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2 years ago

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Two blocks, of mass m and 2m, are initially at rest on a horizontal frictionless surface. A force F is exerted individually on e

ach block, as shown above. The graph shows how F varies with time t. Which block has the greatest average power provided to it between t = 0 s and t = 3 s?

1 answer:

Anon25 [30]2 years ago

7 0

The block has the greatest average power provided is bock m.

<h3>What is instantaneous power?</h3>

This is the product of force and velocity exerted on an object.

Mathematically instantaneous power is calculated as;

P = Fv

where;

F is the applied force
v is the velocity

Both blocks (m and 2m) will experience the same force but different velocity.

The smaller block (m) will experience greater velocity.

Thus, the block has the greatest average power provided is bock m.

Learn more about instantaneous power here: brainly.com/question/8893970

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A wire carries a current of 4.1 A. How many electrons per second are passing any cross sectional area of the wire? Enter your an

aivan3 [116]

To solve this problem it is necessary to apply the concepts related to Current and Load.

The current in terms of the charge of an electron can be expressed as

$i = \frac{q}{t}$

Where,

q = Charge

t = time

At the same time the Charge is the amount of electrons multiplied by the amount of these, that is

q = ne

Replacing in the first equation we have to

$i = \frac{q}{t}$

$i = \frac{ne}{t}$

Clearing n,

$n = \frac{it}{e}$

Here the time is one second then

$n = \frac{i}{e}$

$n = \frac{4.1}{1.6*10^{-9}}$

$n = 2.56*10^{19}electrons$

Therefore the number of electrons per second are passing any cross sectional area of the wire are $2.56*10^{19}electrons$

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3 years ago

Driving home from school one day, you spot a ball rolling out into the street (Figure 5-21). You brake for 1.20 s, slowing your

frosja888 [35]

A ) v = v o + a t ( the acceleration will be negative )
9.50 = 16.0 + a * 1.2
a * 1.2 = -16.0 + 9.50
a * 1.2 = - 6.5
a = - 6.5 : 1.2
a = - 5.4167 m/s²
F = m * a = 950 kg * 5.4167 m/s²
F = 5,145.8 N ( the average force exerted on a car during braking )
b ) d = v o - a t² / 2
d = 16.0 * 1.2 - ( 5.4167 * 1.2² / 2 ) =
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3 years ago

A boy throws a ball vertically up it returns the ground after 10 seconds find the maximum height reached by the ball

Akimi4 [234]

Answer:

Approximately $122.625\; {\rm m}$ (assuming that $g = 9.81\; {\rm m\cdot s^{-2}}$ , the ball was launched from ground level, and that the drag on the ball is negligible.)

Explanation:

Let $v_{0}$ denote the velocity at which the ball was thrown upward.

If the drag (air friction) on the ball is negligible, the ball would land with a velocity of exactly $(-v_{0})$ . The velocity of the ball would be changed from $v$ to $(-v_{0})\!$ (such that $\Delta v = (-v_{0}) - v_{0} = (-2\, v_{0})$ ) within $t = 10\; {\rm s}$ .

Also because the drag on the ball is negligible, the acceleration of the ball would be $a = -g = -9.81\; {\rm m\cdot s^{-2}}$ . Thus:

$\Delta v = a\, t = 10\; {\rm s} \times (-9.81\; {\rm m\cdot s^{-2}}) = -98.1\; {\rm m\cdot s^{-1}}$ .

Since $\Delta v = (-2\, v_{0})$ :

$-2\, v_{0} = \Delta v = -98.1\; {\rm m\cdot s^{-1}$ .

$\begin{aligned}v_{0} &= \frac{-98.1\; {\rm m\cdot s^{-1}}}{-2}= 49.05\; {\rm m \cdot s^{-1}}\end{aligned}$ .

The ball reaches maximum height when its velocity is $v_{1} = 0\; {\rm m\cdot s^{-1}}$ . Apply the SUVAT equation $x = ({v_{1}}^{2} - {v_{0}}^{2}) / (2\, a)$ to find the displacement $x$ between the original position (ground level, where $v_{0} = 49.05\; {\rm m\cdot s^{-1}}$ ) and the max-height position of the ball (where $v_{1} = 0\; {\rm m\cdot s^{-1}}$ .)

$\begin{aligned}x &= \frac{(0\; {\rm m\cdot s^{-1}})^{2} - (49.05\; {\rm m\cdot s^{-1}})^{2}}{2 \times (-9.81\; {\rm m\cdot s^{-2}})} \\ &\approx 122.625\; {\rm m\cdot s^{-1}}\end{aligned}$ .

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2 years ago

A negatively charged balloon has 4 μC of charge. How many excess electrons are on this bal- loon? The elemental charge is 1.6 ×

bagirrra123 [75]

Data:

The charge of a body depends on the amount of electrons it gains or loses. Q = n * e, where "Q" is charge, "n" is the number of plus or minus electrons, and "e" is the fundamental charge of an electron $1,6 * 10 ^{-19}C$ <span>. To know if the body has gained or lost, we look at the signal of its charge, remembering that the electron is negative. The charge of the body is 4 μC (positive), so there is a lack of electrons!

Q = 4 </span>μC → $Q = 4*10^{-6}$
$e = 1,6 * 10 ^{-19}C$
$n = ?$ <span>

We have:
</span> $Q = n*e$
$n = \frac{Q}{e}$
$n = \frac{4*10^{-6}}{1,6 * 10 ^{-19}}$
$n = 2,5*10^{-6-(-19)}$
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