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Musya8 [376]
1 year ago
5

A coin is dropped from a height of 421 m. calculate the velocity of the coin after 3 s​

Physics
1 answer:
Keith_Richards [23]1 year ago
7 0

Answer:

29.4 m.s

Explanation:

Vf = vo + at       v o = original velocity = 0 in this case

Vf = at

   = 9.81 m/s^2 * 3 = 29.4 m/s

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24 A uniform electric field of magnitude 1.1×104 N/C is perpendicular to a square sheet with sides 2.0 m long. What is the elect
Tatiana [17]

Answer:

44,000 Nm^2/C

Explanation:

The electric flux through a certain surface is given by (for a uniform field):

\Phi = EA cos \theta

where:

E is the magnitude of the electric field

A is the area of the surface

\theta is the angle between the direction of the field and of the normal to the surface

In this problem, we have:

E=1.1\cdot 10^4 N/C is the electric field

L = 2.0 m is the side of the sheet, so the area is

A=L^2=(2.0)^2=4.0 m^2

\theta=0^{\circ}, since the electric field is perpendicular to the surface

Therefore, the electric flux is

\Phi =(1.1\cdot 10^4)(4.0)(cos 0^{\circ})=44,000 Nm^2/C

4 0
3 years ago
An object has a mass of 2,000kg. what is its weight on earth? show your work​
artcher [175]

Answer:

F=m*g is the formula and the answer is 19,620 kg

Explanation:

Since the formula is F=m*g and Earth's gravity is 9.81 m/s^2 all you need to do is multiply 2,000 by 9.81

6 0
3 years ago
Read 2 more answers
A block, M1=10kg, slides down a smooth, curved incline of height 5m. It collides elastically with another block, M2=5kg, which i
erma4kov [3.2K]

Answer:

2.86 m

Explanation:

Given:

M₁ = 10 kg

M₂ = 5 kg

\mu_k = 0.5

height, h = 5 m

distance traveled, s = 2 m

spring constant, k = 250 N/m

now,

the initial velocity of the first block as it approaches the second block

u₁ = √(2 × g × h)

or

u₁ = √(2 × 9.8 × 5)

or

u₁ = 9.89 m/s

let the velocity of second ball be v₂

now from the conservation of momentum, we have

M₁ × u₁ = M₂ × v₂

on substituting the values, we get

10 × 9.89 = 5 × v₂

or

v₂ = 19.79 m/s

now,

let the velocity of mass 2 when it reaches the spring be v₃

from the work energy theorem,  we have

Work done by the friction force = change in kinetic energy of the mass 2

or

0.5\times5\times9.8\times2 = \frac{1}{2}\times5\times( v_3^2-19.79^2)

or

v₃ = 20.27 m/s

now, let the spring is compressed by the distance 'x'

therefore, from the conservation of energy

we have

Energy of the spring =  Kinetic energy of the mass 2

or

\frac{1}{2}kx^2=\frac{1}{2}mv_3^2

on substituting the values, we get

\frac{1}{2}\times250\times x^2=\frac{1}{2}\times5\times20.27^2

or

x = 2.86 m

8 0
3 years ago
How much momentum does a 10 kg object have when is moving 30 m/s
nikklg [1K]
P = m x v

P = 30 x 10
=300
3 0
3 years ago
What type of energy is involved when a river moves sediment and erodes its banks?
ziro4ka [17]
<span>First question: The type of energy involved when a river moves sediment and erodes its banks is: option d. Kinetic energy. Kinetic energy is the energy associated with motion. A body (in this case the water) that moves has an energy associated with its motion that is proportional to the speed (exactly to the square of the speed). When the water collides with the banks it is the kinetic energy of the river that erodes it Second question: the answer is the option d. As gravity pulls water down a slope potential energy changes to knietic energy. This is the, water loses altitude and gains velocity. The potential energy. which is proportional to the height, decreases and the kinetic energy, which is proportional to the square of the speed, increases.</span>
8 0
3 years ago
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