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3 years ago

B. How could you determine the instantaneous speed during the trip? Be specific.

Physics

1 answer:

alex41 [277]3 years ago

6 0

Answer:

Speedometer.

Explanation:

If you look at the speedometer in a car, you will see your instantaneous speed. Speedometers tell you your speed of that moment and thats what instantaneous speed.

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Lilit [14]

Answer:

if i do give me brainliest ok ok

Explanation:

7 0

3 years ago

A 60 cm diameter wheel accelerates from rest at a rate of 7 rad/s2. What is the tangential acceleration (in m/s2) of a point on

wolverine [178]

Answer:

2.1 m/s²

Explanation:

By definition, the angular acceleration, is equal to the rate of change of the angular velocity, ω:

α = Δω / Δt (1)

By definition of the angular velocity, we can express the linear velocity, v, as follows:

v = ω*r⇒ Δv = Δω*r (2)

Replacing Δω, from (1) in (2), we get:

Δv = α*Δt*r⇒ Δv/Δt = α*r (3)

By definition of linear acceleration, we can write the following expression;

a = α*r

For a point on the edge of the wheel, the linear acceleration is tangent to the rim, and is equal to the product of the angular acceleration times the distance to the center, which for a point on the edge of the wheel, is just the radius:

⇒ a = 7 rad/sec²*0.3m = 2.1 m/s²

3 0

4 years ago

while the woods hole martha's vineyard ferry boat steamed along at 8 mph through calm seas, passenger casey exercised by walking

mojhsa [17]

Answer:

If Casey is going in the boat's direction, her speed relative to the water is 12 mph in the direction of the boat.

If she is going oposite the boat's direction, her speed relative to the water is 4 mph in the direction of the boat.

If she is going perpendicular to the boat, her speed relative to the water is 8.94 mph in a direction that forms an angle of about 26º with the direction the boat is moving.

Explanation:

The speed of Casey with respect with the water depends on the direction she's moving in the deck. The deck has four sides, so there's four possible directions for Casey to move.

Since the water is calm, it's velocity is 0 mph so it doesn't modify the speeds of either Casey or the boat.

Let's consider the boat is moving to the right. That means it's speed with respecto to the water is 8 mph. If we were floating on the water watching from the side the boat and Casey pass by, when she's moving from left to right she seems faster beacuse she's moving in the same direction as the boat and her speed is added to the boats, so it would seem she's walking at 8 mph + 4 mph = 12 mph.

When Casey is moving from riht to left on the deck, she would seem to be moving from left to right at a speed of 4 mph from the water, because the boats speed is higher than hers and is in the opposite direction.

Finally, when she is moving either upwards or downwards, her direction forms a 90º angle with the direction the boat is moving, so her total speed is a vector with horizontal component of 8 mph (going from left to right, the boats speed) and vertical component of 4 mph (with the direction she's moving, either upwards or downwards). By Pythagoras theorem, her velocity would be $\sqrt{(8 mph)^2+(4mph)^2}=8.94 mph$ and her direction would form an angle of $arccotangent(\frac{4}{8})=0.464 rads$ with the direction the boat is moving, which is about 26º.

4 0

3 years ago

What is an example of total internal reflection at work?

Vesnalui [34]

$\huge\underline{\underline{\boxed{{ \mathbb{SOLUTION:}}}}}$

$\leadsto$ Total internal reflection occurs when a ray of light is traveling through a medium and, when that medium changes, it does not refract into the second medium but instead it reflects back into the first medium.

By Snell's law, we know that

$\longrightarrow \sf{n_1 sin \emptyset_1 = n_2 sin \emptyset_2}$

If this happens, then $\sf{n_2 \angle n_1}$ , and the ray does not lose intensity due to refraction.

$\huge\underline{\underline{\boxed{\mathbb{ANSWER:}}}}$

$\large \bm{ A.}$ A ray of light has the same intensity both entering and exiting a fiber optic cable.

6 0

2 years ago

which has a higher acceleration:a 10kg object acted upon with a net force of 20N or an 18kg object acted on by a net force of 20

MA_775_DIABLO [31]

Answer: The acceleration of 10 kg object is greater than that of 18 kg object.

Explanation:
According to Newton's Second law:
F = ma --- (A)

Let's find the acceleration for both 10 kg and 18 kg objects!
The net force on both of these masses = F = 20N

(1) Acceleration of 10 kg object
Mass = m = 10 kg
Plug in the values in equation (A):
20 = 10 * a
Acceleration = a = 2 m/s^2

(2) Acceleration of 18 kg object
Mass = m = 18 kg
Plug in the values in equation (A):

20 = 18 * a
Acceleration = a = 1.11 m/s^2

2 > 1.11; therefore, 10 kg object has the higher acceleration compared to the acceleration of the 18 kg object.

7 0

4 years ago

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