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Ronch [10]
3 years ago
15

A small 20-kilogram canoe is floating downriver at a speed of 2 m/s. What is the canoe's kinetic energy?

Physics
2 answers:
MariettaO [177]3 years ago
8 0
Ek=1/2mv^2 = (20)(1/2)(2x2) =40
Novosadov [1.4K]3 years ago
5 0

Answer: 40 Joules

Explanation: Kinetic energy is the energy possessed by a body by virtue of its motion.

Kinetic energy depends on the mass and speed of the object.

K.E=\frac{1}{2}\times mv^2

m= mass of canoe = 20 kg

v= speed of canoe = 2 m/s

K.E=\frac{1}{2}\times 20kg\times (2m/s)^2

K.E=40kgm^2s^-{2}=40Joules

(1kgm^2s^{-2}= 1Joule)

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A car speedometer has a 4% uncertainty. What is the range of possible speeds (in km/h) when it reads 110 km/h?
Kruka [31]
4% of 110 is 4.4. So the possible range of speeds is the interval from 110-4.4 till 110+4.4.
105.6 till 114.4
4 0
3 years ago
Argue as to why drivint on icy or snowy roads can be dangerous
Vika [28.1K]
Icy/Snowy roads have less friction than normal roads. This means that the wheels are less likely to stay positioned because of traction, and you will spin out of control
6 0
3 years ago
First right answer gets brainly<br><br>What is the speed of a man who runs 14 miles in 144 minutes?
andreev551 [17]

Answer:

10 mph

Explanation:

7 0
3 years ago
The coefficient of kinetic friction between an object and the surface upon which it is sliding is 0.40 the weight of the object
Ann [662]

The frictional force of an object is the product of the normal force and coefficient of kinetic friction. Here the frictional force acting on the object  is 16.4 N.

<h3>What is frictional force?</h3>

Frictional force is a kind of force acting on a body to resist it from motion. Thus, the direction of the force will be in negative with the magnitude. Frictional force is the product of coefficient of friction and the normal force.

The normal  force acting on the object of mass 4.2 Kg is N = mg

N = 4.2 Kg × 9.8 m/s² = 41.16 N

Frictional force = ц N

                         = 0.40 × 41.16 N

                         = 16.4 N.

Therefore, the frictional force acting between the surface of the object and the floor is 16.4 N

To find more on frictional force, refer here:

brainly.com/question/1714663

#SPJ1

Your question is incomplete. But your complete question probably was:

The coefficient of kinetic friction between an object and the surface upon which it is sliding is 0.40 the weight of the object is 4.2 kg. What is the frictional force of the object?

3 0
1 year ago
Two balls of clay, with masses M1 = 0.49 kg and M2 = 0.47 kg, are thrown at each other and stick when they collide. Mass 1 has a
malfutka [58]

Answer:

a) p_i=1.568\hat{i}+0.752 \hat{j}

b) v_{fx}=1.668\ m.s^{-1}

c) v_{fy}=0.7999\ m.s^{-1}

Explanation:

Given masses:

m_1=0.49\ kg

m_2=0.47\ kg

Velocity of mass 1, v_1=3.2 \hat{i}\ m.s^{-1}

Velocity of mass 2, v_2=1.6 \hat{j}\ m.s^{-1}

a)

Initial momentum:

p_i=m_1.v_1+m_2.v_2

p_i=0.49\times 3.2 \hat{i}+0.47\times 1.6 \hat{j}

p_i=1.568\hat{i}+0.752 \hat{j}

b)

magnitude of initial momentum:

p_i=\sqrt{1.568^2+0.752 ^2}

p_i=1.739\ kg.m.s^{-1}

From the conservation of momentum:

p_f=p_i

m_f.v_f=1.739

v_f=\frac{1.739}{0.49+0.47}

v_f=1.85\ m.s^{-1} is the magnitude of final velocity.

Direction of final velocity will be in the direction of momentum:

tan\theta=\frac{0.752 }{1.568}

\theta=25.62^{\circ}

\therefore v_{fx}=1.85\ cos25.62^{\circ}

v_{fx}=1.668\ m.s^{-1}

c)

Vertical component of final velocity:

v_{fy}=1.85\ sin 25.62^{\circ}

v_{fy}=0.7999\ m.s^{-1}

6 0
4 years ago
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