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3 years ago

Which of the following cools the air in a household refrigerator?

1 answer:

7 0

The answer is A since refrigeration occurs when the normal room temperature heat is drawn away from the inside of the unit through the evaporation of a liquid. Through the process of entropy then, the air instantly cools to the right temperature to preserve food.

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In the auditory system, the unit for measuring the amplitude of sound is _____.

nasty-shy [4]

B decibel is the answer

6 0

4 years ago

Read 2 more answers

Three people are pushing a 500 kg of box in the same direction. applied forces are 30 n, 20 n, and 10 n respectively. if the acc

GREYUIT [131]

The total force applied by the three people is:
$F=30 N+20 N+10 N=60 N$
This force is pushing toward the direction of the motion, while the frictional force $F_f$ points in the opposite direction.
We can write Newton's second law applied to the block: the resultant of the two forces must be equal to the product between the block's mass and its acceleration
$F-F_f = ma$
We know the mass of the block, m=500 kg, and the acceleration, $a=0.02 m/s^2$ , so we can find the friction:
$F_f = F-ma=60 N - (500 kg)(0.02 m/s^2)=50 N$

8 0

3 years ago

a body of a mass 1kg suspended from a spring is found to stretch the spring by 10cm. (A) what is the spring constant? what is pe

Nikitich [7]

Answer:

(A) The spring constant is 98 N/m.

(B) The period of oscillation is 0.635 s.

Explanation:

Given:

Mass of the body is, $m=1\ kg$

Extension length of the spring is, $x=10\ cm=0.1\ m$

Now, let 'k' be the spring constant.

The force acting on the body is due to gravity only and is equal to its weight.

So, weight of the body is given as:

$F_g=mg\\F_g=1\times 9.8\\F_g=9.8\ N$

Now, we know that for a spring-mass system, the net force acting on the body is equal to the product of the spring constant and extension length. Therefore,

$F_g=kx\\9.8=k(0.1)\\k=\frac{9.8}{0.1}=98\ N/m$

Hence, the spring constant is 98 N/m.

(B)

Period of oscillation of a body of spring-mass system in SHM is given as:

$T=2\pi\sqrt{\frac{m}{k}}$

Plug in the given values and solve for period 'T'. This gives,

$T=2\pi\sqrt{\frac{1}{98}}\\T=2\pi\times 0.101\\T=0.635\ s$

Therefore, the period of oscillation is 0.635 s.

5 0

4 years ago

An electric drill

Lera25 [3.4K]

Answer:

Incomplete question. Complete question is: An electric drill starts from rest and rotates with a constant angular acceleration. After the drill has rotated through a certain angle, the magnitude of the centripetal acceleration of a point on the drill is twice the magnitude of the tangential acceleration. Determine the angle through which the drill rotates by this point.

The answer is : Δ θ = 1 rad

Explanation:

Ok, so the condition involves the centripetal acceleration and the tangential acceleration, so let’s start by writing expressions for each:

Ac= centripetal acceleration At= tangential acceleration

Ac = V² / r At = r α

Because we have to determine the angle ultimately, therefore we should convert the linear velocity into angular velocity in the expression for centripetal acceleration

V = r ω

Ac = (r ω)² / r = r² ω² / r

Ac = r ω²

now that we have expressions for the centripetal and tangential acceleration, we can write an equation that expresses the condition given: The magnitude of the centripetal acceleration is twice the magnitude of the tangential acceleration.

Ac = 2 At

That is,

r ω² = 2 r α

it is equivalent to;

ω² = 2 α

now we have the relation between angular speed and angular acceleration, but we also need to determine the angular displacement as well. Therefore choose a kinematics equation that doesn’t involve time because time is not mentioned in the question. Thus,

ω² – ω°² = 2 α Δ θ

such that ω° = 0

and ω² = 2 α

therefore;

2 α - 0 = 2 α Δ θ

2 α = 2 α Δ θ

So the angle will be : Δ θ = 1 rad

7 0

3 years ago

A person throws a ball with a force of 140 N that accelerates at 15 m/s? What is the mass of the

mr_godi [17]

Newton's 2nd law

$\tt \sum F=m.a$

input the value:

$\tt m=\dfrac{\Sum F}{a}=\dfrac{140~N}{15~m/s^2}=9.3~kg$

6 0

3 years ago