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Marta_Voda [28]
3 years ago
6

An alpha particle travels at a velocity of magnitude 550 m/s through a uniform magnetic field of magnitude 0.045T. (An alpha par

ticle has a charge of charge of +3.2 x 10-19 C and a mass 6.6 x 10-27 kg) The angle between the particle's direction of motion and the magnetic field is 52°. What happens to the speed? The answer is that the speed remains the same and I want to know why and how would we know if the speed changes ( there are two questions before this about the force and acceleration force is 6.2x10^-18 and acceleration is 9.5x10^8 )
Physics
1 answer:
creativ13 [48]3 years ago
3 0

Answer:

the force is perpendicular to the speed, it is a type of force that changes the direction of the speed, as in the uniform circular motion te, but does not change its modulus.

Explanation:

The magnetic force is given by the expression

    F = q v x B

The bold are vectors, where v is the velocity and B is the magnetic field, the product is the cross product whose result is a vector perpendicular to the two vectors (v and B)

From the above, the force is perpendicular to the speed, it is a type of force that changes the direction of the speed, as in the uniform circular motion te, but does not change its modulus.

   

Even when the change in direction is real and is caused by a centripetal force

For there to be a change in the velocity modulus there must be a force parallel to the velocity direction, generally a force in electrical

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___ acceleration occurs when an object speeds up
AleksandrR [38]

Explanation:

Acceleration is defined as the change in velocity over time.

When there is an increment or increase in the magnitude of velocity of a moving body then it is known as positive acceleration.

Whereas when there is a decrease in magnitude of velocity of a moving body then it is known as negative acceleration.

Thus, we can conclude that positive acceleration occurs when an object speeds up.

3 0
3 years ago
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A "moving sidewalk" in an airport terminal moves at 1.0 m/s and is 35.0 m long. If a woman steps on at one end and walks at 1.5
pishuonlain [190]

Answer:

a.14 s

b.70 s

Explanation:

a.Let the sidewalk moving in positive x- direction.

Speed  of sidewalk relative to ground=v_s=1m/s

Speed of women relative to sidewalk=v=1.5m/s

The speed of women relative to the ground

v_w=v_s+v=1+1.5=2.5m/s

Distance=35 m

Time=\frac{distance}{speed}

Using the formula

Time taken by women to reach the opposite end if she walks in the same direction the sidewalk is moving=\frac{35}{v_w}=\frac{35}{2.5}=14s

b.If she gets on at the end opposite the end in part (a)

Then, we take displacement negative.

Speed  of sidewalk relative to ground=v_s=1m/s

Speed of women relative to sidewalk=v=-1.5 m/s

The speed of women relative to the ground=v_w=v_s+v=1-1.5=-0.5m/s

Time=\frac{-35}{-0.5}=70 s

Hence, the women takes 70 s to reach the opposite end if she walks in the opposite direction the sidewalk is moving.

3 0
3 years ago
How to derive the fourth equation of motion?
Leya [2.2K]

Answer:

To derive the fourth equation of motion, first we have to consider the equation for acceleration and then to rearrange it. or v2 = u2 + 2as and this equation of motion can be used to find the final velocity or the distance travelled if the other values are given.

Explanation:

v= u + at

s =( u + v ) t /2

s = ut + at2/2

v2 = u2 + 2as

6 0
3 years ago
A bus travels east for 3 km the north for 4 km what is its final displacement
lesya [120]
Im about to the math for this right now.
8 0
3 years ago
HELP PLEASE
4vir4ik [10]

Recall this kinematic equation:

a = \frac{Vi+Vf}{Δt}

This equation gives the acceleration of the object assuming it IS constant (the velocity changes at a uniform rate).

a is the acceleration.

Vi is the initial velocity.

Vf is the final velocity.

Δt is the amount of elapsed time.


Given values:

Vi = 0 m/s (the car starts at rest).

Vf = 25 m/s.

Δt = 10 s


Substitute the terms in the equation with the given values and solve for a:

a = \frac{0+25}{10}

<h3>a = 2.5 m/s²</h3>
7 0
4 years ago
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