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kirill115 [55]
3 years ago
15

Which of the following is a factor that determines the composition of magma?

Physics
1 answer:
Lina20 [59]3 years ago
5 0
The answer is C. I hope I helped!
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If a source of waves produces 30 waves per second, what is the frequency in hertz?
sukhopar [10]

Answer:

30hz is the answer for the question

3 0
2 years ago
Suppose that you'd like to find out if a distant star is moving relative to the earth. The star is much too far away to detect a
rjkz [21]

Answer:

The speed will be "18km/s". A further explanation is given below.

Explanation:

According to the question, the values are:

Wavelength,

\lambda = 656.46 \ nm

\Delta \lambda = 0.04

c=3\times 10^8

As we know,

⇒  \frac{\Delta \lambda}{\lambda} =\frac{v}{c}

On substituting the values, we get

⇒  \frac{656.46}{0.04} =\frac{v}{3\times 10^8}

⇒         v=\frac{656.46}{0.04} (3\times 10^8)

⇒            =16411.5\times 3\times 10^8

⇒            =18280 \ m/s

or,

⇒            =18 \ km/s

8 0
3 years ago
A 150 g sample of brass at 100 °C is placed in a Styrofoam cup of water containing 120 mL of water at 10 °C. No heat is lost to
fredd [130]

Answer:

≈19.144°C.

Explanation:

all the details are in the attachment.

Note, that c₁, m₁, t₁ are the parameters of the sample of brass; c₂, m₂ and t₂ are  the parameters of the sample of water.

P.S. change the provided design according Your requirements.

4 0
2 years ago
Name two objects that have a high density.
Sonbull [250]

Answer:

Iron and stone

Explanation:

4 0
3 years ago
As a quality test of ball bearings, you drop bearings (small metal balls), with zero initial velocity, from a height of 1.94 m i
11Alexandr11 [23.1K]

Answer:

The average acceleration of the bearings is 0.77\times10^{3}\ m/s^2

Explanation:

Given that,

Height = 1.94  m

Bounced height = 1.48 m

Time interval t=14.86\times10^{-3}\ s

Velocity of the ball bearing just before hitting the steel plate

We need to calculate the velocity

Using conservation of energy

mgh=\dfrac{1}{2}mv^2

Put the value into the formula

9.8\times1.94=\dfrac{1}{2}\times v^2

v=\sqrt{2\times9.8\times1.94}

v=6.166\ m/s

Negative as it is directed downwards

After bounce back,

We need to calculate the velocity

Using conservation of energy

mgh=\dfrac{1}{2}mv^2'

Put the value into the formula

9.8\times1.48=\dfrac{1}{2}\times v^2'

v'=\sqrt{2\times9.8\times1.48}

v'=5.38\ m/s

We need to calculate the average acceleration of the bearings while they are in contact with the plate

Using formula of acceleration

a=\dfrac{v-v'}{t}

Put the value into the formula

a=\dfrac{5.38-(-6.166)}{14.86\times10^{-3}}

a=776.98\ m/s^2

a=0.77\times10^{3}\ m/s^2

Hence,The average acceleration of the bearings is 0.77\times10^{3}\ m/s^2

6 0
3 years ago
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