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Sidana [21]
2 years ago
13

A person runs 400 meters in a straight line. What is their distance and displacement.

Physics
1 answer:
suter [353]2 years ago
7 0
Distance = 400 m, Displacement = 400 m in the direction of the straight line.
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A 5 kg object near Earth's surface is released from rest such that it falls a distance of 10 m. After the object falls 10 m, it
makkiz [27]

Answer:D

Explanation:

Given

mass of object m=5 kg

Distance traveled h=10 m

velocity acquired v=12 m/s

conserving Energy at the moment when object start falling and when it gains 12 m/s velocity

Initial Energy=mgh=5\times 9.8\times 10=490 J

Final Energy=\frac{1}{2}mv^2+W_{f}

=\frac{1}{2}\cdot 5\cdot 12^2+W_{f}

where W_{f} is friction work if any

490=360+W_{f}

W_{f}=130 J

Since Friction is Present therefore it is a case of Open system and net external Force is zero

An open system is a system where exchange of energy and mass is allowed and Friction is acting on the object shows that system is Open .

4 0
3 years ago
To move a mass you need a force true or false
monitta

Answer: True

Explanation:

A force must be applied to set a stationary object in motion.

4 0
3 years ago
A ball is thrown horizontally at a speed of 24 meters per second from the top of a cliff. if the ball hits the ground 4.0 second
Viktor [21]

Answer:

78.4 m

Explanation:

To obtain the height of the cliff;

We can use the Relation to obtain the final velocity, v

v = u + at

a = acceleration due to gravity = 9.8m/s²

v = 0 + (9.8*4)

v = 0 + 39.2

v = 39.2 m/s

To obtain the Height, S

v² = u² + 2aS

39.2^2 = 0 + 2(9.8)S

39.2^2 = 0 + 19.6S

1536.64 = 19.6S

S = 1536.64 / 19.6

S = 78.4 m

8 0
2 years ago
A Porsche sports car can accelerate at 8.8 m/s^2. Determine its acceleration in km/h^2.
finlep [7]

Answer:

The acceleration expressed in the new units is 114.048 Km/h^{2}

Explanation:

To convert from m/s^{2} to Km/h^{2} it is necessary to remember that there are 1000 meters in 1 kilometer and 3600 seconds in 1 hour:

Then by means of a rule of three it is get:

8.8\frac{m}{s^{2}}.(\frac{1Km}{1000m}).(\frac{3600s}{1h})^{2}

8.8\frac{m}{s^{2}}.(\frac{1Km}{1000m}).(\frac{12960000s^{2}}{1h^{2}})

Hence, the units of meters and seconds will cancel. Notice the importance of square the ratio 3600s/1h, so that way they can match with the other units:

114.048 Km/h^2

So the acceleration expressed in the new units is 114.048 Km/h^2.

6 0
3 years ago
in this model, the velocity of the spacecraft at position 2 is A.) equal to B.) greater than C.) less than the velocity of the c
finlep [7]
<h2>1. Right answer: the velocity of the spacecraft at position 2 is <u>greater than</u> the velocity of the craft at position 4. </h2>

This is due the gravity field of the planet (The Earth in this case) is used to accelerate the craft. This is true when in a specific point the direction of the movement of the craft is the same direction of the movement of the planet.

In this case the craft will be “catched” by the Earth’s gravitational field, making the craft  to enter a circular orbit.

<h2>2. Right answer: At position 1, the direction of the spacecraft changes because of <u>the gravitational force between Earth and the spacecraft. </u></h2>

As explained in the prior answer, this is the exact and correct point where the trajectory of the spacecraft enters into a circular orbit because of the attraction due gravity of the Earth and therefore changes its direction.


<h2>3. Right answer: Position 3 represents <u>the orbital path or velocity of Earth </u></h2>

Being this the orbital path of the Earth and considering the trajectory of the craft, the condition of accelerating the craft is accomplished.

If the orbital path of the Earth were the opposite from the shown in the figure, the effect on the craft would be braking.

Note all of these is related to the <u>gravitational assistance. </u>

<u>Gravitational assistance</u> is the maneuver in which the energy of the gravitational field of a planet or satellite is used to obtain an acceleration or braking of the probe changing its trajectory.

This maneuver is also called <em>slingshot effect, swing-by</em> or <em>gravity assist</em>. It is a common technique in space for the outer Solar System missions , in order to save costs in the launch rocket and thrusters.


6 0
3 years ago
Read 2 more answers
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