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Evgesh-ka [11]
3 years ago
11

In tests on earth a lunar surface exploration vehicle (mass = 5.57 × 103 kg) achieves a forward acceleration of 0.207 m/s2. to a

chieve this same acceleration on the moon, the vehicle's engines must produce a drive force of 1.60 × 103 n. what is the magnitude of the frictional force that acts on the vehicle on the moon?
Physics
1 answer:
spayn [35]3 years ago
5 0
The answer is 9000         or b                          
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Answer: 1.1 x 10^-10 C

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Would You Rather...Carry a 50 lb box up 3 flights of stairs or a 25 lb box up 5 flights of stairs? Whichever choice you make, ju
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I would carry a 25 pound box up 5 flights, because it is lighter. You will wear out faster carrying a 50 pound box up 3 flights.

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3 years ago
Lithium has 3 protons. how many neutrons are in the isotope lithium -8
Oksi-84 [34.3K]

Answer:

5 neutrons

Explanation:

Both protons and neutrons have a mass of about 1 atomic mass unit (amu). The number in the notation for an isotope is its mass in amu. That means the number is essentially the sum of protons and neutrons in the isotope (electrons don't contribute significantly to the mass).

If lithium has 3 protons, lithium-8 has 8 - 3 = 5 neutrons.

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3 years ago
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A highway patrol car traveling a constant speed of 105 km/h is passed by a speeding car traveling 140 km/h. Exactly 1.00 s after
vodka [1.7K]

Answer:

The elapsed time from when the speeder passes the patrol car until it is caught is 9.24 s.

Explanation:

Hi there!

The position of the patrol car at a time "t" can be calculated using this equation:

x = x0 + v0 · t + 1/2 · a · t²

Where:

x = position of the patrol car at a time "t"

x0 = initial position.

v0 = initial velocity.

t = time.

a = acceleration.

For the speeding car, the equation is the same only that the acceleration is zero. Then, the equation gets reduced to this:

x = x0 + v · t

Where "v" is the constant velocity.

First, let´s convert the velocity units into m/s:

140 km/h · 1000 m / 1 km · 1 h / 3600 s = 38.9 m/s

105 km/h · 1000 m / 1 km · 1 h / 3600 s = 29.2 m/s

We have to find how much time it takes the patrol car to catch the speeder after the speeder passes the patrol car.

When the patrol car catches the speeder, the position of both cars is the same:

position of the patrol car = position of the speeder

x0 + v0 · t + 1/2 · a · t² = x0 + v · t

if we place the origin of the frame of reference at the point where the patrol car starts accelerating (1 s after the speeder passes the patrol car) then, the initial position of the patrol car will be zero, while the initial position of the speeder will be the traveled distance in 1 s:

x = v · t

x = 38.9 m/s · 1 s = 38.9 m

When the patrol car accelerates, the speeder is 38.9 m ahead of it. Then:

x0 + v0 · t + 1/2 · a · t² = x0 + v · t

0 + 29.2 m/s · t + 1/2 · 3.50 m/s² · t² = 38.9 m + 38.9 m/s · t

Let´s agrupate terms and equalize to zero:

-38.9 m - 38.9 m/s · t + 29.2 m/s · t + 1.75 m/s² · t² = 0

-38.9 m - 9.70 m/s · t + 1.75 m/s² · t² = 0

Solving the quadratic equation for t using the quadratic formula:

t = 8.24 s  (the other solution is discarded because it is negative)

The elapsed time from when the speeder passes the patrol car until it is caught is (8.24 s + 1.00) 9.24 s.

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What activity level should participants be in when exercising in their target heart rate zone
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La frecuencia cardíaca objetivo durante las actividades de intensidad moderada es aproximadamente del 50 al 70% de la frecuencia cardíaca máxima, mientras que durante la actividad física intensa es de, aproximadamente, entre el 70 y el 85% del valor máximo.

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