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MAXImum [283]
3 years ago
12

The forklift exerts a 1,500.0 N force on the box and moves it 3.00 m forward to the stack. How much work does the forklift do ag

ainst the force of gravity?
A) 1.50 × 10^3 J

B) 0.00 J

C) 500 J

D) 4.50 × 10^3 J

I think it is D, but I can also justify B, please help!
Physics
1 answer:
Deffense [45]3 years ago
7 0
The answer is D using the work formula
W= F•d but if it was against gravity, it would be 0 if gravity is exerting the same amount, I would pick D using the formula, but I'm not so sure sorry
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If we start with 1.000 g of cobalt-60, 0.675 g will remain after 3.00 yr. this means that the of is _____
Tju [1.3M]
<span>Cobalt-60 is undergoing a radioactivity decay.

The formula of the decay is n=N(1/2)</span>∧(T/t).
<span>Where N </span>⇒ original mass of cobalt
<span>           n </span>⇒ remaining mass of cobalt after 3 years
          T ⇒ decaying period
           t ⇒ half-life of cobalt.

So,
0.675 = 1 × 0.5∧(3/t)
log 0.675 = log 0.5∧(3/t)
3/t = log 0.675 ÷log 0.5
 3/t= 0.567

t = 3÷0.567
  = 5.290626524

the half-life of Cobalt-60 is 5.29 years. 

<span>           
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</span>
8 0
3 years ago
A 4.40-kilogram hoop starts from rest at a height 1.70 m above the base of an inclined plane and rolls down under the influence
Anestetic [448]

Answer:

The linear velocity is  v=4.08m/s

Explanation:

According to the law of conservation of energy

   The potential energy possessed by the  hoop at the top of the inclined plane is converted to the kinetic energy at the foot of the inclined plane

        The kinetic energy can be mathematically represented as

                    KE = \frac{mv^2}{2} + \frac{Iw}{2}

Where I is the moment of inertia possessed by the hoop  which is mathematically represented as

                 I = mr^2

Here R is the radius of the hoop

         w is the angular velocity which the hoop has at the bottom of the lower part of the inclined plane which is mathematically represented as

                          w = \frac{v}{r}

Where v linear speed of the hoop's center of mass just as the hoop leaves the incline and rolls onto a horizontal surface

      Now expressing the above statement mathematically

            potential \ energy = \frac{mv^2}{y} + \frac{Iw^2}{2}

               mgh = \frac{mv^2}{y} + \frac{Iw^2}{2}

=>            mgh =\frac{mv^2}{2} + \frac{(mr^2)(\frac{v}{r})^2 }{2}  

=>          mgh = \frac{mv^2}{2} + \frac{mv^2}{2}

=>           mgh = mv^2

=>              v = \sqrt{gh}

Substituting values

                v = \sqrt{9.81 * 1.7}

                  v=4.08m/s

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