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soldi70 [24.7K]
2 years ago
11

If a box with a mass of 8.0 kg is sitting on a frictionless surface and experiences an acceleration of 4.0 m/s2 to the right, wh

at is the net force acting on the box? (1 point)
A) 2.0 N
B) 12 N
C) 32 N
D) 45 N
Physics
1 answer:
mr Goodwill [35]2 years ago
4 0

The net force acting on a box of mass 8.0kg that experiences an acceleration of 4.0m/s² is 32N. Details about net force can be found below.

<h3>How to calculate net force?</h3>

The net force of a body can be calculated by multiplying the mass of the body by its acceleration as follows:

Force = mass × acceleration

According to this question, a box with a mass of 8.0 kg is sitting on a frictionless surface and experiences an acceleration of 4.0 m/s2 to the right.

Net force = 8kg × 4m/s²

Net force = 32N

Therefore, the net force acting on a box of mass 8.0kg that experiences an acceleration of 4.0m/s² is 32N.

Learn more about net force at: brainly.com/question/18031889

#SPJ1

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Why is density and the melting point both substances of mass
maw [93]
Density is mass over volume. Density is also a charecteristic property wich defines a substance and two different substance cant have the same charasteristic property.
3 0
3 years ago
A diver springs upward from a board that is 4.40 m above the water. At the instant she contacts the water her speed is 13.5 m/s
Yakvenalex [24]

The diver has the initial velocity, both (a) magnitude is 9.8 m/s and (b) direction is  73.5°.

<h3>What is free falling?</h3>

When an object is released from rest in free air considering no friction, the motion is depend only on the acceleration due to gravity, g.

If we drop an object of mass m near the Earth surface from a height h, it has initial mechanical energy(P.E)

U =mgh

When the object strikes the ground, all the potential energy converted into kinetic energy.

K.E = 1/2mv²

where v is the speed just before hitting the ground.

A diver springs upward from a board that is 4.40 m above the water. At the instant, she contacts the water her speed is 13.5 m/s and her body makes an angle of 78.1 ° with respect to the horizontal surface of the water.

(a)

From energy conservation principle, initial and final mechanical energy are equal.

1/2mu² + mgh = 1/2mv²

where, u is the initial velocity of the diver.

u = sq rt  (v² - 2gh)

u = sq rt (13.5² - 2x9.81x4.4)

u = 9.798 m/s or 9.8 m/s

Thus, the velocity of the diver is 9.8 m/s.

(b)

The horizontal component of velocity will remain constant.

The horizontal component of acceleration is zero.

Then,

ucosθ = vcosΦ

θ = cos⁻¹ [ (13.5 x cos 78.1)/9.8 ]

θ = 73.5°

Thus, the direction of velocity is  73.5°.

Learn more about free falling.

brainly.com/question/13299152

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5 0
2 years ago
Use Hooke's Law to determine the work done by the variable force in the spring problem. Nine joules of work is required to stret
natima [27]

Answer:

29.16 J

Explanation:

From Hook's law,

W = 1/2(ke²)..................... Equation 1

Where W = work done, k = Spring constant, e = extension.

Given: W = 9 J, e = 0.5 m.

Substitute into equation 1

9 = 1/2(k×0.5²)

Solve for k

k = 18/0.5²

k = 72 N/m.

The work done required to stretch the spring by additional 0.4 m is

W = 1/2(72)(0.4+0.5)²

W = 36(0.9²)

W = 29.16 J.

6 0
3 years ago
If the battery of your phone can provide 2 mA of current to your phone and holds a charge of 130 C, how long will it take a full
Naya [18.7K]

Answer: 65000 seconds

Explanation:

Given that,

Current (I) = 2 mA

(Since 1 mA = 1 x 10^-3A

2 mA = 2 x 10^-3A)

Charge (Q) = 130 C

Time taken for a fully charged phone to die (T) = ?

Recall that the charge is the product of current and time taken.

i.e Q = I x T

130C = 2 x 10^-3A x T

T = 130C / (2 x 10^-3A)

T = 65000 seconds (time will be in seconds because seconds is the unit of time)

Thus, it will take a fully charged phone 65000 seconds to die

5 0
3 years ago
There are two parallel conductive plates separated by a distance d and zero potential. Calculate the potential and electric fiel
taurus [48]

Answer:

The total electric potential at mid way due to 'q' is \frac{q}{4\pi\epsilon_{o}d}

The net Electric field at midway due to 'q' is 0.

Solution:

According to the question, the separation between two parallel plates, plate A and plate B (say)  = d

The electric potential at a distance d due to 'Q' is:

V = \frac{1}{4\pi\epsilon_{o}}.\frac{Q}{d}

Now, for the Electric potential for the two plates A and B at midway between the plates due to 'q':

For plate A,

V_{A} = \frac{1}{4\pi\epsilon_{o}}.\frac{q}{\frac{d}{2}}

Similar is the case with plate B:

V_{B} = \frac{1}{4\pi\epsilon_{o}}.\frac{q}{\frac{d}{2}}

Since the electric potential is a scalar quantity, the net or total potential is given as the sum of the potential for the two plates:

V_{total} = V_{A} + V_{B} = \frac{1}{4\pi\epsilon_{o}}.q(\frac{1}{\frac{d}{2}} + \frac{1}{\frac{d}{2}}

V_{total} = \frac{q}{4\pi\epsilon_{o}d}

Now,

The Electric field due to charge Q at a distance is given by:

\vec{E} = \frac{1}{4\pi\epsilon_{o}}.\frac{Q}{d^{2}}

Now, if the charge q is mid way between the field, then distance is \frac{d}{2}.

Electric Field at plate A, \vec{E_{A}} at midway due to charge q:

\vec{E_{A}} = \frac{1}{4\pi\epsilon_{o}}.\frac{q}{(\frac{d}{2})^{2}}

Similarly, for plate B:

\vec{E_{B}} = \frac{1}{4\pi\epsilon_{o}}.\frac{q}{(\frac{d}{2})^{2}}

Both the fields for plate A and B are due to charge 'q' and as such will be equal in magnitude with direction of fields opposite to each other and hence cancels out making net Electric field zero.

3 0
3 years ago
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