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ryzh [129]
2 years ago
5

Please expert solve this​

Physics
1 answer:
777dan777 [17]2 years ago
5 0

Explanation:

I am not expect I am ambitious

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3. Force = _____ x _____?
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Explanation:

Force = mass * acceleration.

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In a LRC circuit, a second capacitor is connected in parallel with the capacitor previously in the circuit. What is the effect o
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3 years ago
A 10.0-kg box starts at rest on a level floor. An external, horizontal force of 2.00 × 102 N is applied to the box for a distanc
Harman [31]

Answer:

vf = 11.2 m/s

Explanation:

m = 10 Kg

F = 2*10² N

x = 4.00 m

μ = 0.44

vi = 0 m/s

vf = ?

We can apply Newton's 2nd Law

∑ Fx = m*a   (→)

F - Ffriction = m*a  ⇒  F - (μ*N) = F - (μ*m*g) = m*a   ⇒  a = (F - μ*m*g)/m

⇒    a = (2*10² N - 0.44*10 Kg*9.81 m/s²)/10 Kg = 15.6836 m/s²

then , we use the equation

vf² = vi² + 2*a*x    ⇒    vf = √(vi² + 2*a*x)

⇒   vf = √((0)² + 2*(15.6836 m/s²)*(4.00m)) = 11.2 m/s

7 0
3 years ago
Very large forces are produced in joints when a person jumps from some height to the ground. Calculate the force produced if an
krok68 [10]

Answer:

A) 31 kJ

B)  1.92 KJ

C) 40 ,  2.48

Explanation:

weight of person ( m ) = 79 kg

height of jump ( h ) = 0.510 m

Compression of joint material ( d ) = 1.30 cm ≈  0.013 m

A) calculate the force

Fd = mgh

F = mgh / d

W = mg

F(net) = W + F  = mg ( 1 + \frac{h}{d} )

          =  79 * 9.81 ( 1 + (0.51 / 0.013) )

          = 774.99 ( 40.231 ) ≈ 31 KJ

B) calculate the force when the stopping distance = 0.345 m

d = 0.345 m

Fd = mgh  hence  F = mgh / d

F(net) = W + F  = mg ( 1 + \frac{h}{d} )

          =  79 * 9.81 ( 1 + (0.51 / 0.345) )

          = 774.99 ( 2.478 ) = 1.92 KJ

C) Ratio of force in part a with weight of person

=  31000 / ( 79 * 9.81 ) = 31000 / 774.99 = 40

  Ratio of force in part b with weight of person

= 1920 / 774.99 = 2.48

4 0
3 years ago
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