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Tanya [424]
2 years ago
7

A battery causes a current of 2.0 A to flow through a lamp. The power used by the lamp is 12 watts. What is the voltage?

Physics
1 answer:
Marina CMI [18]2 years ago
7 0

Answer:

6 v

Explanation:

You might be interested in
A wave passes along the surface of the water in a ripple tank. Describe the motion of a molecule on the surface as the water pas
laila [671]

Answer:

Explanation:

Water waves are generally a transverse wave which do not cause permanent displacement of molecules of the medium. Transverse waves are waves in which the direction of propagation of the wave is perpendicular to the direction of vibration of the particles of the medium.

As the wave propagates from one point to another on the surface of water transferring energy, a molecule of water on its surface vibrates upwards and downwards. Its motion is perpendicular to the direction of propagation of the wave. After the vibration, it comes back to its initial position.

8 0
2 years ago
Rock X is released from rest at the top of a cliff that is on Earth. A short time later, Rock Y is released from rest from the s
frosja888 [35]

Answer:

C) True. S increases with time, v₁ = gt and v₂ = g (t-t₀)  we see that for the same t v₁> v₂

Explanation:

You have several statements and we must select which ones are correct. The best way to do this is to raise the problem.

Let's use the vertical launch equation. The positive sign because they indicate that the felt downward is taken as an opponent.

Stone 1

    y₁ = v₀₁ t + ½ g t²

    y₁ = 0 + ½ g t²

Rock2

It comes out a little later, let's say a second later, we can use the same stopwatch

     t ’= (t-t₀)

    y₂ = v₀₂ t ’+ ½ g t’²

    y₂ = 0 + ½ g (t-t₀)²

    y₂ = + ½ g (t-t₀)²

Let's calculate the distance between the two rocks, it should be clear that this equation is valid only for t> = to

    S = y₁ -y₂

    S = ½ g t²– ½ g (t-t₀)²

    S = ½ g [t² - (t²- 2 t to + to²)]  

    S = ½ g (2 t t₀ - t₀²)

    S = ½ g t₀ (2 t -t₀)

This is the separation of the two bodies as time passes, the amount outside the Parentheses is constant.

For t <to.  The rock y has not left and the distance increases

For t> = to.  the ratio (2t/to-1)> 1 therefore the distance increases as time

passes

Now we can analyze the different statements

A) false. The difference in height increases over time

B) False S increases

C) Certain s increases with time, v₁ = gt and V₂ = g (t-t₀) we see that for the same t   v₁> v₂

3 0
3 years ago
0/2 File Limit
slamgirl [31]

Answer:

Speed at which it will reach the ground is given as

v_f = 46.8 m/s

Total time for which it will remain in air is given as

t = 6.3 s

Explanation:

As we know that the object is projected upwards with speed

v_i = 15 m/s

g = - 9.81 m/s^2

now when it will reach the ground then we have

y = v_y t + \frac{1}{2} at^2

so we have

-100 = 15 t - \frac{1}{2}(-9.81) t^2

4.905 t^2 - 15 t - 100 = 0

so we have

t = 6.3 s

Now speed of the object when it reaches the ground is given as

v_f = v_i + at

v_f = -15 + (9.81)(6.3)

v_f = 46.8 m/s

8 0
3 years ago
Write the SI unit of time and temperature
dsp73

Answer:

The SI unit of time is second (s) and temperature is Kelvin (K)

Explanation:

hope it is helpful to you

7 0
3 years ago
A runner moves 2.88 m/s north. She accelerates at 0.350 m/s^2 at a -52.0 angle. At the point in the motion where she is running
MrRissso [65]

The runner has initial velocity vector

\vec v_0=\left(2.88\dfrac{\rm m}{\rm s}\right)\,\vec\jmath

and acceleration vector

\vec a=\left(0.350\dfrac{\rm m}{\mathrm s^2}\right)(\cos(-52.0^\circ)\,\vec\imath+\sin(-52.0^\circ)\,\vec\jmath)

so that her velocity at time t is

\vec v=\vec v_0+\vec at

She runs directly east when the vertical component of \vec v is 0:

2.88\dfrac{\rm m}{\rm s}+\left(0.350\,\dfrac{\rm m}{\mathrm s^2}\right)\sin(-52.0^\circ)\,t=0\implies t=10.4\,\rm s

It's not clear what you're supposed to find at this particular time... possibly her position vector? In that case, assuming she starts at the origin, her position at time t would be

\vec x=\vec v_0t+\dfrac12\vec at^2

so that after 10.4 s, her position would be

\vec x=(10.1\,\mathrm m)\,\vec\imath+(17.2\,\mathrm m)\,\vec\jmath

which is 19.9 m away from her starting position.

8 0
3 years ago
Read 2 more answers
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