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Alex
3 years ago
13

Pls help! What is the average speed of an airplane that travels from New York to Los Angeles, a total distance of 4800 km, in 6.

0 hours?
A.
800 km/hr
B.
80 km/hr
C.
0.00125 km/hr
D.
28,800 km/hr
Physics
1 answer:
Goshia [24]3 years ago
3 0

Answer:

800 meters per hour

Explanation:

800 meters per hour is the average speed of an airplane that travels from New York to Los Angeles, a total distance of 4800 km, in 6.0 hours.

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An electronic line judge camera captures the impact of a 57.0-g tennis ball traveling at 32.2 m/s with the side line of a tennis
mote1985 [20]

Answer:

average acceleration is 1.365 × 10^{4} m/s²

Explanation:

given data

initila speed  u = -32.2 m/s

final speed v = 21.6 m/s

time taken t = 0.00394 s

solution

we get here average acceleration that will be express as

v = u + at    ..........................1

put here value and we get

21.6 = -32.2 + a × 0.00394

solve it we get

a = 1.365 × 10^{4} m/s²

so average acceleration is 1.365 × 10^{4} m/s²

3 0
3 years ago
b) The distance of the red supergiant Betelgeuse is approximately 427 light years. If it were to explode as a supernova, it woul
Nat2105 [25]

Answer:

b) Betelgeuse would be \approx 1.43 \cdot 10^{6} times brighter than Sirius

c) Since Betelgeuse brightness from Earth compared to the Sun is \approx 1.37 \cdot 10^{-5} } the statement saying that it would be like a second Sun is incorrect

Explanation:

The start brightness is related to it luminosity thought the following equation:

B = \displaystyle{\frac{L}{4\pi d^2}} (1)

where B is the brightness, L is the star luminosity and d, the distance from the star to the point where the brightness is calculated (measured). Thus:

b) B_{Betelgeuse} = \displaystyle{\frac{10^{10}L_{Sun}}{4\pi (427\ ly)^2}} and B_{Sirius} = \displaystyle{\frac{26L_{Sun}}{4\pi (26\ ly)^2}} where L_{Sun} is the Sun luminosity (3.9 x 10^{26} W) but we don't need to know this value for solving the problem. ly is light years.

Finding the ratio between the two brightness we get:

\displaystyle{\frac{B_{Betelgeuse}}{B_{Sirius}}=\frac{10^{10}L_{Sun}}{4\pi (427\ ly)^2} \times \frac{4\pi (26\ ly)^2}{26L_{Sun}} \approx 1.43 \cdot 10^{6} }

c) we can do the same as in b) but we need to know the distance from the Sun to the Earth, which is 1.581 \cdot 10^{-5}\ ly. Then

\displaystyle{\frac{B_{Betelgeuse}}{B_{Sun}}=\frac{10^{10}L_{Sun}}{4\pi (427\ ly)^2} \times \frac{4\pi (1.581\cdot 10^{-5}\ ly)^2}{1\ L_{Sun}} \approx 1.37 \cdot 10^{-5} }

Notice that since the star luminosities are given with respect to the Sun luminosity we don't need to use any value a simple states the Sun luminosity as the unit, i.e 1. From this result, it is clear that when Betelgeuse explodes it won't be like having a second Sun, it brightness will be 5 orders of magnitude smaller that our Sun brightness.

4 0
3 years ago
A 7.0kg object rests on a horizontal frictionless surface. What is the magnitude of the horizontal
sukhopar [10]

Answer:

16.1 N

Explanation:

From the question,

F = ma.............................. Equation 1

Where F = horizontal force, m = mass of the object, a = acceleration .

Given: m = 7.0 kg, a = 2.3 m/s²

Substitute this values into equation 1

F = (7.0×2.3)

F = 16.1 N.

Hence the magnitude of the horizontal force is 16.1 N

6 0
3 years ago
B. Find the centripetal force needed to keep an 60.0 kg rider traveling in a circle in the ride.
schepotkina [342]

Answer:

loloololo

lol

Explanation:

lolloolololo

8 0
3 years ago
A 4.33 kg cat has 41.7 J of KE How fast is the cat moving?
balandron [24]

Answer:

The answer to your question is:

Explanation:

Data

mass = 4.33 kg

E = 41.7 J

v = ?

Formula

Ke = (1/2)mv²

Clear v from the equation

v = √2ke/m

Substitution

v = √2(41.7)/4.33

v = 19.26 m/s          Result

7 0
3 years ago
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