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3 years ago

What kind of energy is used in baking?A. LightB. SoundC. HeatD. Mechanical

Physics

1 answer:

noname [10]3 years ago

6 0

Answer:

Explanation:

Thermal energy is used to cook food. Thermal energy is heat. It's converted from either electrical potential energy or chemical potential energy, depending on the cooking appliance.

sorry if I'm wrong

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Right hand rule exercise

saveliy_v [14]

Answer:

A yxiigxih5dd8yixc99uf8g

6 0

3 years ago

A balloon filled with helium gas has an average density of rhob = 0.22 kg/m3. The density of the air is about rhoa = 1.23 kg/m3.

MissTica

Answer:

$a=g\left(\frac{\rho_a}{\rho_b}-1\right)$

45.03681 m/s²

Explanation:

$F_b$ = Buoyant force

W = Weight of the balloon

$\rho_a$ = Density of air = 1.23 kg/m³

$\rho_b$ = Density of balloon = 0.22 kg/m³

$v_a$ = Volume of air

$v_b$ = Volume of balloon

$F_b=\rho_av_bg$

$W=\rho_bv_bg$

g = Acceleration due to gravity = 9.81 m/s²

The net force acting on the balloon is

$F=F_b-W\\\Rightarrow F=\rho_av_bg-\rho_bv_bg\\\Rightarrow \rho_bv_ba=\rho_av_bg-\rho_bv_bg\\\Rightarrow \rho_bv_ba=v_bg(\rho_a-\rho_b)\\\Rightarrow a=\frac{g}{\rho_b}(\rho_a-\rho_b)\\\Rightarrow a=g\left(\frac{\rho_a}{\rho_b}-1\right)$

The equation is $a=g\left(\frac{\rho_a}{\rho_b}-1\right)$

$a=g\left(\frac{\rho_a}{\rho_b}-1\right)\\\Rightarrow a=9.81\times \left(\frac{1.23}{0.22}-1\right)\\\Rightarrow a=45.03681\ m/s^2$

The acceleration of the balloon is 45.03681 m/s²

8 0

3 years ago

You can make a solute dissolve more quickly in a solvent by

Gekata [30.6K]

Answer:Stirring.

Explanation:Stirring a solute into a solvent speeds up the rate of dissolving because it helps distribute the solute particles throughout the solvent. For example, when you add sugar to iced tea and then stir the tea, the sugar will dissolve faster.

3 0

3 years ago

At a particular instant the magnitude of the momentum of a planet is 2.60 × 10^29 kg·m/s, and the force exerted on it by the sta

aleksley [76]

Answer:

F=(-4.8*10^22,0,0) N

Explanation:

<u>Given :</u>

We are given the magnitude of the momentum of the planet and let us call this momentum (p_now) and it is given by p_now = 2.60 × 10^29 kg·m/s. Also, we are given the force exerted on the planet F = 8.5 × 10^22 N. and the angle between the planet and the star is Ф = 138°

Solution :

We are asked to find the parallel component of the force F The momentum here is not constant, where the planet moving along a curving path with varying speed where the rate change in momentum and the force may be varying in magnitude and direction. We divide the force here into two parts: a parallel force F to the momentum and a perpendicular force F' to the momentum.

The parallel force exerted to the momentum will speed or reduce the velocity of the planet and does not change its moving line. Let us apply the direction cosines, we could obtain the parallel force as next

F=|F|cosФp (1)

Where the parallel force F is in the opposite direction of p as the angle between them is larger than 90°. Now we can plug our values for 0 and I F I into equation (1) to get the parallel force to the planet

F=|F|cosФp

=-4.8*10^22 N*p

<em>As this force is in one direction, we could get its vector as next </em>

F=(-4.8*10^22,0,0) N

F=(0,-4.8*10^22,0) N

F=(0,0-4.8*10^22) N

The cosine of 138°, the angle between F and p is, is a negative number, so F is opposite to p. The magnitude of the planet's momentum will decrease.

8 0

3 years ago

The magnetic field inside a 5.0-cm-diameter solenoid is 2.0 T and decreasing at 5.00 T/s. Part A) What is the electric field str

Veronika [31]

Answer:

(A). The electric field strength inside the solenoid at a point on the axis is zero.

(B). The electric field strength inside the solenoid at a point 1.50 cm from the axis is $3.75\times10^{-2}\ V/m$ .

Explanation:

Given that,

Magnetic field = 2.0 T

Diameter = 5.0 cm

Rate of decreasing in magnetic field = 5.00 T/s

(A). We need to calculate the electric field strength inside the solenoid at a point on the axis

Using formula of electric field inside the solenoid

$E=\dfrac{r}{2}|\dfrac{dB}{dt}|$

Electric field on the axis of the solenoid

Here, r = 0

$E=\dfrac{0}{2}\times5.00$

$E = 0$

The electric field strength inside the solenoid at a point on the axis is zero.

(B). We need to calculate the electric field strength inside the solenoid at a point 1.50 cm from the axis

Using formula of electric field inside the solenoid

$E=\dfrac{r}{2}|\dfrac{dB}{dt}|$

$E=\dfrac{1.50\times10^{-2}}{2}\times|5.00|$

$E=0.0375= 3.75\times10^{-2}\ V/m$

Hence, (A). The electric field strength inside the solenoid at a point on the axis is zero.

(B). The electric field strength inside the solenoid at a point 1.50 cm from the axis is $3.75\times10^{-2}\ V/m$ .

4 0

4 years ago

What kind of energy is used in baking?A. LightB. SoundC. HeatD. Mechanical​

What kind of energy is used in baking?A. LightB. SoundC. HeatD. Mechanical