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3 years ago

________________________can turn igneous rock into sediment

2 answers:

7 0

Igneous rock can change into sedimentary rock or into metamorphic rock

Assoli18 [71]3 years ago

4 0

Weathering and erosion can turn any type of rock into sediments.

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If 5J of work are done on a spring, compressing it by 12cm, what is the spring constant?

laiz [17]

Answer:

Explanation:

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2 years ago

How do we work out the world time zone differences

Virty [35]

To find the time zone in hours of a particular location, you can take the longitude -- in degrees -- and divide it by 15. So, for example, 75° E would be 75/15 which equals 5. That translates to the time zone being 5 hours ahead of UTC or GMT time, which can also be labeled as UTC+5. i hope this helped if not then sorry

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2 years ago

PHYSICAL SCIENCE HELP PLEASE !!!!! Picture below is for question 4

Eva8 [605]

Question 1) excited and moving around a lot.

Question 2) heat

Question 3) specific heat

Question 4) As temperature increases, energy transferred increases.

Hope these answers help!

3 0

3 years ago

A girl (mass M) standing on the edge of a frictionless merry-go-round (radius R, rotational inertia I) that is not moving. She t

vladimir1956 [14]

a) $\omega=\frac{-mvR}{I+MR^2}$

b) $v=\frac{-mvR^2}{I+MR^2}$

Explanation:

Since there are no external torques acting on the system, the total angular momentum must remain constant.

At the beginning, the merry-go-round and the girl are at rest, so the initial angular momentum is zero:

$L_1=0$

Later, after the girl throws the rock, the angular momentum will be:

$L_2=(I_M+I_g)\omega +L_r$

where:

$I$ is the moment of inertia of the merry-go-round

$I_g=MR^2$ is the moment of inertia of the girl, where

M is the mass of the girl

R is the distance of the girl from the axis of rotation

$\omega$ is the angular speed of the merry-go-round and the girl

$L_r=mvR$ is the angular momentum of the rock, where

m is the mass of the rock

v is its velocity

Since the total angular momentum is conserved,

$L_1=L_2$

So we find:

$0=(I+I_g)\omega +mvR\\\omega=\frac{-mvR}{I+MR^2}$

And the negative sign indicates that the disk rotates in the direction opposite to the motion of the rock.

The linear speed of a body in rotational motion is given by

$v=\omega r$

where

$\omega$ is the angular speed

r is the distance of the body from the axis of rotation

In this problem, for the girl, we have:

$\omega=\frac{-mvR}{I+MR^2}$ is the angular speed

$r=R$ is the distance of the girl from the axis of rotation

Therefore, her linear speed is:

$v=\omega R=\frac{-mvR^2}{I+MR^2}$

5 0

2 years ago

2. A carpenter tosses a shingle off a 9.4 m high roof, giving it an initial horizontal

Aleksandr [31]

Sounds like the shingle/ball is thrown from the roof horizontally, so that the distance it travels x after time t horizontally is

x = (7.2 m/s) t

The object's height y at time t is

y = 9.4 m - 1/2 gt²

where g = 9.80 m/s² is the magnitude of the acceleration due to gravity, and its vertical velocity is

v = -gt

(a) The object hits the ground when y = 0:

0 = 9.4 m - 1/2 gt²

t² = 2 * (9.4 m) / (9.80 m/s²)

t ≈ 1.92 s

at which time the object's vertical velocity is

v = -g (1.92 s) = -18.8 m/s ≈ -19 m/s

(b) See part (a); it takes the object about 1.9 s to reach the ground.

(c) The object travels a horizontal distance of

x = (7.2 m/s) * (1.92 s) ≈ 13.8 m ≈ 14 m

8 0

2 years ago