{Total\:Time}

\\ \rm\longrightarrow Avg\:Velocity=\dfrac{85}{5}

\\ \rm\longrightarrow Avg\:Velocity= 17m/s

3 0
3 years ago
Read 2 more answers
What is the restoring force of a spring stretched 0.35 meters with a spring contact of 55 newtons?
Vesna [10]
F=-kx = -19,25 answer D
6 0
4 years ago
Read 2 more answers
A 20~\mu F20 μF capacitor has previously charged up to contain a total charge of Q = 100~\mu CQ=100 μC on it. The capacitor is t
sertanlavr [38]

Explanation:

The given data is as follows.

       C = 20 \times 10^{-6} F

        R = 100 \times 10^{3} ohm

        Q_{o} = 100 \times 10^{-6} C

          Q = 13.5 \times 10^{-6} C

Formula to calculate the time is as follows.

          Q_{t}  = Q_{o} [e^{\frac{-t}{\tau}]

       13.5 \times 10^{-6} = 100 \times 10^{-6} [e^{\frac{-t}{2}}]

               0.135 = e^{\frac{-t}{2}}

         e^{\frac{t}{2}} = \frac{1}{0.135}

                         = 7.407

           \frac{t}{2} = ln (7.407)

                      t = 4.00 s

Therefore, we can conclude that time after the resistor is connected will the capacitor is 4.0 sec.

4 0
3 years ago
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