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uranmaximum [27]
1 year ago
6

what is the magnitude of a force?(1 point) responses the motion of the force the motion of the force the direction of the force

the direction of the force the type of force the type of force the amount of force the amount of force
Physics
1 answer:
mel-nik [20]1 year ago
6 0

The term "unbalanced force" in Newton's first law refers to a force that is not entirely counterbalanced (or cancelled) by the other independent forces. Basically, magnitude can be thought of as simply the "value" or "amount" of any physical quantity. It is a scalar quantity at all times.

Force has both a magnitude and a direction because it is a vector quantity. The amount that encapsulates the force's strength is known as its magnitude.

Consider the following scenario: the force is 10 N in the east. The direction is indicated by "towards east," while the force is indicated by "10."

Unless acted upon by an unbalanced force, an object at rest tends to stay at rest, and an object in motion tends to stay in motion with the same speed and direction.

The term "unbalanced force" in Newton's first law refers to a force that is not entirely counterbalanced (or cancelled) by the other independent forces. An uneven force exists if either all the vertical forces (up and down) or all the horizontal forces do not cancel each other. Looking at the free-body diagram for a particular situation makes it easy to recognize whether an unbalanced force is present.

To know more about unbalanced force click on the link:

brainly.com/question/19054208

#SPJ4

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Does uniform circular motion have constant acceleration?
AysviL [449]
Yes. It's (speed squared)/(radius of the circle).
7 0
3 years ago
An archer shoots an arrow with a mass of 45.0 grams from bow pulled
Sladkaya [172]

Answer:

The force the archer need to pull in order to achieve the height is approximately 101.8 N

Explanation:

By energy conservation principle, puling an elastic bow with a force, for a given distance, performs work which is converted to the potential energy of the arrow at height

The given parameters are;

The mass of the arrow, m = 45.0 grams = 0.045 kg

The distance the elastic bow is pulled, d = 65.0 cm = 0.65 m

The height at which the arrow is reaches, h = 150.0 meters

Let 'F', represent the force the archer need to pull in order to achieve the height

Work done, W = Force × Distance moved in the direction of the force

Therefore;

The work done in pulling the arrow, W = F × d

By energy conservation, we have;

The work done in pulling the arrow, W = The potential energy gained by the arrow, P.E.

W = P.E.

The potential energy gained by the arrow, P.E. = m·g·h

Where;

m = The mass of the arrow

g = The acceleration due to gravity = 9.8 m/s²

h = The height the arrow reaches

∴ by plugging in the values, P.E. = 0.045 kg ×9.8 m/s² × 150 m = 66.15 J

W = F × d = F × 0.065 m

Also, W = P.E. = 66.15 J

∴ W = F × 0.065 m = 66.15 J

F × 0.065 m = 66.15 J

F = 66.15 J/(0.65 m) = 1323/13 N ≈ 101.8 N

The force the archer need to pull in order to achieve the height, F ≈ 101.8 N.

3 0
3 years ago
. Calculate the efficiency of a bicycle if the input work to turn the pedal is 45J and the output work is 20J. * 1 point 2.25 2.
cestrela7 [59]

20/45=0.4*100= 44.4 so the answer is..................................................

Answer: 44.4%

8 0
3 years ago
(11%) Problem 5: A submarine is stranded on the bottom of the ocean with its hatch 25 m below the surface. In this problem, assu
V125BC [204]

Answer:

F = 1.24*10^4 N

Explanation:

Given

Depth of the ship, h = 25 m

Density of water, ρ = 1.03*10^3 kg/m³

Diameter of the hatch, d = 0.25 m

Pressure of air, P(air) = 1 atm

Pressure of water =

P(w) = ρgh

P(w) = 1.03*10^3 * 9.8 * 25

P(w) = 2.52*10^5 N/m²

P(net) = P(w) + P(air) - P(air)

P(net) = P(w)

P(net) = 2.52*10^5 N/m²

Remember,

Pressure = Force / Area, so

Force = Area * Pressure

Area = πr² = πd²/4

Area = 3.142 * 0.25²/4

Area = 3.142 * 0.015625

Area = 0.0491 m²

Force = 0.0491 * 2.52*10^5

F = 12373 N

F = 1.24*10^4 N

5 0
3 years ago
Read 2 more answers
If you could shine a very powerful flashlight beam toward the Moon, estimate the diameter of the beam when it reaches the Moon.
grin007 [14]

To develop this problem it is necessary to apply the Rayleigh Criterion (Angular resolution)criterion. This conceptos describes the ability of any image-forming device such as an optical or radio telescope, a microscope, a camera, or an eye, to distinguish small details of an object, thereby making it a major determinant of image resolution.  By definition is defined as:

\theta = 1.22\frac{\lambda}{d}

Where,

\lambda= Wavelength

d = Width of the slit

\theta= Angular resolution

Through the arc length we can find the radius, which would be given according to the length and angle previously described.

The radius of the beam on the moon is

r = l\theta

Relacing \theta

r = l(\frac{1.22\lambda}{d})

r = 1.22\frac{l\lambda}{d}

Replacing with our values we have that,

r = 1.22*(\frac{(384*10^3km)(\frac{1000m}{1km})(550*10^{-9}m)}{7*10^{{-2}}})

r = 3680.91m

Therefore the diameter of the beam on the moon is

d = 2r

d = 2 * (3690.91)

d = 7361.8285m

Hence, the diameter of the beam when it reaches the moon is 7361.82m

8 0
3 years ago
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