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loris [4]
3 years ago
8

A wave with low frequency would have relatively ________.

Physics
2 answers:
N76 [4]3 years ago
3 0
Answer A, low energy and long wavelength.
Goshia [24]3 years ago
3 0
The answer is
A. low energy and a long wavelength.
Waves with low frequency have a longer wavelength, thus having lower energy..
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A stunt car traveling at 20 m/s flies horizontally off a cliff and lands 39.2 m from the base of the cliff. How tall is the clif
Arada [10]

Answer:

18.82 m

Explanation:

The following data were obtained from the question:

Initial velocity (u) = 20 m/s

Horizontal distance (s) = 39.2 m

Height (h) of the cliff =?

Next, we shall determine the time taken for the car to get to the ground. This can be obtained as follow:

Initial velocity (u) = 20 m/s

Horizontal distance (s) = 39.2 m

Time (t) taken to reach the ground =?

s = ut

39.2 = 20 × t

Divide both side by 20

t = 39.2 / 20

t = 1.96 s

Finally, we shall determine the height of the cliff as follow:

Acceleration due to gravity (g) = 9.8 m/s²

Time (t) taken to reach the ground = 1.96 s

Height (h) of the cliff =?

h = ½gt²

h = ½ × 9.8 × 1.96²

h = 4.9 × 3.8416

h = 18.82 m

Therefore, the cliff is 18.82 m high

7 0
3 years ago
A wave has wavelength of 8m and a speed of 360 m/s. What is the frequency of the wave?
Naily [24]

Answer:

45~ \text{Hz}.

Explanation:

\text{Frequency,}~ f = \dfrac{v}{\lambda}\\\\\\~~~~~~~~~~~~~~~~~~=\dfrac{360~ \text{ms}^{-1}}{8~ \text m}\\\\\\~~~~~~~~~~~~~~~~~~=45~ \text s^{-1}\\\\\\~~~~~~~~~~~~~~~~~~~=45~ \text{Hz}

8 0
2 years ago
Which of the following is the most likely global climate change?a.a decrease in the overall temperature of the Earthb.an increas
andrew11 [14]
I believe the answer is C. an increase in the number of cold climates
5 0
3 years ago
Read 2 more answers
Object A is moving due east, while object B is moving due north. They collide and stick together in a completely inelastic colli
serious [3.7K]

Answer:

pf = 198.8 kg*m/s

θ = 46.8º N of E.

Explanation:

  • Since total momentum is conserved, and momentum is a vector, the components of the momentum along two axes perpendicular each other must be conserved too.
  • If we call the positive x- axis to the W-E direction, and the positive y-axis to the S-N direction, we can write the following equation for the initial momentum along the x-axis:

       p_{ox} = p_{oAx} + p_{oBx}  (1)

  • We can do exactly the same for the initial momentum along the y-axis:

       p_{oy} = p_{oAy} + p_{oBy}  (2)

  • The final momentum along the x-axis, since the collision is inelastic and both objects stick together after the collision, can be written as follows:

       p_{fx} =  (m_{A} + m_{B} ) * v_{fx}  (3)

  • We can repeat the process for the y-axis, as follows:

       p_{fy} =  (m_{A} + m_{B} ) * v_{fy}  (4)

  • Since (1) is equal to (3), replacing for the givens, and since p₀Bₓ = 0, we can solve for vfₓ as follows:

       v_{fx} = \frac{p_{oAx}}{(m_{A}+ m_{B)}} = \frac{m_{A}*v_{oAx} }{(m_{A}+ m_{B)}} =\frac{17.0kg*8.00m/s}{46.0kg} =  2.96 m/s (5)

  • In the same way, we can find the component of the final momentum along the y-axis, as follows:

       v_{fy} = \frac{p_{oBy}}{(m_{A}+ m_{B)}} = \frac{m_{B}*v_{oBy} }{(m_{A}+ m_{B)}} =\frac{29.0kg*5.00m/s}{46.0kg} =  3.15 m/s (6)

  • With the values of vfx and vfy, we can find the magnitude of the final speed of the two-object system, applying the Pythagorean Theorem, as follows:

      v_{f} = \sqrt{v_{fx} ^{2} + v_{fy} ^{2}} = \sqrt{(2.96m/s)^{2} + (3.15m/s)^{2}} = 4.32 m/s (7)

  • The magnitude of the final total momentum is just the product of the combined mass of both objects times the magnitude of the final speed:

       p_{f} = (m_{A} + m_{B})* v_{f}  = 46 kg * 4.32 m/s = 198.8 kg*m/s (8)

  • Finally, the angle that the final momentum vector makes with the positive x-axis, is the same that the final velocity vector makes with it.
  • We can find this angle applying the definition of tangent of an angle, as follows:

       tg \theta = \frac{v_{fy}}{v_{fx}} = \frac{3.15 m/s}{2.96m/s} = 1.06 (9)

       ⇒ θ = tg⁻¹ (1.06) = 46.8º N of E

8 0
3 years ago
A cardboard box sits on top of a concrete sidewalk where the coefficient of friction between the surfaces is 0.4. The mass of th
Sloan [31]

Answer:

Fg = 98.1 [N]; N = 98.1 [N]; Ff = 39.24 [N]; a = 2.076[m/^2]

Explanation:

To solve this problem, we must make a free body diagram and interpret each of the forces acting on the box. In the attached diagram we can find the free body diagram.

The gravitational force is equal to:

Fg = (10 * 9.81) = 98.1 [N]

Now by summing forces on the Y axis equal to zero, we can find the normal force exerted by the surface.

N - Fg = 0

N = Fg

N = 98.1 [N]

The friction force is defined as the product of normal force by the coefficient of friction.

Ff = N * μ

Ff = 98.1 * 0.4

Ff = 39.24 [N]

By the sum forces on the x-axis equal to the product of mass by acceleration (newton's second law), we can find the value of acceleration.

60 - Ff = m * a

60 - 39.24 = 10 * a

a = 2.076[m/^2]

6 0
3 years ago
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