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Vinil7 [7]
3 years ago
10

You are standing on the SCALE The scale is on the floor. What are the forces on the SCALE?

Physics
2 answers:
svlad2 [7]3 years ago
8 0
Normal and gravitational force
ehidna [41]3 years ago
3 0

Answer:

The Normal and Gravitational Force

Explanation:

The normal force pushes up and is between the ground and the scale. The gravitational force is the force exerted on the ground.

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What is the acceleration of this object? The object's mass is 60 kg.
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First determine the net force. Let's say the downwards force is negative and the upwards force is positive.
Since the forces act in opposite directions, the net force would be:
400N - 600N = -200N

Since I said negative is downwards, this translates to the net force being 200N downwards.

Force = mass*acceleration
200N = 60kg * acceleration
acceleration = 3.33 m/s^2
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Which change will cause an increase in the electric current produced through electromagnetic induction?using more wire loops in
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In a standard tensile test, a steel rod of 7 8-in. diameter is subjected to a tension force of 17 kips. Knowing that ν = 0.30 an
natali 33 [55]

Answer:

(a) Elongation of the rod==5.61×10⁻⁹m

(b) Change in diameter=1.640×10⁻⁸m

Explanation:

Given data

Diameter d=78 in=1.9812 m

Cross Area is:

A=(\pi /4)d^{2} \\A=(\pi /4)(1.9812m)^{2}\\A=3.08m^{2}

Applied Load P=17 KN=17×10³N

E=29 × 106 psi=1.99947961×10¹¹Pa

Stress and Strain in x direction

Stress in x direction

σ=P/A

=\frac{17*10^{3}N }{3.08m^{2} }\\ =5517.25Pa

σ=5517.25 Pa

Strain in x direction

ε=σ/E

=\frac{5517.25}{1.99947961*10^{11} } \\=2.76*10^{-8}

ε=2.76×10⁻⁸

Part (a)

Elongation of the rod=Lε

=(0.2032)(2.76×10⁻⁸)

Elongation of the rod==5.61×10⁻⁹m

Part(b) Change in diameter

Strain in y direction

ε₁= -vε

ε₁= -(0.30)(2.76×10⁻⁸)

ε₁=-8.28×10⁻⁹

Change in diameter=d×ε₁

Change in diameter=(1.9812m)×(-8.28×10⁻⁹)

Change in diameter=1.640×10⁻⁸m

4 0
3 years ago
In a 350-m race, runner A starts from rest and accelerates at 1.6 m/s^2 for the first 30 m and then runs at constant speed. Runn
kifflom [539]

Answer:

B can take 0.64 sec for the longest nap .

Explanation:

Given that,

Total distance = 350 m

Acceleration of A = 1.6 m/s²

Distance = 30 m

Acceleration of B = 2.0 m/s²

We need to calculate the time for A

Using equation of motion

s=ut+\dfrac{1}{2}at_{A}^2

Put the value in the equation

30=0+\dfrac{1}{2}\times1.6\times t_{A}^2

t_{A}=\sqrt{\dfrac{30\times2}{1.6}}

t_{A}=6.12\ sec

We need to calculate the time for B

Using equation of motion

s=ut+\dfrac{1}{2}at_{B}^2

Put the value in the equation

30=0+\dfrac{1}{2}\times2.0\times t_{B}^2

t_{B}=\sqrt{\dfrac{30\times2}{2.0}}

t_{B}=5.48\ sec

We need to calculate the time for longest nap

Using formula for difference of time

t'=t_{A}-t_{B}

t'=6.12-5.48

t'=0.64\ s

Hence, B can take 0.64 sec for the longest nap .

4 0
3 years ago
Suppose a current flows through a copper wire. Which two things occur?
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I’m pretty sure the answer is c and d hope this helps and good luck
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