All categories

3 years ago

How much heat is required to heat the temperature of 338g of aluminum by?

1 answer:

3 0

Specific heat is another physical property of matter. ... we con now ask the following question: by how much will the temperature of an .

You might be interested in

As an object moves, the distance it travels increases with time. Agree Disagree

Mnenie [13.5K]

Agree

Hope this helps

4 0

3 years ago

Read 2 more answers

How much momentum will a dumb-bell of mass 10 kg transfer

frosja888 [35]

We want to find how much momentum the dumbbell has at the moment it strikes the floor. Let's use this kinematics equation:

Vf² = Vi² + 2ad

Vf is the final velocity of the dumbbell, Vi is its initial velocity, a is its acceleration, and d is the height of its fall.

Given values:

Vi = 0m/s (dumbbell starts falling from rest)

a = 10m/s² (we'll treat downward motion as positive, this doesn't affect the result as long as we keep this in mind)

d = 80×10⁻²m

Plug in the values and solve for Vf:

Vf² = 2(10)(80×10⁻²)

Vf = ±4m/s

Reject the negative root.

Vf = 4m/s

The momentum of the dumbbell is given by:

p = mv

p is its momentum, m is its mass, and v is its velocity.

Given values:

m = 10kg

v = 4m/s (from previous calculation)

Plug in the values and solve for p:

p = 10(4)

p = 40kg×m/s

6 0

3 years ago

A lamp is labelled '230 V, 100 W'. How many joules of electrical energy is changed to thermal energy and light if the lamp is sw

viva [34]

Given;

V = 230V

Power, P = 100W

time, t = 2hrs = 7200s

from,

P = IV

and Energy, E= Pt

E = 100*7200

E = 720000 Joules

E = 720KJ

4 0

2 years ago

Read 2 more answers

Look at the picture above and scientifically explain how eating a hamburger can cause energy transformation in three ways within

sergejj [24]

Answer:

CLAIM: When you eat the hamburger, your body uses it and breaks it down which is chemical energy than use that food for energy for mechanical energy which makes thermal energy

EVIDENCE: This happens every day to people across the world and also it is scientifically proven

REASONING: The reason why this evidence supports it is because as I said it happens everyday and when you feel energized after and apple after you were lazy that just happened

Explanation:

7 0

3 years ago

In a historical movie, two knights on horseback start from rest 86 m apart and ride directly toward each other to do battle. Sir

Harlamova29_29 [7]

Answer:

Relative to Sir George's starting point, the knights collide at a distance of 38.43 m from Sir George's starting point.

Explanation:

Let the distance covered by Sir George be $S_{1}$

and the distance covered by Sir Alfred be $S_{2}$

Since the knights collide, hence they must have traveled for the same amount of time just before collision

From one of the equations of motion for linear motion

$S = ut + \frac{1}{2}at^{2}$

Where $S$ is the distance traveled

$u$ is the initial velocity

$a$ is the acceleration

and $t$ is the time

For Sir George,

$S = S_{1}$

$u$ = 0 m/s (Since they start from rest)

$a =$ 0.21 m/s²

Hence,

$S = ut + \frac{1}{2}at^{2}$ becomes

$S_{1} = (0)t + \frac{1}{2}(0.21)t^{2}\\S_{1} = 0.105 t^{2}\\$

$t^{2} = \frac{S_{1}}{0.105}$

Now, for Sir Alfred

$S = S_{2}$

$u$ = 0 m/s (Since they start from rest)

$a =$ 0.26 m/s²

Hence,

$S = ut + \frac{1}{2}at^{2}$ becomes

$S_{2} = (0)t + \frac{1}{2}(0.26)t^{2}\\S_{2} = 0.13 t^{2}\\$

$t^{2} = \frac{S_{2}}{0.13}$

Since, they traveled for the same time, $t$ just before collision, we can write

$\frac{S_{1}}{0.105}= \frac{S_{2}}{0.13}$

Since, the two nights are 86 m apart, that is, the sum of the distances covered by the knights just before collision is 86 m. Then we can write that

$S_{1} + S_{2} = 86$ m

Then, $S_{2} = 86 - S_{1}$

Then,

$\frac{S_{1}}{0.105}= \frac{S_{2}}{0.13}$ becomes

$\frac{S_{1}}{0.105}= \frac{86 -S_{1}}{0.13}$

$0.13{S_{1}}= 0.105({86 -S_{1}})\\0.13{S_{1}}= 9.03 - 0.105S_{1}}\\0.13{S_{1}} + 0.105S_{1}}= 9.03 \\0.235{S_{1}} = 9.03\\{S_{1}} =\frac{9.03}{0.235}$

$S_{1} = 38.43$ m

∴ Sir George covered a distance of 38.43 m just before collision.

Hence, relative to Sir George's starting point, the knights collide at a distance of 38.43 m from Sir George's starting point.

6 0

3 years ago