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Irina-Kira [14]
3 years ago
12

The magnitude of the force of the bottom block on the top block is _____ the magnitude of the force of the earth on the top bloc

k.
Physics
1 answer:
ra1l [238]3 years ago
6 0

Hello. This question is incomplete. The full question is:

Two blocks are stacked on top of each other on the floor of an elevator. For each of the following situations, select the correct relationship between the magnitudes of the two forces given.

The elevator is moving downward at a constant speed.

The magnitude of the force of the bottom block on the top block is _____ the magnitude of the force of the earth on the top block.

Answer:

The magnitude of the force of the bottom block on top block is equal to the magnitude of the force of the top block on bottom block.

Explanation:

As the elevator is descending, there is only a normal force being applied to the lower surface of the block. This force has a magnitude equal to the force of the upper block, because the only acceleration that is acting in this case is the force of gravity. From that force, the resulting force is zero.

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Two light bulbs are 2.0 m apart. From what distance can these light bulbs be marginally resolved by a small telescope with a 4.5
andrezito [222]

Answer:

R = 1.2295 10⁵  m

Explanation:

After reading your problem they give us the diameter of the lens d = 4.50 cm = 0.0450 m, therefore if we use the Rayleigh criterion for the resolution in the diffraction phenomenon, we have that the minimum separation occurs in the first minimum of diffraction of one of the bodies m = 1 coincides with the central maximum of the other body

            θ = 1.22 λ / D

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          θ = y / R

where y is the separation of the two bodies (bulbs) y = 2 m and R the distance from the bulbs to the lens

            \frac{y}{R} = 1.22 \frac{ \lambda}{D}

            R = \frac{ y \ D}{1.22 \lambda}

let's calculate

            R = \frac{ 2 \ 0.045}{ 1.22 \ 600 \ 10^{-9}}

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3 0
3 years ago
Can some answer 7 a and b please
pav-90 [236]

Answer:

a.After 15 second Mr Comer's speed =  1.7 m/s

b.Distance travelled by Mr.Comer in 15 seconds  =   18.75\ m

Explanation:

a. Lets recall our first equation of motion V_{f} =V_{i} + at

Now we know that V_{i} = 0.8m/s , a = 0.06\ m/s^2 and

t = 15 \ sec

Plugging the values we have.

V_{f} =V_{i} + at

V_{f} =0.8 + 0.06 \times 15

V_{f} =0.8+ 0.9

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Then Mr.Comer's speed after 15 sec = 1.7ms^-1

b.

Lets find the distance and recall our third equation of motion.

V_{f} ^2-V{i}^2 = 2as

So s = distance covered.

Dividing both sides with 2a we have.

\frac{V_{f} ^2-V{i}^2}{2a} = 2as

Plugging the values.

\frac{(1.7)^2-(0.8)^2}{2\times0.06} = 2as

s= 18.75\ m

So Mr.Comer will travel a distance of s= 18.75\ m.

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