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Makovka662 [10]

1 year ago

11

During a NASCAR race a car goes 58 m/s around a curved section of track that has a radius of 260 m. What is the car's accelerati

on?

1 answer:

Alex777 [14]1 year ago

6 0

Answer:

12.9m/s^2

Explanation:

As, a=(v^2)/r

=(58^3)/260

=12.9

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Consider an oil droplet of mass m and charge q. We want to determine the charge on the droplet in a Millikan-type experiment. We

Katen [24]

Answer:

$q=\frac{mg}{E_o}$

Explanation:

Given:

Charge = <em>q</em>

Electric field strength = $E_o$

weight of the droplet = <em>mg</em>

The charge is suspended motionless. This is because the electric force on the charge is balanced by the weight of the droplet.

electric force on charged droplet, $F=qE_o$

This is balanced by the weight, $mg$

Equating the two:

$qE_o=mg\\\Rightarrow q=\frac{mg}{E_o}$

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3 years ago

What is the definition of cell ?

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2 years ago

A new ride being built at an amusement park includes a vertical drop of 71.6 meters. Starting from rest, the ride vertically dro

Allisa [31]

Answer: 2.6x107

Explanation:

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2 years ago

A spaceship ferrying workers to Moon Base I takes a straight-line path from the earth to the moon, a distance of 384,000 km. Sup

Tema [17]

Answer:

a) v = 19,149.6 m/s

b) f = 95%

c) t = 346.5min

Explanation:

First put all values in metric units:

$15.8 min*\frac{60s}{1min}=948s$

The equation of motion you need is:

$v_f = a*t+v_0$

where $v_f$ is the final velocity, a is acceleration and t is time in hours.

Since the spaceship starts from 0 velocity:

$v_f = a*t = 20.2*948 = 19,149.6 m/s$

Next, you need to calculate the distances traveled on each interval, considering that both starting and final intervals travel the same distance because the acceleration and time are equal. For this part you need the next motion equation:

$x=\frac{v_0+v_f}{2}t$

solving for first and last interval:

Since the spaceship starts and finish with 0 velocity:

$x=\frac{v}{2}t=\frac{19,149.6}{2}948=9,076,910.4m=9,076.9104km$

Then the ship traveled $384,000-9,076.9104*2 = 361,846.1792km$ at constant speed, which means that it traveled:

$f_{constant_speed} =\frac{ x_{constant_speed}}{x_total} =\frac{361,846.1792}{380,000} =0.95$

Which in percentage is 95% of the trip.

to calculate total time you need to calculate the time used during constant speed:

$t = \frac{361,846,179.2}{19,149.6} = 18,895.75s = 314min$

That added to the other interval times:

$t_{total} = t_1+t_2+t_3=15.8+314.93+15.8=346.5min$

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2 years ago

A block with a mass of 8.7 kg is dropped from rest from a height of 8.7 m, and remains at rest after hitting the ground. 1)If we

Harlamova29_29 [7]

To solve this problem we will apply the concepts related to gravitational potential energy.

This can be defined as the product between mass, gravity and body height.

Mathematically it can be expressed as

$\Delta P = mgh$

$\Delta P = (8.7)(9.8)(3)$

$\Delta P = 255.78J$

Therefore the change in the internal energy of the system is 255.78

7 0

3 years ago