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klemol [59]
3 years ago
10

Work equals force the times

Physics
1 answer:
soldier1979 [14.2K]3 years ago
6 0
<span>Work equals force times distance. When you move an object, you are exerting a force onto it. By exerting a force on the object, you are actually displacing it from its initial position. You cannot apply force to the object without altering its position. Keep in mind that when you exert work, you are exerting energy too. So the work must have a unit in joules in SI units. Force is in newtons or kilogram meter per second squared. And distance is in meters. So you will have newton-meter or joules. </span>
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If the half-life of niobium-91 is 680 years, how long will it take for the decay described in the previous question?
Svet_ta [14]
2040

15.4+2.2/2 until it equals 2.2
( divide by 3)

680(years)*3 devisions = 2040
3 0
4 years ago
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1. Calculate the charge in a circuit which carries a current of 30 amperes for 5 seconds
sweet-ann [11.9K]

Answer:

1. 150C.

2. 50sec

3.1.5a

Explanation:

1. I = Q/T

Q= 30x5

=150c

2.applying the formulae, I = Q/T

T= Q/I

=500/10

=50sec.

3. using the formulae i=q/t

i= 120/80

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6 0
3 years ago
A rock is projected upward from the surface of the moon, at time t = 0.0 s, w a velocity of 30 m/s. The acceleration due to grav
Vinvika [58]
<h2>Answer: 277.777 m</h2>

Explanation:

The situation described here is parabolic movement. However, as we are told that the rock was<u> projected upward from the surface</u>, we will only use the equations related to the Y axis.

In this sense, the movement equations in the Y axis are:

y-y_{o}=V_{o}.t+\frac{1}{2}g.t^{2}    (1)

V=V_{o}-g.t    (2)

Where:

y  is the rock's final position

y_{o}=0  is the rock's initial position

V_{o}=30\frac{m}{s} is the rock's initial velocity

V is the final velocity

t is the time the parabolic movement lasts

g=1.62\frac{m}{s^{2}}  is the acceleration due to gravity at the surface of the moon

As we know y_{o}=0 , equation (2) is rewritten as:

y=V_{o}.t+\frac{1}{2}g.t^{2}    (3)

On the other hand, the maximum height  is accomplished when V=0:

V=V_{o}-g.t=0    (4)

V_{o}-g.t=0    

V_{o}=g.t    (5)

Finding t:

t=\frac{V_{o}}{g}    (6)

Substituting (6) in (3):

y=V_{o}(\frac{V_{o}}{g})+\frac{1}{2}g(\frac{V_{o}}{g})^{2}    (7)

y_{max}=\frac{{V_{o}}^{2}}{2g}    (8)  Now we can calculate the maximum height of the rock

y_{max}=\frac{{(30m/s)}^{2}}{(2)(1.62m/s^{2})}   (9)

Finally:

y_{max}=277.777m  

4 0
3 years ago
During the period that the moon’s phases are changing from new to full, the moon is _____. waxing exhibiting retrograde motion w
Sindrei [870]
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3 0
3 years ago
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Unless you're a frequent user of this type of event, i think it would be economically more efficient if you pay the resort on one-day price
8 0
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