How is heat transferred through radiation?

Physics

1 answer:

g100num [7]4 years ago

5 0

A. Through the direct contact of particles

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PLEASE HELP!!!!

Goryan [66]

Answer:

The braking distance would be about nine times as long (assuming that acceleration during braking stays the same.)

Explanation:

Let $u$ denote the initial velocity of the vehicle ( $20\; \text{mph}$ or $60\; \text{mph}$ ) and let $v$ denote the velocity of the vehicle after braking ( $0\; \text{mph}$ ). Let $x$ denote the braking distance.

Assume that the acceleration during braking are both constantly $a$ in both scenarios. The SUVAT equations would apply. In particular:

$\begin{aligned} x &= \frac{v^{2} - u^{2}}{2\, a}\end{aligned}$ .

Since $v = 0$ <em> </em>(the vehicle has completely stopped), the equation becomes $x = (-u^{2}) / (2\, a)$ .

Assuming that $a$ (braking acceleration) stays the same, the braking distance $x$ would be proportional to $u^{2}$ , the square of the initial velocity.

Hence, increasing the initial speed from $20\; \text{mph}$ to $60\; \text{mph}$ would increase the braking distance by a factor of $3^{2} = 9$ .

7 0

1 year ago

Read 2 more answers

An engineer weighs a sample of mercury (ρ = 13.6 × 103 kg/m3 ) and finds that the weight of the sample is 6.0 n. what is the sam

Amanda [17]

Given:
ρ = 13.6 x 10³ kg/m³, density of mercury
W = 6.0 N, weight of the mercury sample
g = 9.81 m/s², acceleration due to gravity.

Let V = the volume of the sample.
Then
W = ρVg
or
V = W/(ρg)
= (6.0 N)/[(13.6 x 10³ kg/m³)*(9.81 m/s²)]
= 4.4972 x 10⁻⁵ m³

Answer: The volume is 44.972 x 10⁻⁶ m³

5 0

3 years ago

an unknown charge exerts an attractive force of 4.23 E 13 N. If these charges are separated by 8cm, what is the magnitude of the

mash [69]

Answer:

Since there is attraction force between two charges so the other charge must be - 3C

Explanation:

As we know that the force between two charges is given by formula

$F = \frac{kq_1q_2}{r^2}$