A 70 kg box is pulled by a 400 N force at an angle of 30° to the horizontal. The force of kinetic friction is 75 N.

How much energy is required to raise the temperature of 50.0 grams of water 10.0 degree C? (Explain yourself answer in joules!)

maria [59]

The amount of energy needed is 2093 J

Explanation:

The amount of energy needed to increase the temperature of a substance by $\Delta T$ is given by the equation

$Q=mC\Delta T$

where

m is the mass of the substance

C is its specific heat capacity

$\Delta T$ is the increase in temperature

For the water in this problem, we have

m = 50.0 g = 0.050 kg

$C=4186 J/g^{\circ}C$ (specific heat capacity of water)

$\Delta T=10.0^{\circ}C$

Therefore, the amount of energy needed is

$Q=(0.050)(4186)(10)=2093 J$

Learn more about specific heat capacity:

brainly.com/question/3032746

brainly.com/question/4759369

#LearnwithBrainly

4 0

3 years ago

What is biasing and doping

Flura [38]

Answer:

Biasing - Biasing means being unfair to someone or something, and liking the other one or helping them.

Doping - Doping is where a person use drug to get better at sporting and get stronger muscles, such as getting stronger, getting faster and so on. It is mostly used by athletics.

5 0

3 years ago

Read 2 more answers

PLEASE HELP ME ASAP!!!!!!!!!

NNADVOKAT [17]

Answer:

letter C. velocity hope this helps

7 0

3 years ago

If a marble is released from a height of 10 meters how long would it take to hit the ground?

zlopas [31]

S = ut + 0.5at^2

10 = 0 + 0.5(9.81)t^2 {and if g = 10 then t^2 = 2 so t ~1.414} 

t^2 ~ 2.04 

t ~ 1.43 seconds

3 0

3 years ago

The component of the external magnetic field along the central axis of a 78-turn circular coil of radius 34.0 cm decreases from

grigory [225]

Answer:

Induced current, I = 18.88 A

Explanation:

It is given that,

Number of turns, N = 78

Radius of the circular coil, r = 34 cm = 0.34 m

Magnetic field changes from 2.4 T to 0.4 T in 2 s.

Resistance of the coil, R = 1.5 ohms

We need to find the magnitude of the induced current in the coil. The induced emf is given by :

$\epsilon=-N\dfrac{d\phi}{dt}$

Where

$\dfrac{d\phi}{dt}$ is the rate of change of magnetic flux,

And $\phi=BA$

$\epsilon=-NA\dfrac{dB}{dt}$

$\epsilon=-78\times \pi (0.34)^2\dfrac{(0.4-2.4)}{2}$

$\epsilon=28.32\ V$

Using Ohm's law, $\epsilon=I\times R$

Induced current, $I=\dfrac{\epsilon}{R}$

$I=\dfrac{28.32}{1.5}$

I = 18.88 A

So, the magnitude of the induced current in the coil is 18.88 A. Hence, this is the required solution.

5 0

3 years ago