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3 years ago

What is the kinetic energy of a 3.2 kg pike swimming at 0.5 m/s?

Physics

1 answer:

lisabon 2012 [21]3 years ago

6 0

Answer:

0.4 J

Explanation:

Because the formula is Ek= $\frac{1}{2}mv^{2}$

So Ek= $\frac{1}{2}$ ×3.2×0.5²

Then Ek=1/2×3.2×0.25

After Ek=1.6×0.25

In the end Ek= 0.4

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The softest known mineral is talc. It has a hardness of 1. What is the hardest known mineral

Vinvika [58]

Answer:

diamond

Explanation:

diamond has a hardness of 10, and is the hardest mineral

3 0

3 years ago

A 1,000 kg ball traveling at 5 m/s would have kinetic energy

MatroZZZ [7]

Answer:

12500 J = 12.5 kJ

Explanation:

Kinetic energy is the energy possessed by a moving object solely due to its motion.

You can get the K.E. of an object using the equation,

K.E. = (1/2)mv²

So you get, for ball

K.E. = (1/2)×1000×5² = 12500 J = 12.5 kJ

6 0

4 years ago

There are Z protons in the nucleus of an atom, where Z is the atomic number of the element. An α particle carries a charge of +2

qaws [65]

Answer:

$r = 2.84 \times 10^{-14} m$

Explanation:

As per energy conservation we know that the electrostatic potential energy of the charge system is equal to the initial kinetic energy of the alpha particle

So here we can write it as

$\frac{1}{2}mv^2 = \frac{k(2e)(ze)}{r}$

now we know that

$m = 1.67 \times 10^{-27} kg$

$e = 1.6 \times 10^{-19} C$

z = 79

here kinetic energy of the incident alpha particle is given as

$KE = 6.4 \times 10^{-13} J$

now we have

$6.4 \times 10^{-13} = \frac{(9\times 10^9)(1.6 \times 10^{-19})^2(79)}{r}$

now we have

$r = 2.84 \times 10^{-14} m$

7 0

4 years ago

A test charge of +2μC is placed halfway between a charge of +6μC and another of +4μC separated by 10 cm. (a) What is the magnitu

Nady [450]

Answer:

Part a)

F = 14.4 N

Part b)

away from +6μC charge

Explanation:

Force on the charge placed midway between two charges is given by

$F = \frac{kq_1q}{r^2} - \frac{kq_2q}{r^2}$

here we know that

$q = 2\mu C$

$q_1 = 6 \mu C$

$q_2 = 4\mu C$

here the charge is placed at mid point so we have

$r = 5 cm$

now we have

$F = \frac{(9\times 10^9)(6\mu C)(2\mu C)}{0.05^2} - \frac{(9\times 10^9)(2\mu C)(4\mu C)}{0.05^2}$

$F = 43.2 - 28.8 = 14.4 N$

Part b)

since +6μC charge is more in magnitude so the force due to this charge will be more so the net force on it is away from +6μC charge

8 0

4 years ago

My fifth time asking this, but can someone please help me with these four????

givi [52]

I think 3 is 1.5...i kinda hope that helps for one

3 0

3 years ago