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1 year ago

15

In a clear cup there are three substances. Their densities are 3, 1, and 2. From top of the cup to the bottom, what would the or

der of substances be sitting inside the cup?

1 answer:

Vesnalui [34]1 year ago

4 0

Answer:

1, 2 and 3

Explanation:

The most dense substance will settle at the bottom of the cup

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Suppose you were asked to find the torque about point p due to the normal force n in terms of given quantities. which method of

igomit [66]

Answer:

T=Lnsin $\alpha$

Please check the attached

Explanation:

The torque can simply be calculated by multiplying the length of the rod by the perpendicular force n as shown in the attached figure.

Note that sin90=1

T=Lsin $\alpha$ (nsin90)

T=Lsin $\alpha$ xn

T=Lnsin $\alpha$

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2 years ago

1. Black Panther ran 567.5 ks to the east in 2.3 hours then tumed around and ran 2218 km to

mario62 [17]

Answer:Average Velocity of the trip=98.77km/h

Explanation:

Given that:

Black Panther started his journey with

Distance ran towards east = 567.5km

Time ran towards east = 2.3 hours

and then turned back,

Distance ran towards west, = 2218 km

Time taken towards west, = 1.2 hours

Let us treat the westward movement as negative movement of the east

So total displacement =567.5 km east + (-2218km east )= 1,650.5km

But it seems the value of the distance ran west is wrong , because i do not think he can run 2218km under 1.2hours considering he ran 567.5m due east at 2.3hours .

So Let we use 221.8 km as distance ran due west.

So that Total Displacement becomes = 567.5 km east + (-221.8km east )= 345.7km

Total time = 2.3 + 1.2 = 3.5hrs.

Average velocity = Total displacement / total time

.345.7km / 3.5hr = 98.77km/h

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2 years ago

What is the net force on a car if the force of friction is 15 N and the forward force due to the engine is 20 N?

anygoal [31]

D. 35n forwards....................

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2 years ago

You're driving a vehicle of mass 1350 kg and you need to make a turn on a flat road. The radius of curvature of the turn is 71 m

sergey [27]

Answer:

$v=12.65\ m.s^{-1}$

Explanation:

Given:

mass of vehicle, $m=1350\ kg$

radius of curvature, $r=71\ m$

coefficient of friction, $\mu=0.23$

<u>During the turn to prevent the skidding of the vehicle its centripetal force must be equal to the opposite balancing frictional force:</u>

$m.\frac{v^2}{r} =\mu.N$

where:

$\mu=$ coefficient of friction

$N=$ normal reaction force due to weight of the car

$v=$ velocity of the car

$1350\times \frac{v^2}{71} =0.23\times (1350\times 9.8)$

$v=12.65\ m.s^{-1}$ is the maximum velocity at which the vehicle can turn without skidding.

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2 years ago

How does the height from which you drop the ball relate to the height that the ball bounces back up?

Stels [109]

The higher you go the more potential energy there is, and the lower it is the more kinetic energy there is, so the more kinetic energy there is the higher the ball will bounce.

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2 years ago