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soldi70 [24.7K]

2 years ago

14

While being thrown, a net force of 132 N acts on a baseball (mass = 140g) for a period of 4.5 x 10^-2 sec. what is the magnitude

of the change in momentum of the ball?

1 answer:

Vera_Pavlovna [14]2 years ago

5 0

Answer:

5.94 N·s

Explanation:

F = 132 N

t = 0.045 s

Impulse = Ft = (132 N)(0.045 s) = 5.94 N·s

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A 38-kg canoe breaks free from its dock and floats downriver at a speed of 2.2 m/s. What is the momentum of the canoe? 48.0 m/s

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Geologists apply various methods to study the layers of the Earth. Which of the following is NOT a method used to study the Eart

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Answer:

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2 years ago

A light bulb emits light that travels uniformly in all directions. Detailed measurements show that at a distance of 56 m from th

vladimir1956 [14]

Complete question:

A light bulb emits light that travels uniformly in all directions. Detailed measurements show that at a distance of 56 m from the bulb, the amplitude of the electric field is 3.78 V/m. What is the average intensity of the light?

Answer:

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Explanation:

Given;

Amplitude of the electric field, E₀ = 3.78 V/m

The average intensity of the light is calculated as follows;

$I_{avg} = \frac{c\epsilon_0 E_0^2}{2}$

where;

$I_{avg}$ is the average intensity of the light

c is speed of light = 3 x 10⁸ m/s

$I_{avg} = \frac{(3\times 10^8)(8.85 \times 10^{-12}) (3.78)^2}{2} \\\\I_{avg} = 0.01897 \ W/m^2\\\\I_{avg} = 0.02 \ W/m^2$

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4 0

2 years ago

A rod extending between x = 0 and x = 13.0 cm has uniform cross-sectional area A = 8.00 cm2. Its density increases steadily betw

mezya [45]

Answer:

The mass is $m = 3.45408 kg$

Explanation:

From the question we are told that

The extension of the rod is $from , \ x_1 = 0 \to x_2 = 13.0$

The area is $A = 8.0 cm^2$

The density increase as follows $from \ \rho_1 =2.5 g/cm^2 \to \rho_2= 19.0 g/cm^3$

The equation $\rho = B + Cx$

at $x_1= 0$ $\rho_1 =2.5 g/cm^2$

So

$2.5 = B + 0$

=> $B =2.5$

So at $x_2 = 13.0$ , $\rho_2= 19.0 g/cm^3$

So

$19.0 = 2.5 + C(13)$

=> $C = 1.27$

Now

$m = 8 \int\limits^{13}_{0} {2.5 + 1.27x} \, dx$

$m = 8 [{2.5 +\frac{ 1.27x^2}{2} } ]\left | 13} \atop {0}} \right.$

$m = 8 [{2.5 +\frac{ 1.27(13)^2}{2} } ]$

$m = 3454.08 g$

$m = 3.45408 kg$

8 0

2 years ago