The question is on the attached image<br> Help please :(

Q: Riri wants to bake a cake. She adds flour, sugar, egg, baking soda, and yeast into a bowl and mixed them together. After all

babunello [35]

Answer:

I don't know what you mean about which changes occurred in this process but if its why the dough starts rising then its caused by the carbon dioxide in baking soda and yeast which is a fungus

8 0

3 years ago

Read 2 more answers

A river flowing steadily at a rate of 175 m3/s is considered for hydroelectric power generation. It is determined that a dam can

Alex_Xolod [135]

Answer:

137200000 watts or 137200 kilowatts

Explanation:

The formula for power is P= dhrg

Where P = Power in watts

d = density of water (1000 kg/m^3)

h = height in meters

r = flow rate in cubic meters per second,

g = acceleration due to gravity of 9.8 m/s^2,

Plugging in the known values,

we get

P = 1000 kg/m^3 * 80 m * 175 m^3/s * 9.8 m/s^2

P = 80000 kg/m^2 * 175 m^3/s * 9.8 m/s^2

P = 14000000 kg m/s * 9.8 m/s^2

P = 137200000 kg m^2/s^3

P = 137200000 watts or 137200 kilowatts

The above figure assumes 100% efficiency which is impossible. A good efficiency would be 90% so the actual power available would be close to 0.90 * 137200 = 123480 kilowatts

6 0

3 years ago

If a box is pulled with a force of 100 N at an angle of 25

love history [14]

The X and Y components of the force are 90.63 Newton and 42.26 Newton respectively.

<u>Given the following data:</u>

Force = 100 Newton.

Angle of inclination = 25°

To determine the X and Y components of the force:

<h3>The horizontal component (X) of a force:</h3>

Mathematically, the horizontal component of a force is given by this formula:

$F_x = Fcos \theta\\\\F_x =100 \times cos25\\\\F_x =100 \times 0.9063$

Fx = 90.63 Newton.

<h3>The vertical component (Y) of tensional force:</h3>

Mathematically, the vertical component of a force is given by this formula:

$F_y = Fsin \theta\\\\F_y =100 \times sin25\\\\F_y =100 \times 0.4226$

Fy = 42.26 Newton.

Read more on horizontal component here: brainly.com/question/4080400

6 0

2 years ago

Compare the gravitational acceleration on the following objects compared to the Sun using:

arsen [322]

The gravitational acceleration of White dwarf compared to Sun is 13,675.86.

The gravitational acceleration of Neutron star compared to Sun is 6.79 x 10⁻²⁴.

The gravitational acceleration of Star Betelgeuse compared to Sun is 8.5 x 10¹⁰.

<h3>Mass of the planets</h3>

Mass of sun = 2 x 10³⁰ kg

Mass of white dwarf = 2.765 x 10³⁰ kg

Mass of Neutron star = 5.5 x 10¹² kg

Mass of star Betelgeuse = 2.188 x 10³¹ kg

<h3>Radius of the planets</h3>

Radius of sun = 696,340 km

Radius of white dwarf = 7000 km

Radius of Neutron star = 11 km

Radius of star Betelgeuse = 617.1 x 10⁶ km

<h3>Gravitational acceleration of White dwarf compared to Sun</h3>

$\frac{g(star)}{g(sun)} = \frac{M(star)}{M(sun)} \times [\frac{R(sun)}{R(star)} ]^2\\\\\frac{g(star)}{g(sun)} = \frac{2.765 \times 10^{30}}{2\times 10^{30}} \times [\frac{696,340,000}{7,000,000} ]^2\\\\\frac{g(star)}{g(sun)} = 13,675.86$

<h3>Gravitational acceleration of Neutron star compared to Sun</h3>

$\frac{g(star)}{g(sun)} = \frac{M(star)}{M(sun)} \times [\frac{R(sun)}{R(star)} ]^2\\\\\frac{g(star)}{g(sun)} = \frac{5.5 \times 10^{12}}{2\times 10^{30}} \times [\frac{11,000}{7,000,000} ]^2\\\\\frac{g(star)}{g(sun)} = 6.79\times 10^{-24}$

<h3>Gravitational acceleration of Star Betelgeuse compared to Sun</h3>

$\frac{g(star)}{g(sun)} = \frac{M(star)}{M(sun)} \times [\frac{R(sun)}{R(star)} ]^2\\\\\frac{g(star)}{g(sun)} = \frac{2.188 \times 10^{31}}{2\times 10^{30}} \times [\frac{617.1 \times 10^9}{7,000,000} ]^2\\\\\frac{g(star)}{g(sun)} = 8.5\times 10 ^{10}$

Learn more about acceleration due to gravity here: brainly.com/question/88039

3 0

3 years ago

Two spheres have identical charges and are 75 cm apart. the force between them is +0.30 n. what is the magnitude of the charge o

ella [17]

Answer:

$4.33\cdot 10^{-6}C$ , charges are both positive or both negative

Explanation:

The electrostatic force between the two spheres is given by

$F=k\frac{q_1 q_2}{r^2}$

where

k is the Coulomb's constant

q1 and q2 are the charges on the two spheres

r is the distance between the centres of the two spheres

In this problem, we have

$F=+0.30 N$ is the force

$r=75 cm=0.75 m$ is the distance between the spheres

$q_1 =q_2 =q$ because the two spheres have identical charge

Solving the formula for q, we find

$q=\sqrt{\frac{Fr^2}{k}}=\sqrt{\frac{(+0.30 N)(0.75 m)^2}{9\cdot 10^9}}=4.33\cdot 10^{-6}C$

And the two charges have the same sign (so, both positive or both negative), since the sign of the force is positive (+0.30 N), so it is a repulsive force.

5 0

3 years ago

The question is on the attached image Help please :(