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AlladinOne [14]

3 years ago

8

You plan to take a trip to the moon. Since you do not have a traditional spaceship with rockets, you will need to leave the eart

h with enough speed to make it to the moon. Some information that will help during this problem: mearth = 5.9742 x 1024 kg rearth = 6.3781 x 106 m mmoon = 7.36 x 1022 kg rmoon = 1.7374 x 106 m dearth to moon = 3.844 x 108 m (center to center) G = 6.67428 x 10-11 N-m2/kg2

1 answer:

hichkok12 [17]3 years ago

7 0

Answer:

$v = 3.5 \times 10^5 m/s$

Explanation:

At some distance from the Earth the force of attraction due to moon is balanced by the force due to Moon

so we will have

$\frac{GM_em}{r^2} = \frac{GM_m}{(d-r)^2}$

now we have

$\frac{d - r}{r} = \sqrt{\frac{M_m}{M_e}}$

$\frac{3.844\times 10^8 - r}{r} = \sqrt{\frac{7.36 \times 10^{22}}{5.9742\times 10^{24}}}$

so we will have

$r = 3.46 \times 10^8 m$

Now by energy conservation

$-\frac{GM_e}{R_e} - \frac{GM_m}{d - (R_e + R_m)} + \frac{1}{2}v^2 = -\frac{GM_e}{r} - \frac{GM_m}{d - r}$

$-6.26 \times 10^{8} - 13046 + \frac{1}{2}v^2 = -1.15 \times 10^6 - 1.28 \times 10^5$

$v = 3.5 \times 10^5 m/s$

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Sam, whose mass is 60 Kg, is riding on a 5.0 kg sled initially traveling at 8.0 m/s. He

umka2103 [35]

<h3>Answer: 130 newtons</h3>

===============================================================

Explanation:

We'll need the acceleration first.

The initial speed (let's call that Vi) is 8.0 m/s
The final speed (Vf) is 0 m/s since Sam comes to a complete stop at the end.
This happens over a duration of t = 4.0 seconds

The acceleration is equal to the change in speed over change in time

a = acceleration

a = (change in speed)/(change in time)

a = (Vf - Vi)/(4 seconds)

a = (0 - 8.0)/4

a = -8/4

a = -2

The acceleration is -2 m/s^2, meaning that Sam slows down by 2 m/s every second. Negative accelerations are often associated with slowing down. The term "deceleration" can be used here.

Here's a further break down of Sam's speeds at the four points of interest

At 0 seconds, he's going 8 m/s
At the 1 second mark, he's slowing down to 8-2 = 6 m/s
At the 2 second mark, he's now at 6-2 = 4 m/s
At the 3 second mark, he's at 4-2 = 2 m/s
Finally, at the 4 second mark, he's at 2-2 = 0 m/s

Next, we'll apply Newton's Second Law of motion

F = m*a

where,

F = force applied
m = mass
a = acceleration

We just found the acceleration, and the mass is fairly easy as all we need to do is add Sam's mass with the sled's mass to get 60+5.0 = 65 kg

So the force applied must be:

F = m*a

F = 65*(-2)

F = -130 newtons

This force is negative to indicate it's pushing against the sled's momentum to slow Sam down.

The magnitude of this force is |F| = |-130| = 130 newtons

8 0

3 years ago

A force F acts in the forward direction on a cart of mass m. A friction force ff opposes this motion. Part A Use Newton's second

stellarik [79]

Answer:

a = F-ff/m

Explanation:

According to Newton's second law of motion which states that "the rate of change in momentum of a body is directly proportional to the applied force F and acts in the direction of the force.

Mathematically;

F = ma

Since two forces acts on the cart i.e the moving force F and the frictional force Ff , we will take the sum of the forces.

∑F = ma where

m is the mass of the cart

a is its acceleration

∑F = F+(-ff )(since frictional force is an opposing force)

F - ff = ma

Dividing both sides by mass m

a = F-ff/m

5 0

3 years ago

A school bus moves at speed of 35 mi/hr for 20 miles. How long will it take the bus to get to school

ludmilkaskok [199]

Answer:

Explanation:

Assuming school is at the end of the 20 mile route, then

20 mi / 35 mi/hr = 0.57142...hr

which is about 34 minutes 17 seconds

6 0

2 years ago

Here on Earth you hang a mass from a vertical spring and start it oscillating with amplitude 1.9 cm. You observe that it takes 3

Vlada [557]

Answer:

T = 3.23 s

Explanation:

In the simple harmonic movement of a spring with a mass the angular velocity is given by

w = √ K / m

With the initial data let's look for the ratio k / m

The angular velocity is related to the frequency and period

w = 2π f = 2π / T

2π / T = √ k / m

k₀ / m₀ = (2π / T)²

k₀ / m₀ = (2π / 3.0)²

k₀ / m₀ = 4.3865

The period on the new planet is

2π / T = √ k / m

T = 2π √ m / k

In this case the amounts are

m = 6 m₀

k = 10 k₀

We replace

T = 2π√6m₀ / 10k₀

T = 2π √6/10 √m₀ / k₀

T = 2π √ 0.6 √1 / 4.3865

T = 3.23 s

5 0

3 years ago

Two transverse waves travel along the same taut string. Wave 1 is described by y1(x, t) = A sin(kx - ωt), while wave 2 is descri

Vadim26 [7]

Answer:

6) Wave 1 travels in the positive x-direction, while wave 2 travels in the negative x-direction.

Explanation:

What matters is the part $kx \pm \omega t$ , the other parts of the equation don't affect time and space variations. We know that when the sign is - the wave propagates to the positive direction while when the sign is + the wave propagates to the negative direction, but <em>here is an explanation</em> of this:

For both cases, + and -, after a certain time $\delta t$ ( $\delta t >0$ ), the displacement <em>y</em> of the wave will be determined by the $kx\pm\omega (t+\delta t)$ term. For simplicity, if we imagine we are looking at the origin (x=0), this will be simply $\pm \omega (t+\delta t)$ .

To know which side, right or left of the origin, would go through the origin after a time $\delta t$ (and thus know the direction of propagation) we have to see how we can achieve that same displacement <em>y</em> not by a time variation but by a space variation $\delta x$ (we would be looking where in space is what we would have in the future in time). The term would be then $k(x+\delta x)\pm\omega t$ , which at the origin is $k \delta x \pm \omega t$ . This would mean that, when the original equation has $kx+\omega t$ , we must have that $\delta x>0$ for $k\delta x+\omega t$ to be equal to $kx+\omega\delta t$ , and when the original equation has $kx-\omega t$ , we must have that $\delta x$ for $k\delta x-\omega t$ to be equal to $kx-\omega \delta t$

<em>Note that their values don't matter, although they are a very small variation (we have to be careful since all this is inside a sin function), what matters is if they are positive or negative and as such what is possible or not .</em>

<em />

In conclusion, when $kx+\omega t$ , the part of the wave on the positive side ( $\delta x>0$ ) is the one that will go through the origin, so the wave is going in the negative direction, and viceversa.

4 0

3 years ago