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PtichkaEL [24]
3 years ago
12

1. How much heat must be absorbed by 375 grams of water to raise its

Physics
1 answer:
antiseptic1488 [7]3 years ago
5 0

Answer:

39225J

Explanation:

Given parameters:

Mass of water  = 375grams of water

Change in temperature  = 25°C

Specific heat capacity of water  = 4.184J/g°C

Unknown:

Amount of heat absorbed  = ?

Solution:

To solve this problem, we use the expression below:

      H  = m c Ф  

H is the heat absorbed

m is the mass

c is the specific heat capacity

Ф is the change in temperature

  Insert the parameters and solve;

      H  = 375 x 4.184 x (25) = 39225J

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What is the force on a 1000kg elevator that is falling freely at 9.8 m/sec^2
Advocard [28]
From Newton's second law, we know F = ma, where a is the acceleration and m is the mass in kg.

F = 1000kg * 9.8m/s = 9800N

F = 9800 N

Hope this helps!
3 0
3 years ago
If your car runs out of gas and you must push it 250 m to the nearest gas station. How much work is done on the car if it is pus
tankabanditka [31]

Answer:

1750 Joules.

Explanation:

Work done = force * distance

= 7 * 250

= 1750 Joules.

6 0
3 years ago
You do 25 J of work to move a 4N object 5 meters. Find your efficiency and write down the formula.
daser333 [38]
4*5=20;  ration of 20:25, 
20/25= %80 
3 0
3 years ago
A system of inertia 0.47 kg consists of a spring gun attached to a cart and a projectile. The system is at rest on a horizontal
Vlad [161]

Answer:

v'=0.83m/a and v=10.2m/s

Explanation:

The information that we have is:

m_1=0.050kg\\m_2=0.47kg\\\theta = 40\\h=2.2m

The maximum height of the projectile is given by the equation

h=\frac{v^2sin^2\theta}{2g}

So, rearrange for the velocity,

v=\sqrt{\frac{2gh}{sin^2\theta}}\\v=\sqrt{\frac{2*9.8*2.2}{sin^2(40)}}\\v=10.2m/s

Apply the conservation of momentum,

mvcos(\theta)=m'v'

Then rearrange the recoil speed,

v'=\frac{mvcos\theta}{m_2}\\v'=\frac{0.05*10.2*cos40}{0.47}\\v'=0.83m/s\\

4 0
3 years ago
A horizontal insulating rod of length 11.8-cm and charge 19 nC is in a plane with a long straight vertical uniform line charge.
SCORPION-xisa [38]

Answer:

11.962337 × 10^-4 N

Explanation:

Given the following :

Length L = 11.8

Charge = 29nC = 29 × 10^-9 C

Linear charge density λ = 1.4 × 10^-7 C/m

Radius (r) = 2cm = 2/100 = 0.02 m

Using the relation:

E = 2kλ/r ; F =qE

F = 2kλq/L × ∫dr/r

F = 2*k*q*λ/L × (In(0.02 + L) - In(0.02))

2*k*q*λ/L = [2 × (9 * 10^9) * (29 * 10^9) * (1.4 * 10^-7)]/ 0.118] = 6193.2203 × 10^(9 - 9 - 7) = 6193.2203 × 10^-7 = 6.1932203 × 10^-4

In(0.02 + 0.118) - In(0.02) = In(0.138) - In(0.02) = 1.9315214

Hence,

(6.1932203 × 10^-4) × 1.9315214 = 11.962337 × 10^-4 N

3 0
3 years ago
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