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Alexeev081 [22]
2 years ago
14

What are the uses of Electricity in our every day life?

Physics
1 answer:
neonofarm [45]2 years ago
7 0
Calling
Heating Food
Going on the internet
Video Games 
Weather
Heating up pools 
You might be interested in
When the tube is filled with mercury vapor, as in this case, a sharp drop in the collected current is observed when the accelera
Nata [24]

Answer:

The energy absorbed by the atomic electrons in the mercury atom is 7.84 \times 10^{-19} J

Explanation:

Given:

Potential V = 4.9 V

According to the conservation law,

   Loss in kinetic energy = Gain in potential energy

Here, energy absorbed by the atomic electrons is given by,

    E = eV

Where e = 1.6 \times 10^{-19} C       ( charge of electron )

    E = 1.6 \times 10^{-19 } \times 4.9

    E = 7.84 \times 10^{-19} J

Therefore, the energy absorbed by the atomic electrons in the mercury atom is 7.84 \times 10^{-19} J

5 0
2 years ago
When a planet orbits the Sun, the
Anna [14]
I’m not sure I think it’s A
6 0
2 years ago
Read 2 more answers
You throw a ball from the balcony onto the court in the basketball arena. You release the ball at a height of 7.00 m above the c
Mariana [72]

Answer:

Your friend has to wait 0.26 s after you throw the ball to start running.

Explanation:

The equation that gives the position vector of the ball is as follows:

r = (x0 + v0 · t · cos α, y0 + v0 · t ·sin α + 1/2 · g · t²)

Where:

x0 = initial horizontal positon

v0 = initial velocity

t = time

α = throwing angle

y0 = initial vertical position

g = acceleration due to gravity

The equation of displacement of your friend is as follows:

x = x0 + v0 · t + 1/2 · a · t²

Where:

x = position of your friend at time t

x0 = initial position

v0 = initial velocity

t = time

a = acceleration

Please, see the attached figure for a description of the situation. Notice that the frame of reference is located at the throwing point.

Let´s find the time of flight of the ball. We know that at the final time, the y-component of the vector r has to be -6.00 m (1 m above the ground). Then:

y = y0 + v0 · t ·sin α + 1/2 · g · t²

-6.00 m = 0 m + 9.00 m/s · t · sin 33.0° - 1/2 · 9.8 m/s² · t²

0 = -4.9 m/s² · t² + 9.00 m/s · sin 33.0° · t + 6.00 m

Solving the quadratic equation:

t = 1.71 s

Now that we have the time of flight, we can calculate the x-component of the vector r (the horizontal distance traveled by the ball):

x= x0 + v0 · t · cos α

x = 0m + 9.00 m/s · 1.71 s · cos 33°

x = 12.9 m

Then, your friend will have to run (12.9 m - 11.0 m) 1.9 m to catch the ball 1 m above the ground.

Let´s see, how much time it takes your friend to run that distance:

x = x0 + v0 · t + 1/2 · a · t²      (x0 = 0, v0 = 0)

x = 1/2 · a · t²

1.9 m = 1/2 · 1.80 m/s² · t²

Solving for t

t = 1.45 s

Then, since the time of flight of the ball is 1.71 s, your friend has to wait

1.71 s - 1.45 s = 0.26 s after you throw the ball to start running.

6 0
2 years ago
By means of a rope whose mass is negligible, two blocks are suspended over a pulley, as the drawing shows, with m1 = 12.1 kg and
puteri [66]

Answer:

14.8 kg

Explanation:

We are given that

m_1=43.7 kg

m_2=12.1 kg

g=9.8 m/s^2

a=\frac{1}{2}(9.8)=4.9 m/s^2

We have to find the mass of the pulley.

According to question

T_2-m_2 g=m_2 a

T_2=m_2a+m_2g=m_2(a+g)=12.1(9.8+4.9)=177.87 N

T_1=m_1(g-a)=43.7(9.8-4.9)=214.13 N

Moment of inertia of pulley=I=\frac{1}{2}Mr^2

(T_2-T_1)r=I(-\alpha)=\frac{1}{2}Mr^2(\frac{-a}{r})=\frac{1}{2}Mr(-4.9)

Where \alpha=\frac{a}{r}

(177.87-214.13)=-\frac{1}{2}(4.9)M

-36.26=-\frac{1}{2}(4.9)M

M=\frac{36.26\times 2}{4.9}=14.8 kg

Hence, the mass of the pulley=14.8 kg

6 0
3 years ago
Need some help with these two physics problems!
Juliette [100K]

The force that keeps the puck moving is 0.25 N while the velocity of the puck is  3.7 m/s.

<h3>What is the centripetal force?</h3>

We know that the centripetal force is the force that acts on a body that is moving along a circular path. In this case, we are told that the puck is moving along a circular path hence it is acted upon by the centripetal force that acts on it.

The centripetal force in this case would be supplied by the weight of the object that is moving in the circular path. Thus we can write in our equation that;

Centripetal force = Weight of object = mg

m = mass of the object

g = acceleration due to gravity

Then;

W = 0.026 Kg * 9.8 m/s^2

W = 0.25 N

To obtain the velocity of the object;

FT = mv^2/r

v = √ FT r/m

v =  √0.25 * 1.4/0.026

v = 3.7 m/s

Learn more about centripetal force:brainly.com/question/11324711

#SPJ1

5 0
1 year ago
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