All categories

3 years ago

Please help me . i need to answer it asap thankyou

Physics

1 answer:

alex41 [277]3 years ago

8 0

The amount of heat needed to melt the block of lead is $5.4\cdot 10^4 J$

Explanation:

The amount of heat needed to melt a solid substance already at its melting point is given by the equation

$Q=m \lambda_f$

where

m is the mass of the substance

$\lambda_f$ is the specific latent heat of fusion

In this problem, we have a block of lead already at its melting point, $327^{\circ}C$ . Its mass is

m = 200 g = 0.2 kg

and its specific latent heat of fusion is

$\lambda_f = 2.7\cdot 10^5 J/kg$

Therefore, the amount of heat needed for the block to completely melt is

$Q=(0.2)(2.7\cdot 10^5)=5.4\cdot 10^4 J$

Learn more about specific heat:

brainly.com/question/3032746

brainly.com/question/4759369

#LearnwithBrainly

You might be interested in

If a bowler rolls the bowling ball down the alleyway with a force of 1,000N. The ball exerts a reaction force again the alleyway

exis [7]

It’s 1,000 because it stays the same

3 0

3 years ago

Read 2 more answers

What are the cues for a forehand stroke

MaRussiya [10]

Answer:

huh

Explanation:

8 0

3 years ago

Explain how are position time, velocity, and acceleration related to each<br> other?

pogonyaev

I had this question before I think was it A or B.

6 0

3 years ago

The temperature of water in a beaker is 45°C. What does this measurement represent?

Tasya [4]

Answer:

the average kinetic energy of water particles - C.

8 0

3 years ago

In measuring the width of a hair sample, a light of wavelength 500 nm is used. The hair sample is 40 um in radius. With the scre

sergejj [24]

Answer:

The distance of the second dark band away from the central bright spot be located is $5.625\times10^{-2}\ m$

Explanation:

Given that,

Wave length = 500 nm

Radius $d= 40\ \mu m$

Distance from the hair sample D= 6 m

We need to calculate the distance of the second dark band away from the central bright spot be located

$\sin\theta=\dfrac{y}{D}$

$\sin\theta=\dfrac{y}{6}$

Using formula for dark fringe

$(n-\dfrac{1}{2})\lambda=2d\sin\theta$

Put the value into the formula

$(2-\dfrac{1}{2})\times500\times10^{-9}=2\times40\times10^{-6}\times\dfrac{y}{6}$

$y=\dfrac{(2-\dfrac{1}{2})\times500\times10^{-9}\times6}{2\times40\times10^{-6}}$

$y=0.05625\ m$

$y=5.625\times10^{-2}\ m$

Hence, The distance of the second dark band away from the central bright spot be located is $5.625\times10^{-2}\ m$

6 0

3 years ago

Please help me . i need to answer it asap thankyou​

Please help me . i need to answer it asap thankyou