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4 years ago

What is velocity change divided by time change?

1 answer:

5 0

Acceleration: is the change in velocity divided by the time it takes for the change to occur. . Acceleration is the change in verity divided by the time it takes to make the change, Acceleration has direction.

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A car is traveling at 114 m/s and changes its velocity to 77 m/s in 9 sec.

suter [353]

Answer:

Given -

Initial Velocity, u = 114 m/s
Final velocity, v = 77 m/s.
Time taken, t = 9 sec.

To find - 

Acceleration of the car.

Solution -

Here, using the equation of motion v = u + at we can find the acceleration easily.

★ Here,

V = Final velocity
U = Initial Velocity
A = Acceleration
T = Time.

Substituting the values -

→ 77 = 114 + a(9)

→ 9a = 114 - 77

→ 9a = 37

→ a = 37/9

→ a = 4.1 m/s

Therefore, the acceleration of the car will be 4.1 m/s.

5 0

3 years ago

A layer of ethyl alcohol (n = 1.361) is on top of water (n = 1.333). To the nearest degree, at what angle relative to the normal

AleksAgata [21]

Answer:

a. 78 degree

Explanation:

According to Snell's Law, we have:

(ni)(Sin θi) = (nr)(Sin θr)

where,

ni = Refractive index of medium on which light is incident

ni = Refractive index of ethyl alcohol = 1.361

nr = Refractive index of medium from which light is refracted

nr = Refractive index of ethyl alcohol = 1.333

θi = Angle of Incidence

θr = Angle of refraction

So, the Angle of Incidence is know as the Critical Angle (θc), when the refracted angle becomes 90°. This is the case of total internal reflection. That is:

θi = θc

when, θr = 90°

Therefore, Snell's Law becomes:

(1.361)(Sin θc) = (1.333)(Sin 90°)

Sin θc = 1.333/1.361

θc = Sin⁻¹ (0.9794)

θc = 78.35° = 78° (Approximately)

Therefore, correct answer will be:

a. 78 degree

8 0

3 years ago

A civil engineer wishes to redesign the curved roadway in the example What is the Maximum Speed of the Car? in such a way that a

vlabodo [156]

Answer:

24.3 degrees

Explanation:

A car traveling in circular motion at linear speed v = 12.8 m/s around a circle of radius r = 37 m is subjected to a centripetal acceleration:

$a_c = \frac{v^2}{r} = \frac{12.8^2}{37} = 4.43 m/s^2$

Let α be the banked angle, as α > 0, the outward centripetal acceleration vector is split into 2 components, 1 parallel and the other perpendicular to the road. The one that is parallel has a magnitude of 4.43cosα and is the one that would make the car slip.

Similarly, gravitational acceleration g is split into 2 component, one parallel and the other perpendicular to the road surface. The one that is parallel has a magnitude of gsinα and is the one that keeps the car from slipping outward.

So $gsin\alpha = 4.43cos\alpha$