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ziro4ka [17]
3 years ago
14

If a car travels around a gentle curve on a highway at 60 km/h, does the velocity change

Physics
1 answer:
Oksi-84 [34.3K]3 years ago
4 0
Yes the velocity changes. Because velocity changes with direction. The object is moving around a gentle curve. The curve is not linear it is curve the direction changes a bit so obviously the velocity also changes but not much. Juts a minor change. Depends on how much curve the highway is.
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Make sure everything is organized have a planner it can help

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3 0
4 years ago
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What is one disadvantage of a series circuit?
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6 0
3 years ago
One billiard ball is shot east at 2.00 m/s. A second, identical billiard ball is shot west at 1.00 m/s. The balls have a glancin
dimulka [17.4K]

Answer:

Velocity is 1.73 m/s along 54.65° south of east.

Explanation:

Let unknown velocity be v, mass of billiard ball be m and east direction be positive x axis.

Here momentum is conserved.

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Initial momentum = m x 2i + m x (-1)i = m i

Final momentum = m x v + m x 1.41 j = mv + 1.41 m j

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mi = mv + 1.41 m j

v = i - 1.41 j

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Direction,  

       \theta =tan^{-1}\left ( \frac{-1.41}{1}\right )=-54.65^0             

Velocity is 1.73 m/s along 54.65° south of east.

5 0
3 years ago
A charge is moving in a magnetic field that points to the left. What direction can the charge move and experience no magnetic fo
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3 years ago
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A wire of length 5mm and Diameter 2m extends by 0.25 when a force of 50N was use. calculate the
bazaltina [42]

Answer and Explanation:

Data provided in the question

Force = 50N

Length = 5mm

diameter = 2.0m = 2\times 10^{-3}

Extended by = 0.25mm = 0.25\times 10^{-3}

Based on the above information, the calculation is as follows

a. The Stress of the wire is

= \frac{force\ applied}{area\ of \ circle}

here area of circle = perpendicular to the are i.e cross-sectional  i.e

= \frac{\pi d^{2}}{4}

= \frac{\pi(2\times 10^{-3})^2}{4}

Now place these above values to the above formula

= \frac{4\times 50}{\pi\times 4 \times 10^{-6}} \\\\ = \frac{50}{\pi}

= 15.92 MPa

As 1Pa = 1 by N m^2

So,

MPa = 10^6 N m^2

b. Now the strain of the wire is

= \frac{Change\ in\ length}{initial\ length} \\\\ = \frac{0.25\times 10^{-3}}{5}

= 5 \times 10^{-5}

3 0
3 years ago
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