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7nadin3 [17]
2 years ago
11

A soccer ball is kicked from the top of one building with a height of H1 = 30.2 m to another building with a height of H2 = 12.0

m.
The ball is kicked with a speed of v0 = 15.10 m/s at an angle of θ = 74.1° with respect to the horizontal. The mass of a size 5 soccer ball is m = 450 g. When the ball lands on the other building, its speed is 19.89 m/s.
How much energy was lost to air friction? The ball is kicked without a spin.

Physics
1 answer:
salantis [7]2 years ago
8 0

The amount of energy lost to air friction, given the data is 37.71 J

<h3>How to obtain the initial energy</h3>
  • Initial velocity (u) = 15.1 m/s
  • Mass (m) = 450 g = 450 / 1000 = 0.45 Kg
  • Initial Energy (E₁) = ?

E₁ = ½mu²

E₁ = ½ × 0.45 × 15.1²

E₁ = 51.3 J

<h3>How to obtain the final energy</h3>
  • Final velocity (u) = 19.89 m/s
  • Mass (m) = 450 g = 450 / 1000 = 0.45 Kg
  • Final Energy (E₂) = ?

E₂ = ½mv²

E₂ = ½ × 0.45 × 19.89²

E₂ = 89.01 J

<h3>How to determine the energy lost</h3>
  • Initial Energy (E₁) = 51.3 J
  • Final Energy (E₂) = 89.01 J
  • Energy lost =?

Energy lost = E₂ - E₁

Energy lost = 89.01 - 51.3

Energy lost = 37.71 J

Learn more about energy:

brainly.com/question/10703928

#SPJ1

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The magnetic field in a cyclotron is 1.25 T, and the maximum orbital radius of the circulating protons is 0.40 m. (a) What is th
Darya [45]

Answer:

1.92 x 10⁻¹²J

Explanation:

The magnetic force from the magnetic field gives the circulating protons gives the particle the necessary centripetal acceleration to keep it orbiting round the circular path. And from Newton's second law of motion, the force(F) is equal to the product of the mass(m) of the proton and the centripetal acceleration(a). i.e

F = ma

Where;

a = \frac{v^2}{r}             [v = linear velocity, r = radius of circular path]

=> F = m\frac{v^2}{r}           ------------(i)

We also know that the magnitude of this magnetic force experienced by the moving charge (proton) in a magnetic field is given by;

F = q v B sin θ       ----------(ii)

Where;

q = charge of the particle

v = velocity of the particle

B = magnetic field

θ = the angle between the velocity and the magnetic field.

Combining equations (i) and (ii) gives

m\frac{v^2}{r} = q v B sin θ           [θ = 90° since the proton is orbiting at the maximum orbital radius]

=> m\frac{v^2}{r} = q v B sin 90°

=> m\frac{v^2}{r} = q v B

Divide both side by v;

=> m\frac{v}{r} = qB

Make v subject of the formula

v = \frac{qBr}{m}

From the question;

B = 1.25T

m = mass of proton = 1.67 x 10⁻²⁷kg

r = 0.40m

q = charge of a proton = 1.6 x 10⁻¹⁹C

Substitute these values into equation(iii) as follows;

v = \frac{(1.6*10^{-19})(1.25)(0.4)}{(1.67*10^{-27})}

v = 4.79 x 10⁷m/s

Now, the kinetic energy, K, is given by;

K = \frac{1}{2}mv²

m = mass of proton

v = velocity of the proton as calculated above

K = \frac{1}{2}(1.67*10^{-27} * (4.79 * 10^7)^2 )

K = 1.92 x 10⁻¹²J

The kinetic energy is 1.92 x 10⁻¹²J

8 0
3 years ago
If a ball that is 10 meters above the ground is thrown horizontally at 5.51 meters per second. a. how long will it take for the
GalinKa [24]

Answer:

a. t = 1.43 s

b. d = 7.88 m

Explanation:

a. The time of flight can be found using the following equation:

y_{f} = y_{0} + v_{0_{y}}t - \frac{1}{2}gt^{2}

Where:

y_{f}: is the final height = -10 m

y_{0}: is the initial height = 0

v_{0_{y}}: is the initial speed in the vertical direction = 0

g: is the acceleration due to gravity = 9.81 m/s²

By solving the above equation for "t" we have:

t = \sqrt{\frac{2y_{f}}{g}} = \sqrt{\frac{2*10 m}{9.81 m/s^{2}}} = 1.43 s

Hence, the ball will hit the ground in 1.43 s.

b. The distance in the horizontal direction can be found as follows:

x_{f} = x_{0} + v_{0}t + \frac{1}{2}at^{2}

Where:

x₀: is the initial position in the horizontal direction = 0

a: is the acceleration in the horizontal direction = 0 (it is moving at constant speed)

x_{f} = 5.51 m/s*1.43 s = 7.88 m

Therefore, the ball will travel 7.88 m before it hits the ground.

I hope it helps you!

4 0
3 years ago
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Irina-Kira [14]

The response is False, both bars, iron bars and plastic bars have de same inertia, this characteristic does not depend on the type of material, the inertia depends on his transverse section, since we can estimate in the following formula

<span>Area moment of inertia Ixx = BH3/12</span>

<span>Area moment of inertia Iyy= HB3/12</span>

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Which bright solar feature is shown in the picture above?
Ghella [55]

Answer : (B) Prominence

Explanation :

A large, glittering and gaseous characteristic which is extending outward from the surface of the sun is called <em>Prominence</em>.

<em>Photosphere</em> is one of the layer of sun where the prominence are anchored and then they move into the corona of the sun.

<em>Corona</em> is a region in the surface of the sun which is the constituent of hot ionized gases (plasma).

The prominence consists of colder plasma and this prominence plasma is much more shining and denser as compared to coronal plasma.

Hence, the correct option is (B) Prominence.

6 0
3 years ago
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a scale model of the solar system where 50 cm represents 1.0x10 to the fifth km is actual distance what would be the dimension o
Fofino [41]

The distance between Mars and the Sun in the scale model would be 1140 m

Explanation:

In this scale model, we have:

x_1 = 50 cm represents an actual distance of

d_1 = 1.0\cdot 10^5 km

The actual distance between Mars and the Sun is 228 million km, therefore

d_2=228\cdot 10^6 km

On the scale model, this would corresponds to a distance of x_2.

Therefore, we can write the following proportion:

\frac{x_1}{d_1}=\frac{x_2}{d_2}

And solving for x_2, we find:

x_2=\frac{x_1 d_2}{d_1}=\frac{(50)(228\cdot 10^6)}{1\cdot 10^5}=1.14\cdot 10^5 cm = 1140 m

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4 0
3 years ago
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