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spayn [35]
3 years ago
13

F = ma

Physics
2 answers:
tekilochka [14]3 years ago
6 0
The correct answer is A
k0ka [10]3 years ago
4 0

Answer:

A. 6000N

Explanation:

Force=mass*acceleration

Force=3000kg*2 m/s

6000N=3000kg*2 m/s

hope this helps!

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A ball is dropped from rest out of a high window in a tall building for 5 seconds. Assuming we ignore air resistance and assume
lara [203]

Answer:

what is the upwards force?

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An object travels in a circular path of radius 5.0 meters at a uniform speed of 10. m/s. What is the magnitude of the object's a
vladimir1956 [14]
I think F= mv²/r
And F=ma
So, ma = mv²/r
a = v²/r
a = 100/5
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Find the force required to do 25 joule work when the force causes a displacement of 0.5 m​
Eduardwww [97]

Answer:

<h2>50 N</h2>

Explanation:

The force required can be found by using the formula

f =  \frac{w}{d}  \\

w is the workdone

d is the distance

From the question we have

f =  \frac{25}{0.5}  \\

We have the final answer as

<h3>50 N</h3>

Hope this helps you

6 0
3 years ago
A circuit contains an AC voltage source, a resistor, and another component. The voltage amplitude is held constant as the freque
Temka [501]

Answer:

Capacitor

Explanation:

A capacitor is a device that stores charges. For a sinusoidal voltage circuit which contains a capacitor, the capacitor will alternately charge and discharge at a rate determined by the frequency of the supply.

When an alternating sinusoidal voltage is applied to the plates of a capacitor in an AC circuit, the capacitor is charged firstly in one direction and then in the opposite direction changing its polarity at the same rate as the AC mains voltage.

4 0
3 years ago
A block weighting 400kg rests on a horizontal surface and support on top of it another block of weight 100kg placed on the top o
masha68 [24]

The horizontal force applied to the block is approximately 1,420.84 N

The known parameters;

The mass of the block, w₁ = 400 kg

The orientation of the surface on which the block rest, w₁ = Horizontal

The mass of the block placed on top of the 400 kg block, w₂ = 100 kg

The length of the string to which the block w₂ is attached, l = 6 m

The coefficient of friction between the surface, μ = 0.25

The state of the system of blocks and applied force = Equilibrium

Strategy;

Calculate the forces acting on the blocks and string

The weight of the block, W₁ = 400 kg × 9.81 m/s² = 3,924 N

The weight of the block, W₂ = 100 kg × 9.81 m/s² = 981 N

Let <em>T</em> represent the tension in the string

The upward force from the string = T × sin(θ)

sin(θ) = √(6² - 5²)/6

Therefore;

The upward force from the string = T×√(6² - 5²)/6

The frictional force = (W₂ - The upward force from the string) × μ

The frictional force, F_{f2} = (981 - T×√(6² - 5²)/6) × 0.25

The tension in the string, T = F_{f2} × cos(θ)

∴ T = (981 - T×√(6² - 5²)/6) × 0.25 × 5/6

Solving, we get;

T = \dfrac{5886}{\sqrt{6^2 - 5^2} + 28.8} \approx 183.27

Frictional \ force, F_{f2} = \left (981 -  \dfrac{5886}{\sqrt{6^2 - 5^2} + 28.8}  \times \dfrac{\sqrt{6^2 - 5^2} }{6} \times  0.25 \right) \approx 219.92

The frictional force on the block W₂, F_{f2} ≈ 219.92 N

Therefore;

The force acting the block w₁, due to w₂ F_{w2} = 219.92/0.25 ≈ 879.68

The total normal force acting on the ground, N = W₁ + \mathbf{F_{w2}}

The frictional force from the ground, \mathbf{F_{f1}} = N×μ + \mathbf{F_{f2}} = P

Where;

P = The horizontal force applied to the block

P = (W₁ + \mathbf{F_{w2}}) × μ + \mathbf{F_{f2}}

Therefore;

P = (3,924 + 879.68) × 0.25 + 219.92 ≈ 1,420.84

The horizontal force applied to the block, P ≈ 1,420.84 N

Learn more about friction force here;

brainly.com/question/18038995

3 0
3 years ago
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