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Roman55 [17]
3 years ago
15

How long does light take to travel from the sun to earth? heres the exact question: light travels at 300,000 km.s. The sun is a

total of 148 million kilometres from the earth. How long does it take light to travel from the sun to the earth
Physics
1 answer:
Artist 52 [7]3 years ago
4 0
I've heard it takes around 8 min
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How does nuclear energy work? Explain
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There are many processes to get nuclear energy. Nuclear energy is basically energy from an atom. For example fission is where the nucleus of an atom ( typically radioactive atoms ) gets split then energy is released ( typically heat). And in radioactive decay radiation is released from an radioactive atom. Hope this helps
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3 years ago
How will a current change if the resistance of a circuit remains constant while the voltage across the circuit decreases to half
beks73 [17]

Answer:

1. The current will drop to half of its original value.

Explanation:

The problem can be solved by using Ohm's law:

V=RI

where

V is the voltage across the circuit

R is the resistance of the circuit

I is the current

We can rewrite it as

I=\frac{V}{R}

In this problem, we have:

- the resistance of the circuit remains the same: R' = R

- the voltage is decreased to half of its original value: V'=\frac{V}{2}

So, the new current will be

I'=\frac{V'}{R'}=\frac{V/2}{R}=\frac{1}{2}\frac{V}{R}=\frac{I}{2}

so, the current will drop to half of its original value.

4 0
3 years ago
Is Algae Biotic or Abiotic?
muminat
It’s biotic because it’s a living thing
Abiotic would be like a rock non living things
Answer:Biotic
4 0
3 years ago
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Calculate the linear acceleration (in m/s2) of a car, the 0.310 m radius tires of which have an angular acceleration of 15.0 rad
love history [14]

Answer:

a) The linear acceleration of the car is 4.65\,\frac{m}{s^{2}}, b) The tires did 7.46 revolutions in 2.50 seconds from rest.

Explanation:

a) A tire experiments a general plane motion, which is the sum of rotation and translation. The linear acceleration experimented by the car corresponds to the linear acceleration at the center of the tire with respect to the point of contact between tire and ground, whose magnitude is described by the following formula measured in meters per square second:

\| \vec a \| = \sqrt{a_{r}^{2} + a_{t}^{2}}

Where:

a_{r} - Magnitude of the radial acceleration, measured in meters per square second.

a_{t} - Magnitude of the tangent acceleration, measured in meters per square second.

Let suppose that tire is moving on a horizontal ground, since radius of curvature is too big, then radial acceleration tends to be zero. So that:

\| \vec a \| = a_{t}

\| \vec a \| = r \cdot \alpha

Where:

\alpha - Angular acceleration, measured in radians per square second.

r - Radius of rotation (Radius of a tire), measured in meters.

Given that \alpha = 15\,\frac{rad}{s^{2}} and r = 0.31\,m. The linear acceleration experimented by the car is:

\| \vec a \| = (0.31\,m)\cdot \left(15\,\frac{rad}{s^{2}} \right)

\| \vec a \| = 4.65\,\frac{m}{s^{2}}

The linear acceleration of the car is 4.65\,\frac{m}{s^{2}}.

b) Assuming that angular acceleration is constant, the following kinematic equation is used:

\theta = \theta_{o} + \omega_{o}\cdot t + \frac{1}{2}\cdot \alpha \cdot t^{2}

Where:

\theta - Final angular position, measured in radians.

\theta_{o} - Initial angular position, measured in radians.

\omega_{o} - Initial angular speed, measured in radians per second.

\alpha - Angular acceleration, measured in radians per square second.

t - Time, measured in seconds.

If \theta_{o} = 0\,rad, \omega_{o} = 0\,\frac{rad}{s}, \alpha = 15\,\frac{rad}{s^{2}}, the final angular position is:

\theta = 0\,rad + \left(0\,\frac{rad}{s}\right)\cdot (2.50\,s) + \frac{1}{2}\cdot \left(15\,\frac{rad}{s^{2}}\right)\cdot (2.50\,s)^{2}

\theta = 46.875\,rad

Let convert this outcome into revolutions: (1 revolution is equal to 2π radians)

\theta = 7.46\,rev

The tires did 7.46 revolutions in 2.50 seconds from rest.

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3 years ago
Plzzzzz help me plzzz
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8 miles per hour

(extra space)
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