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3 years ago

T Or F- True or False There three types of cloud's are .

Physics

1 answer:

AURORKA [14]3 years ago

7 0

Answer:

There are three main cloud types.

The foundation consists of 10 major cloud types. In addition to cirrus, stratus, cumulus, and nimbus clouds, there are cirrostratus, cirrocumulus, altostratus, altocumulus, stratocumulus, nimbostratus, and cumulonimbus clouds. The following table places these cloud types into the four major cloud groups.

Explanation:

So false, depends what you have learned and your grade level ig

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is it true that parallel worlds exists? if yes or no, give proper reasons and proof. brainliest to the best

Daniel [21]

Answer:

It is just a theory that multi verse exists. Researchers have found tracers of mirror particles which could only be from other universe.

Explanation:

Its said above.

5 0

3 years ago

An observer in frame S sees lightning simultaneously strike two points 100 m apart. The first strike occurs at xx1 = yy1 = zz1 =

Veseljchak [2.6K]

Answer:

a) 0, = -0.33 us

b) 140m

c) No, The event are not simultaneous i.e they did not occur at the same time, the second even (-0.33 usec) occurs 0.33 usec earlier than the first event.

Explanation:

the lorentz factor expression is written as;

y = 1₀ / √(1 - (v²/c²))

where v is the relative speed of an observer and c is the speed of light

so we were given that relative speed to be o.7c

therefore

y = 1 / √(1 - ((0.7c)² / c²))

y = 1 / √(1 - (0.49c² / c²))

y = 1 / √(1 - 0.49)

y = 1 / 0.7141

y = 1.4

1 - the coordinates of the first event, the s' frame of reference is,

x1 ' = y(x1 - vt1) = 0

y1 ' = y1, z1' = z1 and

t1 ' = y [t1 - v/c²x1]

= 0

2 - the coordinates of the second event, the s ' frame of reference is'

x2 ' = y(x2-vt2)

= 1.4(100m - 0)

= 140m

y2 ' = y2, z2 ' = z2

t2 ' = y [ t2 - v/c²x2 ]

= 1.4 [ 0 - 0.7c/c²(100) ]

using speed of light c as 3*10^8

1.4 [ 0 - (0.7*3*10^8) / (3*10^8)²(100) ]

= -0.33 us

distance between

delltaX' = X2' - X1'

= 140m - 0

= 140m

No, The event are not simultaneous i.e they did not occur at the same time.

the second even (-0.33 us) occurs 0.33 us earlier than the first event.

4 0

3 years ago

An object moving in circular motion has a mass of 15 kg and a centripetal acceleration of 10 m/s2. What is the centripetal force

sweet-ann [11.9K]

Answer:

1) A

2) C

3) B

4) A

5) Incomplete information(picture missing)

6) Incomplete information(picture missing)

7) Incomplete information(picture missing)

8) A

9) C

10) C

Explanation:

1) m = 15kg, a = $10ms^{-2}$ , F = ma = 15*10 = 150N

2) m = 3kg, v = $4ms^{-1}$ , r = 4m, F = $\frac{mv^{2} }{r}$

$\frac{3*4^{2} }{4}$ = 12N

3) a = $10ms^{-2}$ , r = 10m, v=?

F = $\frac{mv^{2} }{r}$ and F = ma

equating the two equations and cancelling a, we have:

$\frac{v^{2} }{r}$ = a

making v the subject of formula, we have:

v = $\sqrt{ar}$

= $\sqrt{100}$

= 10 $ms^{-1}$

4) r = 10m, v = $5ms^{-1}$ , a = ?

F = $\frac{mv^{2} }{r}$

F = ma

equating the above equations and making a subject of formula, we get:

a = $\frac{v^{2} }{r}$

a = 25/10 = 2.5 $ms^{-2}$

5) I can't find the picture associated with this question

6) I can't find the picture associated with this question

7) I can't find the picture associated with this question

8) F = $\frac{mv^{2} }{r}$

assuming m and r is unity, that is the values are 1 respectively, the formula simplifies to:

F = $v^{2}$

Now, if v is tripled

F = $(3v)^{2}$

F = 9 $v^{2}$

We can see that the force will be 9X greater than it was.

9) F = $\frac{mv^{2} }{r}$

assuming m and r is unity, that is the values are 1 respectively, the formula simplifies to:

F = $v^{2}$

Now, if v is doubled

F = $(2v)^{2}$

F = 4 $v^{2}$

We can see that the force will be 4X greater than it was.

10) F = $\frac{mv^{2} }{r}$

assuming m and v is unity, that is the values are 1 respectively

F = 1/r

if r is doubled,

F = 1/2 * 1/r

We can see that the force is 1/2 as big as it was

7 0

3 years ago

A canon is tilled back 30.0 degrees and shoots a cannon ball at 155 m/s. What is the

Umnica [9.8K]

Answer:

$= \frac{(115^2)(\sin 30)^2}{2\times 9,8} \\\\= 168.7m$

Therefore, highest point that the cannon ball reaches is 168.7m

Explanation:

the cannon is fired at an angle 30 o to the horizonatal with a speed of 155 m/s

highest point that the cannon ball reaches?

$H_{max}=\frac{V^2\sin ^2 \theta}{2g}$

g = 9.8m/s2

$= \frac{(115^2)(\sin 30)^2}{2\times 9,8} \\\\= 168.7m$

Therefore, highest point that the cannon ball reaches is 168.7m

6 0

4 years ago

A sample of gold has a mass of 38.6 grams and a volume of 2 cm3. What is the density of gold?

raketka [301]

Density is equal to mass divided by volume; that said, you would divide 38.6 by 2 to get your answer

7 0

4 years ago