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charle [14.2K]
3 years ago
12

Calculate the kinetic energy of a 100.0-kg meteor approaching the Earth at a speed of 10.0 km/s. Remember that 1 km = 1000 m.

Physics
2 answers:
Alex17521 [72]3 years ago
8 0
Kinetic Energy = (1/2) (mass) (speed squared)

Mass = 100 kg
Speed = 10 km/s = 10,000 m/s
Speed squared = 100,000,000 m²/s²

1/2 (M)(V²) = 1/2 (100) (100,000,000) = 5,000,000,000 joules  (5 x 10⁹ J)

That's  5 billion joules.

That's the amount of energy a 100-watt light bulb uses in  579 days.
olchik [2.2K]3 years ago
5 0
KE=1/2 x mass x speed^2
Substitute:

KE=1/2 x 100 x 10000^2
5000000000J of KE
So either <u>5 gigajoules</u>
Or <u>5x10^9J</u>
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Find the instantaneous acceleration at t=ls for an object moving along a straight axis with velocity function:
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The answer is A.
Explanation:
We know that the average acceleration a for an interval of time Δt is expressed as:

a = Δv
Δt
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e formula for the instantaneous acceleration a is almost the same, except that we need to indicate that we're interested in knowing what the ratio of Δv to Δt approaches as Δt approaches zero.

We can indicate that by using the limit notation.

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How remove local action and polarization ​
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Local action is removed by amalgamating zinc rod.

Explanation:

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Shoukd be a base hit

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Which of the following objects cannot make a shadow? Broken piece of glass from a window/wooden pane of the same window? Explain
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broken pieces of glass from a window can't form a shadow.

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Even when the head is held erect, as in the figure below, its center of mass is not directly over the principal point of support
alexandr1967 [171]

We are asked to determine the force required by the neck muscle in order to keep the head in equilibrium. To do that we will add the torques produced by the muscle force and the weight of the head. We will use torque in the clockwise direction to be negative, therefore, we have:

\Sigma T=r_{M\perp}(F_M)-r_{W\perp}(W)

Since we want to determine the forces when the system is at equilibrium this means that the total sum of torque is zero:

r_{M\perp}(F_M)-r_{W\perp}(W)=0

Now, we solve for the force of the muscle. First, we add the torque of the weight to both sides:

r_{M\perp}(F_M)=r_{W\perp}(W)

Now, we divide by the distance of the muscle:

(F_M)=\frac{r_{W\perp}(W)}{r_{M\perp}}

Now, we substitute the values:

F_M=\frac{(2.4cm)(50N)}{5.1cm}

Now, we solve the operations:

F_M=23.53N

Therefore, the force exerted by the muscles is 23.53 Newtons.

Part B. To determine the force on the pivot we will add the forces we add the vertical forces:

\Sigma F_v=F_j-F_M-W

Since there is no vertical movement the sum of vertical forces is zero:

F_j-F_M-W=0

Now, we add the force of the muscle and the weight to both sides to solve for the force on the pivot:

F_j=F_M+W

Now, we plug in the values:

F_j=23.53N+50N

Solving the operations:

F_j=73.53N

Therefore, the force is 73.53 Newtons.

8 0
1 year ago
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