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1 year ago

15

A satellite phone can be used to place a call from virtually anywhere in the world. What type of wave does the satellite phone l

ikely transmit to the satellite orbiting above Earth?
sound wave

visible light

gamma ray

radio wave

2 answers:

SIZIF [17.4K]1 year ago

7 0

Radio wave because the others are wrong

Strike441 [17]1 year ago

6 0

Gamma ray because I believe it makes sense

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Light of wavelength 600 nm passes though two slits separated by 0.22mm and is observed on a screen 1.1m behind the slits. The lo

Ne4ueva [31]

Answer:

$y(m=1)=\frac{(1)(600*10^{-9}m)(1.1m)}{0.22m}=3*10^{-6}m\\\\y(m=1)=\frac{(-1)(600*10^{-9}m)(1.1m)}{0.22m}=-3*10^{-6}m$

Explanation:

We have to take into account the expression for the position of the fringes

$dsen\theta=m\lambda\\y=\frac{m\lambda D}{d}$

where m is the number of the maximum, d is the separation of the slits, D is the distance to the screen.

(a) By replacing we obtain

$y(m=1)=\frac{(1)(600*10^{-9}m)(1.1m)}{0.22m}=3*10^{-6}m\\\\y(m=1)=\frac{(-1)(600*10^{-9}m)(1.1m)}{0.22m}=-3*10^{-6}m$

(b) more information is required to solve this point. Please complete the information.

HOPE THIS HELPS!

4 0

3 years ago

Read 2 more answers

Three people pull simultaneously on a stubborn donkey. Jack pulls directly ahead of the donkey with a force of 77.3 N, Jill pull

DiKsa [7]

Answer:

$F_{net} = 232.8 N$

towards right so it is -15 degree

Explanation:

Net force in forward direction due to all three is given as

$F_x = F_1 + F_2cos45 + F_3cos45$

here we know that

$F_1 = 77.3 N$

$F_2 = 61.7 N$

$F_3 = 147 N$

$F_x = 77.3 + 61.7 cos45 + 147 cos45$

$F_x = 224.9 N$

Similarly in Y direction we will have

$F_y = F_3 sin45 - F_2 sin45$

$F_y = (147 - 61.7)sin45$

$F_y = 60.3 N$

Now the net force on the donkey is given as

$F_{net} = \sqrt{F_x^2 + F_y^2}$

$F_{net} = \sqrt{224.9^2 + 60.3^2}$

$F_{net} = 232.8 N$

Now direction of force is given as

$tan\theta = \frac{F_y}{F_x}$

$tan\theta = \frac{60.3}{224.9}$

$\theta = 15^o$ towards right so it is -15 degree

3 0

2 years ago

What is the source of all magnetism?

ivanzaharov [21]

The answer is electric current

3 0

2 years ago

Train cars are coupled together by being bumped into one another. Suppose two loaded train cars are moving toward one another, t

wolverine [178]

Answer:

final velocity = 0.08585m/s

Explanation:

We are taking train cars as our system. In this system no external force is acting. So we can apply the law of conservation of linear momentum.

The law of conservation of linear momentum states that the total linear momentum of a system remains constant if there is no external force acting on the system. That is total linear momentum before = total linear momentum after

total linear momentum before = linear momentum of first train car + linear momentum of second train car

We know that linear momentum = mv

where,

m = mass

v = velocity

thus,

total linear momentum before = m₁v₁ + m₂v₂

m₁ = mass of first train car = 135,000kg

v₁ = velocity of first train car = 0.305m/s

m₂ = mass of first second car = 100,000kg

v₂ = velocity of second train car = −0.210m/s

Note: Momentum is a vector. So while adding momentum we should take account of its direction too. Here since second train car is moving in a direction opposite to that of the first one, we have taken its velocity as negative.

total linear momentum before = m₁v₁ + m₂v₂

= 135,000x0.305 + 100,000x(−0.210)

= 135,000x0.305 - 100,000x0.210

= 20,175 kgm/s

Now we have to find total linear momentum after bumping. After the bumping both the train cars will be moving together with a common velocity(say v).

Therefore, total linear momentum after = mv

m = m₁ + m₂ = 135,000 + 100,000 = 235,000

total linear momentum before = total linear momentum after

235,000v = 20,175

v = $\frac{20,175}{235,000}$

= 0.08585m/s

8 0

2 years ago

Read 2 more answers

Problem 5 You are playing a game where you drop a coin into a water tank and try to land it on a target. You often find this gam

Dvinal [7]

Answer:

Option D: 1.5in in front of the target

Explanation:

The object distance is $y= 6in$ .

Because the surface is flat, the radius of curvature is infinity .

The incident index is $n_i=\frac{4}{3}$ and the transmitted index is $n_t= 1$ .

The single interface equation is $\frac{n_i}{y}+\frac{n_t}{y^i}=\frac{n_t-n_i}{r}$

Substituting the quantities given in the problem,

$\frac{\frac{4}{3}}{6in}+\frac{1}{y^i}= 0$

The image distance is then $y^i=-\frac{18}{4}in =-4.5in$

Therefore, the coin falls $1.5in$ in front of the target

6 0

2 years ago