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3 years ago

The graph below shows the velocity of a car over time.

Physics

2 answers:

Vadim26 [7]3 years ago

6 0

Answer:

its calculate the area under the graph like the other person said :0

Explanation:

jekas [21]3 years ago

3 0

Answer:

D. Calculate the area under the graph.

Explanation:

The distance made during a particular period of time is calculated as (distance in m) = (velocity in m/s) * (time in s)

You can think of such a calculation as determining the area of a rectangle whose sides are velocity and time period. If you make the time period very very small, the rectangle will become a narrow "bar" - a bar with height determined by the average velocity during that corresponding short period of time. The area is, again, the distance made during that time. Now, you can cover the entire area under the curve using such narrow bars. Their areas adds up, approximately, to the total distance made over the entire span of motion. From this you can already see why the answer D is the correct one.

Going even further, one can make the rectangular bars arbitrarily narrow and cover the area under the curve with more and more of these. In fact, in the limit, this is something called a Riemann sum and leads to the definition of the Riemann integral. Using calculus, the area under a curve (hence the distance in this case) can be calculated precisely, under certain existence criteria.

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g n diffraction, the formula for minima is given by a times s i n (theta )equals m lambda, where a is the width of the slit, the

ki77a [65]

Answer:

θ = 22.2

Explanation:

This is a diffraction exercise

a sin θ = m λ

The extension of the third zero is requested (m = 3)

They indicate the wavelength λ = 630 nm = 630 10⁻⁹ m and the width of the slit a = 5 10⁻⁶ m

sin θ = m λ / a

sin θ = 3 630 10⁻⁹ / 5 10⁻⁶

sin θ = 3.78 10⁻¹ = 0.378

θ = sin⁻¹ 0.378

to better see the result let's find the angle in radians

θ = 0.3876 rad

let's reduce to degrees

θ = 0.3876 rad (180º /π rad)

θ = 22.2º

4 0

2 years ago

A uniform plank of length 5.0 m and weight 225 N rests horizontally on two supports, with 1.1 m of the plank hanging over the ri

lawyer [7]

Answer:

x = 0.6034 m

Explanation:

Given

L = 5 m

Wplank = 225 N

Wman = 522 N

d = 1.1 m

x = ?

We have to take sum of torques about the right support point. If the board is just about to tip, the normal force from the left support will be going to zero. So the only torques come from the weight of the plank and the weight of the man.

∑τ = 0 ⇒ τ₁ + τ₂ = 0

Torque come from the weight of the plank = τ₁

Torque come from the weight of the man = τ₂

⇒ τ₁ = + (5 - 1.1)*(225/5)*((5 - 1.1)/2) - (1.1)*(225/5)*((1.1)/2) = 315 N-m (counterclockwise)

⇒ τ₂ = Wman*x = 522 N*x (clockwise)

then

τ₁ + τ₂ = (315 N-m) + (- 522 N*x) = 0

⇒ x = 0.6034 m

7 0

2 years ago

Can someone explain what streamlining is in a simple way with an example.

Usimov [2.4K]

Imagine a car, and imagine you see little arrows below and beneath it, just around it. <span>The path of a particle that is flowing steadily and without turbulence in a fluid past an object.</span>

6 0

3 years ago

HELP PLEASE <br> Find the net force necessary for a 15 kg object to accelerate at 20 m/s/s.

tensa zangetsu [6.8K]

Answer:

<h3>The answer is 300 N</h3>

Explanation:

The force acting on an object given the mass and acceleration we use the formula

<h3>force = mass × acceleration</h3>

We have

force = 15 × 20

We have the final answer as

Hope this helps you

3 0

2 years ago

What does the rotational speed of the ring module have to be so that an astronaut standing on the outer rim of the ring module f

Usimov [2.4K]

Complete question:

In the movie The Martian, astronauts travel to Mars in a spaceship called Hermes. This ship has a ring module that rotates around the ship to create “artificial gravity” within the module. Astronauts standing inside the ring module on the outer rim feel like they are standing on the surface of the Earth. (The trailer for this movie shows Hermes at t=2:19 and demonstrates the “artificial gravity” concept between t= 2:19 and t=2:24.)

Analyzing a still frame from the trailer and using the height of the actress to set the scale, you determine that the distance from the center of the ship to the outer rim of the ring module is 11.60 m

What does the rotational speed of the ring module have to be so that an astronaut standing on the outer rim of the ring module feels like they are standing on the surface of the Earth?

Answer:

The rotational speed of the ring module have to be 0.92 rad/s

Explanation:

Given;

the distance from the center of the ship to the outer rim of the ring module r, = 11.60 m

When the astronaut standing on the outer rim of the ring module feels like they are standing on the surface of the Earth, then their centripetal acceleration will be equal to acceleration due to gravity of Earth.

Centripetal acceleration, a = g = 9.8 m/s²

Centripetal acceleration, a = v²/r

But v = ωr

a = g = ω²r

$\omega = \sqrt{\frac{g}{r}} = \sqrt{\frac{9.8}{11.6} } = 0.92 \ rad/s$

Therefore, the rotational speed of the ring module have to be 0.92 rad/s

3 0

3 years ago