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3 years ago

14

In the warm up, you reviewed the equation to calculate kinetic energy. What question could you ask about kinetic energy which wi

ll include the variables that affect it?

2 answers:

enot [183]3 years ago

7 0

Answer:

what is the average efficiency of the potensial to kinetic energy on a pendulum

Explanation:

shutvik [7]3 years ago

4 0

Explanation:

HEY PLS DON'T JOIN THE ZOOM CALL OF A PERSON WHO'S ID IS 825 338 1513 (I'M NOT SAYING THE PASSWORD) HE IS A CHILD PREDATOR AND A PERV. HE HAS LOTS OF ACCOUNTS ON BRAINLY BUT HIS ZOOM NAME IS MYSTERIOUS MEN.. HE ASKS FOR GIRLS TO SHOW THEIR BODIES AND -------- PLEASE REPORT HIM IF YOU SEE A QUESTION LIKE THAT. WE NEED TO TAKE HIM DOWN!!! PLS COPY AND PASTE THIS TO OTHER COMMENT SECTIONS!!

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1. The magnitude of the electrostatic force a charge particle exerts on a 5.0nC particle 0.20m away from it is 18N. Find the mag

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The magnitude force a charge of the 16nC particle of 0.40

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2 years ago

If the total mass is m, find the moment of inertia about an axis through the center and perpendicular to the plane of the square

My name is Ann [436]

Question: A thin, uniform rod is bent into a square<span> of... A thin, uniform rod is bent into a </span>square<span> of side length a. </span>If the total mass is M<span>, </span>find the moment of inertia about an axis through the center and perpendicular to the plane of the square<span>. </span>Use the parallel-axis theorem<span>.</span>

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3 years ago

a 5.5 g dart is fired into a block of wood with a mass of 22.6 g. the wood block is initially at rest on a 1.5 m tall post. afte

IgorLugansk [536]

<span>From the problem alone we can say that the dart and the block of wood combined into a single object moving together at the end. With that clue we know that the collision is an inelastic collision. The formula of an inelastic collision is:

$m_{1}v_{1i}+m_{2}v_{2i}=(m_{1}+m_{2})v_{f}$

First let us sort out our given:Mass should be in kg to get the proper answer. Now let's assign m1 as the mass of the dart and m2 as the mass of the block.
m1 = 5.5g

$5.5g$ x $\frac{1kg}{1000g}= 0.0055kg$

m2 = 22.6g

$22.6g$ x $\frac{1kg}{1000g}= 0.0226kg$

So now we settled that we can set our given as:
M1 = .0055 kg
v1i = ?
M2 = 0.0226 kg
v2i = 0 m/s
dx = 2.5 m
dy = -1.5 m

Now you can see that we have 2 unknowns: v1i and vf. We need the vf to solve for the initial velocity of the dart or object 1. We have other given to consider, so we can make use of that to get our missing vf.

Now, vf is the horizontal velocity after the collision. We do this by first using the equations for projectiles considering that we have an x and y dimension to consider. We use the y dimension to get the x.
</span>

dy = -1.5 m

a = 9.8m/s^2

viy = 0 (take note that the initial vertical velocity is 0)

t = ?

<span>We can use the UAM equations to solve for the time in the y-dimension (vertical) to get the horizontal velocity.

$dy = v_{iy}t + \frac{1}{2} at^{2}$ </span>

$1.5 = (0)t+\frac{1}{2} (9.8)t^{2}$

<span> $1.5 = \frac{1}{2} (9.8)t^{2}$

$\frac{(2)(1.5)}{9.8}=t^{2}$

$\frac{(3)}{9.8}=t^{2}$

$\sqrt{0.3061} = \sqrt{t^{2}$

$0.553s = t$

Now using this, we can get the horizontal (x-dimension) velocity using the formula:
$v_{x} =d_{x}t$ and our given earlier for the horizontal distance is 2.5m and we solved for time 0.553s. Let's put that into our equation:
$v_{x} =d_{x}t$
$v_{x} =(2.5m)(0.553s)$
$v_{x} =4.52m/s$

Now we finally have our vf or velocity after the collision. Now let's get back to the equation.

$m_{1}v_{1i}+m_{2}v_{2i}=(m_{1}+m_{2})v_{f}$

From this we can derive the equation for v1i by isolating it.

$v_{1i}= \frac{((m_{1}+m_{2})v_{f})-(m_{2}v_{2i})}{m_{1}}$

Now let's put in all our given and what we solved:

$v_{1i}= \frac{((0.0055kg+0.0226kg)4.52m/s)-((0.0226kg)0m/s)}{0.0055kg}$

$v_{1i}= \frac{(0.0281kg)4.52m/s)}{0.0055kg}$

$v_{1i}= \frac{0.127012kg.m/s}{0.0055kg}$

$v_{1i}= 23.09m/s$

The initial speed of the dart is 23.09 m/s or 23.10 m/s.</span>

7 0

4 years ago

A student throws a water balloon vertically downward from the top of a building. The balloon leaves the thrower's hand with a sp

dusya [7]

from the question you can see that some detail is missing, using search engines i was able to get a similar question on "https://www.slader.com/discussion/question/a-student-throws-a-water-balloon-vertically-downward-from-the-top-of-a-building-the-balloon-leaves-t/"

here is the question : A student throws a water balloon vertically downward from the top of a building. The balloon leaves the thrower's hand with a speed of 60.0m/s. Air resistance may be ignored,so the water balloon is in free fall after it leaves the throwers hand. a) What is its speed after falling for 2.00s? b) How far does it fall in 2.00s? c) What is the magnitude of its velocity after falling 10.0m?

Answer:

(A) 26 m/s

(B) 32.4 m

(C) v = 15.4 m/s

Explanation:

initial speed (u) = 6.4 m/s

acceleration due to gravity (a) = 9.9 m/s^[2}

time (t) = 2 s

(A) What is its speed after falling for 2.00s?

from the equation of motion v = u + at we can get the speed

v = 6.4 + (9.8 x 2) = 26 m/s

(B) How far does it fall in 2.00s?

from the equation of motion $s=ut+0.5at^{2}$ we can get the distance covered

s = (6.4 x 2) + (0.5 x 9.8 x 2 x 2)

s = 12.8 + 19.6 = 32.4 m

c) What is the magnitude of its velocity after falling 10.0m?

from the equation of motion below we can get the velocity

$v^{2} = u^{2} + 2as\\v^{2} = 6.4^{2} + (2x9.8x10)\\V^{2} = 236.96\\v = \sqrt{236.96}$

v = 15.4 m/s

7 0

3 years ago

I need this done pls

Aleonysh [2.5K]

Answer:

all the details are in the attached picture, the answers are marked with colour.

5 0

2 years ago