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3 years ago

Select all of the answers that apply.

2 answers:

5 0

Researchers found the "cosmic microwave background radiation", which is a heat imprint left over from the big bang.

The redshift of light emitted by most galaxies indicates the universe is expanding.

Over [174]3 years ago

5 0

Researchers found the "cosmic microwave background radiation", which is a heat imprint left over from the big bang.

The redshift of light emitted by most galaxies indicates the universe is expandinding.

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A bullet of mass 0.0021 kg initially moving at 497 m/s embeds itself in a large fixed piece of wood and travels 0.65 m before co

alina1380 [7]

Answer:

The force exerted by the wood on the bullet is 399.01 N

Explanation:

Given;

mass of bullet, m = 0.0021 kg

initial velocity of the bullet, u = 497 m/s

final velocity of the bullet, v = 0

distance traveled by the bullet, S = 0.65 m

Determine the acceleration of the bullet which is the deceleration.

Apply kinematic equation;

V² = U² + 2aS

0 = 497² - (2 x 0.65)a

0 = 247009 - 1.3a

1.3a = 247009

a = 247009 / 1.3

a = 190006.92 m/s²

Finally, apply Newton's second law of motion to determine the force exerted by the wood on the bullet;

F = ma

F = 0.0021 x 190006.92

F = 399.01 N

Therefore, the force exerted by the wood on the bullet is 399.01 N

6 0

3 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

PLSS HELLPPP MEEE WILL MARK BRAINLIEST

Lesechka [4]

Answer:

a:s

b:s

c:r

d: r

e: s

f: s

g: s

Explanation:

8 0

3 years ago

How is a frame of reference used to describe motion

Zinaida [17]

In particle kinematics, we measure the acceleration of the particle. For this, an observer is taken as a reference point. Generally we consider the observer as in origin of a co-ordinate system. The place where observer is placed is known as frame of reference.
Newton's laws only work in the inertial frame of reference. Inertial frame is a frame of reference whose acceleration is zero. But, if you are observing the motion of an object from non-inertial frame, you need to add and extra force on the system under observation which is known as pseudo force.

5 0

4 years ago

Your little sister (mass 25 kg) is sitting in her little red wagon (mass 8.5 kg) at rest. You begin pulling her forward, acceler

Nata [24]

Answer:

Your little sister (mass 25.0 ) is sitting in her little red wagon (mass 8.50 at rest. You begin pulling her forward and continue accelerating her with a constant force for 2.35 at the end of which time she's moving at a speed of 1.80 .

(a) Calculate the impulse you imparted to the wagon and its passenger. (b) With what force did you pull on the wagon?

4 0

3 years ago