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3 years ago

A 70 kg man is walking at a speed of 2 m/s. What is his Kinetic Energy?

Physics

2 answers:

Ilya [14]3 years ago

8 0

Kinetic energy = 1/2 (mass) (speed squared)

KE = 1/2 (70kg) (2m/s)^2

KE = (35 kg) (4 m^2/s^2)

KE = (140) kg-m^2/s^2

That's 140 Newton-meter

140 Joules

den301095 [7]3 years ago

5 0

140 Jules. Is the correct answer

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Two parallel plates that are initially uncharged are separated by 1.7 mm, have only air between them, and each have surface area

yaroslaw [1]

Answer:

$5.63\cdot 10^{-6} C$

Explanation:

The capacitor of a parallel-plate capacitor is given by:

$C=\epsilon_0 \frac{A}{d}$

where

A is the area of each plate

d is the separation between the plates

$\epsilon_0$ is the vacuum permittivity

The energy stored in a capacitor instead is given by

$U=\frac{1}{2}\frac{Q^2}{C}$

where

Q is the charge stored in each plate

Substituting the expression we found for C inside the last formula,

$U=\frac{1}{2}\frac{Q^2 d}{\epsilon_0 A}$

And re-arranging it

$Q=\sqrt{\frac{2U\epsilon_0 A}{d}}$

Now if we substitute

$d=1.7 mm=0.0017 m\\A=16 cm^2 = 16\cdot 10^{-4} m^2\\U = 1.9 J$

We find the charge stored on the capacitor:

$Q=\sqrt{\frac{2(1.9)(8.85\cdot 10^{-12})(16\cdot 10^{-4})}{0.0017}}=5.63\cdot 10^{-6} C$

7 0

3 years ago

The items in a mixture can be returned to their original form. True False

Grace [21]

I believe it’s false.

4 0

3 years ago

Read 2 more answers

A 1502.7 kg car is traveling at 33.1 m/s when

Bond [772]

By definition of average acceleration,

a = (20 m/s - 33.1 m/s) / (4.7 s) ≈ -2.78 m/s²

Vertically, the car is in equilibrium, so the net force is equal to the friction force in the direction opposite the car's motion:

∑ F = (1502.7 kg) (-2.78 m/s²) ≈ -4188.38 N ≈ -4200 N

If you just want the magnitude, drop the negative sign.

5 0

3 years ago

A 25 kg block is held against a compressed spring and then the spring is allowed to decompress giving the block a velocity. The

Alex787 [66]

Answer:

h=18.05 cm

Explanation:

Given that

m= 25 kg

K= 1300 N/m

x=26.4 cm

θ= 19.5 ∘

When the block just leave the spring then the speed of block = v m/s

From energy conservation

$\dfrac{1}{2}Kx^2=\dfrac{1}{2}mv^2$

$Kx^2=mv^2$

$v=\sqrt{\dfrac{kx^2}{m}}$

By putting the values

$v=\sqrt{\dfrac{kx^2}{m}}$

$v=\sqrt{\dfrac{1300\times 0.264^2}{25}}$

v=1.9 m/s

When block reach at the maximum height(h) position then the final speed of the block will be zero.

We know that

$V_f^2=V_i^2-2gh$

By putting the values

$0^2=1.9^2-2\times 10\times h$

h=0.1805 m

h=18.05 cm

4 0

3 years ago

A fluid in an aquifer is 23.6 m above a reference datum, the fluid pressure (in gage pressure) is 4390 n/m2 and the flow velocit

Phoenix [80]

As per Bernuolli's Theorem total energy per unit mass is given as

$\frac{P}{\rho} + \frac{1}{2}v^2 + gH = E$

now from above equation

$P = 4390 N/m^2$

$\rho = 0.999 * 10^3$

$v = 7.22 * 10^{-4} m/s$

$H = 23.6 m$

now by above equation

$\frac{4390}{0.999*10^3} + \frac{1}{2}*(7.22*10^{-4})^2 + 9.8*23.6 = E$

$E = 235.7 J/kg$

Part B)

Now energy per unit weight

$U = \frac{E}{g}$

$U = \frac{235.7}{9.8}$

$U = 24 m$

7 0

3 years ago