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3 years ago

Write down formula of power

Physics

2 answers:

r-ruslan [8.4K]3 years ago

8 0

There u go -> P=f/a

zzz [600]3 years ago

5 0

Answer:

Power is a measure of the amount of work that can be done in a given amount of time. So, the formula is the energy or work divided by time:

P = W ÷ T, where p is power , w is work and t is time.

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The engine in an imaginary sports car can provide constant power to the wheels over a range of speeds from 0 to 70 miles per hou

Inessa05 [86]

Answer:

a) 4.40 s

b) 2.20 s

Explanation:

Given parameters are:

At constant power ,

initial speed of the car, $v_0=0$

final speed of the car, $v=32$ mph

At full power,

initial speed of the car, $v_0=0$

final speed of the car, $v=64$ mph

At constant power, $KE = \frac{1}{2} mv^2$

At full power, $KE = \frac{1}{2} m(2v)^2$

So $KE_f = 4KE_i$

So, time to reach 64 mph speed is 4 times more than the initial time

$t = 4*1.10 =4.40$ s

$v=v_0+at\\a=\frac{v-v_0}{t}=\frac{32-0}{1.1/3600}=104727.27$ $miles/hours^2$

For final 64 mph speed,

$v=v_0+at\\t=\frac{v-v_0}{a}=\frac{64-0}{104727.27} = 6.111*10^{-4}$ $hours$ = $6.111*10^{-4}*3600=2.20$ s

7 0

3 years ago

A 30 g horizontal metal bar, 13 cm long, is free to slide up and down between two tall, vertical metal rods that are 13 cm apart

natita [175]

Answer:

Terminal speed, v = 6901.07 m/s

Explanation:

It is given that,

Mass of the horizontal bar, m = 30 g = 0.03 kg

Length of the bar, l = 13 cm = 0.13 m

Magnetic field, $B=5.5\times 10^{-2}\ T$

Resistance, R = 1.2 ohms

We need to find the terminal speed oat which the bar falls. When terminal speed is reached,

Force of gravity = magnetic force

$mg=ilB$ ..................(1)

i is the current flowing

l is the length of the rod

Due to the motion in rods, an emf is induced in the coil which is given by :

$E=Blv$ , v is the speed of the bar

$iR=Blv$

$i=\dfrac{Blv}{R}$

Equation (1) becomes,

$mg=\dfrac{B^2l^2v}{R}$

$v=\dfrac{mgR}{B^2l^2}$

$v=\dfrac{0.03\times 9.8\times 1.2}{(5.5\times 10^{-2})^2(0.13)^2}$

v = 6901.07 m/s

So, the terminal speed at which the bar falls is 6901.07 m/s. Hence, this is the required solution.

5 0

3 years ago

onsider 1000 mL of a 1.00 × 10-4 M solution of a certain acid HA that has a Ka value equal to 1.00 × 10-4. Water was added or re

solmaris [256]

Answer:

The volume of the final solution, V = 0.0305L

Explanation:

Number of moles = Concentration * volume

Concentration of HA = 1.00 * 10⁻⁴M

Volume of HA = 1000mL = 1 L

Number of moles of HA = 1.00 * 10⁻⁴ * 1

Number of moles of HA = 1.00 * 10⁻⁴ mols

Equation of reaction:

HA → H⁺ + A⁻

If 1 mol of HA produces 1 mol of H⁺ and A⁻, 1.00 * 10⁻⁴ mol of HA will produce 1.00 * 10⁻⁴ mol of H⁺ and A⁻.

Since only 16% dissociation occurs = 0.16

Number of moles of H⁺ produced = 0.16 * 1.00 * 10⁻⁴

Number of moles of H⁺ produced = 1.6 * 10⁻⁵mols

Number of moles of A⁻ produced = 0.16 * 1.00 * 10⁻⁴

Number of moles of A⁻ produced = 1.6 * 10⁻⁵mols

Since 16% of HA dissociated into H⁺ and A⁻, 84% of HA is left

Number of mols of HA left = 0.84 * 1.00 * 10⁻⁴

Number of mols of HA left = 8.4 * 10⁻⁵mols

Concentration = num of moles/volume

Let the volume of the final solution be V

Conc of HA = 8.4 * 10⁻⁵/V

Conc of H⁺ = 1.6 * 10⁻⁵/V

Conc of A⁻ = 1.6 * 10⁻⁵/V

To calculate the dissociation constant

$k_{a} = [H^{+} ][A^{-} ]/[HA]$

$k_{a}= [1.6 * 10^{-5} /V][1.6 * 10^{-5} /V]/[8.4 * 10^{-5} /V]\\k_{a}= 3.05 * 10^{-6} /V\\k_{a} = 1.00 * 10^{-4}\\ 1.00 * 10^{-4} = 3.05 * 10^{-6} /V\\V= 3.05 * 10^{-6}/ 1.00 * 10^{-4}\\V=3.05 * 10^{-2}\\V=0.0305 L$

3 0

3 years ago

A punter drops a 2.0 kg ball from rest vertically 1 meter down onto his foot. The ball leaves the foot with a speed of 18 m/s at

pav-90 [236]

Answer:

Explanation:

Given

mass of drop $m=2 kg$

height of fall $h=1 m$

ball leaves the foot with a speed of 18 m/s at an angle of $55^{\circ}$

Velocity of ball just before the collision with the floor

$u^=2gh$

$u=\sqrt{2gh}$

$u=\sqrt{2\times 9.8\times 1}=4.42 m/s$

Impulse delivered in Y direction

$J_y=m(v\sin (55)-(-u))$

$J_y=2(18\sin (55)+4.42)$

$J_y=38.32 kg-m/s$

Impulse in x direction

$J_x=m\times v\cos (55)$

$J_x=2\times 4.42\cos (55)=20.646$

$J_{net}=\sqrt{J_x^2+J_y^2}$

$J_{net}=\sqrt{(38.32)^2+(20.64)^2}$

$J_{net}=43.52 kg-m/s$

at an angle of $\tan \phi =\frac{J_y}{J_x}=\frac{38.32}{20.64}$

$\phi =tan^{-1}(1.856)$

$\phi =61.7^{\circ}$

7 0

3 years ago

Which type of acceleration would be best if you were merging onto the interstate?

kondaur [170]

Answer:

If merging onto an interstate, it is best to merge at the speed of the moving traffic.

Explanation:

As a driver, you cannot always count on the other drivers to see you wanting to merge onto the interstate, and even if they see you, they might not be willing to stop just so you merge, therefore it is best advised to look for a gap in the traffic and merge at the traffic speed.

4 0

3 years ago

Write down formula of power​

Write down formula of power