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3 years ago

John traveled East at 10 m/s for ten

Physics

1 answer:

NNADVOKAT [17]3 years ago

5 0

Answer:

Option D

130 m

Explanation:

From the concept of speed, distance=speed*time

When traveling East, displacement=10*10=100 m (since he takes 10 seconds while traveling at a speed of 10 m/s)

When traveling West, displacement=5*6=30 m (since John takes 6 seconds to travel at a speed of 5 m/s)

Total displacement=Displacement East+ Displacement West=100+30=130 m

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PLEASE HELP ME What does a moving charge experience when it is near a magnetic field? static friction gravity a stationary charg

attashe74 [19]

A moving charge experiences a force as it moves close to a magnetic field present in a region.

Explanation:

A charged particle when moves near a magnetic field is forced to move in a circular path. If there is a straight moving charged particle enters the region of the magnetic field, it's path is changed from straight path to the curved path.

The motion of the charged path inside or near the magnetic field will be a circular path and this circular path is due to the force experienced by the charged particle in the magnetic field.

The force experienced by the a charged particle as it moves near the magnetic field is termed as the Lorentz force. The expression for the Lorentz force experienced by the charged particle in the magnetic field is given by:

$\boxed{F_m=Q(\overrightarrow{v}\times\overrightarrow{B})}$

Here, $F_m$ is the magnetic force, $Q$ is the amount of charge, $\overrightarrow{v}$ is the velocity vector and $\overrightarrow{B}$ is the magnetic field intensity in the region.

Thus, A moving charge experiences a force as it moves close to a magnetic field present in a region.

Learn More:

1. Which one is true for the electromagnetic spectrum brainly.com/question/1619496

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Answer Details:

Grade: High School

Subject: Physics

Chapter: Electromagnetism

Keywords:

moving, charge, straight, circular, path, magnetic, field, amount, force, velocity, intensity, Lorentz, expression, stationary.

6 0

3 years ago

A firework is launched with a force of 700 N and a momentum of 200 kg-m/s. How much time before is explodes?

8_murik_8 [283]

Answer:

t = 0.28 seconds

Explanation:

Given that,

Force acting on a firework, F = 700 N

The momentum of the firework, p =200 kg-m/s

We need to find the time before it explodes. Ket the time be t. We know that, the rate of change of momentum is equal to external frce. So,

$F=\dfrac{P}{t}\\\\t=\dfrac{P}{F}\\\\t=\dfrac{200}{700}\\t=0.28\ s$

So, the required time is equal to 0.28 seconds.

7 0

3 years ago

A projectile is fired with an initial speed of 60.3 m/s at an angle of 34.2 above the horizontal on a long flat firing range.

amid [387]

Answer:

A.) H = 58.6 m

B.) T = 6.92 s

C.) 345.12 m

D.) V = 22.13 m/s

E.) Ø = 32.1 degree

Explanation:

Given that the

initial speed U = 60.3 m/s

Angle Ø = 34.2 degree

A.) At maximum height, final velocity V is equal to zero.

Using the third equation of motion under gravity.

V^2 = U sin Ø^2 - 2gH

Substitute for U and g. Where g = 9.8 m/s^2

0 = (60.3 sin 34.2)^2 - 2 × 9.8 × H

1148.78 = 19.6 H

H = 1148.78/19.6

H = 58.6 m

B.) To Determine the total time in the air, let us use the formula

V = UsinØ - gt

At maximum height, V = 0

t = UsinØ/g

Total time T = 2t

Therefore, T = 2UsinØ/g

T = (2 × 60.3 × sin 34.2)/9.8

T = 67.79/9.8

T = 6.92 s

C.) To determine the total horizontal distance covered which is the range, we will use second equation of motion.

S = UcosØT - 1/2gt^2

Where S = range R

g = 0, since the range is not a vertical distance

T = total time

Substitute all the parameters into the formula

R = 60.3 cos 34.2 × 6.92

R = 345.12 m

D.) After 1.2 s firing,

V = UsinØ - gt

Where t = 1.2 s

Substitute into the formula

V = 60.3 × sin34.2 - 9.8 × 1.2

V = 33.89 - 11.76

V = 22.13 m/s

Therefore the speed of the projectile 1.20 s after firing is 22.13 m/s

E.) The direction will be determined by using the formula

t = VsinØ/ g

Cross multiply

VsinØ = gt

Make SinØ the subject of formula

SinØ = gt/V

SinØ = (9.8×1.2)/22.13

Sin Ø = 11.76/22.13

Sin Ø = 0.53

Ø = sin^-1( 0.53 )

Ø = 32.1 degree

3 0

3 years ago