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Verdich [7]
3 years ago
15

Cori uses 475 J of energy from her muscles to push a bar 1 m on a weight machine at the gym. Between the bar’s motion , the heat

generated by the machine, and all other types of work done by machine, ( blank) J of output work will be done.
Physics
2 answers:
Softa [21]3 years ago
8 0

Answer:

Cori uses 475 J of energy from her muscles to push a bar 1 m on a weight machine at the gym. Between the bar’s motion , the heat generated by the machine, and all other types of work done by machine, ( 475 ) J of output work will be done.

Explanation:

on edge

dimulka [17.4K]3 years ago
5 0

Answer:

475

Explanation:

Cori does not exert any more force than 475 J, so 475 is the answer.

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Answer:

A

Explanation:

what more to say.. :| it's the distance from x to a crest or trough

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Akimi4 [234]

the machine shown in the diagram is called a tramp.

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Water flows straight down from an open faucet. The cross-sectional area of the faucet is 2.4 × 10-4m2 and the speed of the water
Ksenya-84 [330]

To solve this problem it is necessary to apply the continuity equations in the fluid and the kinematic equation for the description of the displacement, velocity and acceleration.

By definition the movement of the Fluid under the terms of Speed, acceleration and displacement is,

v_2^2 = v_1^2 + 2gh

Where,

V_i = Velocity in each state

g= Gravity

h = Height

Our values are given as,

A_1 = 2.4*10^{-4} m^2

v_1 = 0.8 m/s

h = 0.11m

Replacing at the kinetic equation to find V_2 we have,

v_2 = \sqrt{v_1^2 + 2gh}

v_2 = \sqrt{(0.8 m/s)^2 + 2(9.80 m/s2)(0.11 m)}

v_2= 1.67 m/s

Applying the concepts of continuity,

A_1v_1 = A_2v_2

We need to find A_2 then,

A_2= \frac{A_1v_1 }{v_2}

So the cross sectional area of the water stream at a point 0.11 m below the faucet is

A_2= \frac{A_1v_1 }{v_2}

A_2= \frac{(2.4*10^{-4})(0.8)}{(1.67)}

A_2= 1.14*10^{-4} m2

Therefore the cross-sectional area of the water stream at a point 0.11 m below the faucet is 1.14*10^{-4} m2

8 0
3 years ago
You have two contentious friends, Chris and Pat, and you’ve quickly discovered that they need you to resolve arguments they have
pshichka [43]

Answer:

v_average = (d₂-d₁) / Δt

this average velocity is not necessarily the velocity of the extreme points,

Explanation:

To resolve the debate, it must be shown that the two have part of the reason, the space or distance between the two points divided by time is the average speed between the points.

             v_average = (d₂-d₁) / Δt

this average velocity is not necessarily the velocity of the extreme points, in the only case that it is so is when there is no acceleration.

Therefore neither of them is right.

3 0
3 years ago
Calculate the magnitude of the linear momentum for the following cases. (a) a proton with mass 1.67 10-27 kg, moving with a spee
abruzzese [7]

Answer:

<h2>8.0995×10^-21 kgms^-1</h2>

Explanation:

Mass of proton :

m_P=1.67\times 10^-^2^7\:kg\\

Speed of Proton:

v_P=4.85\times 10^6

Linear Momentum of a particle having mass (m) and velocity (v) :

-> p =m->v\:\:\: (1)

Magnitude of momentum :

p=mv\:\:\: (2)

Frome equation (2), magnitude of linear momentum of the proton :

p_P=m_P\:v_P\\\\p_P=1.67\times 10^-^2^7 \:kg\times4.85\times 10^6\:ms^-^1\\\\p_P= 8.0995\times 10^-^2^1\:kgms^-^1

7 0
2 years ago
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