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evablogger [386]

3 years ago

5

An automobile moving along a straight track changes its velocity from 40 m/s to 80 m/s in a distance of 200 m. What is the (cons

tant) acceleration of the vehicle during this time? Group of answer choices

1 answer:

Nata [24]3 years ago

8 0

Answer:

Dear Kaleb

Answer to your query is provided below

Acceleration of the vehicle is 12m/s^2

Explanation:

Explanation for the same is attached in image

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Answer:

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3 years ago

What is ozone? How and where is it formed in the atmosphere?

mrs_skeptik [129]

Ozone gas is made up of oxygen molecules that have three atoms. it exists in polluted air and ozone layer. Ozone layer is formed in stratosphere(part of atmosphere).

4 0

3 years ago

A 4 V power source is connected in parallel to four resistors: 1, 2, 2, and 3. Find the total current.

Keith_Richards [23]

Answer:

Explanation:

Equivalent resistance is 1 / ((1/1) + (1/2) + (1/2) + (1/3)) = 3/7 Ω

I = V/R = 4(7/3) = 28/3 = 9.3 A

4 0

2 years ago

Casey is playing with her yo-yo. She performs an "around the world" trick, in which the yo-yo travels up and around in a circle

USPshnik [31]

Answer:

It has the least potential energy at the bottom of its circular path.

Explanation:

It has the least potential energy at the bottom of its circular path.

Remember the equation

U = m*g*h

where U is the potential energy

m is the mass of the yo-yo

h is the height

4 0

3 years ago

Suppose that the hatch on the side of a Mars lander is built and tested on Earth so that the internal pressure just balances the

Kaylis [27]

Answer:

The value is $F_{net} = 4444 lb$

The force will be outward

Explanation:

From the question we are told that

The diameter of the disk is $d = 50.0 \ cm = \frac{50}{100} = 0.5 \ m$

The external pressure on Mars is $P = 650 \ N/m^2$

From the question we are told that

Internal pressure = External pressure

Generally the external Force on earth is

$F_E = P_{atm} * A$

Here $P_{atm}$ is the atmospheric pressure with value $P_{atm} = 1.013*10^{5}\ Pa$

So

$F_E = 1.013 *10^{5} * \pi * \frac{d^2}{4}$

=> $F_E = 1.013 *10^{5} *3.142 * \frac{0.50 ^2}{4}$

=> $F_E = 19893 \ N$

Generally the external Force on Mars is

$F= P * A$

$F = 650 * \pi * \frac{d^2}{4}$

=> $F = 650 *3.142 * \frac{0.5^2}{4}$

=> $F = 127.6 \ N$

Net force is mathematically represented as

$F_{net} = F_E -F$

=> $F_{net} = 19893 -127.6$

=> $F_{net} = 19765.6 \ N$ converting to pounds

$F_{net} = \frac{19765.6}{4.448}$

=> $F_{net} = 4444 lb$

Given that that the value is positive then the force will be outward

7 0

3 years ago