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3 years ago

When it comes to graphs I am no genius lol

Physics

1 answer:

Lunna [17]3 years ago

5 0

Answer:

Option B

Explanation:

Displacement = Area under the velocity-time graph

So to find the displacement of a particle from 7 to 8 seconds we take the triangle and calculate its area.

$Area\ of\ a\ right\ angled\ triangle = \frac{1}{2}*(base)*(height) \\\\Area\ of\ a\ right\ angled\ triangle = \frac{1}{2}*(1)*(3) \\\\Area\ of\ a\ right\ angled\ triangle = 1.5$

So the displacement is 1.5 meters

I have an uploaded image so you can understand it better hope it helps

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Helpppp!!!

Korvikt [17]

Answer:

A) 0.25 Hz

B) 6 m

C) 1.5 m/s

Explanation:

The number of waves that pass a certain spot every 20 seconds = 5 waves

The distance between the crest = 6 m

A) The frequency of the wave, f = The number of cycles per second = 5/20 = 0.25 per second = 0.25 Hz

B) The wavelength of the wave, λ = The distance between crests = 6 m

C) The velocity of the wave, v = f × λ = 0.25/s × 6 m = 1.5 m/s.

5 0

3 years ago

An ideal spring is fixed at one end. A variable force F pulls on the spring. When the magnitude of F reaches a value of 49.1 N,

balu736 [363]

When the spring is stretched by 15.2 cm = 0.152 m, the spring exerts a restorative force with magnitude (due to Hooke's law)

$F = kx$

where $k$ is the spring constant. Solve for $k$ .

$49.1\,\mathrm N = k (0.152\,\mathrm m) \implies k \approx 323 \dfrac{\rm N}{\rm m}$

The amount of work required to stretch or compress a spring by $x\,\mathrm m$ from equilibrium length is

$W = \dfrac12 kx^2$

Then the work needed to stretch the spring by 15.2 cm is

$W_1 = \dfrac12 \left(343\dfrac{\rm N}{\rm m}\right) (0.152\,\mathrm m)^2 \approx 3.73\,\mathrm J$

and by 15.2 + 13.7 = 28.9 cm is

$W_2 = \dfrac12 \left(343\dfrac{\rm N}{\rm m}\right) (0.289\,\mathrm m)^2 \approx 13.5\,\mathrm J$

so the work needed to stretch from 15.2 cm to 28.9 cm from equilibrium is

$\Delta W = W_2 - W_1 \approx \boxed{9.76\,\mathrm J}$

4 0

2 years ago

How fast would the car need to go to double its kinetic energy? 12 m/s?

S_A_V [24]

The car is traveling at v=12m/s. Kinetic energy is given by this formula:
$E_k=\frac{mv^2}{2}$
We can see that if the kinetic energy depends on the speed quadratically. This means that if you want to increase the kinetic energy x times you would have to increase speed $\sqrt{x}$ times. We conclude from this that car would have to go:
$v'=\sqrt2v=\sqrt2\cdot 12=16.97\frac{m}{s}$

7 0

3 years ago

Two identical soccer balls are rolled toward each other. What will be true after they collide head-on

Juliette [100K]

no because there is no chance that it will possibly hit eachother

6 0

2 years ago

During the day, susan notices that the wind is blowing onshore at the beach. What is this called? land breeze land breeze sea br

In-s [12.5K]

The correct option is (b) Sea breeze

During the day, susan notices that the wind is blowing onshore at the beach called sea breeze.

What is sea breeze?

Any wind that flows from a big body of water onto or onto a landmass is called a sea breeze or an onshore breeze.
Sea breezes form as a result of changes in air pressure brought on by the different heat capacities of water and dry land. Sea breezes are therefore more confined than prevailing winds.
A sea wind is frequently seen along coasts after sunrise because land warms up far more quickly than water does when exposed to solar radiation.
The sea wind front is significant because it can serve as a catalyst for afternoon thunderstorms and provide welcome cooling along the coast.

Learn more about the sea breeze with the help of the given link:

brainly.com/question/13015619

#SPJ4

8 0

2 years ago

When it comes to graphs I am no genius lol​

When it comes to graphs I am no genius lol