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tekilochka
2 months ago
9

Ronnie kicks a playground ball with an initial velocity of 16 m/s at an angle of 40° relative to the ground. What is the approxi

mate horizontal component of the initial velocity?
Physics
2 answers:
Softa [3K]2 months ago
7 0
The calculation for the horizontal component is performed as follows:
Vhorizontal = V · cos(angle)

For your instance, Vhorizontal = 16 · cos(40) equates to 12.3 m/s

Conclusion: 12.3 m/s
serg [3.5K]2 months ago
4 0

Answer:

Concluding answer: D 12.3 m/s

Explanation:

I just completed the quiz.

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Ellen does an experiment by releasing a ball from a height of 1 m above each floor in a tall building. She records the time it t
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The question Ellen is likely exploring is "In what way does distance influence the gravitational force acting on objects?"

Explanation:

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1 month ago
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The hot glowing surfaces of stars emit energy in the form of electromagnetic radiation. It is a good approximation to assume tha
Maru [3345]

Answer:

A) 5.1*10^10m B) 5.4*10^6m

Explanation:

Utilizing the formula for surface radiation P (energy per second in Watts) = emissivity constant * surface area * Stefan-Boltzmann constant * Temperature in Kelvin^4 *

2.7*10^31 = 1* 5.67*10^-8*A*11000^4

Rearranging to solve for A = 2.7*10^31 / (5.67*10^-8*1.46*10^16) = 0.3261*10^23m^2

Assuming the shape is spherical, the surface area is = 4πR^2 (radius of Rigel)

R = √(0.3261*10^23 / 4*π) = 5.1 * 10^10m

B) repeating the same calculation

2.1 *10^23 = 1*A*5.67*10^-8*10000^4 where A is the surface area of Procyon

Rearranging gives A = 2.1*10^23/(5.67*10^-8*10^16)

A = 0.37*10^15

Assuming the star is spherical;

A = 4πR^2 where R is Procyon's radius

R = √(0.37*10^15/4π) = 5.4*10^6m

4 0
15 days ago
An ideal gas is allowed to expand isothermally from 2.00 l at 5.00 atm in two steps:
Sav [3153]

Heat supplied to the gas = Q = 743 Joules

Work applied to the gas = W = -743 Joules

\texttt{ }

Additional explanation

The Ideal Gas Law that should be remembered is:

\large {\boxed {PV = nRT} }

P = Pressure (Pa)

V = Volume (m³)

n = number of moles (moles)

R = Gas Constant (8.314 J/mol K)

T = Absolute Temperature (K)

Now, let’s proceed with the problem!

\texttt{ }

Given:

Initial volume of the gas = V₁ = 2.00 L

Initial pressure of the gas = P₁ = 5.00 atm

Unknown:

Work done on the gas = W =?

Heat supplied to the gas = Q =?

Solution:

Step A:

An ideal gas expands isothermally:

P_1V_1 = P_2V_2

5.00 \times 2.00 = 3.00 \times V_2

V_2 = 10 \div 3

V_2 = 3\frac{1}{3} \texttt{ L}

\texttt{ }

Next, we will determine the work performed on the gas:

W_A = -P_2(V_2 - V_1)

W_A = -3.00(3\frac{1}{3} - 2.00)

W_A = \boxed{-4 \texttt{ L.atm}}

\texttt{ }

Step B:

By utilizing the methodology mentioned earlier:

P_2V_2 = P_3V_3

3.00 \times 3\frac{1}{3} = 2.00 \times V_3

V_3 = 10 \div 2

V_3 = 5 \texttt{ L}

\texttt{ }

Next, we will ascertain the work completed on the gas:

W_B = -P_3(V_3 - V_2)

W_B = -2.00(5 - 3\frac{1}{3})

W_B = \boxed{-3\frac{1}{3} \texttt{ L.atm}}

\texttt{ }

Ultimately, we can calculate the total work done and heat supplied as follows:

W = W_A + W_B

W = -4 + (-3\frac{1}{3})

W = -7\frac{1}{3} \texttt{ L.atm}

W = -7\frac{1}{3} \times 101.33 \texttt{ J}

\boxed{W \approx -743 \textt{ J}}

\texttt{ }

\Delta U = Q + W

0 = Q + (-743)

\boxed{Q = 743 \texttt{ J}}

\texttt{ }

Learn more

  • Minimum Coefficient of Static Friction:
  • The Pressure In A Sealed Plastic Container:
  • Effect of Earth’s Gravity on Objects:

\texttt{ }

Answer details

Grade: High School

Subject: Physics

Chapter: Pressure

5 0
29 days ago
Which contributions did Galileo make to the model of the solar system? Select two options.a mathematical model for the orbits of
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Galileo's contributions to the solar system model include: Data indicating that planets reflect sunlight like the Moon, and his observations of Jupiter's moons orbiting the gas giant. With the assistance of an early telescope that he constructed, Galileo made these two significant discoveries.
7 0
1 month ago
Suppose that an owner of the same dog breed has also taken some measurements. They notice that the surface area of the dog has i
inna [3103]

Answer:

The surface area of the dog changes from A to 3A

Explanation:

It is stated that the dog's surface area has increased by a factor of 3 over four years.

We need to calculate the change in the relative surface area of the dog over this timeframe.

Let’s assume the initial surface area is A.

Since the surface area has been multiplied by 3,

it follows that the surface area after four years is equal to 3×A = 3A.

Thus, the dog's surface area transitions from A to 3A.

7 0
1 month ago
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