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3 years ago

If the kinetic energy of the 40kg box is 784 J, what is the velocity before it strikes the ground?

Physics

1 answer:

mamaluj [8]3 years ago

4 0

Answer:

Explanation:

$KE=\frac{1}{2}mv^2$

$784=\frac{1}{2}(40)v^2$

$784=20v^2$

$39.2=v^2$

$v=6.26m/s$

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2 years ago

which fire extinguisher is most appropriate to put out a fire that involves a stack of burning newspapers

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Monoammonium phosphate effectively smothers the fire, while sodium bicarbonate induces a chemical reaction which extinguishes the fire. Fire extinguishers with a Class C rating are suitable for fires in “live” electrical equipment.

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2 years ago

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Light strikes a 5.0-cm thick sheet of glass at an angle of incidence in air of 50°. the sheet has parallel faces and the glass h

mestny [16]

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3 years ago

Consider the 65 N light fixture supported as in the figure. Find the tension in the supporting wires.

ASHA 777 [7]

By using Lami's theorem formula, the tension in the supporting wires is 48.6 Newtons

TENSION

Tension is also a force having Newton as S.I unit.

The tension in the wire will be the same.

This question can be solved by using either vector diagram or by using Lami's theorem.

The sum of two given angles = 42 + 42 = 84 degrees

The third angle = 180 - 84 = 96 degrees.

Below is the Lami's theorem formula

$\frac{T}{sin\alpha } = \frac{T}{sin\beta } = \frac{W}{sinY}$

Where

$\alpha = \beta$ = 42 + 90 = 132 degrees

Y = 96 degrees

W = 65 N

By using the formula, we have

$\frac{T}{sin\alpha } = \frac{W}{sinY}$

T/sin 132 = 65/sin96

Cross multiply

T = 0.743 x 65.57

T = 48.56 N

Therefore, the tension in the supporting wires is 48.6 Newtons approximately.

Learn more about Tension here: brainly.com/question/24994188

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2 years ago

Betty weighs 400 N and she is sitting on a playground swing seat that hangs 0.21 m above the ground. Tom pulls the swing back an

disa [49]

Answer:

4.15 m/s

Explanation:

As the total energy must be conserved (neglecting air resistance) the change in gravitational potential energy, must be equal to the change in kinetic energy:

ΔE = ΔK + ΔU =0

If we take as a zero reference level for the gravitational potential energy, the height of the swing seat above the ground, (which is equal to 0.21 m), we can find the initial gravitational energy, considering the height of the point where the seat is released, regarding this point:

h₀ = 1.09 m -0.21 m = 0.88 m

⇒ U₀ = m*g*h₀ = 400 N*0.88 m = 352 J

As Uf = 0, ΔU = Uf -U₀ = -352 J

As the swing starts from rest, K₀=0, so we can say:

ΔK = Kf = $\frac{1}{2} *m*vf^{2}$ (1)

As ΔK = -ΔU ⇒ ΔK = 352 J (2)

From (1) and (2) we can solve for vf, as follows:

$vf = \sqrt{\frac{2*352J}{40.8kg}} = 4.15 m/s$

So, when the swing passes through its lowest position, Betty moves at 4.15 m/s.

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3 years ago