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Mashcka [7]
3 years ago
6

What types of changes in motion cause acceleration?

Physics
1 answer:
xxTIMURxx [149]3 years ago
6 0

Answer:

Change in velocity and direction over a specific period of time.

Explanation:

In physics, acceleration can be defined as the rate of change of the velocity of an object with respect to time.

This simply means that, acceleration is given by the subtraction of initial velocity from the final velocity all over time.

Hence, if we subtract the initial velocity from the final velocity and divide that by the time, we can calculate the acceleration of an object.

Mathematically, acceleration is given by the equation;

Acceleration (a) = \frac{final \; velocity  -  initial \; velocity}{time}

a = \frac{v  -  u}{t}

Where,

a is acceleration measured in ms^{-2}

v and u is final and initial velocity respectively, measured in ms^{-1}

t is time measured in seconds.

Hence, the types of changes in motion that cause acceleration is a change in velocity and direction over a specific period of time.

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A 20000 kg railroad car is rolling at 1.00 m/s when a 1000 kg load of gravel is suddenly dropped in. part a what is the car's sp
Talja [164]
Using conservation of energy and momentum we get m1*v1=(m1+m2)*v2 so rearranging for v2 and plugging the given values in we get:
(200000kg*1.00m/s)/(21000kg)=.952m/s
8 0
3 years ago
"What is the magnifying power of an astronomical telescope using a reflecting mirror whose radius of curvature is 5.9 m and an e
notka56 [123]

Answer:

The Magnifying power of a telescope is M = 109.26

Explanation:

Radius of curvature R = 5.9 m = 590 cm

focal length of objective f_{objective} = \frac{R}{2}

⇒ f_{objective} = \frac{590}{2}

⇒ f_{objective} = 295 cm

Focal length of eyepiece f_{eyepiece} = 2.7 cm

Magnifying power of a telescope is given by,

M = \frac{f_{objective} }{f_{eyepiece} }

M = \frac{295}{2.7}

M = 109.26

therefore the Magnifying power of a telescope is M = 109.26

4 0
3 years ago
If the error in the angle is 0.50, the error in sin 90o is​
Sindrei [870]

At the given erro in angle, the error in the measurement of sin 90 degrees would be 0.001.

<h3>Percentage error</h3>

The percentage error of any measurement is obtained from the ratio of the error to the actual measurement.

The error of sin 90 degrees is calculated as follows;

sin 90 = 1

error in measurement = sin(90 - 0.5)

error in measurement = sin(89.5) = 0.999

<h3>Error in sin 90 degrees</h3>

Error in sin 90 degrees = 1 - 0.999

Error in sin 90 degrees = 0.001

Thus, at the given erro in angle, the error in sin 90 degrees would be 0.001.

Learn more about error in measurement here: brainly.com/question/26668346

4 0
2 years ago
Write down at least 3 equations as you solve them
Step2247 [10]

Answer:

3+7=          7x3=                                21➗7=          

Explanation:

3+7= 10              7 8 9 10

7x3= 21                  7 14 21

21➗7=3        21 14 7

                 mark me brainliest                      please

4 0
2 years ago
] A new coal-fired 750 MWe power plant with a thermal efficiency of 42% burns 9000 Btu/lb coal, which contains 1.1% sulfur. a. I
Valentin [98]

Answer:

The net emissions rate of sulfur is 1861 lb/hr

Explanation:

Given that:

The power or the power plant = 750 MWe

Since the power plant with a thermal efficiency of 42% (i.e. 0.42) burns 9000 Btu/lb coal, Then the energy released per one lb of the coal can be computed as:

\mathtt{=( 0.42\times 9000\times 1055.06) J}

= 3988126.8  J

= 3.99 MJ

Also, The mass of the burned coal per sec can be calculated by dividing the molecular weight of the power plant by the energy released per one lb.

i.e.

The mass of the coal that is burned per sec =\dfrac{750}{3.99}

The mass of the coal that is burned per sec = 187.97 lb/s

The mass of sulfur burned  = \dfrac{1.1}{100} \times 187.97 \  lb/s

= 2.067 lb/s

To hour; we have:

= 7444 lb/hr

However, If a scrubber with 75% removal efficiency is utilized,

Then; the net emissions rate of sulfur is (1 - 0.75) × 7444 lb/hr

= 0.25 × 7444 lb/hr

= 1861 lb/hr

Hence, the net emissions rate of sulfur is 1861 lb/hr

8 0
2 years ago
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