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2 years ago

A net force of 2800 N is applied to a 1400 kg automobile at rest. The speed of the automobile after 10 s will be:

Physics

1 answer:

Svetach [21]2 years ago

6 0

The speed after 10 s is 20 m/s

<h3>What is the speed after 10 s?</h3>

Given that we have the following;

force = 2800 N

Mass = 1400 kg

time = 10 s

Recall that;

F = ma

F = force

m = mass

a = acceleration

a = F/m

a = 2800 N/1400 kg

a = 2 m/s^2

Now;

v = u + at

v = ??

u = 0 m/s

a= 2 m/s^2

t = 10 s

v = at

v = 2 * 10

v = 20 m/s

Learn more about the speed:brainly.com/question/28224010

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A reconnaissance plane flies 560 km away from its base at 602 m/s, then flies back to its base at 903 m/s.

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Answer:

Approximately $722\; \rm m\cdot s^{-1}$ .

Explanation:

The average speed of a vehicle is calculated as:

$\displaystyle \text{average speed} = \frac{\text{total distance}}{\text{total time}}$ .

In this question, the total distance is $2 \times 560\; \rm km = 1120\; \rm km$ .

The unit of the speeds in this question is meters per second, while the unit of distance is kilometers. Convert the unit of distance to meters:

$560 \; \rm km = 560 \times 10^{3} \; \rm m = 5.6 \times 10^{5}\; \rm m$ .

$1120 \; \rm km = 1120 \times 10^{3} \; \rm m = 1.12 \times 10^{6}\; \rm m$ .

Time required for the first part of this trip:

$\displaystyle \frac{5.60 \times 10^{5}\; \rm m}{602\; \rm m\cdot s^{-1}} \approx 930\; \rm s$ .

Time required for the second part of this trip:

$\displaystyle \frac{5.60 \times 10^{5}\; \rm m}{903\; \rm m\cdot s^{-1}} \approx 620\; \rm s$ .

The time required for the entire trip would be approximately $930 + 620 = 1550\; \rm s$ .

Calculate the average speed of this plane:

$\begin{aligned} \text{average speed} &= \frac{\text{total distance}}{\text{total time}} \\ &\approx \frac{1.12\times 10^{6}\; \rm m}{1550\; \rm s} \approx 722\; \rm m \cdot s^{-1}\end{aligned}$ .

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3 years ago

Two objects of the same mass, but made of different materials, are initially at the same temperature. Equal amounts of heat are

adelina 88 [10]

Answer:

The final temperature of the two objects is the same.

Explanation:

The expression for the heat energy in terms of mass, specific heat and the change in the temperature is as follows:

$Q=mc(T_{f} -T_{i})$

Here, Q is the heat energy, m is the mass of the object, c is the specific heat and $T_{f},T_{i}$ are the final temperature and initial temperature.

According to the given question, Two objects of the same mass, but made of different materials, are initially at the same temperature. Equal amounts of heat are added to each object.

$Q=mc(T_{f}-T_{i})$ ............(1)

$Q=mc(T'_{f}-T_{i})$ .............(2)

From (1) and (2),

$mc(T_{f}-T_{i})=mc(T'_{f}-T_{i})$

$T_{f}-T_{i}=T'_{f}-T_{i}$

$T_{f}=T'_{f}$

Therefore, the final temperature of the two objects is the same.

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3 years ago