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asambeis [7]
3 years ago
6

In which block of the periodic table is uradium (U) found?

Physics
1 answer:
aleksandr82 [10.1K]3 years ago
3 0

Answer: fBlock

Explanation:

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A compact car, mass 660 kg, is moving at 1.00 ✕ 102 km/h toward the east. The driver of the compact car suddenly applies the bra
Maslowich

Answer:59.43\times 10^3 kg-m/s

Explanation:

Given

mass of car m=660 kg

Initial velocity of car =102 km/h\approx 28.33 m/s towards east

Time taken to stop t=2.1 s

Force exerted F_{avg}=4.8\times 10^3 N

change in momentum is given  by impulse imparted to the car

Impulse(J)=-F\cdot t

J=-4.8\times 10^3\times 2.1

J=-59.49\times 10^3 kg-m/s

negative Sign indicates that impulse is imparted opposite to the direction of motion

magnitude of momentum J=59.49\times 10^3

6 0
3 years ago
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Science please help!<br>​
IceJOKER [234]

Explanation:

spherical lenses which are curved outward are CONVEX lenses

7 0
3 years ago
Dina has a mass of 50 kilograms and is waiting at the top of a ski slope that’s 5.0 meters high.
Over [174]
All of Dina's potential energy Ep is converted into kinetic energy Ek so Ep=Ek, where Ep=m*g*h and Ek=(1/2)*m*v². m is the mass of Dina, h is the height of ski slope, g=9.8 m/s² and v is the maximal velocity. 

So we solve for v:

m*g*h=(1/2)*m*v², masses cancel out,

g*h=(1/2)*v², we multiply by 2,

2*g*h=v² and take the square root to get v

√(2*g*h)=v, we plug in the numbers and get:

v=9.9 m/s. 

So Dina's maximum velocity on the bottom of the ski slope is v=9.9 m/s.
8 0
3 years ago
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How does the matter and (solar) energy move through the sphere?
Tema [17]
When the Sun's energy moves through space, it reaches Earth's atmosphere and finally the surface. This radiant solar energy warms the atmosphere and becomes heat energy. This heat energy is transferred throughout the planet's systems in three ways: by radiation, conduction, and convection.
5 0
3 years ago
Two polarizing sheets have their transmission axes crossed so that no light is transmitted. A third sheet is inserted so that it
jek_recluse [69]

Answer:

a)    I= I₀ (cos²θ - cos⁴θ)    b) 75.5º

Explanation:

a) For this exercise we must use Malus's law

         I = I₀ cos² θ

where tea is the angle between the two polarizers.

We apply this expression to our case

* Polarizer 1 suppose that it is vertical and polarizer 2 (intermediate) is at an angle θ with respect to the vertical

         I₁ = I₀ cos² θ

* We analyze for the polarity 2 and the last polarizer 3 which indicate that it must be at 90º from the first one, therefore it must be horizontal.

The angle of polarizers 2 and 3 is θ' measured from the horizontal, if we measure with respect to the vertical

              θ₂ = 90- θ’ = θ

fiate that in the exercise we must take a reference system and measure everything with respect to this system.

          I = I₁ cos² θ'

       

we substitute

         I = (I₀ cos² tea) cos² (θ - 90)

        cos (θ -90) = cos θ cos 90 + sin θ sin 90 = sin θ

         I = Io cos² θ sin² θ

        1= cos²θ+ sin²θ

       sin²θ = 1 - cos²θ

        I= I₀ (cos²θ - cos⁴θ)

b) to find when the intensity is maximum,

we can use that we have an extreme point when the drift is zero

          \frac{dI}{d \theta} = 0

          \frac{dI}{d \theta}= Io (2 cos θ - 4 cos³θ) = 0

whereby

            cos θ - 2 cos³ θ = 0

            cos θ ( 1 - 2 cos² θ) = 0  

The zeros of this function are in

           θ = 90º

           1-2cos²θ =0       cos θ = 0.25  θ =  75.5º

Let's analyze this two results for the angle of 90º the intnesidd is zero with respect to the first polarizer, so it is not an acceptable solution.

Consequently, the angle that allows the maximum intensity to pass is 75.5º

5 0
2 years ago
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