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4 years ago

The ends of a magnet where thee forces are the strongest are called magnets

Physics

1 answer:

ladessa [460]4 years ago

7 0

They are called magnets poles.

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As a skateboarder moves downhill, some of the energy of the skateboarder is

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As the skateboard rolls down the ramp it loses potential energy and gains kinetic energy.

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4 years ago

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What is the name of the force that acts between any two objects because of<br>their masses?

svetoff [14.1K]

Answer:

Explanation:

The force of attraction between 2 masses.

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4 years ago

1. What is the efficiency of a heat engine that does 1250 J of work and gives off 5250 J of heat to

vladimir2022 [97]

answer: in pic

explanation: found it here http://gaaq.weebly.com/uploads/7/8/8/4/78844490/chapter_11_q_as.pdf

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3 years ago

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I was having trouble with this problem, and problems like it: A 3.2 kg pelican, with a 1.73 kg fish in its mouth, is flying 1.52

Oduvanchick [21]

Answer:

28.1 m/s

Explanation:

$u_x$ = Initial velocity of the fish = 1.52 m/s

y = Height of the bird = 40 m

$a_y$ = Acceleration in y axis = $9.81\ \text{m/s}^2$

$u_y$ = Initial velocity in y axis = 0

$y=u_yt+\dfrac{1}{2}a_yt^2\\\Rightarrow 40=0+\dfrac{1}{2}\times 9.81t^2\\\Rightarrow t=\sqrt{\dfrac{40\times 2}{9.81}}\\\Rightarrow t=2.86\ \text{s}$

$v_y=u_y+a_yt\\\Rightarrow v_y=0+9.81\times 2.86\\\Rightarrow v_y=28.057\ \text{m/s}$

The final velocity in x direction will remain the same as the initial velocity as there is no acceleration in the x direction $u_x=v_x=1.52\ \text{m/s}$

Resultant velocity is given by

$v=\sqrt{v_x^2+v_y^2}\\\Rightarrow v=\sqrt{1.52^2+28.057^2}\\\Rightarrow v=28.1\ \text{m/s}$

The fish is moving at a velocity of 28.1 m/s when it hits the water.

6 0

3 years ago

object carries a charge of -8.1 ÂµC, while another carries a charge of -2.0 ÂµC. How many electrons must be transferred from the

LenKa [72]

Number of electrons transferred: $1.91\cdot 10^{13}$

Explanation:

The charge on the first object is

$Q_1 = -8.1\mu C$

while the charge on the 2nd object is

$Q_2=-2.0 \mu C$

When they are in contact, the final charge on each object will be

$Q=\frac{Q_1+Q_2}{2}=\frac{-8.1+(-2.0)}{2}=-5.05 \mu C$

So, the amount of charge (electrons) transferred from the 1st object to the 2nd object is

$\Delta Q = Q_1 - Q = -8.1 -(5.05)=-3.05 \mu C = -3.05\cdot 10^{-6}C$

The charge of one electron is

$e=-1.6\cdot 10^{-19}C$

Therefore, the number of electrons transferred is

$N=\frac{Q}{e}=\frac{-3.05\cdot 10^{-6}}{-1.6\cdot 10^{-19}}=1.91\cdot 10^{13}$

Learn more about electrons:

brainly.com/question/2757829

#LearnwithBrainly

8 0

3 years ago