Ask question

Ask question

All categories

3 years ago

5

Some areas of the Earth receive more solar radiation than others. Which of the following results from the Sun's uneven heating o

f the Earth?

1 answer:

GuDViN [60]3 years ago

7 0

The core of the earth lets out radiation from under the ground which make the core responce of how volcano's errupt.

You might be interested in

Identify five common business uses for electronic spreadsheets

Sloan [31]

For the answer to the question above
Forecasting how a business might do in the future.
Calculating tax.
Doing basic payrolls.
Calculating Revenues.
Producing charts.

--Going past 5--

Inventory tracking
Very (VERY) basic CRM for small businesses
I hope my answer helped you.

4 0

3 years ago

Read 2 more answers

An ammeter with a resistance of 5.0 ohm is connected in series with a 3.0V cell and a lamp rated at 300 mA, 3V. Calculate the cu

Alex

Answer:

I = 0.2 A

Explanation:

Lamp is rated at 300 mA

I_lamp = 0.3 A

Voltage is; V = 3V

Thus; Resistance is given by;

R = V/I

R = 3/0.3

R = 10 ohms

Now, since the ammeter of 5 ohms is connected in series with the lamp. Thus equivalent resistance;

R_eq = 10 + 5

R_eq = 15 ohms

Ammeter current will be;

I = V/R_eq

I = 3/15

I = 0.2 A

3 0

3 years ago

I can’t figure it out! <br> Please help I have a test tomorrow

nignag [31]

it's 20m/s i.e. no. (2)

7 0

3 years ago

The yoga term asanas are which of the following?

frutty [35]

Answer:

i guss its A

Explanation:

3 0

3 years ago

A 5kg object moving horizontally at 3m/s collides with a stationary 3kg object. After the collision, the 5kg object is deflected

gavmur [86]

Answer:

The velocity of each ball after the collision are 2.19 m/s and 2.58 m/s.

Explanation:

Given that,

Mass of object = 5 kg

Speed = 3 m/s

Mass of stationary object = 3 kg

Moving object deflected = 30°

Stationary object deflected = 31°

We need to calculate the velocity of each ball after collision

Using conservation of momentum

Along x-axis

$m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}\cos\theta+m_{2}v_{2}\cos\theta$

Put the value into the fomrula

$5\times3+0=5\times v_{1}\cos30+3\times v_{2}\cos45$

$15=5v_{1}\times\dfrac{\sqrt{3}}{2}+3v_{2}\times\dfrac{1}{\sqrt{2}}$

$15=\dfrac{5\sqrt{3}}{2}v_{1}+\dfrac{3}{\sqrt{2}}v_{2}$ ....(I)

Along y -axis

$m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}\sin\theta+m_{2}v_{2}\sin\theta$

Put the value into the formula

$0+0=5\times v_{1}\sin30-3\times v_{2}\sin45$

$\dfrac{5}{2}v_{1}-\dfrac{3}{\sqrt{2}}v_{2}=0$ ...(II)

From equation (I) and (II)

$v_{1}=\dfrac{15\times2}{5\sqrt{3}+5}$

$v_{1}=2.19\ m/s$

Put the value of v₁ in equation (I)

$\dfrac{5}{2}\times2.19-\dfrac{3}{\sqrt{2}}v_{2}=0$

$v_{2}=\dfrac{5.475\times\sqrt{2}}{3}$

$v_{2}=2.58\ m/s$

Hence, The velocity of each ball after the collision are 2.19 m/s and 2.58 m/s.

3 0

3 years ago