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a_sh-v [17]
2 years ago
10

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?
Physics
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

3 0
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 ) =
= 19.20 - 3.90 = 15.30 m
7 0
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}.

7 0
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)}
n = 2,5*10^{-6+19}
\boxed{n = 2,5*10^{13}electrons}
7 0
3 years ago
What is carried by a wave?
Nata [24]
Waves carry energy from one place to another. Because waves carry energy, some waves are used for communication, eg radio and television waves and mobile telephone signals. ... Some types of waves need to be transmitted through matter, either a solid, liquid or a gas. For example, water waves have to travel in water.
3 0
3 years ago
Read 2 more answers
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