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Zigmanuir [339]
3 years ago
15

A small 22 kilogram canoe is floating downriver at a speed of 2 m/s. What is the canoe's kinetic energy

Physics
2 answers:
VladimirAG [237]3 years ago
4 0
Kinetic energy is simply 1/2 *mass* velocity^2 
so since you know both mass and velocity you should be able to calculate the kinetic energy. 
Anna [14]3 years ago
4 0
A student on a raft floating behind the canoe says the canoe has NO kinetic energy.
A teacher standing on the riverbank when the canoe floats by says it has 44 joules of KE.
They're both correct.
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A cold block of metal feels colder than a block of wood at the same temperature. Why? A hot block of metal feels hotter than a b
tino4ka555 [31]

: In general, metals feel colder or hotter to the touch than other materials at the same temperature because they're good thermal conductors. This means they easily transfer heat to colder objects or absorb heat from warmer objects

5 0
3 years ago
Read 2 more answers
A uniform electric field exists in the region between two oppositely charged plane parallel plates. A proton is released from re
valentinak56 [21]

Answer:

a) 17.33 V/m

b) 6308 m/s

Explanation:

We start by using equation of motion

s = ut + 1/2at², where

s = 1.2 cm = 0.012 m

u = 0 m/s

t = 3.8*10^-6 s, so that

0.012 = 0 * 3.8*10^-6 + 0.5 * a * (3.8*10^-6)²

0.012 = 0.5 * a * 1.444*10^-11

a = 0.012 / 7.22*10^-12

a = 1.66*10^9 m/s²

If we assume the electric field to be E, and we know that F =qE. Also, from Newton's law, we have F = ma. So that, ma = qE, and E = ma/q, where

E = electric field

m = mass of proton

a = acceleration

q = charge of proton

E = (1.67*10^-27 * 1.66*10^9) / 1.6*10^-19

E = 2.77*10^-18 / 1.6*10^-19

E = 17.33 V/m

Final speed of the proton can be gotten by using

v = u + at

v = 0 + 1.66*10^9 * 3.8*10^-6

v = 6308 m/s

5 0
3 years ago
Once a ball leaves a table at 10 m/s, how long will it take to hit the floor
Zarrin [17]

<u>Answer:</u>

The time taken for the ball to hit the floor as 1.02 seconds

<u>Explanation:</u>

As per the given question, the ball leaves at a speed from the table with an initial velocity of 10 m/s, we have the equation

\mathrm{V}_{\mathrm{f}}=\mathrm{V}_{\mathrm{i}}+\mathrm{at}

where Vf represents the final velocity

 Vi represents the initial velocity  

a represents the acceleration and

t represents the time

\mathrm{V}_{\mathrm{f}}=\mathrm{V}_{\mathrm{i}}+\mathrm{at} after rearranging

 \mathrm{t}=\frac{\mathrm{vf}-\mathrm{vi}}{a}

  \mathrm{t}=\frac{0-10}{9.8}  = 1.022 seconds

6 0
4 years ago
Gravity and magnetism are both
ella [17]

Answer:

Explanation:

Comment

You have to read this carefully enough that you don't mix up energy and forces.

Gravity is a force. If you don't believe me try jumping off a building. Which way are you going to go and why? Down because gravity attracts your mass.

So Magnetism must be a force as well. It acts in one direction, but not a specific one the way gravity acts). It also either attracts or repulses (pushes an object away)

Answer A

3 0
2 years ago
Read 2 more answers
Two blocks with rough surfaces are stacked on top of a slippery horizontal surface. The top block has a mass of m, and the botto
lapo4ka [179]

Answer:

Please refer to the attached schematic for free-body diagrams of the blocks.

The contact forces N1 = mg, N2 = 3mg.

The friction force fs = mg/2.

The tension force T = 3mg/2.

The distance y = (gt^2)/4

Explanation:

Starting with the third block which is declining a time t. Newton's Second Law implies that

F = (3m)a\\3mg - T = 3ma\\T = 3mg - 3ma

There are two unknowns in one equation: T and a. Therefore, we need to find more equations to solve these unknown variables.

Let's look at free-body diagram 1:

Newton's Second Law:

F = ma\\f_s = ma

That is another equation with one more unknown: fs.

Free-body diagram 2:

F = (2m)a\\T - f_s = 2ma

That is our third equation. We now have three equations with three unknowns. Combining equation 2 and 3 gives:

T - ma = 2ma\\T = 3ma

Plugging this equation into the first equation gives

T = 3mg - 3ma\\3ma = 3mg - 3ma\\3mg = 6ma\\a = g/2

Following this, we can find the other unknowns:

T = 3ma = 3m(g/2) = \frac{3mg}{2}

f_s = ma = mg/2

The contact forces are N1 and N2.

N_1 = mg\\N_2 = 2mg + N1 = 3mg

Finally, using the acceleration and equation of kinematics, we can find the distance the hanging weight drops in a time t.

y - y_0 = v_0t + \frac{1}{2}at^2\\y = \frac{1}{2}(\frac{g}{2})t^2 = \frac{gt^2}{4}

6 0
4 years ago
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