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marysya [2.9K]
3 years ago
13

Can someone help with this question please :) will mark brainliest

Physics
1 answer:
maria [59]3 years ago
5 0

Answer:

The answer should be south

Explanation:

Because it has more force to the south then to the north, west and east are the same so (40N South)

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A charged particle moves in a circular path in a uniform magnetic field.Which of the following would increase the period of the
Bond [772]

Answer:

Increasing its charge

Increasing the field strength

Explanation:

For a charged particle moving in a circular path in a uniform magnetic field, the centripetal force is provided by the magnetic force, so we can write:

qvB = m\frac{v^2}{r}

where

q is the charge

v is the velocity

B is the magnetic field

m is the mass

r is the radius of the orbit

The period of the motion is

T=\frac{2\pi r}{v}

Re-arranging for r

r=\frac{Tv}{2\pi}

And substituting into the previous equation

qvB = m \frac{Tv^3}{2\pi}

Solving for T,

T=\frac{2\pi q B}{m v^2}

So we see that the period is:

- proportional to the charge and the magnetic field

- inversely proportional to the mass and the square of the speed

So the following will increase the period of the particle's motion:

Increasing its charge

Increasing the field strength

4 0
3 years ago
True or False: Amplitude can increase or decrease wavelength
Marat540 [252]

Answer:

A. False, frequency can increase or decrease wavelength.

For example: a high frequency would mean there are shorter wavelengths that occur in a period. Meanwhile, a low frequency would indicate that the wavelengths are longer and in longer periods.

4 0
3 years ago
Comets travel around the sun in elliptical orbits with large eccentricities. If a comet has speed 1.6×104 m/s when at a distance
xz_007 [3.2K]

Answer:

v₂ = 7.6 x 10⁴ m/s

Explanation:

given,

speed of comet(v₁) = 1.6 x 10⁴ m/s

distance (d₁)= 2.7 x 10¹¹ m

to find the speed when he is at distance of(d₂) 4.8 × 10¹⁰ m

v₂ = ?

speed of planet can be determine using conservation of energy

K.E₁ + P.E₁ = K.E₂ + P.E₂

\dfrac{1}{2}mv_1^2-\dfrac{GMm}{r_1} = \dfrac{1}{2}mv_2^2-\dfrac{GMm}{r_2}

\dfrac{1}{2}v_1^2-\dfrac{GM}{r_1} = \dfrac{1}{2}v_2^2-\dfrac{GM}{r_2}

v_2^2= v_1^2 + \dfrac{2GM}{r_2}-\dfrac{2GM}{r_1}

v_2= \sqrt{v_1^2 +2GM(\dfrac{1}{r_2}-\dfrac{1}{r_1})}

v_2= \sqrt{(1.6\times 10^4)^2 +2\times 6.67 \times 10^{-11}\times 1.99 \times 10^{30}(\dfrac{1}{4.8\times 10^{10}}-\dfrac{1}{2.7\times 10^{11}})}

v₂ = 7.6 x 10⁴ m/s

3 0
3 years ago
a 700k/g race car slowed down from 30m/s to 15m/s what is the impulse and what was the average force applied by the brakes
posledela
So impulse is a change in momentum.

Mass*(final velocity - initial velocity)

I dont think you will be able to find the average force with the given info because you need to know the time it takes for the car to slow down.
5 0
3 years ago
When Bella pushes a box with a mass of 5.25 kilograms with a force of 15.75 newtons, it accelerates at a rate of 2.5 meters/seco
Mariana [72]

Sure !

Start with Newton's second law of motion:

                     Net Force = (mass) x (acceleration) .

This formula is so useful, and so easy, that you really
should memorize it.

Now, watch:

The mass of the box is 5.25 kilograms, and the box is
accelerating at the rate of  2.5 m/s² .
What's the net force on the box ?

                    Net Force = (mass) x (acceleration)

                                     = (5.25 kilograms) x (2.5 m/s²)

                     Net force =       13.125 newtons .

But hold up, hee haw, whoa !  Wait a second !
Bella is pushing with a force of 15.75 newtons, but the box
is accelerating as if the force on it is only 13.125 newtons.
What happened to the rest of Bella's force ? ?

==>  Friction is pushing the box in the opposite direction,
and cancelling some of Bella's force.

How much ?

            (Bella's 15.75 newtons) minus (13.125 that the box feels)

           =      2.625 newtons backwards, applied by friction.


5 0
3 years ago
Read 2 more answers
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