What is a limitation of the Big Bang model?

Why, on a sunny day, it is normally hot inside a greenhouse

Dahasolnce [82]

Because of the greenhouse gases inside the greenhouse, and the gases trap the heat from the sun so the plant don't freeze, hope this helps

5 0

2 years ago

What is the formula of finding displacement

xenn [34]

Displacement is the final position of the object minus the initial position of the object.
Xf - Xi. Displacement is not the distance of the object. If you go to the right 10m and to the left another 10m, your displacement is 0m. But your distance is 20m

8 0

3 years ago

in the circuit diagram even in suppose the resistor R1 R2 and R3 have the value of 5 ohm 10 ohm 30 ohm respectively which have b

Anuta_ua [19.1K]

<h3><u>Given</u> :</h3>

Three identical resistors of resistances 5Ω, 10Ω and 30Ω are connected with a battery of 12V $.$

We have to find current through the each resistor and equivalent resistance of circuit $.$

<h3><u>SoluTion</u> :</h3>

➝ Equivalent resistance of series connection is given by

<u>R = R1 + R2 + R3</u>

➝ We know that, Equal current flow through each resistor in series connection.

➝ As per ohm's law, Current flow through a conductor is directly proportional to the applied potential difference.

<u>V = IR</u>

◈ <u>Equivalent resistance</u> :

⇒ Req = R1 + R2 + R3

⇒ Req = 5 + 10 + 30

⇒ <u>Req = 45Ω</u>

◈ <u>Current flow in circuit</u> :

⇒ V = IReq

⇒ 12 = I × 45

⇒ <u>I = 0.27A</u>

፨ Therefore, 0.27A current will flow through each resistor.

5 0

2 years ago

The temperature of the cooling water as it leaves the hot engine of an automobile is 240 °F. After it passes through the radiato

djverab [1.8K]

Answer:571.09 kJ

Explanation:

Given

Temperature of cooling water from engine exit $=240^{\circ} F\approx 115.55^{\circ}C$

After Passing through the radiator its temperature decreases to $175^{\circ}F\approx 79.44^{\circ}F$

specific heat of water $=4.184 J/g^{\circ}C$

Volume of water $= 1 gallon\approx 3.78 L$

density of water $\rho =1 gm/mL$

Thus mass of water $=\rho \times V=3.78\times 1=3.78 kg$

Heat transferred to the surrounding is equal to heat absorbed by cooling water

$Q=m\cdot c\cdot \Delta T$

$Q=3.78\times 4.184\times 1000\times (115.55-79.44)$

$Q=3.78\times 4.184\times 1000\times (36.11)$

$Q=571.09 kJ$

4 0

2 years ago

Capacitor 2 has half the capacitance and twice the potential difference as capacitor 1. What is the ratio (U_{\rm C})_1/\,(U_{\r

IrinaK [193]

Answer:

1/2

Explanation:

The energy stored in a capacitor is given by

$U=\frac{1}{2}CV^2$

where

C is the capacitance

V is the potential difference

Calling $C_1$ the capacitance of capacitor 1 and $V_1$ its potential difference, the energy stored in capacitor 1 is

$U=\frac{1}{2}C_1 V_1^2$

For capacitor 2, we have:

- The capacitance is half that of capacitor 1: $C_2 = \frac{C_1}{2}$

- The voltage is twice the voltage of capacitor 1: $V_2 = 2 V_1$

so the energy stored in capacitor 2 is

$U_2 = \frac{1}{2}C_2 V_2^2 = \frac{1}{2}\frac{C_1}{2}(2V_1)^2 = C_1 V_1^2$

So the ratio between the two energies is

$\frac{U_1}{U_2}=\frac{\frac{1}{2}C_1 V_1^2}{C_1 V_1^2}=\frac{1}{2}$

4 0

2 years ago