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2 years ago

What is the kinetic energy of a 150kg object that is moving with a speed of 15 m/s?

Physics

1 answer:

otez555 [7]2 years ago

8 0

$ke = \frac{1}{2} \times m {v}^{2}$

so you :

$\frac{1}{2} \times 150 \times {15}^{2}$

= 16875 ( J )

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The type of graph used to show how a part of something relates to the whole is which of the following?

Slav-nsk [51]

Answer:

Circle or pie graph is used to show how a part of something relates to the whole.

Explanation:

Pie graphs are easy to read and can present a very clear picture of the relationships.

4 0

2 years ago

Read 2 more answers

A particle with mass 1.09 kg oscillates horizontally at the end of a horizontal spring. A student measures an amplitude of 0.985

e-lub [12.9K]

Answer:

a) $f = 0.598\,hz$ , b) $v_{max} = 3.701\,\frac{m}{s}$ , c) $k = 15.385\,\frac{N}{m}$ , d) $U = 1.081\,J$ , e) $K = 6.382\,J$ , f) $v\approx 3.422\,\frac{m}{s}$

Explanation:

a) The frequency of oscillation is:

$f = \frac{76}{127\,hz}$

$f = 0.598\,hz$

b) The angular frequency is:

$\omega = 2\pi \cdot f$

$\omega = 2\pi \cdot (0.598\,hz)$

$\omega = 3.757\,\frac{rad}{s}$

Lastly, the speed at the equilibrium position is:

$v_{max} = \omega \cdot A$

$v_{max} = (3.757\,\frac{rad}{s} )\cdot (0.985\,m)$

$v_{max} = 3.701\,\frac{m}{s}$

c) The spring constant is:

$\omega = \sqrt{\frac{k}{m}}$

$k = \omega^{2}\cdot m$

$k = (3.757\,\frac{rad}{s} )^{2}\cdot (1.09\,kg)$

$k = 15.385\,\frac{N}{m}$

d) The potential energy when the particle is located 38.1 % of the amplitude away from the equilibrium position is:

$U = \frac{1}{2}\cdot (15.385\,\frac{N}{m} )\cdot (0.375\,m)^{2}$

$U = 1.081\,J$

e) The maximum potential energy is:

$U_{max} = \frac{1}{2}\cdot (15.385\,\frac{N}{m} )\cdot (0.985\,m)^{2}$

$U_{max} = 7.463\,J$

The kinetic energy when the particle is located 38.1 % of the amplitude away from the equilibrium position is:

$K = U_{max} - U$

$K = 7.463\,J - 1.081\,J$

$K = 6.382\,J$

f) The speed when the particle is located 38.1 % of the amplitude away from the equilibrium position is:

$K = \frac{1}{2}\cdot m \cdot v^{2}$

$v = \sqrt{\frac{2\cdot K}{m} }$

$v = \sqrt{\frac{2\cdot (6.382\,J)}{1.09\,kg} }$

$v\approx 3.422\,\frac{m}{s}$

4 0

3 years ago

Calculate the angular velocity of the neptune (orbital period of 165 earth period) in its orbit around the sun.

ludmilkaskok [199]

The correct answer is: Angular velocity = $1.208 * 10^{-9}$ rad/s

Explanation:
The angular velocity is given as:
ω = $\frac{2\pi}{T}$ --- (1)

Where T = 165 * (365 days) * (24 hours/day) * (60 minutes/hour) * (60 seconds/minute) = 5203440000 s

Plug in the value in (1):
ω = $\frac{2\pi}{5203440000} = 1.208 * 10^{-9}$ rad/s

7 0

2 years ago

Which of the following is NOT an advantage of using a wheel and axle?

kakasveta [241]

Answer:

the first one

Explanation:

4 0

3 years ago

What is the least amount of time required for a given point on this wave to move from y = 0 to y = 12cm?

ale4655 [162]

Time t = ?

When wave is moving from y = 0 to y =12 cm

By using the formula,

y = 15cos [(π/12) t)] = 0,

cos [(π/12) t)] = 0 = cos (π/2), so,

(π/12)t = π/2,

t = (π/2) (12/π)

t = 12/2

t = 6 sec

so 6 sec is the least amount of time required

4 0

3 years ago