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3 years ago

What would the speed of this wave be?

Physics

1 answer:

valentinak56 [21]3 years ago

7 0

Answer:

c) 100,000 m/s

Explanation:

You need to take the same wave length from the top graph and bottom one, so let's take half a wave length then in the top one that is 0.005, but in the bottom one it's 2000/4 = 500 because they are smaller and there are 4 half waves before you get to 2000, whereas in the top one there is 1 half wave before you get to 0.005 on the graph.

Now use speed = distance / time

speed = 500 / 0.005 = 100 000 m/s

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find the time taken, if the speed of a train increased from 72 km/hr to 90 km/hr for 234 km. leave your answer in seconds

Airida [17]

Answer:

Time taken = 10400 s

Explanation:

Given:

Initial speed of the train, $u=72\textrm{ km/h}=72\times \frac{5}{18}=20\textrm{ m/s}$

Final speed of the train, $v=90\textrm{ km/h}=90\times \frac{5}{18}=25\textrm{ m/s}$

Displacement of the train, $S=234\textrm{ km}=234\times 1000=234000\textrm{ m}$

Using Newton's equation of motion,

$v - u = at\\a=\frac{v-u}{t}$

Now, using Newton's equation of motion for displacement,

$v^{2}-u^{2}=2aS$

Now, plug in the value of $a=\frac{v-u}{t}$ in the above equation. This gives,

$v^{2}-u^{2}=2\times \frac{v-u}{t}\times S\\(v+u)(v-u)=\frac{2(v-u)S}{t}\\t=\frac{2(v-u)S}{(v+u)(v-u)}\\t=\frac{2S}{v+u}$

Now, plug in 234000 m for $S$ , 25 m/s for $v$ and 20 m/s for $u$ . Solve for $t$ .

$t=\frac{2S}{v+u}\\t=\frac{2\times 234000}{25+20}\\t=\frac{468000}{45}=10400\textrm{ s}$

Therefore, the time taken by the train is 10400 s.

3 0

3 years ago

how fast will and in what direction will a 20kg object accelerate if one force pushes at a 30 degree angle and another pushes at

DiKsa [7]

Answer:

$|a|=2.83\ m/s^2$

$\theta=75^o$

Explanation:

<u>Net Force And Acceleration </u>

The Newton's second law relates the net force applied on an object of mass m and the acceleration it aquires by

$\vec F_n=m\vec a$

The net force is the vector sum of all forces. In this problem, we are not given the magnitude of each force, only their angles. For the sake of solving the problem and giving a good guide on how to proceed with similar problems, we'll assume both forces have equal magnitudes of F=40 N

The components of the first force are

$\vec F_1=$

$\vec F_1=\ N$

The components of the second force are

$\vec F_2=$

$\vec F_2=\ N$

The net force is

$\vec F_n=$

$\vec F_n=\ N$

The magnitude of the net force is

$|F_n|=\sqrt{14.64^2+54.64^2}$

$|F_n|=\sqrt{3200}=56.57\ N$

The acceleration has a magnitude of

$\displaystyle |a|=\frac{|F_n|}{m}$

$\displaystyle |a|=\frac{56.57}{20}$

$|a|=2.83\ m/s^2$

The direction of the acceleration is the same as the net force:

$\displaystyle tan\theta=\frac{54.64}{14.64}$

$\theta=75^o$

5 0

2 years ago

A laptop can convert 400J of electrical energy into a 240J of light and sound. What is the efficiency? Where does the rest of th

Blababa [14]

Efficiency = (Wanted) energy out ÷ energy in × 100

Energy in = 400J
Wanted Energy out = 240J

Energy cannot be used up, only transferred, so the remaining energy is most likely to be transferred into unwanted energy (loss of energy) such as heat energy.

Efficiency = 240 ÷ 400 × 100
Efficiency = 0.6 × 100
Efficiency = 60%

8 0

3 years ago

Which will increase the kinetic energy of an object more, increasing its mass or increasing its speed? Why?

agasfer [191]

Increasing its velocity will add to the kinetic energy more as the formula for kinetic energy is 0.5*m*v^2. (The speed will be squared making it greater)

4 0

3 years ago

What is the volume of 45.6 g of silver if the density of silver is 10.5 g/ml?

Leviafan [203]

Volume = mass / density = 45.6/10.5 = .... L

8 0

3 years ago

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