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4 years ago

Which answer best describes thermal energy?

Physics

1 answer:

timofeeve [1]4 years ago

8 0

Thermal energy is the energy produces from the movement of atoms in the substance.

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A sled is on an icy (frictionless) slope that is 30° above the horizontal. When a 40-N force, parallel to the incline and direct

son4ous [18]

Answer: 5.8kg

Explanation:

F - mg sinΦ = ma

Given that

M=?

g= 9.8

Φ= 30

a= 2

F= 40 then

40 - 9.8 * m * sin 30 = 2 * m

40 = 2 * m + 9.8 * m * sin 30

40 = (2 + 9.8 * sin 30) m

m = 40 / (2 + 9.8 * 0.5)

m = 40 / (2 + 4.9)

m = 40 / 6.9

m = 5.797kg

m = 5.8kg

4 0

3 years ago

A coil of 160 turns and area 0.20 m2 is placed with its axis parallel to a magnetic field of initial magnitude 0.40 T. The magne

ser-zykov [4K]

Answer:

The rate at which power is generated in the coil is 10.24 Watts

Explanation:

Given;

number of turns of the coil, N = 160

area of the coil, A = 0.2 m²

magnitude of the magnetic field, B = 0.4 T

time for field change = 2 s

resistance of the coil, R = 16 Ω

The induced emf in the coil is calculated as;

emf = dΦ/dt

where;

Φ is magnetic flux = BA

emf = N (BA/dt)

emf = 160 (0.4T x 0.2 m²)/dt

emf = 12.8 V/s

The rate power is generated in the coil is calculated as;

P = V²/ R

P = (12.8²) / 16

P = 10.24 Watts

Therefore, the rate at which power is generated in the coil is 10.24 Watts

8 0

3 years ago

Consider the following cyclic process carried out in two steps on a gas. Step 1: 44 J of heat is added to the gas, and 20. J of

notka56 [123]

Answer:37 J

Explanation:

Given

Step :1

Heat added Q=44 J

Work done=-20 J

$\Delta E_1=Q+W=44-20=24 J$

Step :2

Heat added Q=-61 J

work done $W_2$

$\Delta E_2=Q+W_2$

$\Delta E_2=61+W_2$

$\Delta E_1+\Delta E_2=0$

as the process is cyclic

$44-20-61+W_2=0$

$W_2=37 J$

work done in compression is 37 J

3 0

3 years ago

A 120 kg box is on the verge of slipping down an inclined plane with an angle of inclination of 47º. What is the coefficient of

Alex_Xolod [135]

Given :

A 120 kg box is on the verge of slipping down an inclined plane with an angle of inclination of 47º.

To Find :

The coefficient of static friction between the box and the plane.

Solution :

Vertical component of force :

$mg\ sin\ \theta = 120\times 10 \times sin\ 47^\circ{}=877.62 \ N$

Horizontal component of force(Normal reaction) :

$mg\ cos\ \theta = 120\times 10 \times cos\ 47^\circ{}=818.40 \ N$

Since, box is on the verge of slipping :

$mg\ sin\ \theta= \mu(mg \ cos\ \theta)\\\\\mu = tan \ \theta\\\\\mu = tan\ 47^o\\\\\mu = 1.07$

Therefore, the coefficient of static friction between the box and the plane is 1.07.

Hence, this is the required solution.

7 0

3 years ago

A planet orbits a star, in a year of length 3.37 x 107 s, in a nearly circular orbit of radius 1.04 x 1011 m. With respect to th

PIT_PIT [208]

Answer

Given,

Time period of star,T = 3.37 x 10⁷ s

Radius of circular orbit,R = 1.04 x 10¹¹ m

a) Angular speed of the planet

$\omega = \dfrac{2\pi}{T}=\dfrac{2\pi}{3.37\times 10^{7}}$

$\omega = 1.864\times 10^{-7}\ rad/s$

b) tangential speed

$v = r \omega = 1.04\times 10^{11}\times 1.864 \times 10^{-7}$

v = 1.94 x 10⁴ m/s

c) centripetal acceleration magnitude

$a = \dfrac{v^2}{r}= \dfrac{(1.94\times 10^4)^2}{1.04\times 10^{11}}$

a = 3.62 x 10⁻³ m/s²

8 0

3 years ago