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3 years ago

The__ of friction is a number that represents the resistance to sliding

Physics

1 answer:

Tresset [83]3 years ago

3 0

Answer:

B. coefficient

Explanation:

i dont have to explain right?

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What is an amu? <br><br>lol I'm asking alot of questions<br><br><br><br><br><br><br>

forsale [732]

Atomic Mass Unit is the answer

6 0

2 years ago

The diagram shows two forces of equal magnitude acting on an object. If the common magnitude of the forces is 3.6 N and the angl

Nuetrik [128]

<h3>Answer</h3>

6.6 N pointing to the right

<h3>Explanation</h3>

Given that,

two forces acting of magnitude 3.6N

angle between them = 48°

To find,

the third force that will cause the object to be in equilibrium

Find the vertical and horizontal components of the two forces

vertical force1 = sin(24)(3.6)

vertical force2= -sin(24)(3.6)

<em>(negative sign since it is acting on opposite direction)</em>

vertical force3 = sin(24)(3.6) - sin(24)(3.6)

= 0

horizontal force1 = cos(24)(3.6)

horizontal force2= cos(24)(3.6)

horizontal force3 = cos(24)(3.6) + cos(24)(3.6)

= 2(cos(24)(3.6))

= 6.5775 N

≈ 6.6 N

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4 0

3 years ago

A child pushes a toy car down a hill. The child has a mass of 20 kg. The car has a mass of 1.6 kg and a speed of 7.4 m/s2. When

Andrew [12]

Answer:

The answer is 73.8 J

Explanation:

5 0

3 years ago

Help me plssss <br> 30 pointsssssss

labwork [276]

Answer:

Explanation:

6 0

3 years ago

Read 2 more answers

A proton is initially moving west at a speed of 1.10 106 m/s in a uniform magnetic field of magnitude 0.281 T directed verticall

kifflom [539]

Answer:

so here it will move in circle with radius 4.06 cm

Explanation:

As we know that proton is moving towards west while the magnetic field is vertically upwards

So here the force on the proton must be perpendicular to the velocity

So here we have

$F = q(\vec v \times \vec B)$

so here we have

$F = qvB sin90$

since force is perpendicular to the velocity so here it must be centripetal force

here we have

$\frac{mv^2}{R} = qvB$

so we have

$R = \frac{mv}{qB}$

$R = \frac{(1.66 \times 10^{-27})(1.10 \times 10^6)}{(1.6 \times 10^{-19})(0.281)}$

$R = 4.06 cm$

so here it will move in circle with radius 4.06 cm

8 0

3 years ago