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german
3 years ago
13

What waves are used in the hospitals to take pictures of bones

Physics
1 answer:
stich3 [128]3 years ago
7 0

Answer:

An X-ray is used to take pictures of your bones. The waves that are used are known as radiation waves.

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What type of device forms images through reflection?<br> Answer: Any mirror
inysia [295]

Answer:

wouldnt a phone be one it takes pictures through reflection like a mirror

6 0
3 years ago
What is the restoring force of a pendulum
salantis [7]

Answer:

the Restoring force causes the vibrating object to go slower going further from the equilibrium position and to go faster as it approaches the equilibrium position. the restoring force is what is causing the vibration The tension force comes from the string tugging on the bob of the pendulum.

Explanation:

3 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
A block with a weight of 9.0 N is at rest on a horizontal surface. A 1.2 N upward force is applied to the block by means of an a
lesya [120]

Answer:

a)   9 - 1.2 = 7.8 N

b) Since the force exerted by the box is it's weight, it acts in a downward direction. So the box will exert a force downward (perpendicular to the horizontal surface).

Explanation:

A 9N block would exert 9N of normal force on the horizontal plane under normal conditions. But in this case, we have a spring taking away some of the force by applying an upward force on the box.

So the force exerted by the box on the surface would now be:

a)   9 - 1.2 = 7.8 N

b) Since the force exerted by the box is it's weight, it acts in a downward direction. So the box will exert a force downward (perpendicular to the horizontal surface).

8 0
3 years ago
A 2000 kg car is moving at 1.5 m/s. What is its kinetic energy?*
Dmitrij [34]
Answer is B 3000 J hope I could help
3 0
3 years ago
Read 2 more answers
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