Ask question

Ask question

All categories

Lady bird [3.3K]

3 years ago

11

As a 3.0 kg bucket is being lowered into a 10 m deepwell, starting from top, the tension in the rope is 9.8 N. theacceleration o

f the bucket will be:____________A) 6.5 m/s2 downwardB) 9.8 m/s2 downwardC) zeroD) 3.3 m/s2 upwardE) 5.6 m/s2 upward

1 answer:

777dan777 [17]3 years ago

8 0

Answer:

A) 6.5 m/s²

Explanation:

Mass of the bucket, m = 3.0 kg

depth of the well, d = 10 m

tension on the rope, T = 9.8 N

The net downward force on the bucket is given as;

T = mg - ma

where;

a is downward acceleration of the bucket

9.8 = (3 x 9.8) - 3a

9.8 = 29.4 - 3a

3a = 29.4 - 9.8

3a = 19.6

a = 19.6 / 3

a = 6.53 m/s² downwards

Therefore, the acceleration of the bucket is 6.53 m/s² downwards

You might be interested in

The engineer of a passenger train traveling at 25.0 m/s sights a freight train whose caboose is 200 m ahead on the same track. T

zaharov [31]

a) The train collide after 22.5 seconds

b) The trains collide at the location x = 537.5 m

c) See graph in attachment

d) The freight train must have a head start of 500 m

e) The deceleration must be smaller (towards negative value) than $-0.25 m/s^2$

f) The two trains avoid collision if the acceleration of the freight train is at least $0.35 m/s^2$

Explanation:

a)

We can describe the position of the passenger train at time t with the equation

$x_p(t)=u_p t + \frac{1}{2}at^2$

where

$u_p = 25.0 m/s$ is the initial velocity of the passenger train

$a=-0.100 m/s^2$ is the deceleration of the train

On the other hand, the position of the freight train is given by

$x_f(t)=x_0 + v_f t$

where

$x_0=200 m$ is the initial position of the freight train

$v_f = 15.0 m/s$ is the constant velocity of the train

The collision occurs if the two trains meet, so

$x_p(t)=x_f(t)\\u_pt+\frac{1}{2}at^2=x_0+v_ft\\25t+\frac{1}{2}(-0.100)t^2=200+15t\\0.050t^2-10t+200=0$

This is a second-order equation that has two solutions:

t = 22.5 s

t = 177.5 s

We are interested in the 1st solution, which is the first time at which the passenger train collides with the freight train, so t = 22.5 seconds.

b)

In order to find the location of the collision, we just need to substitute the time of the collision into one of the expression of the position of the trains.

The position of the freight train is

$x_f(t)=x_0 +v_ft$

And substituting t = 22.5 s, we find:

$x_f(22.5)=200+(15)(22.5)=537.5 m$

We can verify that the passenger train is at the same position at the time of the collision:

$x_p(22.5)=(25.0)(22.5)+\frac{1}{2}(-0.100)(22.5)^2=537.5 m$

So, the two trains collide at x = 537.5 m.

c)

In the graph in attachment, the position-time graph of each train is represented. We have:

The freight train is moving at constant speed, therefore it is represented with a straight line with constant slope (the slope corresponds to its velocity, so 15.0 m/s)
The passenger train has a uniformly accelerated motion, so it is a parabola: at the beginning, the slope (the velocity) is higher than that of the freight train, however later it decreases due to the fact that the train is decelerating

The two trains meet at t = 22.5 s, where the position is 537.5 m.

d)

In order to avoid the collision, the freight train must have a initial position of

$x_0'$

such that the two trains never meet.

We said that the two trains meet if:

$x_p(t)=x_f(t)\\u_p t + \frac{1}{2}at^2=x_0' + v_f t$

Re-arranging,

$\frac{1}{2}at^2+(u_p-v_f)t-x_0'=0\\-\frac{1}{2}at^2+(v_f-u_p)t+x_0'=0$

Substituting the values for the acceleration and the velocity,

$0.05t^2-10t+x_0'=0$

The solution of this equation is given by the formula

$t=\frac{+10\pm \sqrt{10^2-4\cdot 0.05 \cdot x_0'}}{2(0.05)}$

The two trains never meet if the discrimant is negative (so that there are no solutions to the equation), therefore

$10^2-4\cdot 0.05 \cdot x_0'100\\x_0'>500 m$

Therefore, the freight train must have a head start of 500 m.

e)

In this case, we want to find the acceleration $a'$ of the passenger train such that the two trains do not collide.

We solve the problem similarly to part d):

$x_p(t)=x_f(t)\\u_p t + \frac{1}{2}a't^2=x_0 + v_f t$

Re-arranging

$\frac{1}{2}a't^2+(u_p-v_f)t-x_0=0\\-\frac{1}{2}a't^2+(v_f-u_p)t+x_0=0$

Substituting,

$-0.5at^2-10t+200=0$

The solution to this equation is

$t=\frac{+10\pm \sqrt{10^2-4\cdot (-0.5a') \cdot (200)}}{2(0.05)}$

Again, the two trains never meet if the discriminant is negative, so

$10^2-4\cdot (-0.5a') \cdot (200)$

So, the deceleration must be smaller (towards negative value) than $-0.25 m/s^2$

f)

In this case, the motion of the freight train is also accelerated, so its position at time t is given by

$x_f(t)=x_0 + v_f t + \frac{1}{2}a_ft^2$

where $a_f$ is the acceleration of the freight train.

Then we solve the problem similarly to the previous part: the two trains collide if their position is the same,

$x_p(t)=x_f(t)\\u_p t + \frac{1}{2}at^2=x_0 + v_f t+\frac{1}{2}a_ft^2$

Re-arranging,

$\frac{1}{2}(a_f-a)t^2+(v_f-u_p)t+x_0=0\\\\\frac{1}{2}(a_f-0.100)t^2-10t+200=0$

And the solution is

$t=\frac{+10\pm \sqrt{10^2-4\cdot (0.5a_f-0.05) \cdot (200)}}{2(0.5a_f-0.05)}$

Again, the two trains avoid collision if the discriminant is negative, so

$10^2-4\cdot (0.5a_f-0.05) \cdot (200)0.35 m/s^2$

Learn more about accelerated motion:

brainly.com/question/9527152

brainly.com/question/11181826

brainly.com/question/2506873

brainly.com/question/2562700

#LearnwithBrainly

8 0

2 years ago

As you heat a steak on a grill, what kind of energy are you increasing in the steak?

katrin2010 [14]

The correct answer to the question above is Thermal Energy. As you heat a steak on a grill, the kind of energy that you are increasing is Thermal Energy. Thermal energy is the energy that come from heat, since you are you grilling the steak which means there is fire underneath, it has thermal energy.

8 0

2 years ago

Suppose that a person riding on the top of a freight car shines a searchlight beam in the direction in which the train is travel

Snezhnost [94]

Answer:

same

Explanation:

Acc. to Einstien's postulate of special theory of

Relativity , Velocity of the light beam is same in all frames of references

(a) If the freight car is at rest

The frame we can assumed as Non - inertial frame of reference s

In the inertial frame of reference , velocity of the light beam has its own value as : 3 x 10^8 m/s

(b) If the freight car is moving , the frame we can assumed as Non -inertial frame of reference

In thus case also , The velocity of the light beam will also have the same value as ; 3 x 108 m/s

6 0

3 years ago

Why are dark matter and dark energy described as dark

zaharov [31]

Dark matter is called dark because it does not appear to interact with the electromagnetic field, which means it does not absorb, reflect or emit electromagnetic radiation, and is therefore difficult to detect.

3 0

2 years ago

How does the electric force between two charged particles change if one particle's charge is increased by a factor of 2?

Burka [1]

if one of the charge is doubled, the electric potential energy would be doubled too

8 0

3 years ago

Read 2 more answers