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OverLord2011 [107]
3 years ago
14

A 10- kilogram block is pushed across a horizontal surface with a horizontal force of 20 N against a friction force of 10 N. The

acceleration of the block is_______.
Physics
1 answer:
Temka [501]3 years ago
4 0

Answer:

1m/s^2

Explanation:

Mass of block=10 kg

Applied horizontal force =F=20 N

Friction force=f=10 N

We have to find the acceleration of block.

Net force=Applied horizontal force-friction force

ma=F-f

Where F= Horizontal force

f=Friction force

m=Mass of object

a=Acceleration of object

10a=20-10=10

a=\frac{10}{10}=1 m/s^2

Hence, the acceleration of the block=1m/s^2

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3 0
1 year ago
A cross-country skier moves 36 meters eastward, then 44
ivanzaharov [21]

Answer:

Magnitude = 14 metres

Direction = eastward.

Explanation:

A cross-country skier moves 36 meters eastward, then 44meters westward, and finally 22 meters eastward.

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The magnitude and direction will be the displacement of the cross country skier.

Let the east ward be positive and the west ward be negative.

Since the skier moves 36 meters eastward, then 44meters westward, and finally 22 meters eastward. Then, that will be:

36 - 44 + 22 = 14

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3 0
3 years ago
A small sphere is at rest at the top of a frictionless semicylindrical surface. The sphere is given a slight nudge to the right
V125BC [204]

Answer:

vi = 4.77 ft/s

Explanation:

Given:

- The radius of the surface R = 1.45 ft

- The Angle at which the the sphere leaves

- Initial velocity vi

- Final velocity vf

Find:

Determine the sphere's initial speed.

Solution:

- Newton's second law of motion in centripetal direction is given as:

                         m*g*cos(θ) - N = m*v^2 / R

Where, m: mass of sphere

             g: Gravitational Acceleration

             θ: Angle with the vertical

             N: Normal contact force.

- The sphere leaves surface at θ = 34°. The Normal contact is N = 0. Then we have:

                         m*g*cos(θ) - 0 = m*vf^2 / R

                         g*cos(θ) = vf^2 / R    

                         vf^2 = R*g*cos(θ)

                         vf^2 = 1.45*32.2*cos(34)

                        vf^2 = 38.708 ft/s

- Using conservation of energy for initial release point and point where sphere leaves cylinder:

                          ΔK.E = ΔP.E

                          0.5*m* ( vf^2 - vi^2 ) = m*g*(R - R*cos(θ))

                          ( vf^2 - vi^2 ) = 2*g*R*( 1 - cos(θ))

                          vi^2 =  vf^2 - 2*g*R*( 1 - cos(θ))

                          vi^2 = 38.708 - 2*32.2*1.45*(1-cos(34))

                          vi^2 = 22.744

                           vi = 4.77 ft/s

4 0
4 years ago
These are nontoxic, nonflammable chemicals containing atoms of carbon, chlorine, and fluorine that have created a hole in the oz
yarga [219]
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These are compounds developed and improved by Thomas Midgley in the late 1920s. They were used as refrigerants and aerosol propellants.

These compounds created a hole in the ozone layer by the following reactions:

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4 years ago
A car traveling at 30 m/s speeds up to 35 m/s over a period of 5 seconds. What is the acceleration of the car?
Delvig [45]

Answer:

u =30 m/s

v = 35 m/s

t = 5 secs

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a = (v- u)/t

a = (35-30)/5

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PLS MARK BRAINLIEST

8 0
2 years ago
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