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iren2701 [21]
3 years ago
13

Which of the following is always the result when a centripetal force is applied?

Physics
2 answers:
Eduardwww [97]3 years ago
7 0
<span>centripetal acceleration
</span>
fiasKO [112]3 years ago
5 0
Centripetal force produces centripetal acceleration.
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Can somebody re-answer the true and false questions!!!
stiv31 [10]

Answer:

I think no 3 is false

and 4 is true

and the the ones you did are correct

if wrong correct me pls

7 0
3 years ago
Read 2 more answers
A bicyclist starting at rest produces a constant angular acceleration of 1.30 rad/s2 for wheels that are 35.5 cm in radius.
Debora [2.8K]

Answer:

a) 0.462 m/s^2

b) 31.5 rad/s

c) 381 rad

d) 135m

Explanation:

the linear acceleration is given by:

a=\alpha *r\\a=1.30rad/s^2*(35.5*10^{-2}m)\\a=0.462m/s^2

the angular speed is given by:

\omega=\frac{v}{r}\\\\\omega=\frac{11.2m/s}{35.5*10^{-2}m}\\\\\omega=31.5rad/s

to calculate how many radians have the wheel turned we need the apply the following formula:

\theta=\frac{1}{2}\alpha*t^2\\\\t=\frac{\omega}{\alpha}\\\\t=\frac{31.5rad/s}{1.30rad/s^2}\\\\t=24.2s\\\\\theta=\frac{1}{2}*1.30rad/s^2*(24.2s)^2\\\\\theta=381rad

the distance is given by:

d=\theta*r

d=381rad*(35.5*10^{-2}m)\\d=135m

4 0
3 years ago
When light propagates through two adjacent materials that have different optical properties, some interesting phenomena occur at
Veseljchak [2.6K]

Answer:

First, the different indices of refraction must be taken into account (in different media): for example, the refractive index of light in a vacuum is 1 (since vacuum = c).  The value of the refractive index of the medium is a measure of its "optical density":  Light spreads at maximum speed in a vacuum but slower in others  transparent media; therefore in all of them n> 1. Examples of typical values ​​of  are those of air (1,0003), water (1.33), glass (1.46 - 1.66) or diamond (2.42).

The refractive index has a maximum value and a minimum value, which we can calculate the minimum value by means of the following explanation:

The limit or minimum angle, α lim, is defined as the angle of refraction from which  the refracted ray disappears and all the light is reflected. As in the maximum value of  angle of refraction, from which everything is reflected, is βmax = 90º, we can  know the limit angle (the minimum angle that we would have to have to know the minimum index of refraction) by Snell's law:

 βmax = 90º ⇒ n 1x sin α (lim) = n 2 ⇒ sin α lim = n 2 / n 1

Explanation:

When a light ray strikes the separation surface between two media  different, the incident beam is divided into three: the most intense penetrates the second  half forming the refracted ray, another is reflected on the surface and the third is  breaks down into numerous weak beams emerging from the point of incidence in  all directions, forming a set of stray light beams.

4 0
3 years ago
You are given a vector in the xy plane that has a magnitude of 90.0 units and a y component of -61.0 units. A) Assuming the x co
kvasek [131]

Answer:

A) magnitude = sqrrt(147.1^2 + 61^2) = 159.2 units

B)22.5° (clockwise form -ve X axis)

Explanation:

given vector = V1 = x i - 60 j

magnitude of V1 = 90

x - component can be found out by resultant formula

90^2 = x^2 + (-60)^2

x = 67.08 = 67.1 units (3sf)

FOR THE VECTOR 80 UNITS IN -VE X DIRECTION

The X component is -80-------(1)

The Y component is 0 ---------(2)

<u>For the x- component of new added vector:</u>

(1)----------- x + 67.1 = -80

x = -147.1 = -147.1

<u>For the y- component of new added vector:</u>

<u>(</u>2)---------- y - 61  = 0

y = 61.0 (3sf)

the new added vector is  = -147.1 i + 61 j

magnitude = sqrrt(147.1^2 + 61^2) = 159.2 units

direction = arctan (61 / 147.1)

               = 22.5° (clockwise form -ve X axis)

<u />

4 0
3 years ago
An automobile having a mass of 1,000 kg is driven into a brick wall in a safety test. The bumper behaves like a spring with cons
vlada-n [284]

Answer:

v=2.02\frac{m}{s}

Explanation:

Assuming no energy lost, according to the law of conservation of energy, the kinetic energy of the automobile becomes potential energy after the crash:

K=U\\\frac{mv^2}{2}=\frac{kx^2}{2}

Here m is the automobile's mass, v is the speed of the car before impact, k is the "bumper" constant and x is the compression of the bumper due to the collision. Solving for v:

v=x\sqrt\frac{k}{m}\\v=2.63*10^{-2}m\sqrt{\frac{5.9*10^6\frac{N}{m}}{10^3kg}}\\v=2.02\frac{m}{s}

8 0
3 years ago
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