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elena-s [515]
4 years ago
15

1) How much work does it take to accelerate a 8.2 kg object from rest to 19 m/s?

Physics
1 answer:
krek1111 [17]4 years ago
3 0

Explanation:

Work = change in energy

W = ½ mv²

W = ½ (8.2 kg) (19 m/s)²

W ≈ 1500 J

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When two or more forces act on an object at the same time what are they called
Vilka [71]

Unbalanced forces is what they are called

7 0
3 years ago
What is the reaction force of the table with a weight of 558N
Delicious77 [7]

Answer: reaction force = -558N

Explanation:

w = f = 558N

since action force and reaction force are equal in magnitude and opposite in direction,

reaction force = -(f)

reaction force = -558N

if that helps.

8 0
3 years ago
If we represent the solar system on a scale that allows us to walk from the Sun to Pluto in a few minutes, then
aksik [14]

Answer:

b. the planets are marble-size or smaller and the nearest stars are thousands of miles away

Explanation:

The correct answer for the question is option b because if the distance between sun and Pluto is any scale is made equivalent to a walking distance of some minutes then the size of planets will be equivalent to the size of marbles and the nearest stars that is present in Alpha centauri triple star system(4.5 light years away) will be approximately thousands of miles away from us.

4 0
3 years ago
a person can throw a 300 gram ball at 25 m/s the maximum height the person can throw the ball on Earth?​
jolli1 [7]

Answer:

Elevation =31.85[m]

Explanation:

We can solve this problem by using the principle of energy conservation. This consists of transforming kinetic energy into potential energy or vice versa. For this specific case is the transformation of kinetic energy to potential energy.

We need to first identify all the input data, and establish a condition or a point where the potential energy is zero.

The point where the ball is thrown shall be taken as a reference point of potential energy.

E_{p} = E_{k} \\where:\\E_{p}= potential energy [J]\\ E_{k}= kinetic energy [J]

m = mass of the ball = 300 [gr] = 0.3 [kg]

v = initial velocity = 25 [m/s]

E_{k}=\frac{1}{2}  * m* v^{2} \\E_{k}= \frac{1}{2} * 0.3* (25)^{2} \\E_{k}= 93.75 [J]

93.75=m*g*h\\where:\\g = gravity = 9.81 [m/s^2]\\h = elevation [m]\\replacing\\h=\frac{E_{k}}{m*g} \\h=\frac{93.75}{.3*9.81} \\h=31.85[m]

5 0
4 years ago
Review. (c) Assume the dipole is a compass needle-a light bar magnet-with a magnetic moment of magnitude μ . It has moment of in
hammer [34]

The frequency of oscillation is \frac{1}{2\pi }  \sqrt{\frac{\mu B}{I} }.

<h3>What is a magnetic moment?</h3>

The magnetic moment is the magnetism of a magnet or other item that creates a magnetic field, as well as its orientation and strength. Electromagnets, permanent magnets, elementary particles like electrons, different compounds, and a variety of celestial objects are examples of things that have magnetic moments (such as many planets, some moons, stars, etc). The phrase "magnetic moment" typically refers to a system's magnetic dipole moment, which can be represented by an analogous magnetic dipole, which has a magnetic north and south pole that are barely separated from one another. For sufficiently small magnets or at sufficiently wide distances, the magnetic dipole component is adequate. For extended objects, additional terms may be required in addition to the dipole moment, such as the magnetic quadrupole moment.

T = \frac{ Id^{2}\theta }{dt^{2} }

-\mu B\theta=\frac{ Id^{2}\theta }{dt^{2} }

\frac{d^{2}\theta }{dt^{2} } = -(\frac{\mu BI}{\theta} )

By Comparing the above equation with the SHM equation

\frac{d^2 \theta} {dt^{2} } = -\omega^{2} \theta

\omega^{2} =\frac{ \mu B}{I}

Frequency = \frac{\mu}{2\pi }

=\frac{\sqrt{\frac{\mu B}{I} } }{2\pi}

=\frac{1}{2\pi }  \sqrt{\frac{\mu B}{I} }

To learn more about a magnetic moment, visit:

brainly.com/question/17000031

#SPJ4

8 0
1 year ago
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