Surgical repair of a muscle

How much work is done when you push a crate horizontally with 130

Harrizon [31]

Work= Force in the direction of displacement*displacement.

You know the force in the direction of displacement (horizontally) and the displacement. So,

W=130*11=1430

Therefore, the work done is 1,430 Joules

3 0

3 years ago

If the mass of a 1.8 g paperclip was able to be completely converted to energy, how much energy would you obtain?

Anton [14]

Answer:

$E=1.62\times 10^{14}\ J$

Explanation:

Given that,

The mass of the paperclip, m = 1.8 g = 0.0018 kg

We need to find the energy obtained. The relation between mass and energy is given by :

$E=mc^2$

Where

c is the speed of light

So,

$E=0.0018\times (3\times 10^8)^2\\\\E=1.62\times 10^{14}\ J$

So, the energy obtained is $1.62\times 10^{14}\ J$ .

7 0

2 years ago

When the velocity of a mass on a spring is zero, the acceleration is at

Akimi4 [234]

Answer:

A maximum

Explanation:

When displacement is maximum, velocity is Zero and vice versa

When displacement is maximum, acceleration is maximum and when it is zero, acc. Is zero

3 0

3 years ago

Calculate the work done by a force of 30 N in lifting a load of 2 kg to a height of 10 m (g = 10 m/s2)

Mars2501 [29]

Answer:

300 J

Explanation:

Given :

Applied force = 30 N
Mass = 2 kg
Height = 10 m
g = 10 m/s²

Work done by the force :

Work done = Force x Displacement
Work done = mg x h
Work done = 30 N x 10 m
Work done = 300 J

Note :

What you have calculated is the work done by gravitational force on the object (that too, incorrectly)
But in the end, it asks for work done by the force of 30N
Hence, the given answer ~

3 0

2 years ago

A cubical box measuring 1.29 m on each side contains a monatomic ideal gas at a pressure of 2.0 atm How much thermal energy do t

Marrrta [24]

Answer:

a) $U = 652.545\,kJ$ , b) $v \approx 659.568\,\frac{m}{s}$

Explanation:

a) According to the First Law of Thermodinamics, the system is not reporting any work, mass or heat interactions. Besides, let consider that such box is rigid and, therefore, heat contained inside is the consequence of internal energy.

$Q = U$

The internal energy for a monoatomic ideal gas is:

$U = \frac{3}{2} \cdot n \cdot R_{u} \cdot T$

Let assume that cubical box contains just one kilomole of monoatomic gas. Then, the temperature is determined from the Equation of State for Ideal Gases:

$T = \frac{P\cdot V}{n\cdot R_{u}}$