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DiKsa [7]
3 years ago
8

Balloon like cell organelle in the cytoplasm that can store food, water, and wastes, also provides pressure against the cell wal

l for cellular support in plant cells
A
Vacuoles
B
Mitochondrion
C
Ribosomes
D
Nucleus
PLSSSSSS HELP MEEEEE!! :C
Physics
1 answer:
melisa1 [442]3 years ago
8 0

Answer:

Vacuoles......final answer

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g If this combination of resistors were to be replaced by a single resistor with an equivalent resistance, what should that resi
Anettt [7]
<h2>Question:</h2>

In this circuit the resistance R1 is 3Ω, R2 is 7Ω, and R3 is 7Ω. If this combination of resistors were to be replaced by a single resistor with an equivalent resistance, what should that resistance be?

Answer:

9.1Ω

Explanation:

The circuit diagram has been attached to this response.

(i) From the diagram, resistors R1 and R2 are connected in parallel to each other. The reciprocal of their equivalent resistance, say Rₓ, is the sum of the reciprocals of the resistances of each of them. i.e

\frac{1}{R_X} = \frac{1}{R_1} + \frac{1}{R_2}

=> R_{X} = \frac{R_1 * R_2}{R_1 + R_2}             ------------(i)

From the question;

R1 = 3Ω,

R2 = 7Ω

Substitute these values into equation (i) as follows;

R_{X} = \frac{3 * 7}{3 + 7}

R_{X} = \frac{21}{10}

R_{X} = 2.1Ω

(ii) Now, since we have found the equivalent resistance (Rₓ) of R1 and R2, this resistance (Rₓ) is in series with the third resistor. i.e Rₓ and R3 are connected in series. This is shown in the second image attached to this response.

Because these resistors are connected in series, they can be replaced by a single resistor with an equivalent resistance R. Where R is the sum of the resistances of the two resistors: Rₓ and R3. i.e

R = Rₓ + R3

Rₓ = 2.1Ω

R3 = 7Ω

=> R = 2.1Ω + 7Ω = 9.1Ω

Therefore, the combination of the resistors R1, R2 and R3 can be replaced with a single resistor with an equivalent resistance of 9.1Ω

4 0
3 years ago
The difference in outcome between the potential energy in a marshmallow burned over a campfire on the end of a stick and in one
Lapatulllka [165]

Answer:

How fast and efficient the energy is released.

Explanation:

Before burning the marshmallow energy is stored in it in the form of chemical bond energy or chemical potential energy. So upon burning this energy is released. So there will be a difference in energy release from a burned marshmallow and the one we eat straight from package.

5 0
3 years ago
When an object moves in uniform circular motion, the direction of its acceleration is?
Harrizon [31]
Clock wise idk i think you should double check my answer

7 0
3 years ago
A rising pendulum bob gains _____ energy.
puteri [66]

Answer:

B. Kinetic energy.............

4 0
3 years ago
Read 2 more answers
As a city planner, you receive complaints from local residents about the safety of nearby roads and streets. One complaint conce
WINSTONCH [101]

Answer:

a)   x₁ = 290.50 feet ,  x₂ = 169.74 feet , b)  v_max= 41 mph

Explanation:

For this exercise we will work in two parts, the first with Newton's second law to find the acceleration of vehicles

X Axis          fr = m a

Y Axis          N-W = 0

                    N = W = mg

The force of friction has the expression

                  fr = μ N

We replace

                 μ mg = ma

                 a = μ g

                 g = 32 feet / s²

Let's calculate the acceleration for each coefficient and friction

μ              a (feet / s2)

0.599       19.168

0.536       17,152

0.480       15.360

0.350        11.200

These are the acceleration values, for the maximum distance we use the minimum acceleration (a₁ = 11,200 feet / s²) and for the minimum braking distance we use the maximum acceleration (x₂ = 19,168 feet / s²)

                 v² = v₀² - 2 a x

When the speed stops it is zero

                 x₁ = v₀² / 2 a₁

                         

Let's reduce speed

            v₀ = 55mph (5280 foot / 1 mile) (1h / 3600s) = 80,667 feet / s²

Let's calculate the maximum braking distance

            x₁ = 80.667² / (2 11.2)

            x₁ = 290.50 feet

The minimum braking distance

            x₂ = 80.667² / (2 19.168)

            x₂ = 169.74 feet

b) maximum speed to stop at distance x = 155 feet

            0 = v₀² - 2 a x

            v₀ = √2 a x

We calculate the speed for the two accelerations

             v₀₁ = √ (2 11.2 155)

             v₀₁ = 58.92 feet / s

       

             v₀₂ = √ (2 19.168 155)

             v₀₂ = 77.08 feet / s

To stop at the distance limit in the worst case the maximum speed must be 58.92 feet / s = 40.85 mph = 41 mph

5 0
3 years ago
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