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grandymaker [24]
2 years ago
10

How does inertia explain why it is difficult to stop a moving skateboard?

Physics
1 answer:
My name is Ann [436]2 years ago
6 0

Answer:

(Example Person) has to push hard to get the skateboard started, but once it begins moving, it takes much less effort to keep it rolling over the smooth, flat pavement. In fact, if (Example Person) tries to stop the rolling skateboard, it may take as much effort to stop it as it did to start it rolling in the first place.

Explanation:

Hope this helps :)

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A 1 530-kg automobile has a wheel base (the distance between the axles) of 2.70 m. The automobile's center of mass is on the cen
NeTakaya

Answer:

Force on front axle = 6392.85 N

Force on rear axle = 8616.45 N

Explanation:

As we know that the weight of the car is balanced by the normal force on the front wheel and rear wheels

Now we know that

F_1 + F_2 = W

F_1 + F_2 = (1530\times 9.81)

F_1 + F_2 = 15009.3 N

now we know that distance between the axis is 2.70 m and centre of mass is 1.15 m behind front axle

so we can write torque balance about its center of mass

F_1(1.15) = F_2(2.70 - 1.15)

F_1 = 1.35 F_2

now from above equation

F_2 + 1.35F_2 = 15009.3

now we have

F_2 = 6392.85 N

now the other force is given as

F_1 = 8616.45 N

4 0
3 years ago
What is the relationship between the applied force of a hanging mass on a spring and the spring force of the spring?
zaharov [31]

Answer:

elastic force and weight are related to the acceleration of the System.

Explanation:

The relationship between these two forces can be found with Newton's second law.

        F_{e} - W = m a

        K x - m g = m a

We see that elastic force and weight are related to the acceleration of the System.

If a harmonic movement is desired, an extra force that increases the elastic force is applied, but to begin the movement this force is eliminated, in general , if the relationship between this external and elastic force is desired, the only requirement is that it be small for harmonic movement to occur

7 0
3 years ago
A wire is joined to points X and Y in the circuit diagram shown. A diagram of a circuit with a power source on the left. Directl
Sidana [21]

The circuit change when the wire is added will see a short circuit occur and makes bulbs 1 and 2 turn off but keeps bulbs 3 and 4 lit. Option D. This is further explained below.

<h3>How does the circuit change when the wire is added?</h3>

Generally, Electronic circuits consist of a series of interconnected parts that form a closed loop through which electricity may flow.

In conclusion, If two wires are linked together, a short circuit will develop, cutting power to bulbs 1 and 2. But there is no impact on bulbs 3 and 4. There is no problem with bulbs 3 and 4.

Read more about circuit

brainly.com/question/21505732

#SPJ1

7 0
1 year ago
A container with volume 1.64 L is initially evacuated. Then it is filled with 0.226 g of N2N
vaieri [72.5K]

Answer:

0.015 atm

Explanation:

The pressure of the gas can be calculated using Ideal Gas Law:

p = \frac{nRT}{V}

<u>Where:</u>

n: is the number of moles of the gas

R: is the gas constant = 0.082 L*atm/(K*mol)

V: is the volume of the container = 1.64 L

T: is the temperature

We need to find the number of moles and the temperature. The number of moles is:

n = \frac{m}{M}

<u>Where:</u>

M: is the molar mass of the N₂ = 14.007 g/mol*2 = 28.014 g/mol

m: is the mass of the gas = 0.226 g

n = \frac{0.226 g}{28.014 g/mol} = 8.07 \cdot 10^{-3} moles

Now, the temperature can be found using the following equation:

v_{rms} = \sqrt{\frac{3RT}{M}}    

<u>Where:</u>

R: is the gas constant = 0.082 L*atm/K*mol = 8.314 J/K*mol

v_{rms}: is the root-mean-square speed of the gas = 182 m/s

By solving the above equation for T, we have:

T = \frac{v_{rms}^{2}*M}{3R} = \frac{(182 m/s)^{2}*28.014 \cdot 10^{-3} Kg/mol}{3*8.314 J K^{-1}mol^{-1}} = 37.20 K        

Finally, we can find the pressure of the gas:

p = \frac{nRT}{V} = \frac{8.07 \cdot 10^{-3} mol*0.082 L*atm* K^{-1}*mol^{-1}*37.20 K}{1.64 L} = 0.015 atm

Therefore, the pressure of the gas is 0.015 atm.

I hope it helps you!

8 0
3 years ago
Old-fashioned pendulum clocks are powered by masses that need to be wound back to the top of the clock about once a week to coun
Andreyy89

Answer:

Energy, E = 178.36 J

Explanation:

It is given that,

Mass 1, m_1=4\ kg

Mass 2, m_2=4\ kg

Mass 3, m_3=6\ kg

Height from which they are dropped, h = 1.3 m

Let m is the energy used by the clock in a week. The energy is equal to the gravitational potential energy. It is given by :

E=(m_1+m_2+m_3)gh

E=(4+4+6)\times 9.8\times 1.3

E = 178.36 J

So, the energy used by the clock in a week is 178.36 Joules. Hence, this is the required solution.

6 0
2 years ago
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