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Anastaziya [24]
3 years ago
12

How fast is a 180 kg motorcycle traveling if it has 36000 J of kinetic energy??

Physics
1 answer:
leva [86]3 years ago
7 0

Answer:

20 ms⁻¹

360 J

Explanation:

Kinetic energy is the energy possessed by a moving object solely due to its motion.

You can get the K.E. of an object using the equation,

K.E. =  (1/2)mv² where all terms in usual meaning

So you get,

K.E. = 36000 =  (1/2)×180×v²

v = 20 ms⁻¹

Also,

K.E. =  (1/2)×80×3² = 360 J

   

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Suppose that you are headed toward a plateau 55 meters high. If the angle of elevation to the top of the plateau is 40degrees​,
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Answer:

x=65.55m

Explanation:

Let x be the distance to the shore

From trigonometry properties:

tan(40^{o} )=\frac{55m}{x} \\x=\frac{55m}{tan(40^{o} )} \\x=65.55m

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3 years ago
An object traveling at 1.5 rad
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The object's final velocity, given the data is 10.5 rad/s

<h3>What is acceleration? </h3>

This is defined as the rate of change of velocity which time. It is expressed as

a = (v – u) / t

Where

  • a is the acceleration
  • v is the final velocity
  • u is the initial velocity
  • t is the time

<h3>How to determine the final velocity</h3>

The following data were obtained from the question

  • Initial velocity (u) = 1.5 rad/s
  • Acceleration (a) = 0.75 rad/s²
  • Time (t) = 12 s
  • Final velocity (v) = ?

The final velocity can be obtained as follow:

a = (v – u) / t

0.75 = (v – 1.5) / 12

Cross multiply

v – 1.5 = 0.75 × 12

v – 1.5 = 9

Collect like terms

v = 9 + 1.5

v = 10.5 rad/s

Thus, the final velocity of the object is 10.5 rad/s

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If =12a andthe distance from each wire to point p is 0.12m, then what is the magnitude of the magnetic force per unit length on
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The magnitude of the magnetic force per unit length on the top wire is

2×10⁻⁵  N/m

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To calculate the magnitude of the magnetic force per unit length on the top wire, we are using the formula

F= \frac{\mu_0 I_f}{2\pi d}

Here we are given,

\mu_0= magnetic permeability

= 4\pi×10⁻⁷ H m⁻¹

If= 12 A

d= distance from each wire to point.

=0.12m

Now we put the known values in the above equation, we get

F= \frac{\mu_0 I_f}{2\pi d}

Or, F = \frac{4\pi \times 10^{-7}\times  12}{2\pi \times 0.12}

Or, F= 2×10⁻⁵ N/m.

From the above calculation, we can conclude that the magnitude of the magnetic force per unit length on the top wire is 2×10⁻⁵ N/m.

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