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Y_Kistochka [10]
2 years ago
10

1) When making a digital animation of a person running on a sidewalk in a scene, which parameter would be an initial condition?

Physics
1 answer:
lbvjy [14]2 years ago
4 0

Answer:

The position of the person's feet. a running shoe company studying how different surfaces affect the life of a shoe tread

Explanation:

100%

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Please help me find out this answer
Elanso [62]
The direction of work.........

4 0
2 years ago
A 10-kg rock falls from a height of 8-m above the ground. What is the kinetic energy of the rock just before it hits the ground?
pishuonlain [190]

Answer: 800

Explanation:

1/2 x m x v^2 = m x g x h

KE = 10 x 10 x 8

KE= 800

3 0
2 years ago
What is the kinetic energy of a 478 kg object that is moving with the speed of 15 m/s
Tpy6a [65]
Kinetic energy = (1/2)*mass*velocity^2
KE = (1/2)mv^2
KE = (1/2)(478)(15)^2
KE = 53775J
5 0
3 years ago
An isolated charged point particle produces an electric field with magnitude E at a point 2 m away. At a point 1 m from the part
guajiro [1.7K]

Explanation:

The electric field at a distance r from the charged particle is given by :

E=\dfrac{kq}{r^2}

k is electrostatic constant

if r = 2 m, electric field is given by :

E_1=\dfrac{kq}{(2)^2}\\\\=\dfrac{kq}{4}\ .....(1)

If r = 1 m, electric field is given by :

E_2=\dfrac{kq}{r_2^2}\\\\=\dfrac{kq}{1}\ ....(2)

Dividing equation (1) and (2) we get :

\dfrac{E_1}{E_2}=\dfrac{\dfrac{kq}{4}}{kq}\\\\\dfrac{E_1}{E_2}=\dfrac{1}{4}\\\\E_2=4\times E_1

So, at a point 1 m from the particle, the electric field is 4 times of the electric field at a point 2 m.

4 0
2 years ago
How many electrons flow through a point in a wire in 7.00 s if there is a constant current of I = 4.35 A?
OLEGan [10]

Answer:

1.90×10²⁰ Electrons

Explanation:

From the question,

Q = It.................... Equation 1

Where Q = charge flowing through the wire, I = current, t = time

Given: I = 4.35 A, t = 7.00 s

Substitute these values into equation 1

Q = 4.35(7.00)

Q = 30.45 C.

But,

1 electron contains 1.6×10⁻¹⁹ C

therefore,

30.45 C = 30.45/1.6×10⁻¹⁹  electrons

= 1.90×10²⁰ Electrons

8 0
3 years ago
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