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anzhelika [568]
3 years ago
9

A crate of apples weighing 71.4 N sits at rest on a ramp that runs from the ground to the bed of a truck. The ramp is inclined a

t 20.0° to the ground. What is the interaction partner of normal force?
1. If the interaction partner is acting above the surface of the ramp, enter a positive value.
2. If the interaction partner is acting below the surface of the ramp, enter a negative value.
Physics
1 answer:
nika2105 [10]3 years ago
5 0

Answer:

Fint = - 67.094 N

Explanation:

Given

W = 71.4 N

∅ = 20º

Fint = ?

The interaction partner of the Normal Force has equal magnitude and opposite direction to the Normal Force, and it is a contact force exerted by the crate on the ramp, because the interaction parteners are the same force acts on the two object.

Then:    Fint = - N = - Wy' = - W*Cos ∅  

⇒   Fint = - 71.4 N*Cos 20º = - 67.094 N

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alexgriva [62]

Answer:

4 gamma closest thing to this V

Explanation:

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6 0
3 years ago
How do microwaves use the behavior and characteristics of the electromagnetic wave to function?
serg [7]

Answer:

Radiation moves out of the microwave into waves causing heat.

Explanation:

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Hope it helped!

6 0
3 years ago
Vector A has a magnitude of 30 units. Vector B is perpendicular to vector Aand has a magnitude of 40 units. What would the magni
Fudgin [204]

Answer:

|\vec A + \vec B| = 50 units

Explanation:

As we know that magnitude of two vectors is given as

|\vec A + \vec B| = \sqrt{A^2 + B^2 + 2AB cos\theta}

here we know that

A = magnitude of vector A

B = magnitude of vector B

\theta = angle between two vectors

so here we know that

A = 30 units

B = 40 units

angle = 90 degree

so we have

|\vec A + \vec B| = \sqrt{30^2 + 40^2 + 2(30)(40)cos90}

|\vec A + \vec B| = \sqrt{30^2 + 40^2}

|\vec A + \vec B| = 50 units

3 0
3 years ago
An ambulance driver traveling at 31.0 m/s (69.3 mph) honks his horn as he sees a motorist ahead on the highway traveling in the
IrinaVladis [17]

Answer:

fo = 378.52Hz

Explanation:

Using Doppler effect formula:

f'=\frac{C-Vb}{C-Va}*fo

where

f' = 392 Hz

C = 340m/s

Vb = 20m/s

Va = 31m/s

Replacing these values and solving for fo:

fo = 378.52Hz

4 0
3 years ago
A system absorbs 194 kj of heat and the surroundings do 120 kj of work on the system. internal eneergy change
notsponge [240]
We can solve the problem by using the first law of thermodynamics, which states that:
\Delta U = Q-W
where
\Delta U is the change in internal energy of the system
Q is the heat absorbed by the system
W is the work done by the system

In our problem, the heat absorbed by the system is Q=+194 kJ, while the work done is W=-120 kJ, where the negative sign means the work is done by the surroundings on the system. Therefore, the variation of internal energy is
\Delta U= Q-W=+194 kJ - (-120 kJ)=+314 kJ
6 0
3 years ago
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