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3 years ago

How much energy is needed to heat 55 Kg of water from 285 K to 360 K?

Physics

1 answer:

GalinKa [24]3 years ago

8 0

Answer:

17 kJ

Explanation:

q = mCΔT

q = (55 kg) (4.184 kJ/kg/K) (360 K − 285 K)

q = 17,000 J

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What are the 2 factors that increase the electric force between objects?

harkovskaia [24]

The two factors that affect the electric force are the charges and the separation between the objects

Explanation:

The magnitude of the electrostatic force between two objects is given by Coulomb's law:

$F=k\frac{q_1 q_2}{r^2}$

where:

$k=8.99\cdot 10^9 Nm^{-2}C^{-2}$ is the Coulomb's constant

$q_1, q_2$ are the charges of the two objects

r is the separation between the two objects

Looking at the equation, we see that the electric force between objects depends on two factors:

The charges of the objects: the force is directly proportional to the product of the two charges, therefore the greater the charges, the stronger the force
The distance between the two objects: the force is inversely proportional to the square of the distance, therefore the smaller the distance between the objects, the stronger the force

Learn more about electric force:

brainly.com/question/8960054

brainly.com/question/4273177

#LearnwithBrainly

4 0

3 years ago

A car is traveling 50 mph. It leaves Boston, Massachusetts at 5 a.m. and is

aleksandrvk [35]

7 A.M
I’m pretty sure that’s it

6 0

3 years ago

NEED THE ANSWER PLZ What are lanthanides?

geniusboy [140]

It is the the ecrins of light

5 0

3 years ago

Sawyer launches his 180 kg raft on the Mississippi River by pushing on it with a force of 75N. How long must Sawyer push on the

Daniel [21]

Answer: 4.8 s

Explanation:

We have the following data:

$m=180 kg$ the mass of the raft

$F=75 N$ the force applied by Sawyer

$V=2 m/s$ the raft's final speed

$V_{o}=0 m/s$ the raft's initial speed (assuming it starts from rest)

We have to find the time $t$

Well, according to Newton's second law of motion we have:

$F=m.a$ (1)

Where $a$ is the acceleration, which can be expressed as:

$a=\frac{\Delta V}{\Delta t}=\frac{V-V_{o}}{t-t_{o}}$ (2)

Substituting (2) in (1):

$F=m\frac{V-V_{o}}{t-t_{o}}$ (3)

Where $t_{o}=0$

Isolating $t$ from (3):

$t=\frac{m(V-V_{o})}{F}$ (4)

$t=\frac{180 kg(2 m/s-0 m/s)}{75 N}$

Finally:

$t=4.8 s$

6 0

3 years ago

At the instant the traffic light turns green, an automobile starts with a constant acceleration a of 2.5 m/s2. At the same insta

Licemer1 [7]

Answer:

6.96 s

Explanation:

Given:

u = initial speed of the automobile = 0 m/s

a = constant acceleration of the automobile = $2.5\ m/s^2$
v = constant speed of the truck = 8.7 m/s

Assume:

t = time instant at which the automobile overtakes the truck.

At the moment the automobile and the truck both meat each other the distance travel by both vehicles must be the same.

$\therefore \textrm{Distance traveled by the automobile }=\textrm{Distance traveled by the truck}\\\Rightarrow ut+\dfrac{1}{2}at^2=vt\\\Rightarrow (0)t+\dfrac{1}{2}\times 2.5\times t^2=8.7t\\\Rightarrow 1.25t^2=8.7t\\\Rightarrow 1.25t^2-8.7t=0\\\Rightarrow t(1.25t-8.7)=0\\\Rightarrow t = 0\,\,\,or\,\,\, t = \dfrac{8.7}{1.25}\\\Rightarrow t = 0\,\,\,or\,\,\, t = 6.96\\$

Since t = 0 s is the initial condition. So, they both meet again at t = 6.96 s such that the automobile overtakes the truck.

6 0

3 years ago