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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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Name ten scalar quantity and explain them ​
Step2247 [10]

Answer:

A scalar is a quantity that is fully described by a magnitude only. It is described by just a single number. Some examples of scalar quantities include speed, volume, mass, temperature, power, energy, and time.

Examples of scalar quantity are:

Distance.

Speed.

Mass.

Temperature.

Energy.

Work.

Volume.

Area.

Explanation:

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2 years ago
A 0.12 kg bird is flying at a constant speed of 7.8 m/s. what is the birds conetic energy?
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KE=1/2mv^2 - equation for kinetic energy
KE=(1/2)(0.12 kg)((7.8 m/s)^2 - plug it into the formula
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3 years ago
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emmasim [6.3K]

Answer:

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Explanation:

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3 years ago
A 500 W immersion heater is placed in a pot containing 1.00 L of water at 20oC. (a) How long will the water take to rise to the
tatiyna

Answer:

96 s.

Explanation:

(a)

From the question,

Q = cm(t₂-t₁)................... Equation 1

Where Q = heat required to boil the water, c = specific heat capacity of the water, m = mass of the water, t₂ = final temperature of water, t₁ = initial temperature of water

Note: The boiling point of water = 100 °C

Given: c = 4200 J/kg.°C, t₂ = 100 °C, t₁ = 20 °C

mass of water = density×volume

m = D×v, Where D = 1000 kg/m³, v = 1.00 L = 0.001 m³

Hence, m = 1000×0.001 = 1 kg.

Substitute into equation 1

Q = 4200×1(100-20)

Q = 4200×8

Q = 33600 J.

But,

P = Q/t................... Equation 2

make t  the subject of the equation

t = Q/P................. Equation 3

Where P = power, t = time

From the question,

70 % of the available energy is absorbed by water.

P = 0.7×500 = 350 W.

Substitute into equation 2

t = 33600/350

t = 96 s.

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