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3 years ago

Average speed is useful but it might not describe your actual speed

1 answer:

6 0

Average speed is useful in describing the motion of a body, but it does not tell everything about the motion. For instance, a car traveling at an average speed of 100 kilometers per hour during a certain hour could conceivably have traveled at 150 km/h for 40 minutes, Energy, Ch. 3, extension 1 Calculating average speed and velocity 3 then halted for the last 20 minutes as a patrolman gave the driver a ticket (picture above). Its average speed really does not describe that hour’s journey very well. The way to deal with such a situation is to take the average over a shorter interval, and do this more times. In the example given above, if the speed were sampled every 15 minutes instead of over the entire hour, we would list average speeds of 150 km/h at the start (0 to 15 minutes), 150 km/h (15 to 30 minutes), 100 km/h (30 to 45 minutes), and 0 km/h (45 to 60 minutes).

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The energy emitted from the sun is a product of ________.

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Q. The energy emitted from the sun is a product of ________.

A. Fusion

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3 years ago

If you were told an atom was an ion, you would know the atom must have a ______?

andreyandreev [35.5K]

Answer:

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Explanation:

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3 years ago

Which type of bond and how many bonds would occur as carbon dioxide (CO2) is formed from carbon in Group IVA and oxygen in Group

Rus_ich [418]

Answer:

The answer is D. Four covalent bounds

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3 years ago

Radio wave radiation falls in the wavelength region of 10.0 to 1000 meters. What is the energy of radio wave radiation that has

dmitriy555 [2]

Answer:

Explanation:

Given

Wavelength of radiation $\lambda =784\ m$

We know Energy of wave with wavelength $\lambda$ is given by

$E=\frac{hc}{\lambda }$

where h=Planck's constant

c=velocity of light

$\lambda$ =wavelength of wave

$E=\frac{6.626\times 10^{-34}\times 3\times 10^8}{784}$

$E=2.53\times 10^{-28}\ J$

Hence the energy of the wave with wavelength 784 m is $2.53\times 10^{-28}\ J$

7 0

3 years ago

An air-track glider with a mass of 239 g is moving at 0.81 m/s on a 2.4 m long air track. It collides elastically with a 513 g g

HACTEHA [7]

Answer:

Glider it stops just when it reaches the end of the runway

Explanation:

This is a shock between two bodies, so we must use the equations of conservation of the amount of movement, in the instant before the crash and the subsequent instant, with this we calculate the second glider speed, as the shock that elastic is also keep it kinetic energy

Po = pf

Ko = Kf

Before crash

Po = m1 Vo1 + 0

Ko = ½ m1 Vo1²

After the crash

Pf = m1 Vif + Vvf

Kf = ½ m1 V1f² + ½ m2 V2f²

m1 V1o = m1 V1f + m2 V2f (1)

m1 V1o² = m1 V1f² + m2 V2f² (2)

We see that we have two equations with two unknowns, so the system is solvable, we substitute in 1 and 2

m1 (V1o -V1f) = m2 V2f (3)

m1 (V1o² - V1f²) = m2 V2f²

Let's use the relationship (a + b) (a-b) = a² -b²

m1 (V1o + V1f) (V1o -V1f) = m2 V2f²

We divide with 3 and simplify

(V1o + V1f) = V2f (4)

Substitute in 3, group and clear

m1 (V1o - V1f) = m2 (V1o + V1f)

m1 V1o - m2 V1o = m2 V1f + m1 V1f

V1f (m1 -m2) = V1o (m1 + m2)

V1f = V1o (m1-m2 / m1+m2)

We substitute in (4) and group

V2f = V1o + (m1-m2 / m1 + m2) V1o

V2f = V1o [1+ + (m1-m2 / m1 + m2)]

V2f = V1o (2m1 / (m1+m2)

We calculate with the given values

V1f = 0.81 (239-513 / 239 + 513)

V1f = 0.81 (-274/752)

V1f = - 0.295 m/s

The negative sign indicates that the planned one moves in the opposite direction to the initial one

V2f = 0.81 [2 239 / (239 + 513)]

V2f = 0.81 [0.636]

V2f = 0.515 m / s

Now we analyze in the second glider movement only, we calculate the energy and since there is no friction,

Eo = Ef

Where Eo is the mechanical energy at the lowest point and Ef is the mechanical energy at the highest point

Eo = K = ½ m2 vf2²

Ef = U = m2 g Y

½ m2 v2f² = m2 g Y

Y = V2f² / 2g

Y = 0.515²/2 9.8

Y = 0.0147 m

At this height the planned stops, let's use trigonometry to find the height at the end of the track of the track

tan θ = Y / x

Y = x tan θ

The crash occurs in the middle of the track whereby x = 1.2 m

Y = 1.2 tan 0.7

Y = 0.147 m

As the two quantities are equal in glider it stops just when it reaches the end of the runway

7 0

3 years ago