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galben [10]
3 years ago
8

What type of speed looks at a particular point in time?

Physics
1 answer:
bearhunter [10]3 years ago
4 0
The answer is D because instantaneous means at a particular point in time
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The curved section of a speedway is a circular arc having a radius of 190 m. this curve is properly banked for racecars moving a
Anestetic [448]

The banking angle of the curved part of the speedway is determined as 32⁰.

<h3>Banking angle of the curved road</h3>

The banking angle of the curved part of the speedway is calculated as follows;

V(max) = √(rg tanθ)

where;

  • r is radius of the path
  • g is acceleration due to gravity

V² = rg tanθ

tanθ = V²/rg

tanθ = (34²)/(190 x 9.8)

tanθ = 0.62

θ = arc tan(0.62)

θ = 31.8

θ ≈ 32⁰

Learn more about banking angle here: brainly.com/question/8169892

#SPJ1

8 0
2 years ago
Help with the two questions above? Correct answers?
LenKa [72]

(6) first choice: the frequency appears higher and wavelength is shorter.

The car approaches a stationary observer and so the sound will appear to have shorter wavelength. This creates an effect of its siren to sound with higher frequency than it would do if both were stationary.

(7) The Doppler formula for frequency in the case of a stationary observer and source approaching it is as follows:

f_O = \frac{v}{v-v_s}\cdot f= \frac{343\frac{m}{s}}{(343-25)\frac{m}{s}}\cdot 400Hz \approx 431Hz

The wavelength is then

\lambda = \frac{343\frac{m}{s}}{431Hz}\approx 0.80 m

The third choice "0.80m; 431Hz" is correct

7 0
2 years ago
two astronauts are taking a spacewalk outside the International Space Station the first astronaut has a mass of 64 kg the second
Fittoniya [83]

Answer:

Approximately 0.88\; {\rm m \cdot s^{-1}} to the right (assuming that both astronauts were originally stationary.)

Explanation:

If an object of mass m is moving at a velocity of v, the momentum p of that object would be p = m\, v.

Since momentum of this system (of the astronauts) conserved:

\begin{aligned} &(\text{Total Final Momentum}) \\ &= (\text{Total Initial Momentum})\end{aligned}.

Assuming that both astronauts were originally stationary. The total initial momentum of the two astronauts would be 0 since the velocity of both astronauts was 0\!.

Therefore:

\begin{aligned} &(\text{Total Final Momentum}) \\ &= (\text{Total Initial Momentum})\\ &= 0\end{aligned}.

The final momentum of the first astronaut (m = 64\; {\rm kg}, v = 0.8\; {\rm m\cdot s^{-1}} to the left) would be p_{1} = m\, v = 64\; {\rm kg} \times 0.8\; {\rm m\cdot s^{-1}} = 51.2\; {\rm kg \cdot m \cdot s^{-1}} to the left.

Let p_{2} denote the momentum of the astronaut in question. The total final momentum of the two astronauts, combined, would be (p_{1} + p_{2}).

\begin{aligned} & p_{1} + p_{2} \\ &= (\text{Total Final Momentum}) \\ &= (\text{Total Initial Momentum})\\ &= 0\end{aligned}.

Hence, p_{2} = (-p_{1}). In other words, the final momentum of the astronaut in question is the opposite of that of the first astronaut. Since momentum is a vector quantity, the momentum of the two astronauts magnitude (51.2\; {\rm kg \cdot m \cdot s^{-1}}) but opposite in direction (to the right versus to the left.)

Rearrange the equation p = m\, v to obtain an expression for velocity in terms of momentum and mass: v = (p / m).

\begin{aligned}v &= \frac{p}{m} \\ &= \frac{51.2\; {\rm kg \cdot m \cdot s^{-1}}}{64\; {\rm kg}} && \genfrac{}{}{0}{}{(\text{to the right})}{} \\ &\approx 0.88\; {\rm m\cdot s^{-1}} && (\text{to the right})\end{aligned}.

Hence, the velocity of the astronaut in question (m = 58.2\; {\rm kg}) would be 0.88\; {\rm m \cdot s^{-1}} to the right.

5 0
2 years ago
A 40.0 kg box is placed on a 2.00 m tall shelf. If the box falls off the shelf, what is its kinetic energy as it strikes the gro
RUDIKE [14]
M = 40 Kg , g=9.8 m/s² , h = 2 m

PE = m g h

PE = (40) (9.8) (2)

PE = 784 J

KE = PE

½m v² = m g h

½ v² = g h

½ v² = (9.8) (2)

½ v² = 19.6

v² = 19.6×2

v² = 39.2

V = √39.2

V = 6.26 m/s

KE = ½mv²

KE = ½(40) (6.26)²

KE =783.8 J
6 0
2 years ago
50 POINTS
Vikki [24]

Answer:

The correct answer is: B area = area

Explanation: You can solve this problem through the use of laws of Kepler's planetary motion. There are Kepler the laws ( 3 of them) In this exercise, we will use the second law. According to this law, a line segment joining a planet and the sun sweeps out areas is equal to that during the equal intervals of time

3 0
3 years ago
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