All categories

3 years ago

State Newton’s third law of motion.

1 answer:

7 0

Answer:every force has an equal and oppisite reaction. meeaning if you punch a wall wwith 50lbs of force the wall pushes back on your hand with 50 lbs of force

Explanation:

You might be interested in

A force in the +x-direction with magnitude ????(x) = 18.0 N − (0.530 N/m)x is applied to a 6.00 kg box that is sitting on the ho

fiasKO [112]

Answer:

$v_f=8.17\frac{m}{s}$

Explanation:

First, we calculate the work done by this force after the box traveled 14 m, which is given by:

$W=\int\limits^{x_f}_{x_0} {F(x)} \, dx \\W=\int\limits^{14}_{0} ({18N-0.530\frac{N}{m}x}) \, dx\\W=[(18N)x-(0.530\frac{N}{m})\frac{x^2}{2}]^{14}_{0}\\W=(18N)14m-(0.530\frac{N}{m})\frac{(14m)^2}{2}-(18N)0+(0.530\frac{N}{m})\frac{0^2}{2}\\W=252N\cdot m-52N\cdot m\\W=200N\cdot m$

Since we have a frictionless surface, according to the the work–energy principle, the work done by all forces acting on a particle equals the change in the kinetic energy of the particle, that is:

$W=\Delta K\\W=K_f-K_i\\W=\frac{mv_f^2}{2}-\frac{mv_i^2}{2}$

The box is initially at rest, so $v_i=0$ . Solving for $v_f$ :

$v_f=\sqrt{\frac{2W}{m}}\\v_f=\sqrt{\frac{2(200N\cdot m)}{6kg}}\\v_f=\sqrt{66.67\frac{m^2}{s^2}}\\v_f=8.17\frac{m}{s}$

5 0

2 years ago

A test tube of length L and cross-sectional area A is submerged in water with the open end down so that the edge of the tube is

Margaret [11]

Answer:

if we measure the change in height of the gas within the had and obtain a straight line in relation to the depth we can conclude that the air complies with Boye's law.

Explanation:

The air in the tube can be considered an ideal gas,

P V = nR T

In that case we have the tube in the air where the pressure is P1 = P_atm, then we introduce the tube to the water to a depth H

For pressure the open end of the tube is

P₂ = P_atm + ρ g H

Let's write the gas equation for the colon

P₁ V₁ = P₂ V₂

P_atm V₁ = (P_atm + ρ g H) V₂

V₂ = V₁ P_atm / (P_atm + ρ g h)

If the air obeys Boyle's law e; volume within the had must decrease due to the increase in pressure, if we measure the change in height of the gas within the had and obtain a straight line in relation to the depth we can conclude that the air complies with Boye's law.

The main assumption is that the temperature during the experiment does not change

6 0

3 years ago

A heat engine with 0.300 mol of a monatomic ideal gas initially fills a 1000 cm3 cylinder at 500 K . The gas goes through the fo

LuckyWell [14K]

Complete Question:

A heat engine with 0.300 mol of a monatomic ideal gas initially fills a 1000 cm3 cylinder at 500 K . The gas goes through the following closed cycle: - Isothermal expansion to 5000 cm3. - Isochoric cooling to 400 K . - Isothermal compression to 1000 cm3. - Isochoric heating to 500 K .

a) what is the work for one cycle

b) what is the thermal efficiency

Answer:

a) Work done for 1 cycle = 402.13

b) Thermal efficiency = 20%

Explanation:

Number of moles, n = 0.300 mol

Initial Volume, V₁ = 1000 cm³

Temperature, T = 500 K

Isothermal expansion to 5000 cm³

Final volume, V₂ = 5000 cm³

R = 8.314 J/ mol.K

Work done, W = nRT ln(V₂/V₁)

W = (0.3 * 8.314 * 500) * ln(5000/1000)

W = 1247.1 * ln5

W₁ = 2007.13 J

Isochoric cooling

In an Isochoric process, volume is constant i.e. V₂ = V₁ = V

W = nRT ln(V/V)

But ln(V/V) = ln 1 = 0

Work done, W₂ = 0 Joules

Isothermal Compression to 1000 cm³

V₂ = 1000 cm³

V₁ = 5000 cm³

W = nRT ln(V₂/V₁)

W = 0.3 * 8.314 * 400 ln(1000/5000)

W₃ = -1605 J

Isochoric heating to 500 K

Since there is no change in volume, no work is done

W₄ = 0 J

a) Work done for 1 cycle

W = W₁ + W₂ + W₃ + W₄

W = 2007.13 + 0 + 0 -1605+0

W = 402.13 Joules

b) Thermal efficiency

Thermal efficiency = (Net workdone for 1 cycle)/(Heat absorbed)

Heat absorbed = Work done due to thermal expansion = 2007.13 J

Thermal efficiency = 402.13/2007.13

Thermal efficiency = 0.2

Thermal efficiency = 0.2 * 100% = 20 %

3 0

3 years ago

A single-turn circular loop of radius 6 cm is to produce a field at its center that will just cancel the earth's field of magnit

djverab [1.8K]

Answer:

The current is $I = 6.68 \ A$

Explanation:

From the question we are told that

The radius of the loop is $r = 6 \ cm = 0.06 \ m$

The earth's magnetic field is $B_e = 0.7G= 0.7 G * \frac{1*10^{-4} T}{1 G} = 0.7 *10^{-4} T$

The number of turns is $N =1$

Generally the magnetic field generated by the current in the loop is mathematically represented as

$B = \frac{\mu_o * N * I}{2 r }$

Now for the earth's magnetic field to be canceled out the magnetic field generated by the loop must be equal to the magnetic field out the earth

$B = B_e$

=> $B_e = \frac{\mu_o * N * I }{ 2 * r}$

Where $\mu$ is the permeability of free space with value $\mu _o = 4\pi * 10^{-7} N/A^2$

$0.7 *10^{-4}= \frac{ 4\pi * 10^{-7} * 1 * I}{2 * 0.06}$

=> $I = \frac{2 * 0.06 * 0.7 *10^{-4}}{ 4\pi * 10^{-7} * 1}$

$I = 6.68 \ A$

3 0

3 years ago

HELP QUICK!!!WILL GIVE BRAINLIEST!! According to the Sternberg's triarchic theory of intelligence, information synthesis is a st

maxonik [38]

Formulated in the 1980s by Robert Sternberg, the triarchic theory distinguishes three aspects of intelligence: analytic skills, such as the ability to think abstractly and evaluate information; creativity, the ability to invent novel solutions or ideas; and practical skills, which enable one to cope with concrete ...

4 0

3 years ago

Read 2 more answers