Ask question

Ask question

All categories

2 years ago

7

Newton's second law A. describes how an object accelerates when a force is applied B. says that objects eventually stop unless a

force is applied C. objects with mass attract each other D. forces come in action/reaction pairs E. an object will remain in uniform motion unless acted upon by a force F. like charges repel, opposite charges attract

1 answer:

Papessa [141]2 years ago

8 0

Answer:

A. describes how an object accelerates when a force is applied

Explanation:

Newton's second law of motion concerns the behavior of objects that do not have a stability between all established forces. The second law states that an object's acceleration depends on two factors: the net force on the entity and the entity's mass.

You might be interested in

makkiz [27]

Answer:

i done know sorry for not answering ur question

3 0

2 years ago

What must the charge (sign and magnitude) of a particle of mass 1.41 gg be for it to remain stationary when placed in a downward

Yuri [45]

Answer:

$q = 2.067 \times 10^{-5}\ C$

Explanation:

Given,

mass = 1.41 g = 0.00141 Kg

Electric field,E = 670 N/C.

We know,

Force in charge due to Electric field.

F = E q

And also we know

F = m g

Equating both the equation of motion

m g = E q

$q =\dfrac{mg}{E}$

$q =\dfrac{0.00141 \times 9.81}{670}$

$q = 2.067 \times 10^{-5}\ C$

Charge of the particle is equal to $q = 2.067 \times 10^{-5}\ C$

6 0

3 years ago

Can someone please help?

kenny6666 [7]

Answer:

uh with what?

Explanation:

4 0

2 years ago

Bill throws a tennis ball to his dog. He throws the ball at a speed of 15 m/s at an angle of 30° to the horizontal. Assume he th

Sidana [21]

1a) Bill and the dog must have a speed of 13.0 m/s

1b) The speed of the dog must be 22.5 m/s

2a) The ball passes over the outfielder's head at 3.33 s

2b) The ball passes 1.2 m above the glove

2c) The player can jump after 2.10 s or 3.13 s after the ball has been hit

2d) One solution is when the player is jumping up, the other solution is when the player is falling down

Explanation:

1a)

The motion of the ball in this problem is a projectile motion, so it follows a parabolic path which consists of two independent motions:

- A uniform motion (constant velocity) along the horizontal direction

- An accelerated motion with constant acceleration (acceleration of gravity) in the vertical direction

In part a), we want to know at what speed Bill and the dog have to run in order to intercept the ball as it lands on the ground: this means that Bill and the dog must have the same velocity as the horizontal velocity of the ball.

The ball's initial speed is

u = 15 m/s

And the angle of projection is

$\theta=30^{\circ}$

So, the ball's horizontal velocity is

$v_x = u cos \theta = (15)(cos 30)=13.0 m/s$

And therefore, Bill and the dog must have this speed.

1b)

For this part, we have to consider the vertical motion of the ball first.

The vertical position of the ball at time t is given by

$y=u_yt+\frac{1}{2}at^2$

where

$u_y = u sin \theta = (15)(sin 30) = 7.5 m/s$ is the initial vertical velocity

$a=g=-9.8 m/s^2$ is the acceleration of gravity

The ball is at a position of y = 2 m above the ground when:

$2=7.5t + \frac{1}{2}(-9.8)t^2\\4.9t^2-7.5t+2=0$

Which has two solutions: t=0.34 s and t=1.19 s. We are told that the ball is falling to the ground, so we have to consider the second solution, t = 1.19 s.

The horizontal distance covered by the ball during this time is

$d=v_x t =(13.0)(1.19)=15.5 m$

The dog must be there 0.5 s before, so at a time

t' = t - 0.5 = 0.69 s

So, the speed of the dog must be

$v_x' = \frac{d}{t'}=\frac{15.5}{0.69}=22.5 m/s$

2a)

Here we just need to consider the horizontal motion of the ball.

The horizontal distance covered is

$d=98 m$

while the horizontal velocity of the ball is

$v_x = u cos \theta = (34)(cos 30)=29.4 m/s$

where u = 34 m/s is the initial speed.

So, the time taken for the ball to cover this distance is

$t=\frac{d}{v_x}=\frac{98}{29.4}=3.33 s$

2b)

Here we need to calculate the vertical position of the ball at t = 3.33 s.

The vertical position is given by

$y= h + u_y t + \frac{1}{2}at^2$

where

h = 1.2 m is the initial height

$u_y = u sin \theta = (34)(sin 30)=17.0 m/s$ is the initial vertical velocity

$a=g=-9.8 m/s^2$ is the acceleration of gravity

Substituting t = 3.33 s,

$y=1.2+(17)(3.33)+\frac{1}{2}(-9.8)(3.33)^2=3.5 m$

And sinc the glove is at a height of y' = 2.3 m, the difference in height is

y - y' = 3.5 - 2.3 = 1.2 m

2c)

In order to intercept the ball, he jumps upward at a vertical speed of

$u_y' = 7 m/s$

So its position of the glove at time t' is

$y'= h' + u_y' t' + \frac{1}{2}at'^2$

where h' = 2.3 m is the initial height of the glove, and t' is the time from the moment when he jumps. To catch the ball, the height must be

y' = y = 3.5 m (the height of the ball)

Substituting and solving for t', we find

$3.5 = 2.3 + 7t' -4.9t'^2\\4.9t'^2-7t'+12 = 0$

Which has two solutions: t' = 0.20 s, t' = 1.23 s. But this is the time t' that the player takes to reach the same height of the ball: so the corresponding time after the ball has been hit is

$t'' = t -t'$

So we have two solutions:

$t'' = 3.33 s - 0.20 s = 3.13 s\\t'' = 3.33 s - 1.23 s = 2.10 s$

So, the player can jump after 2.10 s or after 3.13 s.

2d)

The reason for the two solutions is the following: the motion of the player is a free fall motion, so initially he jump upwards, then because of gravity he is accelerated downward, and therefore eventually he reaches a maximum height and then he falls down.

Therefore, the two solutions corresponds to the two different part of the motion.

The first solution, t'' = 2.10 s, is the time at which the player catches the ball while he is in motion upward.

On the other hand, the second solution t'' = 3.13 s, is the time at which the player catches the ball while falling down.

Learn more about projectile motion:

brainly.com/question/8751410

#LearnwithBrainly

7 0

3 years ago

Den pushes a desk 400 cm across the floor. He exerts a force of 10 N for 8 s to move the desk.

stellarik [79]

Answer: The correct option is Option b.

Explanation:

Power is defined as the rate of work done by an object.

Mathematically,

$P=\frac{W}{t}$ .....(1)

And work done is the product of force exerted on the object times the displacement covered by that object.

Mathematically,

$W=F.s$

Putting this value in above equation, we get:

$P=\frac{F.s}{t}$

where,

P = power = ?W

F = Force exerted = 10N

s = Displacement = 400cm = 4m (Conversion factor: 1m = 100 cm)

t = Time taken = 8s

Putting values in above equation, we get

$P=\frac{10\times 4}{8}\\\\P=5W$

Hence, the correct option is Option b.

7 0

3 years ago

Read 2 more answers