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3 years ago

What is required to cause acceleration

2 answers:

6 0

Constant force. Basicly in no gravity you have to keep puching smth for it to accelerate and once you stop it just would remain at the same speed. On earth the fact that something is falling faster and faster means it keeps beeing constantly pulled by gravity so the same principle applies

forsale [732]3 years ago

4 0

Acceleration means any change in speed or direction of motion. Any of those requires force.

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True or false? To develop good alternatives, one should brainstorm ideas and consider different perspectives

Contact [7]

From my experience, I would say it is true.

7 0

3 years ago

Having difficulty finding the PE and KE for these values no mass is given. Does anyone know to go solve these?

Alexandra [31]

11) $1.04\cdot 10^7 J$

12) $1.04\cdot 10^7 J$

13) 50.0 m/s

14) 41.6 m/s

Explanation:

11)

The potential energy of an object is the energy possessed by the object due to its position relative to the ground. It is given by

$PE=mgh$

where

m is the mass of the object

g is the acceleration due to gravity

h is the height relative to the ground

Here in this problem, when the train is at the top, we have:

m = 8325 kg (mass of the train + riders)

$g=9.8 m/s^2$ (acceleration due to gravity)

h = 127 m (height of the train at the top)

Substituting,

$PE=(8325)(9.8)(127)=1.04\cdot 10^7 J$

12)

According to the law of conservation of energy, the total mechanical energy of the train must be conserved (in absence of friction). So we can write:

$KE_t + PE_t = KE_b + PE_b$

where

$KE_t$ is the kinetic energy at the top

$PE_t$ is the potential energy at the top

$KE_b$ is the kinetic energy at the bottom

$PE_b$ is the potential energy at the bottom

The kinetic energy is the energy due to motion; since the train is at rest at the top, we have

$KE_t=0$

Also, at the bottom the height is zero, so the potential energy is zero

$PE_b=0$

Therefore, we find:

$KE_b=PE_t=1.04\cdot 10^7 J$

13)

The kinetic energy of an object is the energy of the object due to its motion. Mathematically, it is given by

$KE=\frac{1}{2}mv^2$

where

m is the mass of the object

v is the speed of the object

From question 12), we know that the kinetic energy of the train at the bottom is

$KE=1.04\cdot 10^7 J$

We also know that the mass is

m = 8325 kg

Therefore, we can calculate the speed of the train at the bottom:

$v=\sqrt{\frac{2KE}{m}}=\sqrt{\frac{2(1.04\cdot 10^7)}{8325}}=50.0 m/s$

14)

At the top of the second hill, the total mechanical energy of the train is still conserved.

Therefore, we can write again:

$KE_1 + PE_1 = KE_2 + PE_2$

where

$KE_1$ is the kinetic energy at the top of the 1st hill

$PE_1$ is the potential energy at the top of the 1st hill

$KE_2$ is the kinetic energy at the top of the 2nd hill

$PE_2$ is the potential energy at the top of the 2nd hill

From the previous questions, we know that

$KE_1=0$

and

$PE_1=1.04\cdot 10^7 J$

The height of the second hill is

h = 39 m

So we can also find the potential energy at the second hill:

$PE_2=mgh=(8325)(9.8)(39)=3.2\cdot 10^6 J$

So, the kinetic energy at the second hill is

$KE_2=PE_1-PE_2=1.04\cdot 10^7 - 3.2\cdot 10^6 =7.2\cdot 10^6 J$

And so, the speed is

$v=\sqrt{\frac{2KE_2}{m}}=\sqrt{\frac{2(7.2\cdot 10^6)}{8325}}=41.6 m/s$

4 0

3 years ago

To test the performance of its tires, a car

Rom4ik [11]

The coefficient of static friction is 0.222

Explanation:

In order for the car to remain in circular motion, the frictional force must be able to provide the necessary centripetal force. Therefore, the car will start skidding when the two forces are equal:

$\mu mg=m\frac{v^2}{r}$

where the term on the left is the frictional force, while the term on the right is the centripetal force, and where

$\mu$ is the coefficient of static friction

m is the mass of the car

g is the acceleration of gravity

v is the speed of the car

r is the radius of the track

In this problem, we have:

r = 564 m

v = 35 m/s

$g=9.8 m/s^2$

And re-arranging the equation for $\mu$ , we can find the coefficient of static friction:

$\mu = \frac{v^2}{gr}=\frac{35^2}{(9.8)(564)}=0.222$

Learn more about friction:

brainly.com/question/6217246

brainly.com/question/5884009

brainly.com/question/3017271

brainly.com/question/2235246

#LearnwithBrainly

5 0

3 years ago

I'm pretty sure it's density but I need another opinion.

bixtya [17]

I agree with density

6 0

3 years ago

Read 2 more answers

Longer wavelengths of light, such as _______, have ________ energy than shorter wavelengths, such as _________

Ilya [14]

Answer:

Micro and radio waves.

Lower energy.

Gamma rays.

Explanation:

The electromagnetic spectrum is the range of frequencies of electromagnetic radiation and their respective wavelengths.

Ionising radiation os defined as the energy required of photons of a wave to ionize atoms, causing chemical reactions.

The energy of the wave depends on both the amplitude and the frequency. If the energy of each wavelength is a discrete packet of energy, a high-frequency wave will deliver more of these packets per unit time than a low-frequency wave. In summary, the longer the wavelength, the lower the energy to ionise.

The velocity of a wave is directly proportional to the frequency of that wave.

c = f * lambda

Where,

c = velocity of the wave

f = frequency of the wave = 1/time

Lambda = wavelength.

From the above expression, the longer the wavelength, lambda the shorter the frequency.

Examples of waves with longer wavelengths are, micro and radio waves, while radiations with shorter wavelengths like gamma rays.

8 0

3 years ago