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egoroff_w [7]
3 years ago
11

Two masses (5.3kg and 7.5kg) are fastened together with a small amount of explosive. They are loaded into a spring gun that is t

ilted 60.0 ◦ above the horizontal and launched at a speed of 25m/s. At the top of the trajectory, the explosive detonates (assume the mass of explosive is negligble). After it detonates, the 5.3kg mass isn’t moving at all. What is the velocity of the second mass just after the explosion?
Physics
1 answer:
Vsevolod [243]3 years ago
8 0

Answer:

8.2

Explanation:

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Answer:

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Explanation:

Light waves are faster in vacuum because The refractive index of vacuum is the lowest since there is no barrier in the way of light propagation in vacuum. Demonstrate that the maximum speed of light in vacuum.

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<u>LIGHT WAVES -: </u> Light waves are a form of electromagnetic wave that includes both visible and invisible to the naked eye wavelengths on the electromagnetic spectrum. The average human eye can see light with wavelengths ranging from 390 to 700 nanometers, or nm.

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3 years ago
Finding the area of a trapezoid on a velocity versus time graph will tell you
julsineya [31]
The answer is c. velocity
6 0
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What is supercooling? how can it be minimized?
8090 [49]

<span>Supercooling is lowering the temperature of liquid below its freezing point. This is also known as the cooling of liquid. Supercooling can be minimized by cooling the liquid and steering the liquid fast while it is freezing. Freezers in the refrigerators shows the example activity of supercooling.</span>

3 0
3 years ago
A ski lift has a one-way length of 1 km and a vertical rise of 200 m. The chairs are spaced 20 m apart, and each chair can seat
rusak2 [61]

Power is needed for (1) acceleration and (2) lifting the loaded chairs. These two parts can be calculated separately and then added together.

(1) Power for acceleration:

The final speed of the lift is

V=(10 km/h)(1 h×1000 m60 sec×1 km)=2.887 m/s.

Then the power needed is

Pa=12m(V2−V20)/Δt=12(50×250 kg)(2.778 m/s)2=9.6 kW.

(2) Power for lift

Assume that the acceleration is constant (i.e. power supply is constant), its value will be

a=ΔVΔt=2.778 m/s5 s=0.556 m/s2.

Then the vertical lift during acceleration will be

(12at2)×(2001000)=1.36 m.

Hence, the power needed to increase the potential energy of the lift is

Pg=mgΔhΔt=(50×250 kg) (9.89 m/s2)(1.36 m)/(5 s)=3.41 kW.

Then the total Power required is

Ptotal=Pa+Pg=9.6+34.1=43.7 kW.

Learn more about potential energy at

brainly.com/question/14427111

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4 0
2 years ago
What is the elastic potential energy of a spring that is compressed a distance of 0.35 m and has a spring constant of 71.8 N/m?
Vitek1552 [10]

Answer:

P.E = 4.398 Joules.

Explanation:

Given the following data;

Spring constant, k = 71.8N/m

Displacement, x = 0.35m

To find the elastic potential energy;

The elastic potential energy of an object is given by the formula;

P.E = \frac {1}{2}kx^{2}

Substituting into the equation, we have;

P.E = \frac {1}{2}*71.8 *(0.35)^{2}

P.E = 35.9 * 0.1225

Elastic potential energy = 4.398 Joules.

<em>Therefore, the elastic potential energy of the spring is 4.398 Joules. </em>

3 0
3 years ago
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