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2 years ago

5

Which of the following situations would cause the greatest decrease in the

1 answer:

Ronch [10]2 years ago

4 0

A

Explanation:

hindi ako sure kung tama ba yan

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1. What is meant by half-life?

Bezzdna [24]

1. The time for a radioactive sample to reduce to half of its original mass.
4. 10% of 232 is 23.2 times 6 equals 139.2
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3 years ago

The steps in the heating of a metal seat in a park are listed below: Step 1: Heat travels from the sun to Earth Step 2: Heat tra

I am Lyosha [343]

The steps in the heating of a metal seat in a park are:

Step 1: Heat travels from the sun to Earth

Step 2: Heat travels through atmosphere to the top of the seat

Step 3: Heat from the top of the seat travels through the seat to the lower parts of the seat.

-- Heat is transferred by radiation in Step 1 and Step 2. <em>(B)</em>

-- From the top of the seat to the bottom, heat is transferred by conduction.

There's no convection happening anywhere in the park-hot-seat scenario.

3 0

2 years ago

Read 2 more answers

Parallel rays of monochromatic light with wavelength 571nm illuminate two identical slits and produce an interference pattern on

Natasha2012 [34]

Answer:

$8.8\times 10^{-6} W/m^2$

Explanation:

We are given that

Wavelength, $\lambda=571 nm=571\times 10^{-9} m$

$1 nm=10^{-9} m$

R=75 cm= $\frac{75}{100}=0.75 m$

1 m=100 cm

$d=0.640 mm=0.64\times 10^{-3} m$

$1 mm=10^{-3} m$

$a=0.434 mm=0.434\times 10^{-3} m$

$y=0.830 mm=0.830\times 10^{-3} m$

$I_0=5.00\times 10^{-4}W/m^2$

$tan\theta=\frac{y}{R}$

$\theta=\frac{0.830\times 10^{-3}}{0.75\times 10^{-2}}$

$\theta=1.1\times 10^{-3}rad$

$tan\theta\approx \theta$

Because $\theta$ is small.

$\phi=\frac{2\pi dsin\theta}{\lambda}$

$\sin\theta\approx \theta$ ,

Therefore

$\phi=\frac{2\times\pi\times 0.64\times 10^{-3}\times 1.1\times 10^{-3}}{571\times 10^{-9}}$

$\phi=7.74 rad$

$\beta=\frac{2\pi a\theta}{\lambda}$

$\beta=5.3 rad$

$I=I_0cos^2(\frac{\phi}{2})(\frac{sin\frac{\beta}{2}}{\frac{\beta}{2}})^2$

$I=5\times 10^{-4}cos^2(\frac{7.74}{2})(\frac{sin\frac{5.3}{2}}{\frac{5.3}{2}})^2$

$I=8.8\times 10^{-6} W/m^2$

6 0

2 years ago

One end of a 34-m unstretchable rope is tied to a tree; the other end is tied to a car stuck in the mud. The motorist pulls side

anyanavicka [17]

Answer:

Fc = 89.67N

Explanation:

Since the rope is unstretchable, the total length will always be 34m.

From the attached diagram, you can see that we can calculate the new separation distance from the tree and the stucked car H as follows:

L1+L2=34m

$L1^2=L2^2=L^2=2^2+(H/2)^2$ Replacing this value in the previous equation:

$\sqrt{2^2+H^2/4}+ \sqrt{2^2+H^2/4}=34$ Solving for H:

$H=\sqrt{52}$

We can now, calculate the angle between L1 and the 2m segment:

$\alpha = atan(\frac{H/2}{2})=60.98°$

If we make a sum of forces in the midpoint of the rope we get:

$-2*T*cos(\alpha ) + F = 0$ where T is the tension on the rope and F is the exerted force of 87N.

Solving for T, we get the tension on the rope which is equal to the force exerted on the car:

$T=Fc=\frac{F}{2*cos(\alpha) } = 89.67N$

7 0

2 years ago

Suatu gas memiliki volume 2 m³ dipanaskan dengan kondisi isobarik sehingga volumenya menjadi 5,5 m³ jika tekanan 4 atmosfer hitu

Ivan

Answer:

Sorry if wrong.............

5 0

2 years ago