Ask question

Ask question

All categories

3 years ago

7

Batman chased the joker on his batbike for 20 minutes, traveling at the speed of 30 kilometers per minute. How far did Batman go

?

1 answer:

AURORKA [14]3 years ago

8 0

Answer:

123 kilometer

Explanation:

none

You might be interested in

How many units measure one wavelength?<br><br><br> a<br> 16<br> b<br> 8<br> c<br> 2<br> d<br> 4

pochemuha

B it makes more sense

8 0

3 years ago

If you could shine a very powerful flashlight beam toward the Moon, estimate the diameter of the beam when it reaches the Moon.

grin007 [14]

To develop this problem it is necessary to apply the Rayleigh Criterion (Angular resolution)criterion. This conceptos describes the ability of any image-forming device such as an optical or radio telescope, a microscope, a camera, or an eye, to distinguish small details of an object, thereby making it a major determinant of image resolution. By definition is defined as:

$\theta = 1.22\frac{\lambda}{d}$

Where,

$\lambda$ = Wavelength

d = Width of the slit

$\theta$ = Angular resolution

Through the arc length we can find the radius, which would be given according to the length and angle previously described.

The radius of the beam on the moon is

$r = l\theta$

Relacing $\theta$

$r = l(\frac{1.22\lambda}{d})$

$r = 1.22\frac{l\lambda}{d}$

Replacing with our values we have that,

$r = 1.22*(\frac{(384*10^3km)(\frac{1000m}{1km})(550*10^{-9}m)}{7*10^{{-2}}})$

$r = 3680.91m$

Therefore the diameter of the beam on the moon is

$d = 2r$

$d = 2 * (3690.91)$

$d = 7361.8285m$

Hence, the diameter of the beam when it reaches the moon is 7361.82m

8 0

3 years ago

A vehicle travelling at an initial velocity of 20km/hr,accelerates at 4m/s².calculate its final velocity after 10 seconds.

Nikitich [7]

acceleration = Velocity changes ÷ time of the velocity changes

4 m/s^2 =

4 × 10^(-3) × 3600 km / h =

4 × 3.6 =

14.4 km / h

Thus :

14.4 = V(2) - V(1) / t(2) - t(1)

14.4 = V(2) - 20 / 10

Multiply both sides by 10

10 × 14.4 = 10 × ( V(2) - 20 ) / 10

144 = V(2) - 20

Add both sides 20

144 + 20 = V(2) - 20 + 20

V(2) = 164 Km/h

Thus the final velocity after 10 seconds is 164 Km/h .

4 0

3 years ago

Which considerations are used to calculate a windchill factor? Select two options.

vlada-n [284]

Wind speed and air temperature are used to calculate a windchill factor.

<u>Explanation:</u>

<u></u>

Wind-chill factor is the reduction of body temperature due to the passing flow of lower-temperature air.

The air temperature value is always higher than the wind chill numbers. the heat index will be used if the apparent temperature is higher than the air temperature.So, Wind speed and air temperature are mainly used to calculate a windchill factor.

There are many ways, the surface loses its heat through conduction, evaporation,radiation, and convection.The rate of convection depends on the difference in temperature between the surface and the fluid surrounding the surface and the velocity of that fluid with respect to the surface. The air around the warm surface will be heated, an insulating layer of warm air forms against the surface.The layer becomes a boundary between two. As the wind speed is high the surface cools down rapidly.

7 0

3 years ago

Read 2 more answers

IBM has a fast computer that it calls the Blue Gene/L that can do '136.8

dmitriy555 [2]

Answer:

138.6 megacalculations

Explanation:

This is a pretty straightforward one.

All it needs is to convert the degree of measurement.

Dimensions in physics are attributed names, which state the power to which they're are raised. Just as how

Kilo and Mega means the numbers are raised to the power of 3 and 6 respectively. There also exists the ones that indicates how small, such as milli and micro, which are to the powers of -3 & -6.

The question says the IBM computer calculates at an astonishing 136.8 teracalculations.

Tera in physics means it's raised to the power of 12. Thus, the IBM calculates at an astonishing rate of

136.8*10^12 calculations per second.

We're then asked how many calculations it does in 1 micro second. Like I had highlighted earlier, 1 micro second is 1 raised to the power of -6. Or succinctly put,

1 micro second = 1*10^-6.

If the IBM does

138.6*10^12 = 1 second,

Then it does

x = 1*10^-6 second.

When we cross multiply, we have

138.6*10^12 * 1*10^-6, and that is

138.6*10^6 calculations, or say, 138.6 megacalculations.

The IBM does 138.6 megacalculations in 1 micro second, which is still astonishing, by the way

6 0

3 years ago