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3 years ago

If a wave hits a smooth surface at an angle of incidence of 40 degrees, the angle of reflection is

1 answer:

6 0

<span>So we want to know what is the angle of reflection of the wave if the incident angle is 40 degrees. So when the incident wave hits a surface under some angle, the reflecting angle should be the same as the angle of incidence. So the correct answer is B. 40 degrees.</span>

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What affects the braking distance and thinking distance

Radda [10]

Your speed is one of the only factors that has an effect on both your thinking distance and braking distance. Put simply, the faster you are going, the greater the distance travelled before you apply the brakes (thinking distance) and the vehicle comes to a complete stop (braking distance).

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3 years ago

Read 2 more answers

A certain freely falling object, released from rest, requires 1.95 s to travel the last 23.5 m before it hits the ground. (a) Fi

Ratling [72]

Answer:

(a). The velocity of the object is -2.496 m/s.

(b). The total distance of the object travels during the fall is 23.80 m.

Explanation:

Given that,

Time = 1.95 s

Distance = 23.5 m

(a). We need to calculate the velocity

Using equation of motion

$s = ut+\dfrac{1}{2}gt^2$

Put the value into the formula

$-23.5=u\times1.95+\dfrac{1}{2}\times(-9.8)\times(1.95)^2$

$u=\dfrac{-23.5+4.9\times(1.95)^2}{1.95}$

$u=-2.496\ m/s$

(b). We need to calculate the total distance the object travels during the fall

Using equation of motion

$v = u+gt$

Put the value in the equation

$-2.496=0-9.8\times t$

$t =\dfrac{2.496}{9.8}$

$t=0.254\ sec$

The total time is

$t'=t+1.95$

$t'=0.254+1.95$

$t'=2.204\ sec$

We need to calculate the distance

Using equation of motion

$s = ut+\dfrac{1}{2}gt^2$

Put the value into the formula

$s=0+\dfrac{1}{2}\times9.8\times(2.204)^2$

$s=23.80\ m$

Hence, (a). The velocity of the object is -2.496 m/s.

(b). The total distance of the object travels during the fall is 23.80 m.

4 0

3 years ago

What is the magnitude of the sum of the two vectors A = 36 units at 53 degrees, and B =47 units at 157 degrees.

Thepotemich [5.8K]

Answer:

51.82

Explanation:

First of all, let's convert both vectors to cartesian coordinates:

Va = 36 < 53° = (36*cos(53), 36*sin(53))

Va = (21.67, 28.75)

Vb = 47 < 157° = (47*cos(157), 47*sin(157))

Vb = (-43.26, 18.36)

The sum of both vectors will be:

Va+Vb = (-21.59, 47.11) Now we will calculate the module of this vector:

$|Va+Vb| = \sqrt{(-21.59)^2+(47.11)^2}=51.82$

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4 years ago

A 38.2 kg wagon is towed up a hill inclined at 17.5 ◦ with respect to the horizontal. The tow rope is parallel to the incline an

Tema [17]

Answer:

v = 8.57 m/s

Explanation:

As we know that the wagon is pulled up by string system

So the net force on the wagon along the inclined is due to tension in the rope and component of weight along the inclined plane

So as per work energy theorem we know that

work done by tension force + work done by force of gravity = change in kinetic energy

$F_t . d - (mgsin\theta)(d) = \frac{1}{2}mv^2 - 0$

so we have

$F_t = 129 N$

$\theta = 17.5^o$

m = 38.2 kg

d = 85.4 m

so now we have

$129(85.4) - (38.2)9.8sin17.5 (85.4) = \frac{1}{2}(38.2) v^2$

$v = 8.57 m/s$

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3 years ago

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aivan3 [116]

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3 years ago