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3 years ago

10

John pushes a box with a constant force as shown in the graph below.

1 answer:

andrew11 [14]3 years ago

4 0

From the graph, it can be seen that the constant force that John exerted in order to move the object is 14N. Work is calculated by multiplying the force with the distance to which the object moves in parallel with the direction of the force.
Work = Force x displacement
Work = (14 N) x (8 m)
Work = 112 J
The closest value is 110J. Thus, the answer to this item is the second choice.

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16. The energy absorbed in 10minutes by an electrical heater is 1.5 MJ. The supply voltage is 240 V. calculate: a) The current d

dsp73

Answer:

If by 1.5 MJ you mean 1.5E6 Joules then

W = P t = power X time

W / t = P power

P = 1.5E6 J / 600 sec = 2500 J / s

P = I V

a) I = 2500 J/s / (240 J/c) = 10.4 C / sec = 10.4 amps

b) Q = I t = 10.4 C / sec * 300 sec = 3120 Coulombs

c) E = P * t = 2500 J / sec * 100 hr * 3600 sec / hr = 9.0E8 Joules

5 0

2 years ago

If 100 kg person weighed 400 N on the planet Zorg,

Brut [27]

Answer:

<u>4 m/s²</u>

Explanation:

<u>Formula</u>

Weight = Mass x Gravity due to acceleration

<u>Solving</u> :

400 N = 100 kg x g
g = 400/100
g = <u>4 m/s²</u>

5 0

1 year ago

Read 2 more answers

A flywheel is a solid disk that rotates about an axis that is perpendicular to the disk at its center. Rotating flywheels provid

Sergeeva-Olga [200]

Answer:

$8.37*10^5$ rpm

Explanation:

Given that rotational kinetic energy = $4.66*10^9J$

Mass of the fly wheel (m) = 19.7 kg

Radius of the fly wheel (r) = 0.351 m

Moment of inertia (I) = $\frac{1}{2} mr ^2$

Rotational K.E is illustrated as $(K.E)_{rt} = \frac{1}{2} I \omega^2$

$\omega = \sqrt{\frac{2(K.E)_{rt}}{I} }$

$\omega = \sqrt{\frac{2(KE)_{rt}}{1/2 mr^2} }$

$\omega = \sqrt{\frac{4(K.E)_{rt}}{mr^2} }$

$\omega = \sqrt{\frac{4*4.66*10^9J}{19.7kg*(0.351)^2} }$

$\omega = 87636.04$

$\omega = 8.76*10^4 rad/s$

Since 1 rpm = $\frac{2 \pi}{60} rad/s$

$\omega = 8.76*10^4(\frac{60}{2 \pi})$

$\omega = 836518.38$

$\omega = 8.37 *10^5 rpm$

3 0

2 years ago

A force of 6 N acts on a 33 kg object for 9 seconds. What is the objects change in velocity?

GuDViN [60]

Acceleration = change in velocity/time
By F = ma,
6 = 33 x change in velocity / 9
change in velocity = +1.636 m/s

4 0

2 years ago

Elodia [21]

Ok do 200x2 then divide by 6

4 0

2 years ago