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Vsevolod [243]
3 years ago
6

A 300-n force acts on a 25-kg object. the acceleration of the object is?

Physics
2 answers:
Over [174]3 years ago
8 0
To solve the answer use the equation: a = fnet / m

a = 300 N / 25 kg

300 N / 25 kg = 12m/s

The acceleration of the object is 12m/s

Evgesh-ka [11]3 years ago
4 0
This can be obtained by following the Newton's Second Law of motion. It states that the acceleration of an object which is produced by a net force is directly related to the net force and inversely related to the mass of the object. We use the equation:

a = F / m
a = 300 / 25
a = 12 m/s²
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How much force is needed to accelerate a 1,100 kg car at a rate of 1.5 m/s2?
Anna [14]
Assuming there is no force of friction...

F = ma
F = (1300kg)(1.5m/s^2)
F = 1950N
Just multiply mass by acceleration.
1300 x 1.5 = 1950N.
7 0
3 years ago
Find the quantity of heat needed
krok68 [10]

Answer:

Approximately 3.99\times 10^{4}\; \rm J (assuming that the melting point of ice is 0\; \rm ^\circ C.)

Explanation:

Convert the unit of mass to kilograms, so as to match the unit of the specific heat capacity of ice and of water.

\begin{aligned}m&= 100\; \rm g \times \frac{1\; \rm kg}{1000\; \rm g} \\ &= 0.100\; \rm kg\end{aligned}

The energy required comes in three parts:

  • Energy required to raise the temperature of that 0.100\; \rm kg of ice from (-10\; \rm ^\circ C) to 0\; \rm ^\circ C (the melting point of ice.)
  • Energy required to turn 0.100\; \rm kg of ice into water while temperature stayed constant.
  • Energy required to raise the temperature of that newly-formed 0.100\; \rm kg of water from 0\; \rm ^\circ C to 10\;\ rm ^\circ C.

The following equation gives the amount of energy Q required to raise the temperature of a sample of mass m and specific heat capacity c by \Delta T:

Q = c \cdot m \cdot \Delta T,

where

  • c is the specific heat capacity of the material,
  • m is the mass of the sample, and
  • \Delta T is the change in the temperature of this sample.

For the first part of energy input, c(\text{ice}) = 2100\; \rm J \cdot kg \cdot K^{-1} whereas m = 0.100\; \rm kg. Calculate the change in the temperature:

\begin{aligned}\Delta T &= T(\text{final}) - T(\text{initial}) \\ &= (0\; \rm ^\circ C) - (-10\; \rm ^\circ C) \\ &= 10\; \rm K\end{aligned}.

Calculate the energy required to achieve that temperature change:

\begin{aligned}Q_1 &= c(\text{ice}) \cdot m(\text{ice}) \cdot \Delta T\\ &= 2100\; \rm J \cdot kg \cdot K^{-1} \\ &\quad\quad \times 0.100\; \rm kg \times 10\; \rm K\\ &= 2.10\times 10^{3}\; \rm J\end{aligned}.

Similarly, for the third part of energy input, c(\text{water}) = 4200\; \rm J \cdot kg \cdot K^{-1} whereas m = 0.100\; \rm kg. Calculate the change in the temperature:

\begin{aligned}\Delta T &= T(\text{final}) - T(\text{initial}) \\ &= (10\; \rm ^\circ C) - (0\; \rm ^\circ C) \\ &= 10\; \rm K\end{aligned}.

Calculate the energy required to achieve that temperature change:

\begin{aligned}Q_3&= c(\text{water}) \cdot m(\text{water}) \cdot \Delta T\\ &= 4200\; \rm J \cdot kg \cdot K^{-1} \\ &\quad\quad \times 0.100\; \rm kg \times 10\; \rm K\\ &= 4.20\times 10^{3}\; \rm J\end{aligned}.

The second part of energy input requires a different equation. The energy Q required to melt a sample of mass m and latent heat of fusion L_\text{f} is:

Q = m \cdot L_\text{f}.

Apply this equation to find the size of the second part of energy input:

\begin{aligned}Q_2&= m \cdot L_\text{f}\\&= 0.100\; \rm kg \times 3.36\times 10^{5}\; \rm J\cdot kg^{-1} \\ &= 3.36\times 10^{4}\; \rm J\end{aligned}.

Find the sum of these three parts of energy:

\begin{aligned}Q &= Q_1 + Q_2 + Q_3 = 3.99\times 10^{4}\; \rm J\end{aligned}.

3 0
3 years ago
What is the answer for this question (4x³)²​
Delvig [45]
16x to the power of 6
6 0
3 years ago
Read 2 more answers
For the PE formula, why is the height required for calculations? Why do we need to know the height in order to determine PE? *
Fudgin [204]

Answer:

Answer in Explanation

Explanation:

Whenever we talk about the gravitational potential energy, it means the energy stored in a body due to its position in the gravitational field. Now, we know that in the gravitational field the work is only done when the body moves vertically. If the body moves horizontally on the same surface in the Earth's Gravitational Field, then the work done on the body is considered to be zero. Hence, the work done or the energy stored in the object while in the gravitational field is only possible if it moves vertically. This vertical distance is referred to as height. <u>This is the main reason why we require height in the P.E formula and calculations.</u>

The derivation of this formula is as follows:

Work = Force * Displacement

For gravitational potential energy:

Work = P.E

Force = Weight = mg

Displacement = Vertical Displacement = Height = h

Therefore,

P.E = mgh

5 0
3 years ago
How are step up transformers used in the transmission of electrical energy
tensa zangetsu [6.8K]

Answer:

Transformers are used to increase or decrease the voltage of AC currents

Explanation:

A transformer is a device consisting of two coils (called primary and secondary coil) wrapped at the two sides of a soft iron core. When an AC current is present in the primary coil, it induces a magnetic field inside the core, and the presence of this changing magnetic field induces a voltage (and a current) into the secondary coil.

The voltages in the primary and the secondary coil are related by the transformer equation:

\frac{V_p}{V_s}=\frac{N_p}{N_s}

where

Vp, Vs are the voltages in the primary and secondary coil

Np, Ns are the number of turns in the primary and secondary coil

There are two types of transformers:

- Step-up transformers: these have N_s > N_p, so that V_s > V_p, which means that they increase the voltage. They are used to increase the voltage of the AC current produced by the power plants, before being sent into the transmission lines.

- Step-down transformers: these have N_s < N_p, so that V_s < V_p, which means that they decrease the voltage. They are used at the end of the transmission lines, before the houses, in order to decrease the voltage and allow the household appliances to work properly (in fact, household appliances need lower voltages to work)

8 0
3 years ago
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