Which of the following has kinetic energy? (1 point)

You weigh 716 newtons on Earth. You

Otrada [13]

Answer:

537 N

Explanation:

The force due to gravity of a planet is:

F = GMm / r²

where G is the universal gravitational constant

M is the mass of the planet

m is the mass of the object

and r is the distance between the object and the center of the planet

On Earth, you weigh 716 N, so:

716 N = GMm / r²

On planet X:

F = G (3M) m / (2r)²

F = 3/4 GMm / r²

F = 3/4 (716 N)

F = 537 N

8 0

3 years ago

A plane starting from rest (vo = 0 m/s) when t0 = 0s. The plane accelerates down the runway, and at 29 seconds, its velocity is

IrinaK [193]

Answer:

Acceleration, $a=2.48\ m/s^2$

Explanation:

Given that,

The plane is at rest initially, u = 0

Final speed of the plane, v = 72.2 m/s

Time, t = 29 s

We need to find the average acceleration for the plane. It can be calculated as :

$a=\dfrac{v-u}{t}$

$a=\dfrac{72.2}{29}$

$a=2.48\ m/s^2$

So, the average acceleration for the plane is $2.48\ m/s^2$ . Hence, this is the required solution.

4 0

2 years ago

Solve for work when

BlackZzzverrR [31]

So, <u>the value of the work is approximately 84.65 J</u>.

<h2>Introduction</h2>

Hi ! Here I will help you to discuss the subject about work that caused by force in amount value of angle. Work is affected by the force and displacement.

If related to the magnitude of the force, the amount of work will be proportional to the magnitude of the applied force. Thats mean, if the value of the force that applied on it is greater, then the value of the work will be greater.
If related to the magnitude of shift, the amount of work will be proportional to the magnitude of shift of object. Thats mean, if the value of the shift on it is greater, then the value of the work will be greater.

<h3>Formula Used</h3>

The work done by a moving object can be expressed in the equation:

If the Angle Is Ignored

$\boxed{\sf{\bold{W = F \times s}}}$

If the Angle Effect on Work

$\boxed{\sf{\bold{W = F \times s \times \cos(\theta)}}}$

With the following condition:

W = work that done by object (J)
F = force that applied (N)
s = shift or distance (m)
$\sf{\theta}$ = angle of elevation (°)

<h3>Solution</h3>

We know that :

F = force that applied = $\sf{1.41 \times 10^4}$ N
s = shift or distance = 84.9 m
$\sf{\theta}$ = angle of elevation = 45°

What was asked ?

W = work that done by object = ... J

Step by step :

$\sf{W = F \times s \times \cos(\theta)}$

$\sf{W = (1.41 \cdot 10^4) \times 84.9 \times \cos(45^o)}$

$\sf{W = (1.41 \cdot 10^4) \times 84.9 \times \frac{\sqrt{2}}{2}}$

$\sf{W = 119.709 \times \frac{\sqrt{2}}{2}}$

$\sf{W = 59.8545 \sqrt{2}}$

$\boxed{\sf{W \approx 84.65 \: J}}$

<h3>Conclusion</h3>

So, the value of the work is approximately 84.65 J.

3 0

1 year ago

An object is propelled straight up from ground level with an initial velocity of 48 feet per second. Its height at time t is mod

Ket [755]

Answer:

Explanation:

For a. its max height and when it occurs. First the max height. That's a y-dimension thing, and in the y-dimension we have this info:

v₀ = 48 ft/s

a = -32 ft/s/s

v = 0 (the max height of an object occurs when the final velocity of the object is 0). Use the following equation for this part of the problem:

v² = v₀² + 2aΔx and filling in:

$0=48^2+2(-32)$ Δx and

0 = 2300 - 64Δx and

-2300 = -64Δs so

Δx = 36 feet.

Now for the time it takes to get to this max height. Final velocity is still 0 here, but the equation is a different one for this part of the problem. Use:

v = v₀ + at and filling in:

0 = 48 - 32t and

-48 = -32t so

t = 1.5 sec. That's part a. Onto part b:

The object hits the ground when its displacement, Δx, is 0. Use this equation for this problem:

Δx = v₀t + $\frac{1}{2}at^2$ and filling in:

$0=48t+\frac{1}{2}(-32)t^2$ and

$0=48t-16t^2$ and

0 = 16t(3 - t) so

t = 0 and t = 3. t = 0 is before the object is propelled, so it makes sense that at 0 seconds, the object was still on the ground, right? Then at 3 seconds, it's back on the ground. (Isn't math just perfectly, beautifully sensible!?) Now onto part c:

We are looking for the time interval when the object is >32 feet. So we use the same equation we just used, but with an inequality instead of an equals sign:

$48t+\frac{1}{2}(-32)t^2 >32$ and get everything on one side and factor it again:

$-16t^2+48t-32>0$ and we find that

1 < t < 2 so the time interval is between 1 and 2 seconds that the object is over 32 feet in the air.

8 0

2 years ago

The upper, plastic-like part of Earth that the plates ride on is called the _____.

zhenek [66]

Hey! The upper, plastic-like part of earth that the plates ride on are called the (Asthenosphere) I hope this helps! :)

5 0

3 years ago

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