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3 years ago

I NEEEED HELPPPPPPP PLEASEEEE

Physics

1 answer:

monitta3 years ago

6 0

Answer:

Below

Explanation:

1) To find the average velocity of an object you can use this formula :

v = ∆displacement / ∆time

v = 200 m / 5 s

v = 40 m/s

The average velocity of the turtle is 40 m/s

2) To find the distance an object can travel you can use this formula :

d = (∆velocity)(∆time)

d = (20m/s)(5s)

d = 100 m

The cheetah can run 100 meters in 5 seconds

3) To find the amount of time it takes for an object to move :

t = ∆displacement / ∆velocity

t = 5mi / 20 mph

t = 0.25 h

It would take this car 0.25 h to travel 5 miles at 20 mph

4) To find how far an object can go :

d = (∆velocity)(∆time)

d = (20mph)(3h)

d = 60 mi

The car can drive 60 miles driving 20 mph for 3 hours

5) To find how far an object can go :

d = (∆velocity)(∆time)

d = (20mph)(30min)

d = (20mph)(0.5h)

d = 10 mi

The car can drive 10 miles driving 20mph for 30 minutes

6) To find how long it would take :

t = ∆displacement / ∆velocity

t = 10 m / 50 m/s

t = 0.2 s

It would take the train 0.2 seconds for a train to travel 10 m at 50 m/s

7) To find how far :

d = (∆velocity)(∆time)

d = (670,000,000 mph)(1 yr)

d = (670,000,000 mph)(8760 h)

d = 5869200000000 mi

1 light year is 5869200000000 miles

Hope this helps! Best of luck <3

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In a Rutherford scattering experiment a target nucleus has a diameter of 1.34×10-14 m. The incoming α particle has a mass of 6.6

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Answer:

E = 2.5 x 10⁻¹⁴ J

Explanation:

given,

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mass = 6.64 x 10⁻²⁷ kg

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$\lambda = \dfrac{h}{mv}$

$1.33 \times 10^{-14}= \dfrac{6.626 \times 10^{-34}}{6.64 \times 10^{-27}\times v}$

$v= \dfrac{6.626 \times 10^{-34}}{6.64 \times 10^{-27}\times 1.33 \times 10^{-14}}$

v = 7.5 x 10⁶ m/s

Kinetic energy is equal to

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3 years ago

Suppose the activity of a sample of radioactive material was 100Bq at the start. What would you divide 100Bq by to obtain the ac

myrzilka [38]

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3 years ago

Why are heat transfers associated with phase changes known as latent or "hidden" heats?

pshichka [43]

The best and most correct answer among the choices provided by your question is the first choice or letter A.

Heat transfers associated with phase changes known as latent or "hidden" heats because h<span>eat absorbed or released in a phase change is measured in kJ while temperature is measured in °C.</span>

I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!

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4 years ago

Under favorable circumstances, including reaction time, a motor vehicle with good brakes going 50 miles per hour can be stopped

aleksklad [387]

Answer:

about 229 feet.

Explanation:

According to my research on the information provided by the drivers educational book, It is said that a motor vehicle with good brakes that is going at 50 miles per hour can be stopped within about 229 feet. This is dependent 100% on having good brakes as well as there being normal driving conditions (on pavement with no rain or other weather that may affect driving conditions).

I hope this answered your question. If you have any more questions feel free to ask away at Brainly.

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3 years ago

A 175-kg roller coaster car starts from rest at the top of an 18.0-m hill and rolls down the hill, then up a second hill that ha

Anni [7]

Answer:

The work done by non-conservative forces on the car from the top of the first hill to the top of the second hill is 6574.75 joules.

Explanation:

By Principle of Energy Conservation and Work-Energy Theorem we present the equations that describe the situation of the roller coaster car on each top of the hill. Let consider that bottom has a height of zero meters.

From top of the first hill to the bottom

$m\cdot g \cdot h_{1} = \frac{1}{2}\cdot m\cdot v_{1}^{2} +W_{1, loss}$ (1)

From the bottom to the top of the second hill

$\frac{1}{2}\cdot m\cdot v_{1}^{2} = m\cdot g \cdot h_{2} + \frac{1}{2}\cdot m \cdot v_{2}^{2}+W_{2,loss}$ (2)

Where:

$m$ - Mass of the roller coaster car, in kilograms.

$v_{1}$ - Speed of the roller coaster car at the bottom between the two hills, in meters per second.

$g$ - Gravitational acceleration, in meters per square second.

$h_{1}$ - Height of the first top of the hill with respect to the bottom, in meters.

$W_{1, loss}$ - Work done by non-conservative forces on the car between the top of the first hill and the bottom, in joules.

$v_{2}$ - Speed of the roller coaster car at the top of the second hill, in meters per seconds.

$h_{2}$ - Height of the second top of the hill with respect to the bottom, in meters.

$W_{2, loss}$ - Work done by non-conservative forces on the car bewteen the bottom between the two hills and the top of the second hill, in joules.

By using (1) and (2), we reduce the system of equation into a sole expression:

$m\cdot g\cdot h_{1} = m\cdot g\cdot h_{2} + \frac{1}{2}\cdot m \cdot v_{2}^{2} + W_{loss}$ (3)

Where $W_{loss}$ is the work done by non-conservative forces on the car from the top of the first hill to the top of the second hill, in joules.

If we know that $m = 175\,kg$ , $g = 9.807\,\frac{m}{s^{2}}$ , $h_{1} = 18\,m$ , $h_{2} = 8\,m$ and $v_{2} = 11\,\frac{m}{s}$ , then the work done by non-conservative force is:

$W_{loss} = m\cdot\left[ g\cdot \left(h_{1}-h_{2}\right)-\frac{1}{2}\cdot v_{2}^{2} \right]$

$W_{loss} = 6574.75\,J$

The work done by non-conservative forces on the car from the top of the first hill to the top of the second hill is 6574.75 joules.

8 0

3 years ago