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3 years ago

6

A particle is being accelerated through space by a 10 N force. Suddenly the particle encounters a second of 10 N force in the op

posite direction. The particle with both forces acting?

1 answer:

Lyrx [107]3 years ago

7 0

Answer:

The particle will be rendered still

Explanation:

The particle will be rendered still because of the Newton’s 1st law of motion states that a body will continue to move on forever until contacted by an equal and opposite force. The same case has been produced here and it is justifying the law. If the Force has another direction and magnitude than 10 N then the body will continue to move on in the direction of the greater force.

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2 years ago

The periodic wave in the diagram below has a frequency of 80. hertz.

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Answer:

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2 years ago

The acceleration due to gravity on the surface of Jupiter is about 2.5 times the acceleration due to gravity on Earth’s surface.

Ann [662]

Answer: The correct answer is option C.

Explanation:

Weight = Mass × Acceleration

Let the mass of the space probe be m

Acceleration due to gravity on the earth = g

Weight of the space probe on earth = W

$W=m\times g$

Acceleration due to gravity on the Jupiter = g' = 2.5g

Weight of the space probe on earth = W'

$W'=mg'=m\times 2.5g$

$\frac{W'}{W}=\frac{m\times 2.5g}{m\times g}$

$W'=2.5\times W$

The weight of the space probe on the Jupiter will be 2.5 times the weight of the space probe on earth.

Hence, the correct answer is option C.

6 0

2 years ago

A particle has a charge of q = +4.9 μC and is located at the origin. As the drawing shows, an electric field of Ex = +242 N/C ex

irina1246 [14]

a)

$F_{E_x}=1.19\cdot 10^{-3}N$ (+x axis)

$F_{B_x}=0$

$F_{B_y}=0$

b)

$F_{E_x}=1.19\cdot 10^{-3} N$ (+x axis)

$F_{B_x}=0$

$F_{B_y}=3.21\cdot 10^{-3}N$ (+z axis)

c)

$F_{E_x}=1.19\cdot 10^{-3} N$ (+x axis)

$F_{B_x}=3.21\cdot 10^{-3} N$ (+y axis)

$F_{B_y}=3.21\cdot 10^{-3}N$ (-x axis)

Explanation:

a)

The electric force exerted on a charged particle is given by

$F=qE$

where

q is the charge

E is the electric field

For a positive charge, the direction of the force is the same as the electric field.

In this problem:

$q=+4.9\mu C=+4.9\cdot 10^{-6}C$ is the charge

$E_x=+242 N/C$ is the electric field, along the x-direction

So the electric force (along the x-direction) is:

$F_{E_x}=(4.9\cdot 10^{-6})(242)=1.19\cdot 10^{-3} N$

towards positive x-direction.

The magnetic force instead is given by

$F=qvB sin \theta$

where

q is the charge

v is the velocity of the charge

B is the magnetic field

$\theta$ is the angle between the directions of v and B

Here the charge is stationary: this means $v=0$ , therefore the magnetic force due to each component of the magnetic field is zero.

b)

In this case, the particle is moving along the +x axis.

The magnitude of the electric force does not depend on the speed: therefore, the electric force on the particle here is the same as in part a,

$F_{E_x}=1.19\cdot 10^{-3} N$ (towards positive x-direction)

Concerning the magnetic force, we have to analyze the two different fields:

- $B_x$ : this field is parallel to the velocity of the particle, which is moving along the +x axis. Therefore, $\theta=0^{\circ}$ , so the force due to this field is zero.

$- B_y$ : this field is perpendicular to the velocity of the particle, which is moving along the +x axis. Therefore, $\theta=90^{\circ}$ . Therefore, $\theta=90^{\circ}$ , so the force due to this field is:

$F_{B_y}=qvB_y$

where:

$q=+4.9\cdot 10^{-6}C$ is the charge

$v=345 m/s$ is the velocity

$B_y = +1.9 T$ is the magnetic field

Substituting,

$F_{B_y}=(4.9\cdot 10^{-6})(345)(1.9)=3.21\cdot 10^{-3} N$

And the direction of this force can be found using the right-hand rule:

- Index finger: direction of the velocity (+x axis)

- Middle finger: direction of the magnetic field (+y axis)

- Thumb: direction of the force (+z axis)

c)

As in part b), the electric force has not change, since it does not depend on the veocity of the particle:

$F_{E_x}=1.19\cdot 10^{-3}N$ (+x axis)

For the field $B_x$ , the velocity (+z axis) is now perpendicular to the magnetic field (+x axis), so the force is

$F_{B_x}=qvB_x$

And by substituting,

$F_{B_x}=(4.9\cdot 10^{-6})(345)(1.9)=3.21\cdot 10^{-3} N$

And by using the right-hand rule:

- Index finger: velocity (+z axis)

- Middle finger: magnetic field (+x axis)

- Thumb: force (+y axis)

For the field $B_y$ , the velocity (+z axis) is also perpendicular to the magnetic field (+y axis), so the force is

$F_{B_y}=qvB_y$

And by substituting,

$F_{B_y}=(4.9\cdot 10^{-6})(345)(1.9)=3.21\cdot 10^{-3} N$

And by using the right-hand rule:

- Index finger: velocity (+z axis)

- Middle finger: magnetic field (+y axis)

- Thumb: force (-y axis)

3 0

3 years ago

A bullet is fired horizontally from a 17.7 m high cliff at a speed of 482.0. What is the distance from the cliff that the bullet

Marysya12 [62]

Answer:

Horizontal distance=?m

Explanation:

Horizontal velocity,u=482ms⁻¹

Height of the cliff=17.7m

Horizontal distance,R=?

R=v×√2h/g

5 0

2 years ago