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marin [14]
2 years ago
12

What is 5kg fall from 2.1 meter using g=9.8m/s

Physics
1 answer:
lakkis [162]2 years ago
8 0

Answer:

103.J

Explanation:

Given data

Mass= 5kg

Height= 2.1m

g=9.81m/s^2

Required

The potential energy

PE= mgh

substitute

PE= 5*9.81*2.1

PE=103.J

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:What will be the value of the refractive index of the medium? Critical angle of that medium is 30 degree
earnstyle [38]

Answer:

Let the second medium be air (n₁=1)

The refractive index n₂ of the medium where first medium is air is found (a)

(a) n₂ = 2

Explanation:

Critical angle can be defined as the angle of incidence that provides the angle of refraction of 90°.

Refractive index of a medium can be defined as a number that describes that how fast a light will travel through that medium.

Critical angle and Refractive index are related by:

\theta_{critical}= sin^{-1}(\frac{n_1}{n_2})

sin \theta_{critical}=\frac{n_1}{n_2}

To find refractive index of medium with respect to air, substitute n₁=1 (Refractive index of air is 1)

Also θ(critical)=30°

Find n₂ :

sin30= \frac{1}{n_2}\\0.5=\frac{1}{n_2}\\n_2=\frac{1}{0.5}\\n_2=2

8 0
2 years ago
1) The current in a light bulb is 0.335 Amps. How long does it take for a total charge of 2.76 C to
uysha [10]

Answer:

Time=8.23880597 seconds

Explanation:

Quantity of charge(q)=2.76c

Current(I)=0.335A

Time(t)=?

t=q/I

t=2.76/0.335

t=8.23880597seconds

7 0
3 years ago
A person carries a mass of 10 kg and walks along the +x-axis for a distance of 100m with a constant velocity of 2 m/s. What is t
gizmo_the_mogwai [7]
Since the direction of the force and the direction of the path is perpendicular, the person is not doing any physical work.
3 0
3 years ago
Read 2 more answers
A ball, which has a mass of 1.25 kg, is thrown straight up from the top of a building 225 meters tall with a velocity of 52.0 m/
Anastasy [175]

First we will find the speed of the ball just before it will hit the floor

so in order to find the speed of the cart we will first use energy conservation

KE_i + PE_i = KE_f + PE_f

\frac{1}{2}mv_i^2 + mgh = \frac{1}{2}mv_f^2 + 0

\frac{1}{2}(1.25)(52)^2 + 1.25(9.8)(225) = \frac{1}{2}(1.25)v_f^2

So by solving above equation we will have

v_f = 84.3 m/s

now in order to find the momentum we can use

P = mv

P = 1.25 \times 84.3

P = 105.4 kg m/s

3 0
3 years ago
A spring is used to stop a 50-kg package which is moving down a 20º incline. The spring has a constant k = 30 kN/m and is held b
Elina [12.6K]

Answer:

0.3 m

Explanation:

Initially, the package has both gravitational potential energy and kinetic energy.  The spring has elastic energy.  After the package is brought to rest, all the energy is stored in the spring.

Initial energy = final energy

mgh + ½ mv² + ½ kx₁² = ½ kx₂²

Given:

m = 50 kg

g = 9.8 m/s²

h = 8 sin 20º m

v = 2 m/s

k = 30000 N/m

x₁ = 0.05 m

(50)(9.8)(8 sin 20) + ½ (50)(2)² + ½ (30000)(0.05)² = ½ (30000)x₂²

x₂ ≈ 0.314 m

So the spring is compressed 0.314 m from it's natural length.  However, we're asked to find the additional deformation from the original 50mm.

x₂ − x₁

0.314 m − 0.05 m

0.264 m

Rounding to 1 sig-fig, the spring is compressed an additional 0.3 meters.

8 0
3 years ago
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