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2 years ago

What is the net force on a car with a mass of 1000 kg if its acceleration is 35 m/s^2?

Physics

1 answer:

VashaNatasha [74]2 years ago

7 0

Answer:

3000N

Explanation:

divided to get answer

the force needed to accelerate the 1000kg car by 3m/s2 is 3000N

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Boyle's law balloon was filled to a volume of 2.50 l when the temperature was 30.0∘

aliina [53]

To solve this we assume that the gas inside the balloon is an ideal gas. Then, we can use the ideal gas equation which is expressed as PV = nRT. At a constant pressure and number of moles of the gas the ratio T/V is equal to some constant. At another set of condition of temperature, the constant is still the same. Calculations are as follows:

T1 / V1 = T2 / V2

V2 = T2 x V1 / T1

V2 =284.15 x 2.50 / 303.15

V2 = 2.34 L

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3 years ago

Particles in a ______ have a small amount of energy, vibrate, and are stuck in place?

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Particles in a SOLID have a small amount of energy, vibrate, and are stuck in place?

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3 years ago

A bicycle is traveling at 10 m/s on a flat road. As it reaches a hill, it starts accelerating,

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Answer: why did my brainly points get negative

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3 years ago

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An inductor with an inductance of 2.30H and a resistance of 8.00 Ω isconnected to the terminals of a battery with an emf of 6.00

Basile [38]

Answer:

(a). The initial rate is 2.60 A/s.

(b). The rate of current increases is 0.8658 A/s.

(c). The current is 0.435 A.

(d). The final steady-state current is 0.75 A.

Explanation:

Given that,

Inductance = 2.30 H

Resistance = 8.00 Ω

Voltage = 6.00 V

(a). We need to calculate the initial rate of increase of current in the circuit.

Using formula of initial rate

$V=initial\ rate\times inductance$

$initial\ rate=\dfrac{V}{L}$

Put the value into the formula

$initial \rate=\dfrac{6.00}{2.30}$

$initial\ rate=2.60\ A/s$

The initial rate is 2.60 A/s.

(b). We need to calculate the rate of increase of current at the instant when the current is 0.500

Using formula of rate of increase of current

$rate\ of \ current\ increase=initial\ rate\times e^{\dfrac{-t}{T}}$ ....(I)

Where, $T=\dfrac{L}{R}$

$T=\dfrac{2.30}{8.00}$

$T=0.2875$

Using formula of current

$i=\dfrac{V}{R}(1-e^{\dfrac{-t}{T}})$

$e^{\dfrac{-t}{T}}=1-(i\times\dfrac{R}{V})$

$e^{\dfrac{-t}{T}}=1-(0.500\times\dfrac{8.00}{6.00})$

$e^{\dfrac{-t}{T}}=0.333$

The rate of current increases is

Put the value in the equation (I)

$rate\ of\ current\ increase=2.60\times0.333$

$rate\ of\ current\ increase=0.8658\ A/s$

The rate of current increases is 0.8658 A/s.

(c). We need to calculate the current

Using formula of current

$i=\dfrac{V}{R}(1-e^{\dfrac{-t}{T}})$

Put the value into the formula

$i=\dfrac{6.00}{8.00}\times(1-e^{\dfrac{-0.250}{0.2875}})$

$i=0.435\ A$

The current is 0.435 A.

(d). We need to calculate the final steady-state current

Using formula of steady state

$i=\dfrac{V}{R}$

$i=\dfrac{6.00}{8.00}$

$i=0.75\ A$

The final steady-state current is 0.75 A.

Hence, This is the required solution.

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3 years ago

How much heat energy is required to raise the temperature of 20 kilograms of water from 0°c to 35°c?

saveliy_v [14]

Energy required = Specific Heat × mass × change in temp.
Specific Heat (Cp) water = 4.186kJ/kgC 
Mass = 20kg 
Change in temperature = 35C - 0C = 35C 

Q = 4.186kJ/kgC × 20kg × 35C = 293.02kJ hope this helps! (:

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3 years ago

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