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3 years ago

Which of the follow are types of changes that can happen in the rock cycle

Physics

1 answer:

ollegr [7]3 years ago

3 0

you must have a rock first of all then the cycle continues

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QUESTION 1

Sveta_85 [38]

distance from the Sun of 2.77 astronomical units or about 414 million km 257 million miles and orbiting period of 4.62 years

7 0

3 years ago

A 180 lb crate is on the ground, and a strong rope is attached. You need to move it across the basement floor, which has a coeff

jekas [21]

Answer:

$F = 505.13 N$

Yes it is better to pull the rope rather than push it

Explanation:

Let the force is applied at an angle of 60 degree

so we will have net vertical force on the crate is given as

$F_n + Fsin60 = mg$

here we know

$m = 180 lb$

$m = 81.65 kg$

$F_n = 81.65(9.81) - Fsin60$

$F_n = 801 - 0.866 F$

now friction force on the crate is given as

$F_x = \mu F_n$

$Fcos60 = 0.7(801 - 0.866 F)$

$0.5F + 0.61F = 560.7$

$F = 505.13 N$

Yes it is better to pull the rope rather than push it

6 0

3 years ago

Jonah is trying to move his 22-kg desk by pushing on it with a force of 130 N, but his brother is leaning on it with a downward

Dafna11 [192]

Answer:

0.54

Explanation:

Draw a free body diagram. There are 5 forces on the desk:

Weight force mg pulling down

Applied force 24 N pushing down

Normal force Fn pushing up

Applied force 130 N pushing right

Friction force Fnμ pushing left

Sum of the forces in the y direction:

∑F = ma

Fn − mg − 24 = 0

Fn = mg + 24

Fn = (22)(9.8) + 24

Fn = 240

Sum of the forces in the x direction:

∑F = ma

130 − Fnμ = 0

Fnμ = 130

μ = 130 / Fn

μ = 130 / 240

μ = 0.54

6 0

3 years ago

9. A radioisotope has a half-life of 4.50 min and an initial decay rate of 8400 Bq. What will be

Akimi4 [234]

Answer:

525 Bq

Explanation:

The decay rate is directly proportional to the amount of radioisotope, so we can use the half-life equation:

A = A₀ (½)^(t / T)

A is the final amount

A₀ is the initial amount,

t is the time,

T is the half life

A = (8400 Bq) (½)^(18.0 min / 4.50 min)

A = (8400 Bq) (½)^4

A = (8400 Bq) (1/16)

A = 525 Bq

8 0

3 years ago

A meter stick A hurtles through space at a speed v = 0.25c relative to you, with its length aligned with the direction of motion

yaroslaw [1]

Answer:

$L_0\approx1.0328\ m$

Explanation:

Given:

relativistic length of stick A, $L=1\ m$

relativistic velocity of stick A with respect to observer, $v=0.25c=7.5\times 10^{7}\ m.s^{-1}$

<em>Since the object is moving with a velocity comparable to the velocity of light with respect to the observer therefore the length will appear shorter according to the theory of relativity.</em>

<u> Mathematical expression of the theory of relativity for length contraction:</u>

$L=\frac{L_0}{\gamma}$

where:

L = relativistic length

$L_0=$ original length at rest

$\gamma =$ Lorentz factor $=\frac{1}{\sqrt{1-\frac{v^2}{c^2} } }$

$\Rightarrow 1=\frac{L_0}{\frac{1}{\sqrt{1-\frac{(0.25c)^2}{c^2} } }}$

$L_0=\frac{1}{\sqrt{1-\frac{(0.25c)^2}{c^2} } }$

$L_0\approx1.0328\ m$

4 0

3 years ago