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3 years ago

State the laws of reflection

Physics

1 answer:

dlinn [17]3 years ago

3 0

Answer:

Explanation:

The law of reflection says that the reflected angle (measured from a vertical line to the surface called the normal) is equal to the reflected angle measured from the same normal line.

All other properties of reflection flow from this one statement.

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For which question could a testable hypothesis be developed?

adell [148]

Does wood float in water

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3 years ago

Read 2 more answers

A certain elevator cab has a total run of 198 m and a maximum speed is 317 m/min, and it accelerates from rest and then back to

tester [92]

Answer:

Explanation:

Given

Total run=198 m

maximum speed=317 m/min=5.28 m/s

acceleration $=1.11 m/s^2$

(a)Distance traveled while accelerating

$v^2-u^2=2as$

$(5.28)^2-0=2\times 1.11\times s$

s=12.57 m

time taken

v=u+at

$5.28=0+1.11\times t$

t=4.75 s

(b)Time taken to cover 198 m

if car takes 12.57 m & 4.75 s to reach max speed so it also takes 12.57 m and 4.75 s to stop from max speed

Distance traveled with max speed=198-25.14=172.86 m

time taken $=\frac{172.86}{5.28}=32.73 s$

total time=4.75+32.73+4.75=42.25 s

7 0

3 years ago

The roller-coaster car shown in fig. 6-41 (h1 = 45 m, h2 = 16 m, h3 = 26 m), is dragged up to point 1 where it is released from

kirill [66]

There are many ways to solve this but I prefer to use the energy method. Calculate the potential energy using the point then from Potential Energy convert to Kinetic Energy at each points.

PE = KE

From the given points (h1 = 45, h2 = 16, h<span>3 </span>= 26)

Let’s use the formula:

v2= sqrt[2*Gravity*h1] where the gravity is equal to 9.81m/s2

v3= sqrt[2*Gravity*(h1 - h3 )] where the gravity is equal to 9.81m/s2

v4= sqrt[2*Gravity*(h1 – h2)] where the gravity is equal to 9.81m/s2

Solve for v2

v2= sqrt[2*Gravity*h1]

= √2*9.81m/s2*45m

v2= 29.71m/s

v3= sqrt[2*Gravity*(h1 - h3 )

=√2*9.81m/s2*(45-26)

=√2*9.81m/s2*19

v3=19.31m/s

v4= sqrt[2*Gravity*(h1 – h2)]

=√2*9.81m/s2*(45-16)

=√2*9.81m/s2*(29)

v4=23.85m/s

7 0

3 years ago

Two blocks, with masses M2>M1, are connected by ropes. You pull to the right on a second rope, with external force "T1".The b

Gre4nikov [31]

Answer:

$(M_1 + M_2) a > M_2 a$

Becuase $M_1 +M_2> M_2$

So then we can conclude that:

$T_1 > T_2$

And that makes sense since the force $T_1$ needs to accelerate the two masses and $T_2$ just need to accelerate $M_2$ .

So the best option for this case would be:

a. T1 > T2

See explanation below.

Explanation:

For this case we consider the system as shown on the figure attached.

Since the system is connected the acceleration for both masses are equal, that is $a_{M_1}= a_{M_2} = a$

From the second Law of Newthon we have that the force applied for the mass $M_2$ is $F_{M_2}= M_2 a$ and we know that the force acting on the x axis for the mass 2 is $F_{M_2}= T_2$ so then we have that $T_2= M_2 a$

Now when we consider the system of $M_1 +M_2$ as a whole mass, this system have the same acceleration $a$ and on this case we will see that the only force acting on the entire system would be $T_1$ and then by the second law of Newton we have that: