All categories

3 years ago

What is the equation for finding the acceleration of an object moving in a straight line?

Physics

1 answer:

kiruha [24]3 years ago

4 0

Acceleration can be found using one of the following suvat equations:

$v=u+at\\s=ut+\frac{1}{2}at^2\\s=vt-\frac{1}{2}at^2\\v^2-u^2=2as$

Explanation:

The acceleration of an object is defined as the rate of change of velocity of the object:

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

where

v is the final velocity

u is the initial velocity

t is the time taken for the velocity to change from u to t

There are several equations used to find the acceleration of an object moving in a straight line (they are usually called suvat equation). Depending on which quantities are given in the problem, you can use one of the following equations:

$v=u+at\\s=ut+\frac{1}{2}at^2\\s=vt-\frac{1}{2}at^2\\v^2-u^2=2as$

where

$a$ is the acceleration

$s$ is the distance covered

$t$ is the time

$u$ is the initial velocity

$v$ is the final velocity

Learn more about acceleration and suvat equations:

brainly.com/question/9527152

brainly.com/question/11181826

brainly.com/question/2506873

brainly.com/question/2562700

#LearnwithBrainly

You might be interested in

Determine the voltage ratings of the high-and-low voltage windings for this connection and the MVA rating of the autotransformer

SIZIF [17.4K]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The High Voltage Rating for Auto - Transformer is 86kV

The Low Voltage Rating for Auto - Transformer is 78kV

The MVA rating is 268.75 $MVA$

The efficiency is 99.4%

Explanation:

From the question we are given are given that

The transformer has Mega Volt Amp rating of 25MVA

The frequency is 60-Hz

Voltage rating 8.0kV : 78kV

The short circuit test gives : 453kV,321A,77.5kW

The open circuit test gives : 8.0kV, 39.6A, 86.2kW

This can be represented on a diagram shown on the second uploaded image

From this diagram we can deduce that the The High Voltage Rating for Auto - Transformer is 86kV and the Low Voltage Rating for Auto - Transformer is 78kV

Now to obtain the current flowing through the 8kV coil in the Auto-transformer we have

$\frac{25 \ Mega \ Volt\ Ampere }{8\ Kilo Volt}$

The volt will cancel each other

$\frac{25*10^6}{8*10^3} = 3125\ A$

Now to obtain the required MVA rating we would multiply the value of Power obtained during the open circuit test by the value of the current calculated.we are making use of the power obtain during open circuit testing because the transformer at this point is not under any load.

$MVA \ rating = (86*10^3)(3125) =268.75$

We need to understand that Iron losses is due to open circuit test which has power = 86.2kW

While copper loss is due to short circuit test which has power = 77.5kW

The the current flowing through the secondary coil $I_2$ as shown in the circuit diagram can be obtained as

$I_2 = \frac{25*10^6}{78*10^3} =320.52 A \approx 321$

Now the efficiency can be obtained as thus

$\frac{(operational \ MVA )*(Power factor \pf))}{(operational\ MVA (power factor pf) + copper loss + Iron loss)}*\frac{100}{1}$

=99.941%

8 0

3 years ago

A car is moving with the velocity of 90km/h. If the car come to rest after 10 seconds. Calculate the final velocity and distance

djyliett [7]

Answer:

you can learn from here

https://www.toppr.com/ask/en-bd/question/a-car-is-moving-with-a-velocity-of-10-ms-the-driver-sees-a-wall/

6 0

3 years ago

How much meters is a mile

elena-s [515]

Roughly 1609 meters in one mile

6 0

3 years ago

A 15-ft3 tank contains oxygen initially at 14.7 psia and 80°F. A paddle wheel within the tank is rotated until the pressure insi

Nikolay [14]

Explanation:

Equation for energy balance will be as follows.

$\Delta E_{system} = E_{in} - E_{out}$

$\Delta U = W_{in} - Q_{out}$

Hence, $W_{in} = Q_{out} + mC_{v} (T_{2} - T_{1})$

Therefore, we will calculate the final temperature as follows.

$\frac{P_{1}V}{T_{1}} = \frac{P_{2}V}{T_{2}}$

$T_{2} = \frac{20 psia}{14.7}(638 R)$

= 868.03 R

Now, we will calculate the mass as follows.

m = $\frac{P_{1}V}{RT_{1}}$

= $\frac{14.7 psia \times 15 ft^{3}}{0.3353 psi ft^{3}/lbm R \times 638 R}$

= 1.031 lbm

Hence,

$W_{in} = Q_{out} + mC_{v} (T_{2} - T_{1})$

Putting the values into the above equation as follows.

$W_{in} = Q_{out} + mC_{v} (T_{2} - T_{1})$

$W_{in} = 20 Btu + 1.031 lbm (\frac{0.160 Btu}{lbm R})(735 - 540)R$

$W_{in}$ = 655.2 Btu

Thus, we can conclude that work done by paddle wheel is 655.2 Btu.

6 0

3 years ago

What is difference between kilowatt and kilowatt hour?

Scrat [10]

Those two units can be compared to a 'mile per hour' and a 'mile per hour - hour'.
One is a rate. The other is a quantity, after maintaining a rate for some time.

-- 'Joule' is a unit of energy. It's the amount of work (energy) you do
when you push with a force of 1 newton though a distance of 1 meter.
Lifting 10 pound of beans 3 feet off the floor takes about 40.7 joules of energy.

-- 'Watt' is a rate of using energy . . . 1 joule per second.
If you lift 10 pounds 3 feet off the floor in 1 second, your power is 40.7 watts.

-- 'Watt-second' is the amount of energy used in one second,
at the rate of 1 joule per second . . . 1 joule.

-- 'Watt-hour' is the amount of energy used in one hour,
at the rate of 1 joule per second . . . 3,600 joules.

-- 'Kilowatt' is a bigger rate of using energy . . . 1,000 joules per second.

-- 'Kilowatt - second' is the amount of energy used in one second,
at the rate of 1,000 joules per second . . . 1,000 joules .

-- 'Kilowatt - hour' is the amount of energy used in one hour,
at the rate of 1,000 joules per second . . . 3,600,000 joules .

Depending on where you live, 3,600,000 joules of energy bought
from the electric company costs something between 5¢ and 25¢.

6 0

3 years ago

Read 2 more answers