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blagie [28]
4 years ago
5

What are possible formulas for impulse? Check all that apply. J = Fdeltat J = StartFraction force over change in time EndFractio

n J = Fdeltav J = StartFraction mass over change in time EndFraction J = mdeltav J = deltap
Physics
1 answer:
Alex4 years ago
6 0

<u>The possible formulas for impulse are as follows:</u>

J = FΔt

J = mΔv

J = Δp

Answer: Option  A, E and F

<u>Explanation:</u>

The quantity which explains the consequences of a overall force acting on an object (moving force) is known as impulse. It is symbolised as J. When the average overall force acting on an object than such products are formed and in given duration than the start fraction force over change in time end fraction J = FΔt.

The impulse-momentum theorem explains that the variation in momentum of an object is same as the impulse applied to it: J = Δp J = mΔv if mass is constant J = m dv + v dm if mass changes. Logically, the impulse-momentum theorem is equivalent to Newton second laws of motion which is also called as force law.

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3 years ago
A physics student shoves a 0.50-kg block from the bottom of a frictionless 30.0° inclined plane. The student performs 4.0 j of w
Musya8 [376]

Answer:1.63 m

Explanation:

Given

mass of block m=0.5 kg

inclination \theta =30^{\circ}

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block slides a distance s along the Plane

Work done =change in Potential Energy

Increase in height of block is s\sin \theta

Change in Potential Energy =mg(\sin \theta -0)

\Delta P.E.=0.5\times 9.8\times s\sin 30

4=0.5\times 9.8\times s\sin 30

s=\frac{4}{2.45}      

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7 0
3 years ago
A jet of water squirts out horizontally from a hole on the side of the tank as shown below. 0.579 m 1.45 m h If the hole has a d
klemol [59]

Answer:

The height of the water above the hole in the tank is 58 mm

Explanation:

In order to solve this problem we need to draw a sketch of the dimensions that include the input variables of the problem.

Where:

x = 0.579[m]

y = 1.45 [m]

Using the following kinematic equation we can find the time that takes the water to hit the ground, and then with this time, we can find the velocity of the water in the x-component.

y = y_{o} +v_{o} *t+\frac{1}{2} *g*t^{2} \\

It is necessary to clarify the value of each of the respective variables below

y = - 1.45 [m] "It is negative because this point is below the water outlet"

yo = 0

vo = 0 "The velocity is zero because the component of the speed on the Y-axis does not exist"

therefore:

-1.45=0.5*(-9.81)*t^{2} \\t = \sqrt{\frac{1.45}{0.5*9.81} } \\t = 0.543[s]

The next step is to determine the velocity in component x, knowing the time.

v=\frac{x}{t} \\v=\frac{0.579}{.543} \\v = 1.06[m/s]

Now using torricelli's law we can find the elevation.

v=\sqrt{2*g*h} \\h=\frac{v^{2} }{2*g} \\h=\frac{1.06^{2} }{2*9.81} \\h= 0.057[m] = 57.95[mm]

3 0
3 years ago
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