Bronze is what type of matter

A 330 kg piano slides 3.6 m down a 28o incline and is kept from accelerating by a man who is pushing back on it parallel to the

e-lub [12.9K]

Answer:

a. 652.68N

b. -2349.65J

c. -3116.12J

d. 5465.77J

e. Zero

Explanation:

a. According to equilibrium of forces, the force of gravity is equal to the sum of the frictional force and force exerted by the man in the opposite direction (since they're both resistant forces).

Fg = Fm + Fr

Fm = Fg - Fr

Fm = mgsin(28°) - umgcos(28°)

u = coefficient of frictional force.

Fm = 330*9.8*sin28 - 0.4*330*9.8*cos28

Fm = 1518.27 - 865.59

Fm = 652.68N

b. Work done by man is:

Wm = -Fm * d

Wm = -652.68 * 3.6

Wm = -2349.65J

c. Work done by friction force:

W(Fr) = -Fr * d

W(Fr) = -865.59 * 3.6

W(Fr) = -3116.12J

d. Work done by gravity:

Wg = Fg * d

Wg = 1518.27 * 3. 6

Wg = 5465.77J

e. Net work done on the piano is:

Work done by friction + work done by gravity + work done by man

= -3116.12 + 5464.77 + (-2349.65)

= 0J

7 0

3 years ago

An aluminum wire with a diameter of 0.115 mm has a uniform electric field of 0.235 V/m imposed along its entire length. The temp

tekilochka [14]

Answer with Explanation:

We are given that

Diameter of coil=d=0.115mm

Radius, r= $\frac{d}{2}=\frac{0.115}{2}=0.0575mm=0.0575\times 10^{-3} m$

Using $1mm=10^{-3} m$

Electric field=E=0.235V/m

T=55 degree C

$T_0=20^{\circ} C$

$\rho_0=2.82\times 10^{-8}\Omega m$

$\alpha=3.9\times 10^{-3}/C$

(a).We know that

$\rho=\rho_0(1+\alpha(T-T_0))$

Substitute the values

$\rho=2.82\times 10^{-8}(1+3.9\times 10^{-3}(55-20))$

$\rho=3.2\times 10^{-8}\Omega m$

(b).Current density, $J=\frac{E}{\rho}$

Using the formula

$J=\frac{0.235}{3.2\times 10^{-8}}=7.3\times 10^6A/m^2$

c.Total current,I=JA

Where $A=\pi r^2$

$\pi=3.14$

Using the formula

$I=7.3\times 10^6\times 3.14\times (0.0575\times 10^{-3})^2$

I=0.076A

d.Length of wire=l=2m

$V=El$

Substitute the values

$V=0.235\times 2=0.47 V$

5 0

3 years ago

1. How much heat energy ( Q ) is required to heat 2.0 kg of copper from 30.0 oC to 80.0 oC?

djverab [1.8K]

Answer:

Heat capacity, Q = 38500 Joules

Explanation:

Given the following data;

Mass = 2 kg

Initial temperature, T1 = 30°C

Final temperature, T2 = 80°C

Specific heat capacity of copper = 385 J/Kg°C

To find the quantity of heat required;

Mathematically, heat capacity is given by the formula;

$Q = mcdt$

Where;

Q represents the heat capacity or quantity of heat.

M represents the mass of an object.

C represents the specific heat capacity of water.

dt represents the change in temperature.

dt = T2 - T1

dt = 80 - 30

dt = 50°C

Substituting the values into the formula, we have;

$Q = 2 * 385*50$

Heat capacity, Q = 38500 Joules

4 0

3 years ago

The time lag between education of a muscle cell and the beginning of contraction is known as the ______

Alexus [3.1K]

I think it will be muscle contraction. Hope this helps! Mark brainly please!

6 0

3 years ago

What is the scientific law?

storchak [24]

Answer:

Explanation:

A scientific law is the statement which is based on the repeated experimental trials and observations. The scientific law is abide by the scientific community and major research are based on this. The content or information of scientific law cannot be changed.

7 0

3 years ago