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3 years ago

A small 18 kilogram canoe is floating downriver at a speed of 1 m/s. What is the canoe's kinetic energy?

Physics

1 answer:

In-s [12.5K]3 years ago

8 0

Kinetic energy = (1/2) (mass) (speed²).

A Physicist in the canoe, or on a raft floating downriver next to the canoe, will say that the canoe's kinetic energy is zero.

A Physicist on the riverbank, watching the canoe drift by at 1 m/s, will say that its kinetic energy is 9 Joules.

They're both correct.

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Explain the process of a nuclear fusion reaction using hydrogen. Include the particles that are used to start and maintain the c

Zielflug [23.3K]

Answer it might be this

Explanation:

8 0

3 years ago

The valence electrons of metals are weakly attracted to the parent nuclei, so the electrons break free and float. The moving ele

siniylev [52]

The valence electrons of metals are weakly attracted to the parent nuclei, so the electrons break free and float. The moving electrons form a electron negative blanket that binds the atomic positive nuclei together, forming a metallic bond.

So the answers are { Negative } and { Positive }.

Please vote Brainliest (:

5 0

3 years ago

Read 2 more answers

Canadian geese migrate essentially along a north-south direction for well over a thousand kilometers in some cases, traveling at

Wewaii [24]

Answer:

θ=19.877⁰

Explanation:

Given data

Velocity Va=34.0 km/h

Velocity Va=100 km/h

To find

Angle θ

Solution

We want the bird to fly with velocity Vb=100 km/h with an angle θ relative to the ground so that the bird fly due south relative to the ground.From figure which is attached we got

Sinθ=(Va/Vb)

Sinθ=(34.0/100)

θ=Sin⁻¹(34.0/100)

θ=19.877⁰

5 0

3 years ago

A fisherman notices that his boat is moving up and down periodically, owing to waves on the surface of the water. It takes 2.5 s

gulaghasi [49]

(a) 0.96 m/s

The period of the wave corresponds to the time taken for one complete oscillation of the boat, from the highest point to the highest point again. Since the time between the highest point and the lowest point is 2.5 s, the period is twice this time:

$T=2\cdot 2.5 s=5.0 s$

The frequency of the waves is the reciprocal of the period:

$f=\frac{1}{T}=\frac{1}{5.0 s}=0.20 Hz$

The wavelength instead is just the distance between two consecutive crests, so

$\lambda=4.8 m$

And the wave speed is given by:

$v=\lambda f=(4.8 m)(0.20 Hz)=0.96 m/s$

(b) 0.265 m

The total distance between the highest point of the wave and its lowest point is

d = 0.53 m

The amplitude is just the maximum displacement of the wave from the equilibrium position, so it is equal to half of this distance. So, the amplitude is

$A=\frac{d}{2}=\frac{0.53 m}{2}=0.265 m$

(c) Amplitude: 0.15 m, wave speed: same as before

In this case, the amplitude of the wave would be lower. In fact,

d = 0.30 m

So the amplitude would be

$A=\frac{d}{2}=\frac{0.30 m}{2}=0.15 m$

Instead, the wave speed would not change, since neither the frequency nor the wavelength of the wave have changed.

8 0

3 years ago

(Thought question) At noon On June 21 will the shadow length of 42 degrees north latitude be shorter or longer than it is on Mar

g100num [7]

Answer: Shorter

Explanation: Shadow is formed when an light source is obstructed by an opaque object. The closer the source, shorter is the length of the shadow. In fact, when the source is exactly overhead, no shadow of the object is formed.

June 21 marks the Summer solstice which means the Sun passes directly overhead Tropic of cancer (23.5° N) at noon. March 21 marks the equinox which means sun passes directly overhead equator (0°).

Shadow length of an object at 42° Northern latitude will be shorter on June 21 because the Sun will be closer to this latitude as compared to March 21.

5 0

3 years ago