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Alex777 [14]
3 years ago
8

The acceleration of the object at time t = 0.7 s is most nearly equal to which of the following?

Physics
1 answer:
VladimirAG [237]3 years ago
6 0

We have that for the Question "the acceleration of the object at time t = 0.7 s is most nearly equal to which of the following?"

  • it can be said that the acceleration of the object at time t = 0.7 s is most nearly equal to the slope of the line connecting the origin and the point where the graph where the graph crosses the 0.7s grid line

From the question we are told

the acceleration of the object at time t = 0.7 s is most nearly equal to which of the following?

Generally the equation for the Force  is mathematically given as

F=\frac{F}{dx}

Therefore

F=-kdx

k=600Nm^{-1}

now

K.E=0.5x ds^2

K.E=600*(-0.1^2)

K.E=3J

Therefore

the acceleration of the object at time t = 0.7 s is most nearly equal to the slope of the line connecting the origin and the point where the graph where the graph crosses the 0.7s grid line

For more information on this visit

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adelina 88 [10]

Answer:

<h2>3,343.68kJ </h2>

Explanation:

Heat energy used up can be calculated using the formula:

H = mcΔt

m = mass oof the object (in kg) = 20kg

c = specific heat capacity of water = 4179.6J/kg°C

Δt change in temperature = 80-40 = 40°C

H= 20* 4179.6 * 40

H = 3,343,680Joules

H = 3,343.68kJ

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A father is trying to teach his child to ice skate. As the child stands still, the father pushes him forward with an acceleratio
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Hi there!

We can use Newton's Second Law:

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We are given the mass and acceleration, so:

∑F = 20 · 2 = <u>40 N</u>

4 0
2 years ago
A huge tank of glycerine with a density of 1.260 g/cm3 is vertically stationed on a platform which is 15 m above the ground. The
EleoNora [17]

Answer:

The tank is losing 4.976*10^{-4}  m^3/s

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Explanation:

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We are being informed that both the tank and the hole is being exposed to air :

∴ P₁ = P₂

Also as the tank is voluminous ; we take the initial volume  v_1 ≅ 0 ;

then v_2 can be determined as:\sqrt{[2g (h_1- h_2)]

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v_2 = \sqrt{[2*9.81*(20 - 15)]

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v_2= 9.9 \ m/s  as it leaves the hole at the base.

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b)

How fast is the water from the hole moving just as it reaches the ground?

In order to determine that; we use the relation of the velocity from the equation of motion which says:

v² = u² + 2gh ₂

v² = 9.9² + 2×9.81×15

v² = 392.31

The velocity of how fast the water from the hole is moving just as it reaches the ground is : v_g = \sqrt{392.31}

v_g = 19.81 \ m/s

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