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3 years ago

8

Most of the ________ deposits in Virginia are located in the Appalachian Plateau and the Valley and Ridge provinces. A) coal B)

copper C) gold D) iron

1 answer:

Nesterboy [21]3 years ago

8 0

Answer:d Iron

Explanation:

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Are both intertwined. exercise and you will feel brand new!!!

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3 years ago

When driving on roads that may be slippery: A. Always drive at the maximum speed limit. B. Use cruise control to maintain a stea

erik [133]

When driving on roads that may be slippery, do not make any sudden changes in speed or direction. Option D is correct.

<h3 /><h3>What is a slippery surface?</h3>

The slick road sign serves as a warning. When the road is wet or ice, drivers should use extra caution and reduce their speed. When the weather is bad, avoid making any rapid changes in direction.

When driving on roads that may be slippery, do not make any sudden changes in speed or direction. It may cause accident because the vehicle can lose their balance.

Hence, option D is correct.

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2 years ago

Kinetic Energy - What does it depend on?

Greeley [361]

Answer:

faster; more kinetic energy

Explanation:

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3 years ago

A spring has a natural length of 28.0 cm. If a 23.0-N force is required to keep it stretched to a length of 36.0 cm, how much wo

krek1111 [17]

Answer:

0.23 J

Explanation:

k*(36 - 28) = 23

so k = 23/8 N/cm

W = k(32 - 28)²/2 = 23/8 * 4²/2 = 23 N-cm = 0.23 J

4 0

2 years ago

A 62 kg skier is moving at 6.5 m/s on frictionless horizontal snow-covered plateau when she encounters a rough patch 3.50 m long

Degger [83]

Answer:

(A). The work done by friction in crossing the patch is -637.98 J.

(B). The speed of skier is 10.57 m/s.

Explanation:

Given that,

Mass of skier = 62 kg

Speed = 6.5 m/s

Length = 3.50 m

Coefficient kinetic friction = 0.30

Height = 2.5 m

(A) we need to calculate the work done by friction in crossing the patch

Using formula of work done

$W=-\mu mg\times l$

Put the value into the formula

$W=-0.30\times62\times9.8\times3.50$

$W=-637.98\ J$

The work done by friction in crossing the patch is -637.98 J.

(B) we need to calculate the speed of skier

Using conservation of energy

$K.E_{i}+U_{i}-W_{friction}=K.E_{f}+U_{f}$

$\dfrac{1}{2}mv_{1}^2+mgh-\mu mgl=\dfrac{1}{2}mv_{2}^2+U_{f}$

Final potential energy is zero

So, $\dfrac{1}{2}mv_{1}^2+mgh-\mu mgl=\dfrac{1}{2}mv_{2}^2$

$\dfrac{1}{2}v_{2}^2=\dfrac{1}{2}v_{1}^2+gh-\mu gl$

Put the value into the formula

$\dfrac{1}{2}v_{2}^2=\dfrac{1}{2}\times6.5^2+9.8\times2.5+0.30\times9.8\times3.50$

$v_{2}=\sqrt{2\times55.915}$

$v_{2}=10.57\ m/s$

The speed of skier is 10.57 m/s.

Hence, (A).The work done by friction in crossing the patch is -637.98 J.

(B).The speed of skier is 10.57 m/s.

6 0

3 years ago