Action reaction forces are in which one of newtons laws

The sunlight hits the surface of a lake this will like caused

ziro4ka [17]

A). No. Condensation happens when you take heat out of a gas.

b). No. I'm not sure what transpiration is.

c). Yes. Evaporation happens when you add heat to a liquid.

d). No. Sublimation sometimes happens when you add heat to a solid.

3 0

3 years ago

Read 2 more answers

A toaster using a Nichrome heating element operates on 120 V. When it is switched on at 28 ∘С, the heating element carries an in

sukhopar [10]

Answer:

The final temperature of the element = 262.67°C

The power dissipated in the heating element initially = 163.21 W

The power dissipated in the heating element when the current reaches 1.23 A = 147.60 W

Explanation:

Our given parameters include;

A Nichrome heating element operates on 120 V.

Voltage (V) = 120V

Initial Current (I₁) = 1.36 A

Initial Temperature (T₁) = 28°C

Final Current (I₂) = 1.23 A

Final Temperature (T₂) = unknown ????

Temperature dependencies of resistance is given by:

$R_{T(2)}=R_1[1+\alpha (T_2-T_1)]$ ---------------------- (1)

in which R₁ is the resistance at temperature T₁

$R_{T(2)$ is the resistance at temperature T₂

Given that V= IR

R = $\frac{V}{I}$

Therefore, the resistance at temperature 28°C is;

$R_{28}= \frac{120V}{1.36A}$

= 88.24Ω

$R_{T(2)$ = $\frac{120V}{1.23A}$

= 97.56Ω

From (1) above;

$R_{T(2)}=R_1[1+\alpha (T_2-T_1)]$

97.56 = 88.24 [ 1 + 4.5×10⁻⁴(°C)⁻¹(T₂-28°C)]

$\frac{97.56}{88.24}= 1+(4.5*10^{-4})(T-28^0C)$

1.1056 - 1 = 4.5×10⁻⁴(°C)⁻¹(T₂-28°C)

0.1056 = 4.5×10⁻⁴(T₂-28°C)

$\frac{0.1056}{4.5*10^{-4}}= T-28^0C$

T - 28° C = 234.67

T = 234.67 + 28° C

T = 262.67 ° C

(b)

What is the power dissipated in the heating element initially and when the current reaches 1.23 A

The power dissipated in the heating element initially can be calculated as:

P = I²₁R₂₈

P = (1.36A)²(88.24Ω)

P = 163.209 W

P ≅ 163.21 W

The power dissipated in the heating element when the current reaches 1.23 A can be calculated as:

$P= I^2_2R_{T^0C$

P = (1.23)²(97.56Ω)

P = 147.598524

P ≅ 147.60 W

6 0

2 years ago

Which of the following statements is true regarding the constructive interference diagram shown below? Select all that apply.

yulyashka [42]

Answer:

I believe its B and C

Explanation:

If I'm wrong please tell me so I can correct my answer.

6 0

2 years ago

Read 2 more answers

24 A uniform electric field of magnitude 1.1×104 N/C is perpendicular to a square sheet with sides 2.0 m long. What is the elect

Tatiana [17]

Answer:

$44,000 Nm^2/C$

Explanation:

The electric flux through a certain surface is given by (for a uniform field):

$\Phi = EA cos \theta$

where:

E is the magnitude of the electric field

A is the area of the surface

$\theta$ is the angle between the direction of the field and of the normal to the surface

In this problem, we have:

$E=1.1\cdot 10^4 N/C$ is the electric field

L = 2.0 m is the side of the sheet, so the area is

$A=L^2=(2.0)^2=4.0 m^2$

$\theta=0^{\circ}$ , since the electric field is perpendicular to the surface

Therefore, the electric flux is

$\Phi =(1.1\cdot 10^4)(4.0)(cos 0^{\circ})=44,000 Nm^2/C$

4 0

2 years ago

A bicycle wheel rotates at a constant 25 rev/min. What is true about its angular acceleration?

Nostrana [21]

Answer:

The angular acceleration is zero

Explanation:

When an object is in rotational motion, it has a certain angular velocity, which is the rate of displacement of its angular position.

This angular velocity can change or remain constant - this is given by the angular acceleration, which is:

$\alpha =\frac{\Delta \omega}{\Delta t}$

where

$\Delta \omega$ is the change in angular velocity

$\Delta t$ is the time elapsed

Therefore, the angular acceleration is the rate of change of angular velocity.

In this problem, the bicycle rotates at a constant angular velocity of

$\omega=25 rev/min$

This means that the change in angular velocity is zero:

$\Delta \omega=0$

And so, that the angular acceleration is zero:

$\alpha=0$

8 0

3 years ago