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Svetllana [295]
2 years ago
10

Using a force of 28.0 Newton, a student pulls a 70.0 Newton weight along the tabletop for a distance of 15.0 meters in 3.0 secon

ds. Compute the power developed by the student.
Physics
1 answer:
Musya8 [376]2 years ago
8 0

Answer:

140 watt

Explanation:

We are given that

Force applied by student ,F=28 N

Weight pulled  by students=70 N

Displacement,s=15 m

Time=3 s

We have to find the power developed  by the student.

Work done=w=F\times s

Work done by the student=28\times 15=420 J

Power=\frac{work\;done}{time}

Using the formula

Power=\frac{420}{3}=140watt

Hence, the power developed by the students=140 watt

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Suppose you want to use this human engine to lift a 2.35 kg box from the floor to a tabletop 1.30 m above the floor. How much mu
morpeh [17]

Answer:

The increase in the gravitational potential energy is 29.93 joules.

Explanation:

Given that,

Mass of the box, m = 2.35 kg

It is lifted from the floor to a tabletop 1.30 m above the floor, h = 1.3 m

We need to find the increase the gravitational potential energy. Initial it will placed at ground i.e. its initial gravitational potential is equal to 0. The increase in the gravitational potential energy is given by :

U=mgh

U=2.35\ kg\times 9.8\ m/s^2\times 1.3\ m

U = 29.93 Joules

So, the increase in the gravitational potential energy is 29.93 joules. Hence, this is the required solution.

6 0
3 years ago
Which of the following weather conditions would result in the greatest rate of evaporation from the Earth's surface? A. hot and
lisov135 [29]
A: A hit and dry weather

Hope this helps :)
6 0
2 years ago
A weight lifter is trying to do a bicep curl with a weight of 300 N. At the "sticking point", the moment arm of this weight is 3
lesantik [10]

Answer:

The weight lifter would not get past this sticking point.

Explanation:

Generally torque applied on the weight is mathematically represented as

             T =  F z

To obtain Elbow torque we substitute 4000 N for F (the force ) and 2cm = \frac{2}{100} = 0.02m for z the perpendicular distance

So Elbow Torque is   T_e= 4000 * 0.02

                                   = 80Nm

 To obtain the torque required we substitute 300 N for F and 30cm =\frac{30}{100} = 0.3 m

  So the Required Torque is T_R = 300 *0.3

                                                     =90Nm

Now since   T_e < T_R it mean that the weight lifter would not get past this sticking point

                                   

7 0
3 years ago
What is the wavelength associated with 0.113kg ball traveling with velocity of 43 m/s?
lesya [120]

Answer:

2.73×10¯³⁴ m.

Explanation:

The following data were obtained from the question:

Mass (m) = 0.113 Kg

Velocity (v) = 43 m/s

Wavelength (λ) =?

Next, we shall determine the energy of the ball. This can be obtained as follow:

Mass (m) = 0.113 Kg

Velocity (v) = 43 m/s

Energy (E) =?

E = ½m²

E = ½ × 0.113 × 43²

E = 0.0565 × 1849

E = 104.4685 J

Next, we shall determine the frequency. This can be obtained as follow:

Energy (E) = 104.4685 J

Planck's constant (h) = 6.63×10¯³⁴ Js

Frequency (f) =?

E = hf

104.4685 = 6.63×10¯³⁴ × f

Divide both side by 6.63×10¯³⁴

f = 104.4685 / 6.63×10¯³⁴

f = 15.76×10³⁴ Hz

Finally, we shall determine the wavelength of the ball. This can be obtained as follow:

Velocity (v) = 43 m/s

Frequency (f) = 15.76×10³⁴ Hz

Wavelength (λ) =?

v = λf

43 = λ × 15.76×10³⁴

Divide both side by 15.76×10³⁴

λ = 43 / 15.76×10³⁴

λ = 2.73×10¯³⁴ m

Therefore, the wavelength of the ball is 2.73×10¯³⁴ m.

8 0
3 years ago
In the equation Ca(s) + 2H2O(l) --&gt;Ca(OH)2(aq) + H2(g) list each element used as a reactant, and tell how many atoms of each
Mekhanik [1.2K]

The reactants are <em>Calcium</em>, <em>Hydrogen</em> and <em>Oxygen</em>.

We need 1 atom of <em>Calcium</em>, 4 atoms of <em>Hydrogen</em> and 2 atoms of <em>Oxygen</em>.

Hope this helps.

r3t40

3 0
2 years ago
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