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3 years ago

Average weather conditions in a specific location for a long time

Physics

2 answers:

masya89 [10]3 years ago

7 0

Answer:

B- the climate

Mariulka [41]3 years ago

6 0

Answer:

B - the climate

Explanation:

humidity is the concentration of water vapour present in air.

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Copper has a specific heat of 0.386 J/g°C. How much heat is required to increase 5.00 g of copper from 0.0°C to 10.0°C?

Leto [7]

The answer is 19.3 j

3 0

3 years ago

Read 2 more answers

A 70mm long blockhas cross-section of 50mm by 10mm the block is subjected to forces 60KN (tension) on the 50mm by 10mm face and

sammy [17]

Answer:

970 kN

Explanation:

The length of the block = 70 mm

The cross section of the block = 50 mm by 10 mm

The tension force applies to the 50 mm by 10 mm face, F₁ = 60 kN

The compression force applied to the 70 mm by 10 mm face, F₂ = 110 kN

By volumetric stress, we have that for there to be no change in volume, the total pressure applied by the given applied forces should be equal to the pressure removed by the added applied force

The pressure due to the force F₁ = 60 kN/(50 mm × 10 mm) = 120 MPa

The pressure due to the force F₂ = 110 kN/(70 mm × 10 mm) = 157.142857 MPa

The total pressure applied to the block, P = 120 MPa + 157.142857 MPa = 277.142857 MPa

The required force, F₃ = 277.142857 MPa × (70 mm × 50 mm) = 970 kN

7 0

2 years ago

What is the speed of a car that traveled 500 meter in 30 seconds?

GenaCL600 [577]

<h2>Greetings!</h2>

To find speed, you need to remember the formula:

Speed = distance ÷ time

So plug the given values in:

500 ÷ 30 = 16.66

<h3>So the speed is 16.66m/s (metres per second)</h3>
<h2>Hope this helps!</h2>

3 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

The maximum wavelength that an electromagnetic wave can have and still eject electrons from a metal surface is 542 nm. What is t

Veseljchak [2.6K]

Answer:

2.295 eV

Explanation:

maximum wavelength, λ = 542 nm = 542 x 10^-9 m

The work function of the metal is defined as the minimum amount of energy falling on the metal so that the photo electrons just ejects the surface of metal.

$W_{o}=\frac{hc}{\lambda }$

where, h is the Plank's constant and c be the speed of light

h = 6.634 x 10^-34 Js

c = 3 x 10^8 m/s

$W_{o}=\frac{6.634\times 10^{-34}\times 3\times 10^{8}}{542\times10^{-9} }$

$W_{o}=3.67\times 10^{-19} J=\frac{3.67\times 10^{-19}}{1.6\times 10^{-19}} eV$

Wo = 2.295 eV

Thus, the work function of this metal is 2.295 eV.

5 0

2 years ago