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Reil [10]
3 years ago
6

Hilda pushes her sofa across the room with a force of 40 N. It accelerates at 2 m/s/s. What is the mass of the sofa?

Physics
2 answers:
kicyunya [14]3 years ago
6 0

The correct answer of this question is : A) 20 Kg and B) 3 seconds.

EXPLANATION :

As per the first question, the force applied on sofa F = 40 N.

The acceleration of the sofa a = 2\ m/s^2.

We are asked to calculate the mass of the sofa.

From Newton's second law, we know that external unbalanced force is equal to the product of mass with acceleration of the body.

Mathematically force F = ma.

Hence, mass of the sofa is calculated as -

                              mass of sofa  m = \frac{F}{a}

                                                         = \frac{40}{2}\ Kg

                                                         = 20 Kg.

As per the second question, initial velocity u = 2 m/s due west.

The final velocity of the skateboard v = 8 m/s.

The acceleration of the skate board a = 2\ m/s^2

The time taken [ t ] to reach that velocity is calculated as -

From equation of kinematics, we know that -

                                           v = u + at

                                          ⇒8 = 2 + 2t

                                         ⇒ 2t = 6

                                         ⇒  t = \frac{6}{2}\ s

                                           ⇒t = 3 s.            [ans]


                                                     

GalinKa [24]3 years ago
4 0
The first answer to the first question is 20 kg
the second answer to the second question is 3 seconds
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Komok [63]

Complete question is;

Shoveling snow can be extremely taxing since the arms have such a low efficiency in this activity. Suppose a person shoveling a sidewalk metabolizes food at the rate of 800 W. (The efficiency of a person shoveling is 3%.)

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Answer:

A) P_out = 24 W

B) t = 1470 s

C) Q = 1140.72 KJ

Explanation:

We are given;

Input Power; P_in = 800 W

Efficiency; η = 3% = 0.03

A) Formula for efficiency is;

η = P_out/P_in

Making P_out the subject, we have;

P_out = η•P_in

P_out = 0.03 × 800

P_out = 24 W

B) We know that;

Power = work done/time taken

Thus;

P_out = mgh/t

We are given;

m = 3000 kg

h = 1.20 m

Thus, time is;

t = (3000 × 9.8 × 1.2)/24

t = 1470 s

C) amount of heat wasted is calculated from;

Q = (P_in - P_out)t

Q = (800 - 24) × 1470

Q = 1,140,720 J

Q = 1140.72 KJ

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2 years ago
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Answer:

The two forces that affect on the pendulum are the force of gravity,

Explanation:

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Therefore, I hope this helps!

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3 years ago
A −3.0 nC charge is on the x-axis at x=−9 cm and a +4.0 nC charge is on the x-axis at x=16 cm. At what point or points on the y-
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Answer:

y = 10.2 m

Explanation:

It is given that,

Charge, q_1=-3\ nC

It is placed at a distance of 9 cm at x axis

Charge, q_2=+4\ nC

It is placed at a distance of 16 cm at x axis

We need to find the point on the y-axis where the electric potential zero. The net potential on y-axis is equal to 0. So,

\dfrac{kq_1}{r_1}+\dfrac{kq_2}{r_2}=0

Here,

r_1=\sqrt{y^2+9^2} \\\\r_2=\sqrt{y^2+15^2}

So,

\dfrac{kq_1}{r_1}=-\dfrac{kq_2}{r_2}\\\\\dfrac{q_1}{r_1}=-\dfrac{q_2}{r_2}\\\\\dfrac{-3\ nC}{\sqrt{y^2+81} }=-\dfrac{4\ nC}{\sqrt{y^2+225} }\\\\3\times \sqrt{y^2+225}=4\times \sqrt{y^2+81}

Squaring both sides,

3\times \sqrt{y^2+225}=4\times \sqrt{y^2+81}\\\\9(y^2+225)=16\times (y^2+81)\\\\9y^2+2025=16y^2-+1296\\\\2025-1296=7y^2\\\\7y^2=729\\\\y=10.2\ m

So, at a distance of 10.2 m on the y axis the electric potential equals 0.

8 0
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Tasya [4]

Answer:

I(x)  = 1444×k ×{\pi}

I(y)  = 1444×k ×{\pi}

I(o) = 3888×k ×{\pi}  

Explanation:

Given data

function =  x^2 + y^2 ≤ 36

function =  x^2 + y^2 ≤ 6^2

to find out

the moments of inertia Ix, Iy, Io

solution

first we consider the polar coordinate (a,θ)

and polar is directly proportional to a²

so p = k × a²

so that

x = a cosθ

y = a sinθ

dA = adθda

so

I(x) = ∫y²pdA

take limit 0 to 6 for a and o to \pi /2 for θ

I(x) = \int_{0}^{6}\int_{0}^{\pi/2} y²p dA

I(x) = \int_{0}^{6}\int_{0}^{\pi/2} (a sinθ)²(k × a²) adθda

I(x) = k  \int_{0}^{6}a^(5)  da ×  \int_{0}^{\pi/2}  (sin²θ)dθ

I(x) = k  \int_{0}^{6}a^(5)  da ×  \int_{0}^{\pi/2}  (1-cos2θ)/2 dθ

I(x)  = k ({r}^{6}/6)^(5)_0 ×  {θ/2 - sin2θ/4}^{\pi /2}_0

I(x)  = k × ({6}^{6}/6) × (  {\pi /4} - sin\pi /4)

I(x)  = k ×  ({6}^{5}) ×   {\pi /4}

I(x)  = 1444×k ×{\pi}    .....................1

and we can say I(x) = I(y)   by the symmetry rule

and here I(o) will be  I(x) + I(y) i.e

I(o) = 2 × 1444×k ×{\pi}

I(o) = 3888×k ×{\pi}   ......................2

3 0
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