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Which type of environment results from an equal amount of solute and solvent in and out of the cell?

grigory [225]

The answer to the first question is C isotonic environment. To help me learn this I memorized that isotonic is equal hypotonic is over hydrated while hypertonic is dehydrated. And for the second question is they move from high to low so C

6 0

3 years ago

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Which of the following is a pure substance that can't be separated by physical means, and can't be decomposed into other substan

g100num [7]

Your answer is C. element

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3 years ago

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A 1100 kg car rounds a curve of radius 68 m banked at an angle of 16 degrees. If the car is traveling at 95 km/h, will a frictio

Mariulka [41]

Answer:

Yes. Towards the center. 8210 N.

Explanation:

Let's first investigate the free-body diagram of the car. The weight of the car has two components: x-direction: towards the center of the curve and y-direction: towards the ground. Note that the ground is not perpendicular to the surface of the Earth is inclined 16 degrees.

In order to find whether the car slides off the road, we should use Newton's Second Law in the direction of x: F = ma.

The net force is equal to $F = \frac{mv^2}{R} = \frac{1100\times (26.3)^2}{68} = 1.1\times 10^4~N$

Note that 95 km/h is equal to 26.3 m/s.

This is the centripetal force and equal to the x-component of the applied force.

$F = mg\sin(16) = 1100(9.8)\sin(16) = 2.97\times10^3$

As can be seen from above, the two forces are not equal to each other. This means that a friction force is needed towards the center of the curve.

The amount of the friction force should be $8.21\times 10^3~N$

Qualitatively, on a banked curve, a car is thrown off the road if it is moving fast. However, if the road has enough friction, then the car stays on the road and move safely. Since the car intends to slide off the road, then the static friction between the tires and the road must be towards the center in order to keep the car in the road.

5 0

4 years ago

The gravitational field strength at a distance R from the center of moon is gR. The satellite is moved to a new circular orbit t

3241004551 [841]

Answer:

$g'=\frac{g__R}{4}$

Explanation:

Given:

gravitational field strength of moon at a distance R from its center, $g__R$

Distance of the satellite from the center of the moon, $h=2R$

<u>Now as we know that the value of gravity of any heavenly body is at height h is given as:</u>

$g'=g__{R}} \times \frac{R^2}{(2R)^2}$

$g'=\frac{g__R}{4}$

∴The gravitational field strength will become one-fourth of what it is at the surface of the moon.

6 0

3 years ago

according to newton's third law, of a sledgehammer hits a wall with a force of 1000 n, how much force does the wall hit the sled

astra-53 [7]

The correct option is A: the force exerted on the wall by the sledgehammer is - 1000 N

<h3>Newton's third law</h3>

Newton's third law of motion states that for every action or force, there is an equal but oppositely directed force.

<h3>Application of the newton's third law</h3>

Force with which the sledgehammer hits the wall = 1000 N

From the Newton's third law, an equal but oppositely directed force is exerted on the sledgehammer by the wall.

Thus, the force exerted by the sledgehammer on the wall = -1000 N

The negative sign indicates that the force acts in a direction opposite to that of the sledgehammer.

Thus, the correct option is A: the force exerted on the wall by the sledgehammer is - 1000 N

Learn more about Newton's third law and force at: brainly.com/question/13874955

7 0

2 years ago