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ss7ja [257]
3 years ago
11

What is the acceleration of a 45 kg mass pushed with 15 N of force? What is the acceleration if the force changes to 25 N? What

if the mass changes to 70 kg?
Physics
2 answers:
Rama09 [41]3 years ago
8 0
F = ma
Rearrange this so acceleration is the subject:
A = f/m

Now input your values into this equation for the first question :
A = f/m
A = 15/45
A = 0.33333333 m/s

Then using this same equation change the force of 15 to 25 :
A = f/m
A = 25/45
A = 0.55555556 m/s

For the last question keep the force of 15N but change the mass to 70kg into the same equation :
A = f/m
A = 15/70
A = 0.21 m/s (rounded)
Contact [7]3 years ago
6 0

Answer:

Part a)

a = \frac{1}{3} m/s^2

Part b)

a = \frac{5}{9} m/s^2

Part c)

a = \frac{3}{14} m/s^2

Explanation:

As per Newton's II law we know that

F = ma

Now we know that

m = 45 kg

F = 15 N

so we will have

15 = 45 a

a = \frac{15}{45} = \frac{1}{3} m/s^2

Now we have

m = 45 kg

F = 25 N

now again from Newton's Law

25 = 45 a

a = \frac{25}{45} m/s^2

a = \frac{5}{9} m/s^2

Now if mass is changed to 70 kg

again we have

15 = 70 a

a = \frac{15}{70}

a = \frac{3}{14} m/s^2

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The density of gasoline is 730 kg/m3 at 0°C. Its average coefficient of volume expansion is 9.60 10-4(°C)−1. Assume 1.00 gal of
kipiarov [429]

Answer: 0.4911 kg

Explanation:

We have the following data:

\rho_{0\°C}= 730 kg/m^{3} is the density of gasoline at 0\°C

\beta=9.60(10)^{-4} \°C^{-1} is the average coefficient of volume expansion

We need to find the extra kilograms of gasoline.

So, firstly we need to transform the volume of gasoline from gallons to m^{3}:

V=8.50 gal \frac{0.00380 m^{3}}{1 gal}=0.0323 m^{3} (1)

Knowing density is given by: \rho=\frac{m}{V}, we can find the mass m_{1} of 8.50 gallons:

m_{1}=\rho_{0\°C}V

m_{1}=(730 kg/m^{3})(0.0323 m^{3})=23.579 kg (2)

Now, we have to calculate the factor f by which the volume of gasoline is increased with the temperature, which is given by:

f=(1+\beta(T_{f}-T_{o})) (3)

Where T_{o}=0\°C is the initial temperature and T_{f}=21.7\°C is the final temperature.

f=(1+9.60(10)^{-4} \°C^{-1}(21.7\°C-0\°C)) (4)

f=1.020832 (5)

With this, we can calculate the density of gasoline at 21.7\°C:

\rho_{21.7\°C}=730 kg/m^{3} f=(730 kg/m^{3})(1.020832)

\rho_{21.7\°C}=745.207 kg/m^{3} (6)

Now we can calculate the mass of gasoline at this temperature:

m_{2}=\rho_{21.7\°C}V (7)

m_{2}=(745.207 kg/m^{3})(0.0323 m^{3}) (8)

m_{2}=24.070 kg (9)

And finally calculate the mass difference \Delta m:

\Delta m=m_{2}-m_{1}=24.070 kg-23.579 kg (10)

\Delta m=0.4911 kg (11) This is the extra mass of gasoline

6 0
3 years ago
If a car's mass is 439 kg, what is its weight on earth?
trasher [3.6K]
The correct answer is B.
8 0
3 years ago
The mass of the car?
Nonamiya [84]

Answer:

1050 kg

Explanation:

The formula for kinetic energy is:

KE (kinetic energy) = 1/2 × m × v² where <em>m</em> is the <em>mass in kg </em>and <em>v</em> is the velocity or <em>speed</em> of the object <em>in m/s</em>.

We can now substitute the values we know into this equation.

KE = 472 500 J and v = 30 m/s:

472 500 = 1/2 × m × 30²

Next, we can rearrange the equation to make m the subject and solve for m:

m = 472 500 ÷ (1/2 × 30²)

m = 472 500 ÷ 450

m = 1050 kg

Hope this helps!

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