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Art [367]
3 years ago
7

Anyone want text if yes ill give u my phone number

Physics
2 answers:
Jlenok [28]3 years ago
3 0

Answer:

Explanation:OFC!!!

Tatiana [17]3 years ago
3 0

Answer:

Why would I text a stranger?

Give me 3 good reasons to change my mind.

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A truck is speeding up as it travels on an interstate. The truck's momentum (in kg · m/s) is proportional to the truck's speed (
muminat

Answer:

At the moment, the truck's momentum (in kg · m/s)  is 2200 times as large as the truck's speed (in m/s).

This means that as the truck travels, the truck's momentum (in kg · m/s) is always 2200 times as large as the truck's speed (in m/s).

If the truck is travelling at 16 m/s, the truck's momentum is 35200 kg · m/s.

Explanation:

From the question,

The truck's momentum (in kg · m/s) is proportional to the truck's speed (in m/s).

Let the truck's momentum be P and the truck's speed be v,

Then  we can write that

P∝v

Then,

P = kv

Where k is the proportionality constant

From the question,

At some moment the truck's momentum is 50600 kg · m/s and the truck's speed is 23 m/s,

To determine how many times the truck's speed is as large as the truck's momentum at this moment, we will divide the truck's momentum by the speed, that is

50600 ÷ 23 = 2200

Hence, at the moment, the truck's momentum (in kg · m/s)  is 2200 times as large as the truck's speed (in m/s).

Since, dividing the truck's momentum by the truck's speed gives the proportionality constant k (that is, P/v = k), then

This means that as the truck travels, the truck's momentum (in kg · m/s) is always 2200 times as large as the truck's speed (in m/s).

From

P = kv

Then, k = P/v

At a moment, P = 50600 kg · m/s and v = 23 m/s

∴ k = 50600 kg · m/s ÷ 23 m/s = 2200 kg

k = 2200 kg

To determine the truck's momentum if the truck is traveling at 16 m/s

From

P = kv

k = 2200 kg

v = 16 m/s

∴ P = 2200 kg × 16 m/s

P = 35200 kg · m/s

Hence, if the truck is travelling at 16 m/s, the truck's momentum is 35200 kg · m/s.

3 0
4 years ago
How much force is necessary to stretch a spring 0.5 m when the spring constant is 190N/m
Sauron [17]

Answer:

190×0.5=95N

Explanation:

The spring constant is 190 N/m . Therefore the force is necessary to stretch a spring 0.5 m when the spring constant is 190 N/m is 95 N

4 0
3 years ago
What tension must a 47.0 cm length of string support in order to whirl an attached 1,000.0 g stone in a circular path at 2.55 m/
stealth61 [152]

Answer:

I hope it will help you....

7 0
3 years ago
A satellite is in a circular orbit around the Earth at an altitude of 2.80 3 106 m. Find (a) the period of the orbit, (b) the sp
Katen [24]

Explanation:

Given that,

Altitude h= 2.803\times10^{6}\ m

We need to calculate the radius

r=R+h

Where, R = radius of the earth

h = radius of altitude

Put the value into the formula

r=(6.38\times10^{6}+2.803\times10^{6})

r=9.18\times10^{6}\ m

(a). We need to calculate the period of the orbit,

Using formula of period

T^2=\dfrac{4\pi^2}{GM}r^3

T^2=\dfrac{4\pi^2}{6.67\times10^{-11}\times5.98\times10^{24}}\times(9.18\times10^{6})^3

T^2=76570372.9509\ sec

T=8750.45\ sec

(b). We need to calculate the speed of the satellite

Using formula of speed

v^2=\dfrac{GM}{r}

Put the value into the formula

v^2=\dfrac{6.67\times10^{-11}\times5.98\times10^{24}}{9.18\times10^{6}}

v^2=43449455.3377\ m/s

v=6.59\times10^{3}\ m/s

(c). We need to calculate the acceleration of the satellite

Using formula of acceleration

a_{c}=\dfrac{v^2}{r}

Put the value into the formula

a_{c}=\dfrac{(6.59\times10^{3})^2}{9.18\times10^{6}}

a_{c}=4.73\ m/s^2

Hence, This is the required solution.

6 0
4 years ago
Select all that apply. There MIGHT be more than one.
Stolb23 [73]
I believe the answer to your question is “Lithosphere plate boundaries”

The planet Earth is covered by a layer formed by land and rocks called the earth's crust or lithosphere. This crust is not smooth and uniform, but rather irregular and composed of tectonic plates, also called lithosphere plates. These plates are not fixed as they are under the magma (high temperature molten rock).

Hope this helps!:)
4 0
3 years ago
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