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Bess [88]
3 years ago
7

White dwarfs are too small to see with telescopes true or false

Physics
1 answer:
trasher [3.6K]3 years ago
5 0

Answer:

False

Explanation:

A white dwarf star that is easy to locate and see with small telescopes.

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Average wavelength of radio waves​
Nesterboy [21]

Answer:

Radio waves have frequencies as high as 300 gigahertz(GHz)to as low as 30 hertz(Hz).At 300 GHz the corresponding wavelength is 1mm and at 30Hz is 10,000 km

7 0
3 years ago
An airplane flies 1000 miles in 2 hours. What is its average speed in miles per hour?
OLEGan [10]

Answer:

500km per hour

Explanation:

if in 2 hours the airplane flies 1000 km then 1000 divided by 2 is 500km per hour.

5 0
3 years ago
Boston Red Sox pitcher Roger Clemens could routinely throw a fastball at a horizontal
azamat

Answer: 0.145 seconds

Explanation:

Given that Roger Clemens could routinely throw a fastball at a horizontal speed of 119.7 m/s. How long did the ball take to reach home plate 17.3 m away

Since the speed is horizontal

Using the formula for speed which is

Speed = distance/time

Where

Speed = 119.7 m/s

Distance covered = 17.3 m

Time is what we are looking for

Substitute all the parameters into the formula

119.7 = 17.3/ time

Make time the subject of formula

Time = 17.3 / 119.7

Time = 0.145 seconds.

Therefore, it will take 0.145 seconds to reach the home plates

6 0
3 years ago
A fan connected to a 120-volt electrical system by an extension cord was worn through and exposed the bare, energized conductor,
kari74 [83]

Answer: the ladder

Explanation: since the energized conductor is already in contact with the ladder there by making electric current to flow. The base of the ladder is on the ground there by making the circuit to be complete and causing electrocution.

8 0
3 years ago
Read 2 more answers
An offshore oil well is 2 kilometers off the coast. The refinery is 4 kilometers down the coast. Laying pipe in the ocean is twi
shusha [124]

Answer:

Rectangular path

Solution:

As per the question:

Length, a = 4 km

Height, h = 2 km

In order to minimize the cost let us denote the side of the square bottom be 'a'

Thus the area of the bottom of the square, A = a^{2}

Let the height of the bin be 'h'

Therefore the total area, A_{t} = 4ah

The cost is:

C = 2sh

Volume of the box, V = a^{2}h = 4^{2}\times 2 = 128            (1)

Total cost, C_{t} = 2a^{2} + 2ah            (2)

From eqn (1):

h = \frac{128}{a^{2}}

Using the above value in eqn (1):

C(a) = 2a^{2} + 2a\frac{128}{a^{2}} = 2a^{2} + \frac{256}{a}

C(a) = 2a^{2} + \frac{256}{a}

Differentiating the above eqn w.r.t 'a':

C'(a) = 4a - \frac{256}{a^{2}} = \frac{4a^{3} - 256}{a^{2}}

For the required solution equating the above eqn to zero:

\frac{4a^{3} - 256}{a^{2}} = 0

\frac{4a^{3} - 256}{a^{2}} = 0

a = 4

Also

h = \frac{128}{4^{2}} = 8

The path in order to minimize the cost must be a rectangle.

8 0
3 years ago
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