Explain this statement: Cells are the basic unit of structure and function in organisms.

Two masses sit at the top of two frictionless inclined planes that have different angles,__deleted9917f34947e1359b5705bbac0d9227

Strike441 [17]

Answer:

v = $\sqrt{2gh}$

the speed in the two planes will be the same since it does not depend on the angle of the same

Explanation:

In this exercise we are told that the two inclined planes have no friction force, so we can apply the conservation of energy for each one, we will assume that the initial height in the two planes is the same

starting point. Highest part of each plane

Em₀ = U = m g h

final point. Lowest part of each plane

$Em_{f}$ = K = ½ m v²

as there is no friction, the mechanical energy is preserved

Em₀ = Em_{f}

mg h = ½ m v²

v = $\sqrt{2gh}$

As we can see, the speed in the two planes will be the same since it does not depend on the angle of the same

7 0

3 years ago

Nuclear fusion occurs in stars. True or false

Butoxors [25]

<span>nuclear fusion is observed in stars coming from the combination of two hydrogens to form an atom of helium. Fusion means combination of two or more elements. helium is known to fuel energy which in this case, the great sun which is more than a million Kelvin in temperature already today. hence the answer is true.</span>

8 0

3 years ago

Read 2 more answers

In a car collision, is it better for the passenger if the change in momentum is over a short period of time or a long period of

e-lub [12.9K]

Explanation:

The force on the passenger will be F = ma

Here, m does not change, but a is the variable.

If the cars slows down very fast, the acceleration will be higher, and thus the force will be higher.

If the acceleration is lower, the force will be lower as well, which would be the most desirable scenario for the passenger.

5 0

3 years ago

To find the mass defect for uranium-238 several calculations must be made. Which is the calculation that should be made first? A

BaLLatris [955]

mass number of uranium 238 is sum of number of proton and neutrons in it

While atomic number is just number of protons

So total number of protons in it = 92

total number of neutrons in it = 238 - 92 = 146

mass of proton = 1.0073 amu

mass of neutron = 1.0087 amu

now for the calculation of mass of nuclei is given as

$M = zm_p + (A-z)m_n$

$M = 92(1.0073) + 146(1.0087)$

so option B is correct

3 0

4 years ago

A large crate with mass m rests on a horizontal floor. The static and kinetic coefficients of friction between the crate and the

Mariana [72]

a) $F=\frac{\mu_k mg}{cos \theta - \mu_k sin \theta}$

Here the crate is moving at constant velocity, so no acceleration:

a = 0

Let's analyze the forces acting along the horizontal and vertical direction.

- Vertical direction: the equation of the forces is

$R-Fsin \theta - mg = 0$ (1)

where

R is the normal reaction of the floor (upward)

$F sin \theta$ is the component of the force F in the vertical direction (downward)

mg is the weight of the crate (downward)

- Horizontal direction: the equation of the forces is

$F cos \theta - \mu_k R = 0$ (2)

where

$F cos \theta$ is the horizontal component of the force F (forward)

$\mu_k R$ is the force of friction (backward)

From (1) we get

$R=Fsin \theta +mg$

And substituting into (2)

$F cos \theta - \mu_k (Fsin \theta +mg) = 0\\F cos \theta -\mu _k F sin \theta = \mu_k mg\\F(cos \theta - \mu_k sin \theta) = \mu_k mg\\F=\frac{\mu_k mg}{cos \theta - \mu_k sin \theta}$

b) $\mu_s=cot(\theta)$

In this second case, the crate is still at rest, so we have to consider the static force of friction, not the kinetic one.

The equations of the forces will be:

$R-Fsin \theta - mg = 0$ (1)

$F cos \theta - \mu_s R = 0$ (2)

In this second case, we want to find the critical value of $\mu_s$ such that the woman cannot start the crate: this means that the force of friction must be at least equal to the component of the force pushing on the horizontal direction, $F cos \theta$ .

Therefore, using the same procedure as before,

$R=Fsin \theta +mg$

$F cos \theta - \mu_s (Fsin \theta +mg) = 0$

And solving for $\mu_s$ ,

$F cos \theta = \mu_s (Fsin \theta +mg) \\\mu_s = \frac{F cos \theta}{F sin \theta + mg}$

Now we analyze the expression that we found. We notice that if the force applied F is very large, $F sin \theta >> mg$ , therefore we can rewrite the expression as

$\mu_s \sim \frac{F cos \theta}{F sin \theta}\\\mu_s=cot(\theta)$

So, this is the critical value of the coefficient of static friction.

8 0

3 years ago