Ask question

Ask question

All categories

3 years ago

9

If the latent heat of fusion of ice is 334kj/kg and the specific heat of water is 4.18kj kg-1 k-1 how much heat, in joules, must

6.5g of ice absorb to reach the final temperature in part a?

1 answer:

lord [1]3 years ago

4 0

The answer is 500 because it is

You might be interested in

How is item A different from Item B?

iris [78.8K]

Explanation:

well there is nothing there and it could be different by diffrent objects, idk

8 0

2 years ago

Read 2 more answers

A vehicle starts to move from the rest gets an acceleration of 5 m/s2 within 2

likoan [24]

Answer:

When an object moves in a straight line with a constant acceleration, you can calculate its acceleration if you know how much its velocity changes and how long this takes.

The formula is,

Acceleration = change in velocity / time taken

The equation for acceleration can also be represented as:

a = (v-u) \ t

The change in velocity v – u = 5 – 0 = 5 m/s.

The acceleration = change in velocity ÷ time = 5 m/s ÷ 2 s = 2.5 m/s^2

3 0

2 years ago

Read 2 more answers

In deep space (no gravity), the bolt (arrow)

salantis [7]

t = 0.527 s

<u>It accelerates for 0.527 s.</u>

<u>Explanation:</u>

We use the formula:

v = u+at

Given:

v = 106 m/s

u = 0 (since no gravity)

$a=201 \mathrm{m} / \mathrm{s}^{2}$

So applying the formula,

v = u+at

106 = 0 + 201t

t = 106/201

t = 0.527 s

4 0

3 years ago

An airplane is flying through a thundercloud at a height of 2000 m (This is a very dangerous thing to do because of updrafts, tu

Doss [256]

Answer:

$400000\ \text{N/C}$

Explanation:

$q_1$ = Charge at 3000 m = 40 C

$q_2$ = Charge at 1000 m = -40 C

$r_1$ = 3000 m

$r_2$ = 1000 m

k = Coulomb constant = $9\times10^9\ \text{Nm}^2/\text{C}^2$

Electric field due to the charge at 3000 m

$E_1=\dfrac{k|q_1|}{r_1^2}\\\Rightarrow E_1=\dfrac{9\times 10^9\times 40}{3000^2}\\\Rightarrow E_1=40000\ \text{N/C}$

Electric field due to the charge at 1000 m

$E_2=\dfrac{k|q_2|}{r_2^2}\\\Rightarrow E_2=\dfrac{9\times 10^9\times 40}{1000^2}\\\Rightarrow E_2=360000\ \text{N/C}$

Electric field at the aircraft is $E_1+E_2=40000+360000=400000\ \text{N/C}$ .

7 0

3 years ago

Stephen throws a 0.46 kg football at a speed of 14 m/s and it goes through the goal posts 2.4 m above where he threw it. What is

Dovator [93]

Answer:
$\\ \\ 0.46 \times 14 {2 }^{2} = 90.16 \\$

5 0

2 years ago