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nataly862011 [7]
3 years ago
7

A 2.0-kg object moving 3.0 m/s strikes a 1.0-kg object initially at rest. immediately after the collision, the 2.0-kg object has

a velocity of 1.5 m/s directed 30o from its initial direction of motion. what is the x-component of the velocity of the 1.0-kg object just after the collision?
Physics
1 answer:
Stells [14]3 years ago
5 0
The answer is 1.5 m/s
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Ultraviolet rays from the sun are very strong, therefore if direct contact is made with skin, it could cause sun burn, or in a worser case skin cancer
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Which statement comparing examples of renewable and nonrenewable resources is true?
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I believe the answer is A
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A beam of protons is directed in a straight line along the +z direction through a region of space in which there are crossed ele
IceJOKER [234]

Answer:

B = 7.6 T   direction of + x

Explanation:

For the proton beam to continue in the same direction the electric and magnetic forces must be equal

              F_{m} - F_{e} = 0

              F_{m} = F_{e}

              Fm = q E

             

The electric force is in the direction of the electric field because it is the charge of the positive proton, the electric force goes in the direction of –y, therefore, the magnetic force cancels this force must go in the direction of + y

 The magnetic force is

             F_{m} = q v x B = q v B sin θ

             θ = 90

             B = q E / q v

             B = E / v

             B = 800/105

             B = 7.6 T

To find the direction of the magnetic field we use the right hand rule, the thumb goes in the direction of the proton velocity, the fingers extended in the direction of the magnetic field and the palm is the direction of force, for a positive charge.

Thumb goes in the direction of the + z axis

Palm in the direction of +y

Fingers point in the direction of + x

7 0
3 years ago
A uniform, solid, 1800.0 kgkg sphere has a radius of 5.00 mm. Find the gravitational force this sphere exerts on a 2.30 kgkg poi
iVinArrow [24]

Answer

given,

Mass of the solid sphere = 1800 Kg

radius of the sphere,R = 5 m

mass of the small sphere, m = 2.30 Kg

when the Point is outside the sphere the Force between them is equal to

      F = \dfrac{GMm}{r^2}   when r>R

When Point is inside the Sphere

      F = \dfrac{GMm}{R^2}\ \dfrac{r}{R}  when r<R

where r is the distance where the point mass is placed form the center

Now Force calculation

a) r = 5.05 m

       F = \dfrac{GMm}{r^2}[/tex]

       F = \dfrac{6.67\times 10^{-11}\times 1800\times 2.3}{5.05^2}

               F = 1.082 x 10⁻⁸ N

b) r = 2.65 m

      F = \dfrac{GMm}{R^2}\ \dfrac{r}{R}

      F = \dfrac{6.67\times 10^{-11}\times 1800\times 2.3}{5.05^2}\ \dfrac{2.65}{5.05}

     F = 5.68\times 10^{-9}\ N

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3 years ago
What is a tiltmeter?
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A tiltmeter is an object to measure small movements.
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