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yawa3891 [41]
3 years ago
9

What force is required to accelerate a 1,000 kg car 5 m/s2?

Physics
1 answer:
marysya [2.9K]3 years ago
5 0
F = ma
F = (1000 kg)•(5 m/s^2)
F = 5000 N
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<h3>Why the list is incorrect?</h3>

The list is incorrect because skin is not a reproductive organ and the spines do not attract predators towards each other. Skin protects the body from the outer environment whereas the spines repel the predators from the plant body.

So we can conclude that the skin is NOT a reproductive organ and the spines attract predators are the incorrect statements.

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2 years ago
Write down the condition required for the thermonuclear fusion.​
storchak [24]

Explanation:

The temperature must be hot enough to allow the ions of deuterium and tritium to have enough kinetic energy to overcome the Coulomb barrier and fuse together. The ions must be confined with a high ion density to achieve a suitable fusion reaction rate.

7 0
2 years ago
Read 2 more answers
If a 100-N net force acts on a 50-kg car, what will the acceleration of the car be?
Tema [17]
Newton's 2nd law of motion: 

                             Force = (mass) x (acceleration)

Divide each side by (mass):     

                             Acceleration = (force) / (mass)

                                               =  (100 N) / (50 kg)

                                               =  2 m/s²  


5 0
3 years ago
What do scientist use to determine the temperature of a star
REY [17]

Color is what scientist use to determine the temperature of a star!

hope this helps!

4 0
4 years ago
A cube has a drag coefficient of 0.8. What would be the terminal velocity of a sugar cube 1 cm on a side in air ( = 1.2 kg/mº)?
anzhelika [568]

0.495 m/s

Explanation

the formula for the terminal velocity is given by:

\begin{gathered} v=\sqrt[]{\frac{2mg}{\sigma AC}} \\ \text{where} \\  \end{gathered}

m is the mass

g is 9.81 m/s²

ρ is density

A is area

C is the drag coefficient

then

Step 1

Let's find the mass

\begin{gathered} \sigma=\frac{m}{v} \\ m=\sigma\cdot v \\ \text{mass}=(2\cdot10^3\frac{\operatorname{kg}}{m^3})\cdot(0.01m)^3 \\ \text{mass}=(2\cdot10^3\frac{\operatorname{kg}}{m^3})\cdot(1\cdot10^{-6}) \\ \text{mass}=2\cdot10^{-3}\operatorname{kg} \\ \text{mass}=0.002\text{ kg } \\ \text{Area}=(0.01\text{ m}\cdot0.01m)=0.0001m^2 \end{gathered}

now, replace

\begin{gathered} v=\sqrt[]{\frac{2mg}{\sigma AC}} \\ v=\sqrt[]{\frac{2(0.002kg)(9.81\text{ }\frac{m}{s^2})}{(2\cdot10^3\frac{\operatorname{kg}}{m^3})(0.0001m^2)0.8}} \\ v=\sqrt[]{\frac{0.03924\frac{\operatorname{kg}m}{s^2}}{0.16\frac{\operatorname{kg}}{m^{}}}} \\ v=\sqrt[]{0.2452\frac{m^2}{s^2}} \\ v=0.495\text{ m/s} \end{gathered}

hence, the answer is 0.495 m/s

3 0
1 year ago
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