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prisoha [69]
3 years ago
11

How much work is done if a 50.ON force is applied on an object in the direction of

Physics
1 answer:
jonny [76]3 years ago
4 0

Answer:150Nm

Explanation:

Force=50.0N distance=3.00m

Work=force x distance

Work=50 x 3

Work=150Nm

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You are designing a flywheel. It is to start from rest and then rotate with a constant angular acceleration of 0.200 rev/s^2. Th
Rama09 [41]

Answer:

 I = 8.75 kg m

Explanation:

This is a rotational movement exercise, let's start with kinetic energy

        K = ½ I w²

They tell us that K = 330 J, let's find the angular velocity with kinematics

      w² = w₀² + 2 α θ

as part of rest w₀ = 0

      w = √ 2α θ

let's reduce the revolutions to the SI system

      θ = 30.0 rev (2π rad / 1 rev) = 60π rad

let's calculate the angular velocity

      w = √(2  0.200  60π)

      w = 8.683 rad / s

we clear from the first equation

        I = 2K / w²

let's calculate

        I = 2 330 / 8,683²

        I = 8.75 kg m

4 0
3 years ago
Which letter represents the location of the resister in this diagram? A B C D​
Elodia [21]

Answer:

A

Explanation:

7 0
3 years ago
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Nuclear energy is..
AlexFokin [52]

I am going to say

C.  Energy contained in the nucleus of an atom

4 0
3 years ago
A piston-cylinder device initially contains 1.4 kg saturated liquid water at 200oC. Now heat is transferred to the water until t
postnew [5]

Answer:

Explanation:

Given

mass of saturated liquid water m=1.4\ kg

at 200^{\circ} specific volume is \nu =0.001157\ m^3\kg(From Table A-4,Saturated water Temperature table)

V_1=m\nu _1

V_1=1.4\times 0.001157

V_1=1.6198\times 10^{-3}\ m^3

Final Volume V_2=4V_1

V_2=4\times (1.6198\times 10^{-3})

V_2=6.4792\times 10^{-3}\ m^3

Specific volume at this stage

\nu _2=\frac{V_2}{m}

\nu _2=\frac{6.4792\times 10^{-3}}{1.4}

\nu _2=0.004628\ m^3/kg

Now we see the value and find the temperature it corresponds to specific volume at vapor stage in the table.

T_2=T_1^{*}+\frac{T_2^{*}-T_1^{*}}{\alpha _2^{*}-\alpha _1^{*}}\times (\alpha _2-\alpha _1^{*})

T_2=370^{\circ}+\frac{373.95-370}{0.003106-0.004953}\times (0.004628-0.004953)

T_2=370.7^{\circ} C

4 0
3 years ago
All of the following steps are included in a self-modification program except __________. A. specifying target behavior B. desig
Y_Kistochka [10]

Answer:

C)focusing on one goal at a time

Explanation:

Self-modification programs could be regarded as a program that help individual in managing unwanted as well as dysfunctional behavioral responses whenever they are going through a problem and try to deal with it. For instance the dysfunctional behavioral response for someone with a panic attack is avoidance. Some of the the steps that are involved in in a self-modification program are;

✓ specifying target behavior ✓designing a program

✓ gathering data about target behavior

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3 years ago
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