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3 years ago

How much heat in joules would you need to raise the temperature of 1 kg of water by 5 °C?

1 answer:

7 0

First we need to know the specific heat capacity of water. You can look it up online because it is uniform. The specific heat capacity of water is 4184 kg^-1 °C^-1.

To calculate the energy required to heat a specific thing up, we can use the formula
E = mc△T
Where E is the total energy required, m as mass, c as specific heat capacity and △T

Substitute the numbers in
E = 1 x 4184 x 5
E = 20920J

Therefore the answer is 20920 joules

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Zoe is shown two mystery boxes that are both 8,000cm3. Her teacher tells her that one mystery box is filled with rocks and the o

krek1111 [17]

Answer:

The box of rocks will have depression which can be seen without touching the box.

Explanation:

The density of rocks is very large as compared with napkins. So, the weight of the rocks will be much more greater than that of napkins.

As both boxes have same volume the heavier box will show depression on the lower surface as compared to the lighter box. So, the box of rocks will have depression which can be seen without touching the box.

5 0

3 years ago

A car is moving at a rate of 72 km/hr .How far does car move when it stop after 4 seconds?

Zanzabum

Answer:

Assuming it starts at 72 kmph and hits a dead stop: Divide 72 by 60 for distance per minute. So, 1.2km per minute. 1.2km is 1200m and 4 seconds is one fifteenth of a minute.

Explanation:

5 0

3 years ago

A baseball is thrown at an angle of 20° relative to the ground at a speed of 25 m/s if the ball was caught 50 m from the thrower

scoray [572]

Answer:

2.1 s

Explanation:

The motion of the ball is a projectile motion. We know that the horizontal range of the ball is

$d = 50 m$

And that the initial speed of the ball is

$u=25 m/s$

at an angle of

$\theta=20^{\circ}$

So, the horizontal speed of the ball (which is constant during the entire motion) is

$u_x = u cos \theta = 25 \cdot cos 20^{\circ} = 23.5 m/s$

And since the horizontal range is 50 m, the time taken for the ball to cover this distance was

$t=\frac{d}{u_x}=\frac{50}{23.5}=2.1 s$

which is the time the ball spent in air.

8 0

2 years ago

Have you ever written a bio-data or an application letter? Share your experience in the

Georgia [21]

Answer:

I found the experience tasking

Explanation:

I wouldn't say it was hard, neither was it easy. I'd rather go for something like it being tasking. It's worthy of note that it was my first time, and I think it's very normal especially when one hasn't been doing something of that nature previously. Of course I did my draft, which unsurprisingly happened to be not good enough, and I had to look for templates to guide me through the acceptable way.

I still did it in my own way, but in the right way. Ever since then though, I have never stuttered when writing application letters, as it had since then seem inborn

8 0

3 years ago

What (in N/C) is the electric field 3.00 m from the center of the terminal of a Van de Graaff with an 8.00 mC charge, noting tha

mixer [17]

To solve this problem we will apply the concept related to the electric field defined from the laws of Coulomb. For this purpose we will remember that the electric field is equivalent to the product of the Coulomb constant due to the change of the charge over the squared distance, mathematically this is

$E = \frac{kq}{r^2}$