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Leokris [45]
3 years ago
10

The apparent height of a building 10.5 km away is 0.02 radians. What is the approximate height of the building to the nearest me

ter
Physics
1 answer:
Ksenya-84 [330]3 years ago
8 0

Answer:

Approximate height of the building is 23213 meters.

Explanation:

Let the height of the building be represented by h.

0.02 radians = 0.02 × \frac{180^{o} }{\pi }

                     = 0.02 x (180/\frac{22}{7})

0.02 radians  = 1.146°

10.5 km = 10500 m

Applying the trigonometric function, we have;

Tan θ = \frac{opposite}{adjacent}

So that,

Tan 1.146° = \frac{h}{10500}

⇒ h = Tan 1.146° x 10500

      = 2.21074 x 10500

      = 23212.77

h = 23213 m

The approximate height of the building is 23213 m.

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Two wooden crates rest on top of one another. The smaller top crate has a mass of m1 = 21 kg and the larger bottom crate has a m
Andrews [41]

Answer:

2.35 m/s²

Explanation:

Given that

Mass of the smaller crate, m₁ = 21 kg

Mass of the larger crate, m₂ = 90 kg

Tensión of the rope, T = 261 N

We know that the sum of all forces for the two objects with a force of friction F and a tension T are:

(i) m₁a₁ = F

(ii) m₂a₂ = T - F, where m and a are the masses and accelerations respectively.

1) no sliding can also mean that:

a₁ = a₂ = a

This makes us merge the two equations written above together as:

m₂a = T - m₁a

If we then solve for a, we would have something like this

a = T / (m₁+m₂)

a = 261 / (21 + 90)

a = 261 / 111

a = 2.35 m/s²

Therefore, the needed acceleration of the small crate is 2.35 m/s²

8 0
3 years ago
You are holding the axle of a bicycle wheel with radius
erastova [34]

Answer:

The angular acceleration of the wheel is -6.54 rad/s²

Explanation:

We'll use the equations of motion for this.

w = 2πf

f = 75 rpm = 1.25 rps = 1.25 rev/s

w₀ = initial angular velocity = 2π × 1.25 = 7.85 rad/s

w = final angular velocity = 0 rad/s

t = 1.2 s

α = ?

w = w₀ + αt

0 = 7.85 + 1.2α

α = 7.85/1.2 = - 6.54 rad/s²

6 0
4 years ago
An unstable particle is created in the upper atmosphere from a cosmic ray and travels straight down toward the surface of the ea
german

Answer:

Q1)  Time taking by particle to travel the 40 km wrt. earth = 1.34\times10^{-6} sec.

Q2) The distance traveled by particle where the particle created to surface of earth wrt. particle's frame = 3.84 km.

Q3) The time taking by particle to travel from where it is created to the surface of the earth = 1.285\times10^{-5} sec.

Explanation:

Given :

Speed of particle wrt. earth v=0.99537c

Distance between where particle is created and earth surface = 40 km

we know that,

⇒       v = \frac{x}{t}

Where x = 40\times10^{3} m, v = 0.99537c, we know speed of light c = 3 \times10^{8}

∴      t = \frac{x}{v}

         = \frac{40 \times10^{3} }{0.99537\times3\times10^{8} }

      t = 1.34\times10^{-6} sec

∴ Thus, time taking by particle to travel the 40 km wrt. earth t = 1.34\times10^{-6} sec

According to the lorentz transformation,

⇒    l = l_{o} \sqrt{1-\frac{v^2}{c^2} }

Where l = improper length, l_{o} =proper length (distance measured wrt. rest frame) = 40 km

     l = 40 \sqrt{1-\frac{v^2}{c^2 }

     l = 40 \times 0.096

     l = 3.84 km

∴ Thus, the distance traveled by particle where the particle created to surface of earth wrt. particle's frame = 3.84 km.

According to the time dilation,

   \Delta t = \frac{\Delta t_{o} }{\sqrt{1-\frac{v^2}{c^2} } }

Where \Delta t = improper time (wrt. earth frame time) =1.34\times10^{-6} sec ,  \Delta t _{o} = proper time (wrt. particle frame).

 1.34\times10^{-6} = \frac{ \Delta t_{o}}{0.096}

 \Delta t_{o} = 1.285 \times10^{-5} sec

Thus, the time taking by particle to travel from where it is created to the surface of the earth = 1.285 \times10^{-5} sec.

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tatuchka [14]

Answer:

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Explanation:

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