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2 years ago

Please help with my geology hw

Physics

1 answer:

Dominik [7]2 years ago

3 0

Uhhhh not a question the can be answered sorry :/

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Charges that do not transfer

ratelena [41]

Answer:what is the question exactly

Explanation:

8 0

3 years ago

Which of the following is an example of a force<br> inertia<br> push <br> pull

Whitepunk [10]

A force can be considered a push or pull

hope this helps :)

8 0

3 years ago

A 15kg hoop with a radius of 3 m is rolling at on a level surface when it reaches a 25 degree incline. If it reaches a height of

dimulka [17.4K]

The initial velocity of the hoop is determined as 8.854 m/s.

<h3>Conservation of energy</h3>

The initial velocity of the hoop can be determined from the principle of conservation of energy.

Final potential energy = Initial kinetic energy

P.E = K.E

mgh = ¹/₂mv²

gh = ¹/₂v²

2gh = v²

√2gh = v

√(2 x 9.8 x 4) = v

8.854 m/s = v

Thus, the initial velocity of the hoop is determined as 8.854 m/s.

Learn more about initial velocity here: brainly.com/question/19365526

#SPJ1

4 0

2 years ago

A 50-kg toboggan is coasting on level snow. As it passes beneath a bridge, a 20-kg parcel is dropped straight down and lands in

timama [110]

Answer:

$KE2/KE1=0.71$

Explanation:

By conservation of the linear momentum:

m1*V1 = (m1+m2)*V2

Solving for V2:

$V2 = \frac{m1}{m1+m2}*V1$

The kinetic energies are:

$KE1=1/2*m1*V1^2$

$KE2 = 1/2*(m1+m2)*V2^2$

$KE2 = 1/2*(m1+m2)*(\frac{m1}{m1+m2}*V1)^2$

Simplifying:

$KE2 = 1/2*\frac{m1^2}{m1+m2}*V1^2$

The ratio will be:

$KE2/KE1=\frac{1/2*\frac{m1^2}{m1+m2}*V1^2}{1/2*m1*V1^2} =\frac{m1}{m1+m2}$

$KE2/KE1=0.71$

5 0

3 years ago

What is the minimum coefficient of friction needed for a frightened driver to take the same curve at 20.0 km/h?

Tanya [424]

Answer:

The minimum coefficient of friction is 0.22

Explanation:

Suppose If a car takes a banked curve at less than the ideal speed, friction is needed to keep it from sliding toward the inside of the curve.

We need to calculate the ideal speed to take a 85 m radius curve banked at 15°.

Given that,

Radius = 85 m

Angle = 15°

Speed = 20 km/h

We need to calculate the ideal speed

Using formula of speed

$\tan\theta=\dfrac{v^2}{rg}$

$v=\sqrt{rg\tan\theta}$

Put the value into the formula

$v=\sqrt{85\times9.8\tan15}$

$v=14.9\ m/s$

We need to calculate the minimum coefficient of friction

Using formula for coefficient of friction

$v^2=\dfrac{rg(\sin\theta-\mu\cos\theta)}{\mu\sin\theta+\cos\theta}$

Put the value into the formula

$(5.55)^2=\dfrac{85\times9.8(\sin15-\mu\cos15)}{\mu\sin15+\cos15}$

$\dfrac{30.8025}{85\times9.8}=\dfrac{0.2588-\mu0.966}{\mu0.2588+0.966}$

$0.9754462\ mu-0.223541=0$

$\mu=\dfrac{0.223541}{0.9754462}$

$\mu=0.22$

Hence, The minimum coefficient of friction is 0.22

3 0

2 years ago