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3 years ago

What is the mass of an object that requires 100N of force in order to accelerate it at 10m/s2 ?

Physics

2 answers:

Debora [2.8K]3 years ago

7 0

Answer:

We know that f=ma

Rearranging

M=f÷a so m=10kg

Explanation:

jeka57 [31]3 years ago

3 0

<u>Answer:</u> The mass of an object is 10 kg.

<u>Explanation:</u>

Force is defined as the push or pull on an object with some mass that causes change in its velocity.

It is also defined as the mass multiplied by the acceleration of the object.

Mathematically,

$F=ma$

where,

F = force exerted on the object = 100 N

m = mass of the object = ?

a = acceleration of the object = $10m/s^2$

Putting values in above equation, we get:

$100kg.m/s^2=m\times 10m/s^2\\\\m=\frac{100}{10}=10kg$

Hence, the mass of an object is 10 kg.

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A gas undergoes a process in a piston–cylinder assembly during which the pressure-specific volume relation is pv1.3 = constant.

Galina-37 [17]

Answer:

Change in specific internal Energy $=250\ \rm Btu/lb$

Explanation:

Given:

Mass of the gas, m=0.4 lb
Initial pressure and volume are $p_1=160\ \rm lbf/in^2\ and\ v_1=1\ \rm ft^3\\$

Final pressure and temperature are $p_1=480\ \rm lbf/in^2$
Heat transfer from the gas is 2.1 Btu

Since the process is isotropic we have

$p_1v_1^{1.3}=p_2v_2^{1.3}\\160\times 1^{1.3}=480\times v_2^{1.3}\\v_2=0.43\ \rm ft^3\\$

So the final volume of the gas is calculated.

Work in any isotropic is given by w

$w=\dfrac{p_1v_1-P_2v_2}{n-1}\\\\w=\dfrac{160\times1-480\times0.43}{1.3-1}\\w=-154.67\ \rm Btu\\$

According to the first law of thermodynamics we have

$Q=\Delta U+w\\-2.1=\Delta U-154.67\\\Delta U=152.56\ \rm Btu\\$

So the Specific Internal Change is given by

$\Delta u=\dfrac{\Delta U}{m}\\\Delta u=\dfrac{152.56}{0.4}\\\Delta u=250\ \rm Btu/lb$

So the specific Change in Internal energy is calculated.

6 0

3 years ago

Calculate the average times it took the car to travel 0. 25 and 0. 50 meters. Record the averages, to two decimal places, in Tab

Illusion [34]

The average speed of the given car is 2.22 s and 3.13 s for 0.25 m and 0.50 m distance respectively.

<h3>How to calculate the Average speed?</h3>

The average speed can be calculated by adding the speed of each trial divided by the number of trials,

For 0.25 m the average speed will be:

$S_{avg} = \dfrac{2.24 + 2.21 + 2.23}{ 3}\\\\S_{avg} = 2.22$

For the 0.50 m, the average speed will:

$S_{avg} = \dfrac {3.16 + 3.08 + 3.15} {3 }\\\\S_{avg} = 3.13\rm \ s$

Therefore, the average speed of the given car is 2.22 s and 3.13 s for 0.25 m and 0.50 m distance respectively.

Learn more about Average speed:

brainly.com/question/26386984

6 0

2 years ago

A seagull flies at a velocity of 9.00 m/s straight into the wind. (a) if it takes the bird 20.0 min to travel 6.00 km relative t

enot [183]

Here we will the speed of seagull which is v = 9 m/s

this is the speed of seagull when there is no effect of wind on it

now in part a)

if effect of wind is in opposite direction then it travels 6 km in 20 min

so the average speed is given by the ratio of total distance and total time

$v_{avg} = \frac{6000}{20*60}$

$v_{avg} = 5m/s$

now since effect of wind is in opposite direction then we can say

$V_{net} = v_{bird} - v_{wind}$

$5 = 9 - v_{wind}$

$v_{wind}= 4 m/s$

Part b)

now if bird travels in the same direction of wind then we will have

$v_{net}= v_{bird} + v_{wind}$

$v_{net} = 9 + 4 = 13 m/s$

now we can find the time to go back

$time = \frac{distance}{speed}$

$time = \frac{6000}{13}$

$time = 7.7 minutes$

Part c)

Total time of round trip when wind is present

$T = t_1 + t_2$

$T = 20 + 7.7 = 27.7 min$

now when there is no wind total time is given by

$T = \frac{6000}{9} + \frac{6000}{9}$

$T = 22.22 min$

So due to wind time will be more

4 0

3 years ago

. A milk truck carries milk with density 64.6 lbyft3 in a horizontal cylindrical tank with diameter 6 ft. (a) Find the force exe

Goryan [66]

Answer:

Explanation:

one end of tank will be circular in shape . Area of circle A

= π r² , r is radius of the circle

= 3.14 x 3²

A = 28.26 ft³

To calculate force on the circular area , we first find pressure at the center of the circle which is at depth equal to r

pressure at the center = h d g ' here h = depth = r , d = density of milk

pressure = 3 x 64.6 x 32 poundal / ft²

= 6201.6 poundal / ft²

total force on circular face = pressure at the center x area of circle

= 6201.6 x 28.26

= 175257.21 poundal .

8 0

2 years ago

The size of a balloon increases when the pressure inside it increases. The balloon gets bigger when it is left in the heat from

Masteriza [31]

Explanation:

This happens because the gas inside tend to expand because its temperature gets higher.

This is why the balloon that is put in a freezer for too long tend to gets smaller, because the gas temperature that is inside the balloon decreases.

(you can try it at home)

It is related to the temperature of the gas.

5 0

2 years ago