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3 years ago

A motorbike accelerates from 15m/s to 25m/s in 15 seconds.

Physics

1 answer:

RSB [31]3 years ago

5 0

distance traveled by a uniformly accelerated bike is given as

$d = \frac{v_f + v_i}{2} (t)$

here we know that

$v_f = 25 m/s$

$v_i = 15 m/s$

$t = 15 s$

now we will have from above equation

$d = \frac{15 + 25}{2} (15)$

$d = 20 (15) = 300 m$

so it will cover the total distance of 300 m

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The world energy consumption was about 6*10^22 J. How much area must a parallel plate capacitor need to store this energy Assume

Nastasia [14]

Answer:

$A = 5 \times 10^{32} m^2$

Explanation:

As we know that the energy stored in the capacitor is given as

$Q = \frac{1}{2}CV^2$

here we know that

$Q = 6 \times 10^{22} J$

also we know that

$V = 5 Volts$

now we have

$6 \times 10^{22} = \frac{1}{2}C(5^2)$

$C = 4.8 \times 10^{21} F$

now we know the formula of capacitance

$C = \frac{\epsilon_0 A}{d}$

$4.8 \times 10^{21} = \frac{(8.85 \times 10^{-12})(A)}{1}$

$A = 5 \times 10^{32} m^2$

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3 years ago

Goals can be big or small? Question 1 options: True False

pav-90 [236]

Answer:

Probably big

Explanation:

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2 years ago

As viewed from above in this picture, what direction will the current be in the coil of wire that will cause the loop to rotate

Gala2k [10]

Answer:

When viewed from above, the current in the coil should point towards the top-right corner of the picture.

Explanation:

The current in this coil have only two possible directions: clockwise or counter-clockwise. However, since the diagram shows the coil from above, not from a cross-section, just saying clockwise or counter-clockwise might be ambiguous. The statement that the current is directed towards the top-right corner of the picture is equivalent to saying that when viewed from the lower-right corner of this diagram, the current in the coil is moving clockwise.

Note that at the center of this picture, the current is parallel to the magnetic field- there will be no force on the coil at that position. On the other hand, (also when viewed from above,) at the top-right corner and the lower-left corner of the coil, the current in the coil will be perpendicular to the magnetic field. That's where the force on the coil will be the strongest.

With that in mind, apply the right-hand rule to find the direction of the force on the coil in each of the two possibilities.

Assume that when viewed from above, the current is flowing towards the top-right corner of the picture. Consider the wire near the top-right corner of this coil (as viewed above on this picture.) The current will be going into the picture into the magnetic field. By the right-hand rule, the current on the wire near that point should be pointing towards the bottom of this picture. (Point fingers on the right hand in the direction of the current $I$ . Rotate the right hand such that when curling the fingers, they point in the direction of the magnetic field $B$ . The direction of the right thumb should now point in the direction of the force on the wire $F$ .)

Based on the same assumption, the current in the wires near the bottom left corner of this coil will be pointing out of the picture. By the right hand rule, the magnetic force on the coil in that region should be pointing towards the top of this picture. Combing these two forces, the coil would indeed be rotating around the center of this picture in the direction shown in the diagram.

It can also be shown that if the current points towards the bottom left corner of the picture when viewed from above, the coil will be rotating about the center of this picture in the opposite direction.

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2 years ago

For a series circuit what is the terminal voltage of a battery or power supply equal to in terms of the potential difference or

trapecia [35]

Answer: V=IR

Explanation: for a series circuit connected to a battery supply, the total emf across the circuit is given as

E = I(R + r) and by expanding, we have that E =IR + It

Where r is the internal resistance of the battery

I is the total current flowing in the circuit

R total load resistance in the circuit.

E is the total emf of the circuit.

The total emf is the sum of 2 separate voltages.

"IR" which is the terminal voltage and "Ir" which is the loss voltage.

The teenila voltage is the voltage flowing in the circuit based on the equivalent resistance of the circuit while the loss voltage is the wasted voltage based on the internal resistance of the battery source.

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3 years ago

a 0.0780 kg lemming runs off a 5.36 m high cliff at 4.84 m/s. what is its potential energy (PE) when it is 2.00 m above the grou

zmey [24]

Answer:

1.56 J

Explanation:

The potential energy only depends on the vertical height from the ground level.

We consider the ground level to have zero P.E.

So when it is 2 m above the ground level,

P.E. = mgh

= 0.078×10×2

= 1.56 J

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3 years ago