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strojnjashka [21]
3 years ago
7

Suppose the initial kinetic energy and final potential energy in an experiment are both zero. What can you conclude?

Physics
1 answer:
Kisachek [45]3 years ago
5 0

Initially, the experiment has only potential energy (since total energy is the sum of kinetic and potential energy). And at the end, the experment has only kinetic energy.

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How long will it take to travel 200 km traveling 10 m/s ?
GuDViN [60]
<span>Convert 200 km to meters which equals 200000 meters. Then take 10m/s*200000meters which equals 2000000 seconds.
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5 0
3 years ago
A 6 kilogram block in outer space is moving at -100 m/s (to the left). It suddenly experiences three forces as shown below (in t
Blizzard [7]

Answer:

x = 1474.9 [m]

Explanation:

To solve this problem we must use Newton's second law, which tells us that the sum of forces must be equal to the product of mass by acceleration.

We must understand that when forces are applied on the body, they tend to slow the body down to stop it.

So as the body continues to move to the left, it is slowing down. Therefore we must calculate this deceleration value using Newton's second law. We must perform a sum of forces on the x-axis equal to the product of mass by acceleration. With leftward movement as negative and rightward forces as positive.

ΣF = m*a

10 +12*sin(60)= - 6*a\\a = - 3.39[m/s^{2}]

Now using the following equation of kinematics, we can calculate the distance of the block, before stopping completely. The initial speed must be 100 [m/s].

v_{f}^{2} =v_{o}^{2}-2*a*x

where:

Vf = final velocity = 0 (the block stops)

Vo = initial velocity = 100 [m/s]

a = - 3.39 [m/s²]

x = displacement [m]

0 = 100^{2}-2*3.39*x\\x=\frac{10000}{2*3.39}\\x=1474.9[m]

4 0
3 years ago
Please help me! As quickly as possible
k0ka [10]

Answer:

1. matter

2. kilograms

3. same

4. gravitational

5. gravity

6. space

7. weightlessness

8. Newton

9. weight

10. more

I HOPE THESE ARE CORRECT AND IT HELPS

7 0
2 years ago
The downsprue leading into the runner of a certain mold has a length of 175 mm. The cross-sectional area at the base of the spru
adoni [48]

Answer:

(a) Velocity at bottom is 1.85 m/s

(b) Volume flow rate is 7.4 x 10⁻⁴ m³/s.

(c) The time required to fill the mold is 1.35 s.

Explanation:

(a)

Applying Bernoulli's Equation on both ends of the down sprue, with the assumptions that every point is at atmospheric pressure and the liquid metal at the pouring basin is at zero velocity. The equation then becomes:

V = √2gh

where,

V = velocity at bottom of down sprue

h = height of down sprue = 175 mm = 0.175 m

V = √2(9.8 m/s²)(0.175 m)

<u>V = 1.85 m/s</u>

<u></u>

(b)

The volume flow rate is given as:

Volume Flow Rate = (V)(A)

where,

V = velocity at bottom = 1.85 m/s

A = Area of bottom = 400 mm² = 0.0004 m²

Therefore,

Volume Flow Rate = (1.85 m/s)(0.0004 m²)

<u>Volume Flow Rate = 7.4 x 10⁻⁴ m³/s = 740 cm³/s</u>

(c)

The time required to fill the cavity is given as:

Volume Flow Rate = V/t

where,

V = Volume of mold Cavity = 0.001 m³

t = time required to fill the cavity = ?

Therefore,

t = V/Volume Flow Rate

t = 0.001 m³/7.4 x 10⁻⁴ m³/s

<u>t = 1.35 s</u>

<u></u>

5 0
3 years ago
Which of the following could be used to create an open circuit?
Arturiano [62]
A switch
What are the answers choices
4 0
3 years ago
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