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2 years ago

If you are given force and time, you can determine power if you can know

2 answers:

8 0

The answer would be A, because power is defined as the rate of doing work. Hence power = work / time then you obtain watts. Work is the product of force and displacement (distance). Hence in formula, w = F x s.

Mamont248 [21]2 years ago

4 0

Answer:

joules/energy.

Explanation:

power is defined as the rate of work. Hence power=work/time, then you obtain Watts.

Work is the product of force and displacement.

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The type of friction between the pavement and the tires on a moving vehicle is called.._____friction.

FinnZ [79.3K]

Answer:

A) Kinetic

Explanation:

Kinetic friction is friction caused by motion. Kinetic energy is energy in motion.

5 0

3 years ago

5.00 kg of liquid water is heated to 100.0 °C in a closed system. At this temperature, the density of liquid water is 958 kg/m3.

user100 [1]

Answer:

$1.04\times 10^7\ J.$

Explanation:

In the question given :

Pressure is constant

Therefore, Work done, $W=P\times\Delta V$

Pressure, P=1.01 × 105 Pa.

Final volume, $V_f=8.50\ m^3.$

Initial volume, $V_i=\dfrac{Mass}{density}=\dfrac{5}{958}=5.22\times10^-3\ m^3.$

Therefore, W=8.58\times 10^{5}\ J.

Also, Heat Given, $Q=m\times L=5\times 2.26\times 10^{6}\ J=1.13\times 10^7\ J.$

Also, according to First law of thermodynamics:

$\Delta U=Q-W=(1.13\times 10^7)-(8.58\times 10^5)=1.04\times 10^7\ J.$

Hence, this is the required solution.

8 0

3 years ago

And 8 kg bowling ball is rolling along the frictionless alley

VLD [36.1K]

It will stop eventually

8 0

3 years ago

Why does the current is reduced as electrons move through a conductor

hammer [34]

Because the electrons collide with the particles inside the conductor so are therefore slowed down seen as current is the rate of flow of electrons

3 0

3 years ago

Q = cmAT

castortr0y [4]

Answer: $Q=3000 cal$

Explanation:

We are given the following formula:

$Q=m. c. \Delta T$ (1)

Where:

$Q=3000 cal$ is the amount of heat

$m=300g$ is the mass of water

$c=1 cal/g \°C$ is the specific heat of water

$\Delta T$ is the variation in temperature, which in this case is $\Delta T=30\°C-20\°C=10\°C$

Rewriting equation (1) with the known values at the right side, we will prove the result is $3000 cal$ :

$Q=(300g)(1 cal/g \°C)(10\°C)$ (2)

$Q=3000 cal$ This is the result

8 0

2 years ago