1.What is true about heat capacity and specific heat?

Nanotechnology _____.

mojhsa [17]

Answer:

B

Explanation:

please mark brainliest

3 0

3 years ago

Read 2 more answers

Highlight the correct terms:

larisa [96]

The (more) speed an object has, the (lower) the potential energy and the (higher) the kinetic energy. (I believe that is correct but it’s been a while since I’ve done this)

8 0

2 years ago

During a storm, a tree limb breaks off and comes to rest across a barbed wire fence at a point that is not in the middle between

Whitepunk [10]

Answer:

a. 12.12°

b. 412.04 N

Explanation:

Along vertical axis, the equation can be written as

T_1 sin14 + T_2sinA = mg

T_2sinA = mg - T_1sin12.5 ....................... (a)

Along horizontal axis, the equation can be written as

T_2×cosA = T_1×cos12.5 ......................... (b)

(a)/(b) given us

Tan A = (mg - T_1sin12.5) / T_1 cos12.5

= (176 - 413sin12.5) / 413×cos12.5

A = 12.12 °

(b) T2 cosA = T1 cos12.5

T2 = 413cos12.5/cos12.12

= 412.04 N

4 0

3 years ago

Read 2 more answers

When the plutonium bomb was tested in New Mexico in 1945, approximately 1 gram of matter was converted into energy. Suppose anot

gayaneshka [121]

Answer:

The value is $E = 1.35 *10^{14} \ J$

Explanation:

From the question we are told that

The mass of matter converted to energy on first test is $m = 1 \ g = 0.001 \ kg$

The mass of matter converted to energy on second test $m_1 = 1.5 \ g = 1.5 *10^{-3} \ kg$

Generally the amount of energy that was released by the explosion is mathematically represented as

$E = m * c^2$

=> $E = 1.5 *10^{-3} * [ 3.0 *10^{8}]^2$

=> $E = 1.35 *10^{14} \ J$

7 0

3 years ago

A straight wire of length 0.62 m carries a conventional current of 0.7 amperes. What is the magnitude of the magnetic field made

anyanavicka [17]

Answer:

Magnetic field at point having a distance of 2 cm from wire is 6.99 x 10⁻⁶ T

Explanation:

Magnetic field due to finite straight wire at a point perpendicular to the wire is given by the relation :

$B=\frac{\mu_{0}I }{2\pi R }\times\frac{L}{\sqrt{L^{2}+R^{2} } }$ ......(1)

Here I is current in the wire, L is the length of the wire, R is the distance of the point from the wire and μ₀ is vacuum permeability constant.

In this problem,

Current, I = 0.7 A

Length of wire, L = 0.62 m

Distance of point from wire, R = 2 cm = 2 x 10⁻² m = 0.02 m

Vacuum permeability, μ₀ = 4π x 10⁻⁷ H/m

Substitute these values in equation (1).

$B=\frac{4\pi\times10^{-7}\times 0.7 }{2\pi \times0.02 }\times\frac{0.62}{\sqrt{(0.62)^{2}+(0.02) ^{2} } }$

B = 6.99 x 10⁻⁶ T

3 0

2 years ago