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3 years ago

5

A rightward force of 460 N is applied to a 286-kg crate to accelerate it across the floor

1 answer:

Bess [88]3 years ago

4 0

Answer:

Acceleration = 1.61 m/s²

Explanation:

Given the following data:

Force = 460N

Mass = 286kg

To find the acceleration;

Force = mass * acceleration

Substituting into the equation, we have;

460 = 286 * acceleration

Acceleration = 460/286

Acceleration = 1.61 m/s²

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Based on what you have learned from this unit, construct a tri-fold brochure instructing a new freshman as to the best way to le

Rzqust [24]

<span>A tri-fold brochure has two parallel folds, splitting the brochure into three sections. Even when printed on low-weight paper, tri-folds can stand up easily, which makes them a great choice for displaying at conventions. You can fold both folds inwards so that the left and right sections of the brochure sit on top of one another, or you can have one fold inwards and the other outwards, to create an accordion effect, which looks very attractive.</span>

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3 years ago

Can an argon atom undergo vibrational motion?

azamat

No. the answer to the question if can an argon atom undergo vibrational motion is no. it can not even spin either. the argon atom, or the argon is a chemical element that is the third most abundant gas in the earth's atmosphere. it is ore than twice as abundance as water vapor. Thank you for this question.

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3 years ago

If an electron moves in a circle of radius 21 cm perpendicular to a B field of 0.4 T, what are the speed of the electron and the

kodGreya [7K]

Answer:

a)

$v = 4.048 *10^6$ m/s

b)

Angular frequency = $1.92 * 10^7$

Explanation:

As we know

$v = \frac{qBr}{m}$

q is the charge on the electron = $3.2 * 10^{-19}$ C

B is the magnetic field in Tesla = $0.4$ T

r is the radius of the circle = $0.21$ m

mass of the electrons = $6.64 * 10^{-27}$ Kg

a)

Substituting the given values in above equation, we get -

$v = \frac{3.2 * 10^{-19}*0.4*0.21}{6.64 * 10^{-27}} \\v = 4.048 *10^6$ m/s

b)

Angular frequency =

$\frac{4.048 * 10^6 }{0.21} \\1.92 * 10^7$

8 0

2 years ago

Two long, straight wires are parallel and 20 cm apart. One carries a current of 2.2 A, the other a current of 5.9 A. (a) If the

Leya [2.2K]

Answer: $1.298\times 10^{-5}\ N/m$

Explanation:

Given

Current in the first wire $I_1=2.2\ A$

Current in the second wire $I_2=5.9\ A$

wires are $20\ cm$ apart

Force per unit length between the current-carrying wires is

$\Rightarrow \dfrac{F}{l}=\dfrac{\mu_oI_1I_2}{2\pi r}$

Force exerted by the wires is the same

Put the values

$\Rightarrow \frac{F}{l}=f=\dfrac{4\pi \times 10^{-7}\times 2.2\times 5.9}{2\pi \times 0.2}=1.298\times 10^{-5}\ N/m$

This force will be repulsive in nature as the current is flowing opposite

4 0

3 years ago

A top is a toy that is made to spin on its pointed end by pulling on a string wrapped around the body of the top. The string has

AlladinOne [14]

Given Information:

Angular displacement = θ = 51 cm = 0.51 m

Radius = 1.8 cm = 0.018 m

Initial angular velocity = ω₁ = 0 m/s

Angular acceleration = α = 10 rad/s ²

Required Information:

Final angular velocity = ω₂ = ?

Answer:

Final angular velocity = ω₂ = 21.6 rad/s

Explanation:

We know from the equations of kinematics,

ω₂² = ω₁² + 2αθ

Where ω₁ is the initial angular velocity that is zero since the toy was initially at rest, α is angular acceleration and θ is angular displacement.

ω₂² = (0)² + 2αθ

ω₂² = 2αθ

ω₂ = √(2αθ)

We know that the relation between angular displacement and arc length is given by

s = rθ

θ = s/r

θ = 0.51/0.018

θ = 23.33 radians

finally, final angular velocity is

ω₂ = √(2αθ)

ω₂ = √(2*10*23.33)

ω₂ = 21.6 rad/s

Therefore, the top will be rotating at 21.6 rad/s when the string is completely unwound.

3 0

3 years ago