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3 years ago

A car moved 60 meters west in 2 hours. What is its average velocity?

1 answer:

5 0

Explanation: Velocity is the displacement of an object during a specific unit of time. Two measurements are needed to determine velocity. Displacement and time. Displacement includes a direction, so velocity also includes a direction. Speed with direction. Velocity can be an average velocity or an instantaneous velocity. Units for velocity are the same as for speed: m/s, km/h, and mph. Delta x(Δx) is the symbol used for displacement. Delta (Δ) means to "change in." Δx means to "change in position." Δx is calculated by final position minus initial position. Velocity formula: → v=Δx/t as a fraction.

v=Δx/t

$v=\frac{xf-xi}{t}= \frac{60m}{2}=30m$

<em><u>Final answer is 30.</u></em>

Hope this helps!

Thanks!

Have a great day!

-Charlie

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Lee has been moving a crate that has a mass of 125 kg and required 120 J of work to move. The crate was moved a distance of 20 m

natita [175]

Answer: Output power , P= 4 watts

Explanation:We know that power is defined as the energy supplied per unit time.

Therefore Power, $P=\frac{Energy supplied}{Time taken}$

=> $P=\frac{120}{30} watts = 4 watts$

Thus power output is 4 watts

4 0

4 years ago

A roller-coaster track has six semicircular "dips" with different radii of curvature. The same roller-coaster cart rides through

inessss [21]

Answer:

E > A = B > C > D

Explanation:

On a semicircular curve, the magnitude of force is given by the formula as :

$F=\dfrac{mv^2}{r}$

Here, m is the mass of the vehicle

Case A : R = 60 m v = 16 m/s

$F_1=\dfrac{m(16)^2}{60}=4.27m\ N$

Case B : R = 15 m v = 8 m/s

$F_2=\dfrac{m(8)^2}{15}=4.27m\ N$

Case C : R = 30 m v = 4 m/s

$F_3=\dfrac{m(4)^2}{30}=0.534m\ N$

Case D : R = 45 m v = 4 m/s

$F_4=\dfrac{m(4)^2}{45}=0.356m\ N$

Case E : R = 30 m v = 16 m/s

$F_5=\dfrac{m(16)^2}{30}=8.53m\ N$

The order of the magnitude of the force of the roller-coaster track on the cart from largest to smallest is :

E > A = B > C > D

6 0

3 years ago

PLEASE HELP FOR SCIENCE PLEASE

zloy xaker [14]

1. A. medium

2. B. matter

3. B. straight lines

im sorry i couldnt help for number 4 but hope this helps

3 0

4 years ago

Imagine you are in an open field where two loudspeakers are set up and connected to the same amplifier so that they emit sound w

Scilla [17]

Answer:

d= 5.62 m

Explanation:

In order to have destructive interference, the path difference from the sources to the listener must be an odd multiple of half wavelengths, as follows:

d = (2n+1) * λ/2

In orfer to know which is the wavelength, we can use the relationship between propagation speed (in this case speed of sound), frequency and wavelength:

v= λ*f ⇒ λ = v/f = 344 m/s / 688 1/sec = 0.5 m

So, the path difference must be, at least, λ/2:

d = b-a = λ/2, where b is the distance to the speaker B, and a, the distance to the speaker A.

Applying Pithagorean Theorem, as the perpendicular distance d (which is our unknown) is the same for the triangles defined by the horizontal distance to the listener, and the straight line from the new position to the sources, we can write:

d² = a²- (3.0)²

d² = b²- (3.5)²

As the left sides are equal, so do right sides:

a² - (3.0)² = b² - (3.5)²

⇒ b² - a² = (3.5)² - (3.0)² = 3.25

We can replace (b²- a²) as follows:

b² - a² = (b+a)(b-a) = 3.25 (2)

We know that b-a = λ/2 = 0.25

Replacing in (2), we have:

b+a = 3.25 / 0.25 = 13 m

b-a = 0.25 m

Adding both sides:

2*b = 13.25 m ⇒ b= 13.25 /2 = 6.63 m

⇒ d² = (6.63)² - (3.5)² = 31.6 m²

⇒ d=√31.6 m² = 5.62 m

3 0

3 years ago

Can someone help me?

Leviafan [203]

Car X traveled 3d distance in t time. Car Y traveled 2d distance in t time. Therefore, the speed of car X, is 3d/t, the speed of car Y, is 2d/t. Since speed is the distance taken in a given time.

In figure-2, they are at the same place, we are asked to find car Y's position when car X is at line-A. We can calculate the time car X needs to travel to there. Let's say that car X reaches line-A in t' time.

$V_x .t' = 3d\\ \frac{3d}{t} .t' = 3d\\ t'=t$

Okay, it takes t time for car X to reach line-A. Let's see how far does car Y goes.

$V_y.t = \frac{2d}{t} .t = 2d$

We found that car Y travels 2d distance. So, when car X reaches line-A, car Y is just a d distance behind car X.

4 0

3 years ago