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3 years ago

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What problems do you think might arise if the chromosomes did not align during metaphase?

1 answer:

arlik [135]3 years ago

3 0

What problems do you think might arise if the chromosomes did not align during metaphase?

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A dog running with an initial speed of 7.5 m/s comes to a stop in 5 seconds. What is the

garik1379 [7]

Answer:

4.6

Explanation:

7 0

3 years ago

How is the pressure of a gas related to its concentration of particles?A) Pressure will expand a gas, enlarging its volume and r

Lesechka [4]

Answer:

C) Pressure will compress a gas, reducing its volume and giving it a greater density and concentration of particles.

Explanation:

At constant temperature, pressure and volume are inversely related.

P V = constant

$\Rightarrow P \propto \frac{1}{V}$

As the pressure increases, the gas compresses, the particles come closer reducing the volume of gas.

As we know, with decrease in volume, density increases.

$Density = \frac{Mass}{Volume}$

$Density \propto \frac{1}{Volume}$

Thus, the pressure of a gas is directly related to concentration of particles. Increase in pressure causes increase in concentration of the particles.

8 0

3 years ago

A bowling ball and a baseball both roll across your foot at the same speed. The bowling ball hurts much more.

konstantin123 [22]

Answer:

Newtons second law

Explanation:

Depends on mass

3 0

3 years ago

Read 2 more answers

An object slides on a level surface in the +x direction. It slows and comes to a stop with a constant acceleration of -2.45 m/s2

Stolb23 [73]

Answer:

uk = 0.25

Explanation:

Given:-

- An object comes to stop with acceleration, a = -2.45 m/s^2

Find:-

What is the coefficient of kinetic friction between the object and the floor?

Solution:-

- Assuming the object has mass (m) that slides over a rough surface with coefficient of kinetic friction (uk). There is only Frictional force (Ff) acting in the horizontal axis on the object opposing the motion (-x direction).

- We will apply equilibrium equation on the object in vertical direction.

N - m*g = 0

N = m*g

Where, N : Contact force exerted by the surface on the floor

g : Gravitational acceleration constant = 9.81 m/s^2

- Now apply Newton's second law of motion in the horizontal ( x-direction ):

- Ff = m*a

- The frictional force is related to contact force (N) by the following expression:

Ff = uk*N

- Substitute the 1st and 3rd expressions in the 2nd equation:

uk*m*g = -m*a

uk = a / g

- Plug in the values and solve for uk:

uk = - (-2.45) / 9.81

uk = 0.25

8 0

3 years ago

A cannonball is catapulted toward a castle. The cannonball’s velocity when it leaves the catapult is 51.6 m/s at an angle of 37.

Andrei [34K]

Answer:

$v_{y}=35.21m/s$

Explanation:

From the exercise we know the cannonball's <u>initial velocity</u>, the <u>angle</u> which its released with respect to the horizontal and its <u>initial height</u>

$v_{o}=51.6m/s\\\beta =37.0º\\y_{o}=7m$

If we want to know whats the <u>y-component of velocity</u> we need to use the following formula:

$v_{y}^2=v_{oy}^2+ag(y-y_{o})$

Knowing that $g=-9.8m/s^2$

$v_{y}=\sqrt{((51.6m/s)sin(37))^2-2(9.8m/s^2)(0m-7m)}=35.21m/s$

So, the cannonball's y-component of velocity is $v_{y}=35.21m/s$

6 0

3 years ago