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3 years ago

Which two types of potential energies would be types of mechanical energies?

Physics

2 answers:

Crank3 years ago

3 0

Answer:

D ) Elastic and Gravitational

Explanation:

Mechanical energy is defined as the sum of elastic, potential and kinetic energies.

Therefore potential energy and gravitational or potential energy are contemplated within the definition of mechanical energy.

$E_{M} = E_{e}+E_{k}+E_{p}\\where:\\E_{M}= Mechanical Energy\\E_{e} = elastic energy\\E_{k} = kinetic energy\\E_{p}= potential energy$

The units for energy are given in joules [J] for the SI and Btu for the IP system

Minchanka [31]3 years ago

3 0

<h2>D. Elastic and Gravitational</h2>

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A 2kg block of which material would require 450 joules of thermal energy to increase its temperature by 1 degree Celsius?

12345 [234]

The block is made of A) Tin, as its specific heat capacity is $0.225 J/(g^{\circ}C)$

Explanation:

When an amount of energy Q is supplied to a sample of material of mass m, the temperature of the material increases by $\Delta T$ , according to the following equation :

$Q=mC_s \Delta T$

where $C_s$ is the specific heat capacity of the material.

In this problem, we have:

m = 2 kg = 2000 g is the mass of the unknown material

$Q = 450 J$ is the amount of energy supplied to the block

$\Delta T = 1^{\circ}C$ is the change in temperature of the material

Solving the equation for $C_s$ , we can find the specific heat capacity of the unknown sample:

$C_s = \frac{Q}{m \Delta T}=\frac{450}{(2000)(1)}=0.225 J/(g^{\circ}C)$

And by comparing with tabular values, we can find that this value is approximately the specific heat capacity of tin.

Learn more about specific heat capacity:

brainly.com/question/3032746

brainly.com/question/4759369

#LearnwithBrainly

7 0

3 years ago

Calculate the speed of a proton after it accelerates from rest through a potential difference of 350 V.

AVprozaik [17]

The speed of a proton after it accelerates from rest through a potential difference of 350 V is $25.86 \times 10^4 ~m/s$ .

Initial velocity of the proton $u = 0$

Given potential difference $\Delta V = 350V$

let's assume that the speed of the proton is $v$ ,

Since the proton is accelerating through a potential difference, proton's potential energy will change with time. The potential energy of a particle of charge $q$ when accelerated with a potential difference $\Delta V$ is,

$U = q \Delta V$

Due to Work-Energy Theorem and Conservation of Energy - <em>If there is no non-conservative force acting on a particle then loss in Potential energy P.E must be equal to gain in Kinetic Energy K.E</em> i.e

$\Delta K = \Delta V$

If the initial and final velocity of the proton is $u$ and $v$ respectively then,

change in Kinetic Energy $\implies \Delta K = \frac{1}{2}mv^2 -\frac{1}{2}mu^2 = \frac{1}{2}mv^2 - 0$

change in Potential Energy $\implies \Delta U = q\Delta V$

from conservation of energy,

$v= \sqrt{\frac{2q\Delta V}{m}}$

so, $v = \sqrt{\frac{2\times 350 \times 1.6\times 10^{-19}}{1.67 \times 10^{-27}}$

$= 25.86 \times 10^4 ~m/s$

To read more about the conservation of energy, please go to brainly.com/question/14668053

7 0

1 year ago

An athlete at the gym holds a 3.0 kg steel ball in his hand. His arm is 60 cm long and has a mass of 3.8 kg, with the center of

Serggg [28]

Answer:

(a) τ = 26.58 Nm

(b) τ = 18.79 Nm

Explanation:

(a)

First we find the torque due to the ball in hand:

τ₁ = F₁d₁

where,

τ₁ = Torque due to ball in hand = ?

F₁ = Force due to ball in hand = m₁g = (3 kg)(9.8 m/s²) = 29.4 N

d₁ = perpendicular distance between ball and shoulder = 60 cm = 0.6 m

τ₁ = (29.4 N)(0.6 m)

τ₁ = 17.64 Nm

Now, we calculate the torque due to the his arm:

τ₁ = F₁d₁

where,

τ₂ = Torque due to arm = ?

F₂ = Force due to arm = m₂g = (3.8 kg)(9.8 m/s²) = 37.24 N

d₂ = perpendicular distance between center of mass and shoulder = 40% of 60 cm = (0.4)(60 cm) = 24 cm = 0.24 m

τ₂ = (37.24 N)(0.24 m)

τ₂ = 8.94 Nm

Since, both torques have same direction. Therefore, total torque will be:

τ = τ₁ + τ₂

τ = 17.64 Nm + 8.94 Nm

<u></u>

(b)

Now, the arm is at 45° below horizontal line.

First we find the torque due to the ball in hand:

τ₁ = F₁d₁

where,

τ₁ = Torque due to ball in hand = ?

F₁ = Force due to ball in hand = m₁g = (3 kg)(9.8 m/s²) = 29.4 N

42.42 cm = 0.4242 m

τ₁ = (29.4 N)(0.4242 m)

τ₁ = 12.47 Nm

Now, we calculate the torque due to the his arm:

τ₁ = F₁d₁

where,

τ₂ = Torque due to arm = ?

F₂ = Force due to arm = m₂g = (3.8 kg)(9.8 m/s²) = 37.24 N

d₂ = perpendicular distance between center of mass and shoulder = 40% of (60 cm)(Cos 45°) = (0.4)(42.42 cm) = 16.96 cm = 0.1696 m

τ₂ = (37.24 N)(0.1696 m)

τ₂ = 6.32 Nm

Since, both torques have same direction. Therefore, total torque will be:

τ = τ₁ + τ₂

τ = 12.47 Nm + 6.32 Nm

3 0

2 years ago

a train is moving with an initial velocity of 30 m/s, the brakes are applied so as to produce a uniform acceleration of -1.5 m/s

Pepsi [2]

Answer:

$\boxed{\sf Time \ in \ which \ train \ will \ come \ to \ rest = 20 \ sec}$

Given:

Initial velocity (u) = 30 m/s

Final speed (v) = 0 m/s

Acceleration (a) = - 1.5 m/,s²

To Find:

Time in which train will come to rest (t).

Explanation:

$\sf From \ equation \ of \ motion: \\ \sf \implies \bold{v = u + at} \\ \\ \sf Substituting \ value \ of \ v, \ u \ and \ a: \\ \sf \implies 0 = 30 + ( - 1.5)(t) \\ \sf \implies 0 = 30 - 1.5(t) \\ \sf \implies 30 - 1.5(t) = 0 \\ \\ \sf Subtract \: 30 \: from \: both \: sides: \\ \sf \implies (30 - \boxed{ \sf 30}) - 1.5(t) = \boxed{ \sf - 30} \\ \\ \sf 30 - 30 = 0 : \\ \sf \implies - 1.5(t) = - 30 \\ \\ \sf Divide \: both \: sides \: of \: - 1.5(t) = - 30 \: by \: - 1.5 : \\ \sf \implies \frac{ - 1.5(t)}{ \boxed{ \sf - 1.5}} = \frac{ - 30}{ \boxed{ \sf -1.5 }} \\ \\ \sf \frac{ \cancel{ \sf 1.5}}{\cancel{ \sf 1.5}} = 1 : \\ \sf \implies t = \frac{ - 30}{ - 1.5} \\ \\ \sf \frac{ - 30}{ - 1.5} = \frac{\cancel{ \sf 1.5} \times 20}{\cancel{ \sf 1.5}} = 20 : \\ \sf \implies t = 20 \: sec$

So,

Time in which train will come to rest = 20 seconds

4 0

3 years ago

What is a non-example of gravitational force?

qaws [65]

A non-example of force would be something that stay sill like a balloon in the air..... taste, smell, feel, texture, color, opinion, faith, hope, sincerity, honest, speed, momentum, altitude, volume, loudness, area, length, acidity, obesity, nationalism, current, resistance, viscosity, wavelength, flow, rate, frequency, albedo, diameter, age, temperature, acceleration, body mass index, salinity, specific, specific gravity, consciousness, intelligence, refraction index, mass, time, date rate, switching speed, libido, focal length, and latency are not force. And even there are many other things that also are not force, too.

4 0

3 years ago

Read 2 more answers