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Deffense [45]
4 years ago
14

A weightlifter lifts a set of weights a vertical distance of 2.00m.If a constant net force of 350 N is exerted on the weights,wh

at is the bet work done on the weights?
Physics
1 answer:
MrMuchimi4 years ago
4 0

Answer:

<em>W=700 Joule</em>

Explanation:

<u>Physics Work </u>

Is the dot product of the force vector by the displacement vector

W=\vec F \cdot \vec r

When both the force and the displacements are pointed in the same direction, the formula reduces to its scalar version

W=F.d

The weightlifter is applying a net force of 350 N to lift the weights a distance of 2 m, thus the net work done is

W=350\ N\ .\ 2\ m=700\ Joule

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100kg x bicycle speed = 1400 X 2

bicycle speed = 2800/ 100
bicycle speed = 28 m/s
3 0
4 years ago
PLEASE ITS AN Emergency IF ITS RIGHT I WILL GIVE BRAINLIEST
n200080 [17]

Answer:

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Explanation:

6 0
3 years ago
A student pulls horizontally on a 12 kg box, which then moves horizontally with an acceleration of 0.2 m/s^2. If the student use
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The net force of the object is equal to the force applied minus the force of friction. 
                         Fnet = ma = F - Ff
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The value of Ff is 12.6 N. This force is equal to the product of the normal force which is equal to the weight in horizontal surface and the coefficient of friction.
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7 0
3 years ago
A sample of helium behaves as an ideal gas as energy is
alisha [4.7K]

Answer:

0.0321 g

Explanation:

Let helium specific heat c_h = 5.193 J/g K

Assuming no energy is lost in the process, by the law of energy conservation we can state that the 20J work done is from the heat transfer to heat it up from 273K to 393K, which is a difference of ΔT = 393 - 273 = 120 K. We have the following heat transfer equation:

E_h = m_hc_h \Delta T = 20 J

where m_h is the mass of helium, which we are looking for:

m_h = \frac{20}{c_h \Delta T} = \frac{20}{5.193 * 120} \approx 0.0321 g

4 0
3 years ago
How much total energy is dissipated in 10. seconds
noname [10]

Answer : Total energy dissipated is 10 J

Explanation :

It is given that,

Time. t = 10 s

Resistance of the resistors, R = 4-ohm

Current, I = 0.5 A

Power used is given by :

P=\dfrac{E}{t}

Where

E is the energy dissipated.

So, E = P t.............(1)

Since, P=I^2R

So equation (1) becomes :

E=I^2Rt

E=(0.5\ A)^2\times 4\Omega \times 10\ s

E=10\ J

So, the correct option is (3)

Hence, this is the required solution.

7 0
3 years ago
Read 2 more answers
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