All categories

2 years ago

When a car moves up a hill with constant

Physics

2 answers:

zysi [14]2 years ago

5 0

C gains potential energy

lutik1710 [3]2 years ago

5 0

C is 100 percent the correct answer :) welcome mate

You might be interested in

Sam, whose mass is 60 Kg, is riding on a 5.0 kg sled initially traveling at 8.0 m/s. He

umka2103 [35]

<h3>Answer: 130 newtons</h3>

===============================================================

Explanation:

We'll need the acceleration first.

The initial speed (let's call that Vi) is 8.0 m/s
The final speed (Vf) is 0 m/s since Sam comes to a complete stop at the end.
This happens over a duration of t = 4.0 seconds

The acceleration is equal to the change in speed over change in time

a = acceleration

a = (change in speed)/(change in time)

a = (Vf - Vi)/(4 seconds)

a = (0 - 8.0)/4

a = -8/4

a = -2

The acceleration is -2 m/s^2, meaning that Sam slows down by 2 m/s every second. Negative accelerations are often associated with slowing down. The term "deceleration" can be used here.

Here's a further break down of Sam's speeds at the four points of interest

At 0 seconds, he's going 8 m/s
At the 1 second mark, he's slowing down to 8-2 = 6 m/s
At the 2 second mark, he's now at 6-2 = 4 m/s
At the 3 second mark, he's at 4-2 = 2 m/s
Finally, at the 4 second mark, he's at 2-2 = 0 m/s

Next, we'll apply Newton's Second Law of motion

F = m*a

where,

F = force applied
m = mass
a = acceleration

We just found the acceleration, and the mass is fairly easy as all we need to do is add Sam's mass with the sled's mass to get 60+5.0 = 65 kg

So the force applied must be:

F = m*a

F = 65*(-2)

F = -130 newtons

This force is negative to indicate it's pushing against the sled's momentum to slow Sam down.

The magnitude of this force is |F| = |-130| = 130 newtons

8 0

3 years ago

Bohr's model is sometimes called the ________ model because it resembles a mini-solar system.

elena-14-01-66 [18.8K]

His model was also called the Planetary model

7 0

3 years ago

Read 2 more answers

At the same moment, one rock is dropped and one is theown downwand with an iniial velocily of 29 us frm op of a building that is

Helen [10]

Answer:

The thrown rock will strike the ground $2.42s$ earlier than the dropped rock.

Explanation:

<u>Known Data</u>

$y_{i}=300m$

$y_{f}=0m$
$v_{iD}=0m/s$
$v_{iT}=-29m/s$ , it is negative as is directed downward

<u>Time of the dropped Rock</u>

We can use $y_{f}=y_{i}+v_{iy}t-\frac{gt^{2}}{2}$ , to find the total time of fall, so $0=300m-\frac{(9.8m/s^{2})t_{D}^{2}}{2}$ , then clearing for $t_{D}$ .

$t_{D}=\sqrt[2]{\frac{300m}{4.9m/s^{2}}} =\sqrt[2]{61.22s^{2}} =7.82s$

<u>Time of the Thrown Rock</u>

We can use $y_{f}=y_{i}+v_{iy}t-\frac{gt^{2}}{2}$ , to find the total time of fall, so $0=300-29t_{T}-\frac{(9.8)t_{T}^{2}}{2}$ , then, $0=-4.9t_{T}^{2}-29t_{T}+300$ , as it is a second-grade polynomial, we find that its positive root is $t_{T}=5.4s$