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Triss [41]
3 years ago
8

Which of the following is a science?

Physics
1 answer:
nikitadnepr [17]3 years ago
3 0

Answer:

astrology

Explanation:

eeeeeeeeeeeeeeeeee

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An object has a kinetic energy of 275 j and the momentum of magnitude of 25.0 kg m/s. find the speed and mass of the object
lyudmila [28]
<span>Recall formula for Kinetic energy is: KE = 1/2mv^2, where KE = 275J and momentum (which is 25.0 kg m/s) = m*v Therefore substitute for KE and mv in the equation above to get speed => 275 = 0.5 * 25 * v v = 275/12.5 v = 22m/s to get mass m, recall momentum = m*v => 25= 22*m m= 25/22 = 1.3663kg</span>
6 0
4 years ago
A student pushes on a 8-kg box with a force of 35 N forward. The force of sliding friction is 10 N backward. What is the acceler
Ne4ueva [31]

Answer:

(35 N - 10 N)/8kg = 3.125 m/s^2

Explanation:

The formula for Force is:

Force = Mass*Acceleration

(Force is equal to Mass times Acceleration)

Since we're told to find the acceleration of the box. We make acceleration the subject of the equation:

Acceleration = Force/Mass

(Acceleration equal to Force divided by Mass)

We know that the force are 35 N forward and 10 N backward, and the weight of the box is 8kg.

= (35 N - 10 N)/8kg

The reason that 35 N minus 10 N is because the 10 N is pushing the box backward.

= 25 N/8kg

= 3.125 m/s^2

Hope it helps :DD

3 0
3 years ago
How might a <br> Theory relate to a model
Alexxx [7]
A theory can help create a model
3 0
4 years ago
How do kinetic and potential energy transfer to one throughout a roller coaster ride?
mojhsa [17]

Answer:

As the cars ascend the next hill, some kinetic energy is transformed back into potential energy. Then, when the cars descend this hill, potential energy is again changed to kinetic energy. This conversion between potential and kinetic energy continues throughout the ride.

Explanation:

hope it helps U

6 0
2 years ago
A car is traveling at 50 mi/h when the brakes are fully applied, producing a constant deceleration of 38 ft/s2. what is the dist
e-lub [12.9K]

Convert 38 ft/s^2 to mi/h^2. Then we se the conversion factor > 1 mile = 5280 feet and 1 hour = 3600 seconds.

So now we show it > 38  \frac{ft}{s^2}  x  \frac{1mi}{5280ft} x  \frac{(3600s)^2}{(1h)^2} = 93272.27  \frac{mi}{h^2}

Then we have to use the formula of constant acceleration to determine the distance traveled by the car before it ended up stopping.

Which the formula for constant acceleration would be > v_2^2=v_1^2 + 2as

The initial velocity is 50mi/h (v_1=50)

When it stops the final velocity is (v_2=0)

Since the given is deceleration it means the number we had gotten earlier would be a negative so a = -93272.27

Then we substitute the values in....

0^2 = 50^2 + 2(-93272.27)s&#10;&#10;0 = 2500 - 186544.54s&#10;&#10;Isolate S next.&#10;&#10;185644.54s= 2500&#10;&#10;s =  2500/(185644.54)&#10;&#10;s=0.0134&#10;

So we can say the car stopped at 0.0134 miles before it came to a stop but to express the distance traveled in feet we need to use the conversion factor of 1 mile = 5280 feet in otherwards > 0.0134 mi *  \frac{5280ft}{1mi}  = 70.8 ft
So this means that the car traveled in feet 70.8 ft before it came to a stop.

4 0
3 years ago
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