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Temka [501]
3 years ago
6

How can energy conservation be considered an energy source?

Physics
1 answer:
Brut [27]3 years ago
7 0

Answer:This could be in the form of using fewer energy services or using devices that require less energy.

Explanation:

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A ball is hit so that it accelerates at 10 m/s2. The ball hits a glove with a force of 5 N. What is the mass of the ball?
jenyasd209 [6]

Answer:

0.5 kg

Explanation:

» <u>Concepts</u>

Newton's second law, the Law of Acceleration, states that F = ma, where F = Force in Newtons, m = mass in kg, and a = acceleration in m/s^2.

» <u>Application</u>

We are asked to find the mass of the ball using the equation F = ma. We're also given the force and acceleration, so the equation looks like 5 = 10(m).

» <u>Solution</u>

Step 1: Divide both sides by 10.

  • 5/10=10m/10
  • m=0.5

Thus, the mass of the ball is 0.5 kg.

8 0
2 years ago
What are 3 types of frictions and when does each apply/9029244/29edc63f?utm_source=registration
Anon25 [30]
- Static friction: when an object is not moving
- Kinetic friction: if an object is moving
- Rolling friction: when there is rolling (wheel,..)
5 0
3 years ago
The two ends of a coiled wire are connected to the electrodes of a lightbulb.
barxatty [35]

Answer:

a

Explanation:

The bar magnet moves downward with respect to the wire loop, so that the number of magnetic field lines going through the loop decreases with time. This causes an emf to be induced in the loop, creating an electric current.

in other words, the magnets motion creates a current in the loop

5 0
3 years ago
A)If your torsion balance deflects to an angle of 10° when your two spheres are 40cm apart, what angle will it deflect to when t
svp [43]

Answer:

Explanation:

Given

for \theta=10^{\circ}

Sphere are d=40\ cm

when sphere d_2=10\ cm apart suppose deflection is \theta _2

We know

F=k_t\cdot \theta

Where F=force between charged particle

\theta =Deflection

F=\frac{kQ_1Q_2}{r^2}=k_t\cdot \theta

\theta =\frac{k}{k_t}\times \frac{Q_1Q_2}{r^2}----1

thus \theta \propto \frac{1}{r^2}

for \theta _2

\frac{\theta _1}{\theta _2}=(\frac{r_2}{r_1})^2

\theta _2=16\times \theta _1

\theta _2=160^{\circ}

(b)for 10^{\circ} deflection Potential v_1=8\ kV

Electric Potential is V=\frac{kQ}{r}

Q=\frac{V\cdot r}{k}

where V=voltage

k=constant

r=distance between charges

Put value of Q in equation 1

\theta =\frac{k}{k_t}\times \frac{V^2r^2}{k^2}

\theta =\frac{V^2r^2}{k\cdot k_t}

thus \theta \propto V^2

therefore

\frac{\theta _1}{\theta _2}=(\frac{V_1}{V_2})^2

\frac{10}{\theta _2}=(\frac{8}{4})^2

\theta _2=\frac{10}{4}=2.5^{\circ}

5 0
3 years ago
A fire hose ejects a stream of water at an angle of 35.0° above the horizontal. The water leaves the nozzle with a speed of 25.
erma4kov [3.2K]

Answer:

The fire hose be located 59.80 m away to hit the highest possible fire.

Explanation:

Vertical velocity = 25 sin35 = 14.34 m/s

Acceleration = -9.81m/s²

At maximum height , final vertical velocity = 0 m/s

We have v = u + at

Substituting

           0 = 14.34 - 9.81 x t

           t = 1.46 s

Time of flight of water = 2 x 1.46 = 2.92 s

Horizontal velocity = 25 cos35 = 20.48 m/s

Horizontal displacement = 20.48 x 2.92 = 59.80 m

So, the fire hose be located 59.80 m away to hit the highest possible fire.

5 0
3 years ago
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