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Kipish [7]
3 years ago
11

How much force is needed to accelerate a 1245 kg car at a rate of 4.25 m/sec^2?

Physics
1 answer:
BartSMP [9]3 years ago
3 0

Answer:

F = 5291.25 N

Explanation:

F = Ma so 1245 times 4.25^2 ,, that equals 5291.25 N

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What is the proportionality between pressure and temperature, and the proportionality between atmospheric pressure and tempratur
Alborosie

According to Ideal gasTo solve this problem, the fastest relationship allows us to observe the proportionality between the two variables would be the one expressed in the ideal gas equation, which is

PV= NRT

Here

P = Pressure

V = Volume

N = Number of moles

R = Gas constant

T = Temperature

We can see that the pressure is proportional to the temperature, then

P \propto T

This relationship can be extrapolated to all the scenarios in which these two variables are related. As the pressure increases the temperature increases. The same goes for the pressure in the atmosphere, for which an increase in this will generate an increase in temperature. This variable can be observed in areas of different altitude. At higher altitude lower atmospheric pressure and lower temperature.

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3 years ago
An Olympic high jumper, with a mass of 82 kg, has a
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3 years ago
Celina has a water sample that’s contaminated with salt and microorganisms. Which method should she use to purify the water?
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Celina should perform the following method to get pure water:

Boil the water in a pan, as the water boils it changes into gas (water vapours) these vapours should be cooled down through the process of condensation which changes gas into liquid form, Through this process germs would be killed due to high temperature and the salt will remain in the pan as salt has higher boiling point than water,

7 0
3 years ago
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A wheel starts from rest and rotates with constant angular acceleration to reach an angular speed of 11.1 rad/s in 2.99 s.(a) fi
elena55 [62]
The angular acceleration of a rotating object is given by
\alpha =  \frac{\omega_f - \omega_i}{\Delta t}
where
\omega_f is the final angular speed of the object
\omega_i is its initial angular speed
\Delta t is the time taken to accelerate

For the wheel in our problem, \omega_f=11.1 rad/s, \omega_i = 0 and \Delta t=2.99 s, so its angular acceleration is
\alpha= \frac{11.1 rad/s-0}{2.99 s}=3.71 rad/s^2
8 0
3 years ago
A train, traveling at a constant speed of 22.0 m/s, comes to an incline with a constant slope. While going up the incline, the t
Charra [1.4K]

Answer:

123.30 m

Explanation:

Given

Speed, u = 22 m/s

acceleration, a = 1.40 m/s²

time, t = 7.30 s

From equation of motion,

                       v = u + at

where,

v is the final velocity

u is the initial velocity

a is the acceleration

t is time  

                       V = at + U

using equation  v - u = at to get line equation for the graph of the motion of the train on the incline plane

                       V_{x} = mt + V_{o}      where m is the slope

Comparing equation (1) and (2)

V = V_{x}

a = m    

U = V_{o}

Since the train slows down with a constant acceleration of magnitude 1.40 m/s² when going up the incline plane. This implies the train is decelerating. Therefore, the train is experiencing negative acceleration.

          a = -  1.40 m/s²

Sunstituting a = -  1.40 m/s² and  u = 22 m/s

                        V_{x} = -1.40t + 22

                            V_{x} = -1.40(7.30) + 22

                             V_{x} = -10.22 + 22

                             V_{x} = 11. 78 m/s

The speed of the train at 7.30 s is 11.78 m/s.

The distance traveled after 7.30 sec on the incline is the area cover on the incline under the specific interval.

           Area of triangle +  Area of rectangle

          [\frac{1}{2} * (22 - 11.78) * (7.30)]  + [(11.78 - 0) * (7.30)]

                           = 37.303 + 85.994

                           = 123. 297 m

                           ≈ 123. 30 m

                 

4 0
3 years ago
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