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3 years ago

PLEASE HELP! WILL MARK BRAINLIEST!

Physics

1 answer:

Alborosie3 years ago

6 0

Answer:22.22m/s

Explanation:

Answer: 22.22 m/s

Explanation:

Using the law of conservation of momentum:

Initial momentum of the system = final momentum of the system.

mu+m'u' = mv+m'v'

where u and u' are the initial velocities.

v and v' are the final velocities.

mass of bullet, m = 0.04 kg

mass of the block of wood, m' = 0.5 kg

initial velocity of bullet, u = 300 m/s

initial velocity of block, u' = 0

after collision, block and bullet would move together with velocity v

⇒0.04 kg × 300 m/s + 0 = 0.54 kg × v

⇒ v = 22.22 m/s

Thus, the bullet and the wood block move together as one object with the speed 22.2 m/s.

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Use the graph below to answer the following question: if average acceleration is calculated using the equation, “ change in velo

sergiy2304 [10]

Answer:

$a=9\ cm/s^2$

Explanation:

<u>Average Acceleration </u>

Acceleration is a physical magnitude defined as the change of velocity over time. When we have experimental data, we can compute it by calculating the slope of the line in velocity vs time graph.

Note: <em>We cannot see if the time axis is numbered in increments of 1 second, and we'll assume that. </em>

When $t_2=4\ sec$ , the graph shows a value of $v_2=36\ cm/s$

When $t_1=0\ sec$ , the object is at rest, $v_1=0$

We compute the average acceleration as

$\displaystyle a=\frac{v_2-v_1}{t_2-t_1}$

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3 years ago

A Hall probe, consisting of a rectangular slab of current-carrying material, is calibrated by placing it in a known magnetic fie

Citrus2011 [14]

Answer:

(a) 0.345 T

(b) 0.389 T

Solution:

As per the question:

Hall emf, $V_{Hall} = 20\ mV = 0.02\ V$

Magnetic Field, B = 0.10 T

Hall emf, $V'_{Hall} = 69\ mV = 0.069\ V$

Now,

Drift velocity, $v_{d} = \frac{V_{Hall}}{B}$

$v_{d} = \frac{0.02}{0.10} = 0.2\ m/s$

Now, the expression for the electric field is given by:

$E_{Hall} = Bv_{d}sin\theta$ (1)

And

$E_{Hall} = V_{Hall}d$

Thus eqn (1) becomes

$V_{Hall}d = dBv_{d}sin\theta$

where

d = distance

$B = \frac{V_{Hall}}{v_{d}sin\theta}$ (2)

(a) When $\theta = 90^{\circ}$

$B = \frac{0.069}{0.2\times sin90} = 0.345\ T$

(b) When $\theta = 60^{\circ}$

$B = \frac{0.069}{0.2\times sin60} = 0.398\ T$

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3 years ago

7. Two people are pushing a 40.0kg table across the floor. Person 1 pushes with a force of 490N

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Answer:

20.4 $m/s^{2}$

Explanation:

To start doing this problem, first draw a free body diagram of the table. My teacher always tells us to do this, and I find that it is very helpful. I have attached a free body diagram to this answer- take a look at it.

First, let us see if Net force = MA. To do that, we need to determine whether the object is at equilibrium horizontally. For an object to be at equilibrium, it either needs to be moving at a constant velocity or not moving at all. Also, if an object is at equilibrium, there will not be any acceleration. But we know that there IS acceleration horizontally, so it cannot be in equilibrium. If it is not in equilibrium, we can use the formula ∑F= ma.

Let us determine the net force. Since the object is moving horizontally, we can ignore the weight and normal force, because they are vertical forces. The only horizontal forces we need to worry about are the applied force and force of friction.

Applied force = 1055 N (490 + 565)

Friction force= Unknown

To find the friction force, use the kinetic friction formula, Friction = μkN

μk is the coefficient, which the problem includes- it is 0.613.

N is the normal force, which we have to find.

*To find the normal force, we have to determine if the object is at equilibrium VERTICALLY. Since it has no acceleration vertically (it's not moving up/down), it is at equilibrium. Now, when an object is at equilibrium in one direction, it means that all the forces in that direction are equal. What are our vertical forces? Weight (mg) and Normal force (N). So it means that the Normal force is equal to the Weight.

Weight = mg = (40)(9.8) = 392 N

Normal force = 392 N

Now, plug it back into the formula (μkN): (0.613)(392) = 240.296 N

Friction = 240.296 N

Now that we know the friction, we can find the horizontal net force. Just subtract the friction force, 240.296 from the applied force, 1055 N

Horizontal Net Force: 814.704 N

Now that we know the net force, plug in the numbers for the formula

∑F= ma.

814.704 = (40.0)(a)

*Divide on both sides)

a = 20.3676 m/s^2

Round it to 3 significant figures, to get:

20.4 $m/s^{2}$

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Answer:

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3 years ago