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Softa [21]
3 years ago
13

You have landed on an unknown planet, newtonia, and want to know what objects will weigh there. you find that when a certain too

l is pushed on a frictionless horizontal surface by a 12.7 n force, it moves 16.1 m in the first 2.50 s , starting from rest. you next observe that if you release this tool from rest at 10.3 m above the ground, it takes 2.88 s to reach the ground.

Physics
1 answer:
AleksandrR [38]3 years ago
3 0
Refer to the diagram shown.

Because the surface is frictionless, the resistive for, R, is zero.

Let m = the mass of the object.
Let a =  acceleration due to the applied force.
Therefore
12.7 N = (m kg)*(a m/s²)
a = 12.7/m  m/s²

The object travels  16.1 m in 2.5 s, starting from rest. Therefore
16.1 N = (1/2)*(12.7/m m/s²)*(2.5 s)² = 39.6875/m  N
m = 16.1/39.6875 = 0.4057 kg

For freefall, let g =  acceleration due to gravity.
The time to fall from 10.3 m is 2.88 s, therefore
10.3 m = (1/2)*(g m/s²)*(2.88 s)² = 4.1472g m
g = 10.3/4.1472 = 2.484 m/s²

Answer:
The gravitational acceleration on the planet is 2.5 m/s² (nearest tenth)


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Answer:

d = 1.55 * 10⁻⁶ m

Explanation:

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The fringe number, m = 1 since it is a first order maximum

The wavelength of the green laser pointer, \lambda = 532 nm = 532 * 10⁻⁹ m

Distance between the central maximum and the first order maximum = 1.1 m

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A_{m} = Angle between the incident light and the diffracted light

From the setup shown in the attachment, it is a right angled triangle in which

sin(A_{m}) = \frac{opposite}{Hypotenuse} \\sin(A_{m}) =\frac{1.1}{\sqrt{1.1^{2}+3^{2}}}

sin(A_{m} ) = 0.344\\A_{m} = sin^{-1} 0.344\\A_{m} = 20.14^{0}

Putting all appropriate values into equation (1)

d = \frac{1* 532*10^{-9} }{0.344 }\\d = 0.00000155 m\\d = 1.55 * 10^{-6} m

3 0
3 years ago
A 34-m length of wire is stretched horizontally between two vertical posts. The wire carries a current of 68 A and experiences a
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Answer:

7.28×10⁻⁵ T

Explanation:

Applying,

F = BILsin∅............. Equation 1

Where F = magnetic force, B = earth's magnetic field, I = current flowing through the wire, L = Length of the wire, ∅ = angle between the field and the wire.

make B the subject of the equation

B = F/ILsin∅.................. Equation 2

From the question,

Given: F = 0.16 N, I = 68 A, L = 34 m, ∅ = 72°

Substitute these values into equation 2

B = 0.16/(68×34×sin72°)

B = 0.16/(68×34×0.95)

B = 0.16/2196.4

B = 7.28×10⁻⁵ T

7 0
2 years ago
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