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2 years ago

When magma is cooled, it changes into which of the following?

Physics

2 answers:

Oksanka [162]2 years ago

6 0

Answer: You didnt finish the question. But as magma cools it turns to igneous rock.

vekshin12 years ago

4 0

Igneous rock

Explanation:

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A garbage truck has a mass of 2100 kg. It starts from rest and travels 75 meters in 15 seconds. The truck uniformly accelerates

Ivanshal [37]

Answer: F=(2100)(0.33)=693N

Explanation:

3 0

2 years ago

NO LINKS PLEASE

Trava [24]

Answer:

Explanation:

6 0

2 years ago

You're carrying a 3.2-m-long, 24kg pole to a construction site when you decide to stop for a rest. You place one end of the pole

Aleksandr [31]

Answer:

Tension of 132N

Explanation:

We need to apply Summatory of Force to find the tension in the hand.

We define te tensión in the hand as $F_2$ and the Tension in fence post as $F_1$ , then

$\sum F = 24(9.8)$

$F_1 + F_2= 24(9.8)$

We apply summatory of moments then

$F_2*1.25 = F_1*1.6$

Where the Force 2 is 1.25m from the center of summatory,

We can note that,

$1.6 m - 0.35m=1.25m$

We have two equation and two incognites, then replacing (1) in (2)

$1.6(235.2 -F_2) = 1.25F_2$

$376.32 = F2(1.6+1.25)$

$F_2= \frac{376.32}{2.85}$

$F_2 =132 N$

5 0

2 years ago

1.Predict the frequency of a tuning fork that emits a sound with a wavelength of 0.385 m.

lina2011 [118]

Answer:

1.) Frequency F = 890.9 Hz

2.) Wavelength (λ) = 0.893 m

Explanation:

1.) Given that the wavelength = 0.385m

The speed of sound = 343 m / s

To predict the frequency, let us use the formula V = F λ

Where (λ) = wavelength = 0.385m

343 = F × 0.385

F = 343/0.385

F = 890.9 Hz

2.) Given that the frequency = 384Hz

Using the formula again

V = F λ

λ = V/F

Wavelength (λ) = 343/384

Wavelength (λ) = 0.893 m

The two questions can be solved with the use of formula

3 0

2 years ago

An oscillator consists of a block attached to a spring (k = 427 N/m). At some time t, the position (measured from the system's e

White raven [17]

Answer:

a) 4.49Hz

b) 0.536kg

c) 2.57s

Explanation:

This problem can be solved by using the equation for he position and velocity of an object in a mass-string system:

$x=Acos(\omega t)\\\\v=-\omega Asin(\omega t)\\\\a=-\omega^2Acos(\omega t)$

for some time t you have:

x=0.134m

v=-12.1m/s

a=-107m/s^2

If you divide the first equation and the third equation, you can calculate w:

$\frac{x}{a}=\frac{Acos(\omega t)}{-\omega^2 Acos(\omega t)}\\\\\omega=\sqrt{-\frac{a}{x}}=\sqrt{-\frac{-107m/s^2}{0.134m}}=28.25\frac{rad}{s}$

with this value you can compute the frequency:

$f=\frac{\omega}{2\pi}=\frac{28.25rad/s}{2\pi}=4.49Hz$

the mass of the block is given by the formula:

$f=\frac{1}{2\pi}\sqrt{\frac{k}{m}}\\\\m=\frac{k}{4\pi^2f^2}=\frac{427N/m}{(4\pi^2)(4.49Hz)^2}=0.536kg$

c) to find the amplitude of the motion you need to know the time t. This can computed by dividing the equation for v with the equation for x and taking the arctan:

$\frac{v}{x}=-\omega tan(\omega t)\\\\t=\frac{1}{\omega}arctan(-\frac{v}{x\omega })=\frac{1}{28.25rad/s}arctan(-\frac{-12.1m/s}{(0.134m)(28.25rad/s)})=2.57s$

Finally, the amplitude is:

$x=Acos(\omega t)\\\\A=\frac{0.134m}{cos(28.25rad/s*2.57s )}=0.45m$

5 0

2 years ago