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3 years ago

The part of the computer that provides access to the internet is

Physics

1 answer:

Allisa [31]3 years ago

5 0

The Modem

or physically connecting on the computer the data port is for a cable to connect

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An object accelerates in a direction that is always perpendicular to its motion.

anzhelika [568]

The object will not be able to accelerate perpendicular to direction of motion

7 0

3 years ago

A truck tire rotates at an initial angular speed of 21.5 rad/s. The driver steadily accelerates, and after 3.50 s the tire's ang

erma4kov [3.2K]

Given:

initial angular speed, $\omega _{i}$ = 21.5 rad/s

final angular speed, $\omega _{f}$ = 28.0 rad/s

time, t = 3.50 s

Solution:

Angular acceleration can be defined as the time rate of change of angular velocity and is given by:

$\alpha = \frac{\omega_{f} - \omega _{i}}{t}$

Now, putting the given values in the above formula:

$\alpha = \frac{28.0 - 21.5}{3.50}$

$\alpha = 1.86 m/s^{2}$

Therefore, angular acceleration is:

$\alpha = 1.86 m/s^{2}$

5 0

2 years ago

The map here shows land surface temperatures on earth during one month in 2015

Law Incorporation [45]

Latitude is the answer for apex.

8 0

3 years ago

Read 2 more answers

Katelyn (55 kg) is practicing a drop jump in the biomechanics lab. She steps off a plyometrics box, lands on the force plate, an

suter [353]

Answer:

J = 357.5 kg*m/s

Explanation:

The impulse exerted on Katelyn when she was on the force plate, is equal to the change in her momentum, according to Newton's 2nd Law.
Assuming as the positive direction the upward direction (coincident with the positive y-axis) we can express the initial momentum as follows:

$p_{o} = m*v_{o} = 55 kg * (-3.0 m/s) (1)$

By the same token, the final momentum is as follows:

$p_{f} = m*v_{f} = 55 kg * (3.5 m/s) (2)$

As we have already said, the impulse J is just equal to the change in momentum, i.e., the difference between (2) and (1):

$J = p_{f} - p_{o} = m* (v_{f} -v_{o}) = 55 kg* (3.5m/s- (-3.0m/s)) = 357.5 kg*m/s (3)$

5 0

2 years ago

The energy in the typical thunderstorm is about 1.4×1011j. a typical lightning flash transfers 30 c across a potential differenc

iragen [17]

780 seconds, or 13 minutes.

In the future, please use proper capitalization. There's a significant difference in the meaning between mV and MV. One of them indicated millivolts while the other indicates megavolts. For this problem, I'll make the following assumptions about the values presented. They are:
Total energy = 1.4x10^11 Joules (J)
Current per flash = 30 Columbs (C)
Potential difference = 30 Mega Volts (MV)

First, let's determine the power discharged by each bolt. That would be the current multiplied by the voltage, so
30 C * 30x10^6 V = 9x10^8 CV = 9x10^8 J

Now that we know how many joules are dissipated per flash, let's determine how flashes are needed.
1.4x10^11 / 9x10^8 = 1.56E+02 = 156

Since each flash takes 5 seconds, that means that it will take about 5 * 156 = 780 seconds which is about 780/60 = 13 minutes.

3 0

3 years ago