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Alekssandra [29.7K]
3 years ago
15

A 5,257 kg rocket blasts off to the moon with an acceleration of 76 m/s ^2 what is the net force on the rocket

Physics
1 answer:
frutty [35]3 years ago
7 0

Newton's subsequent law expresses that power is corresponding to what exactly is needed for an object of consistent mass to change its speed. This is equivalent to that item's mass increased by its speed increase.

We use Newtons, kilograms, and meters each second squared as our default units, albeit any proper units for mass (grams, ounces, and so forth) or speed (miles each hour out of every second, millimeters per second², and so on) could unquestionably be utilized also - the estimation is the equivalent notwithstanding.

Hence, the appropriate answer will be 399,532.

Net Force = 399532

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The period of a simple pendulum, defined as the time necessary for one complete oscillation, is measured in time units and is gi
Oksana_A [137]

The task is to show that the right side of the equation has units of [Time], just like the left side has.

The right side of the equation is . . . 2 π √(L/G) .

We can completely ignore the  2π since it has no units at all, so it has no effect on the units of the right side of the equation.  Now the task is simply to find the units of  √(L/G) .

L . . . meters

G . . . meters/sec²

(L/G) = (meters) / (meters/sec²)

(L/G) = (meters) · (sec²/meters)

(L/G) = (meters · sec²) / (meters)

(L/G) = sec²

So  √(L/G) = seconds = [Time]

THAT's what we were hoping to prove, and we did it !

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Suppose a miracle car has a 100% efficient engine and burns fuel having an energy content of 40 MJ/L. If the air resistance and
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Distance traveled in 1 liter = 40 km

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The color red that we see depends upon the :
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A carnival game consists of a two masses on a curved frictionless track, as pictured below. The player pushes the larger object
Harman [31]

Answer:

v₁₀ = 1.90 m / s

Explanation:

In this exercise we are given the maximum height data, with energy we can know how fast the body came out

Final mechanical energy, maximum height

    Em_{f} = U = m g h

Initial mechanical energy, in the lower part of the track

    Em₀ = K = ½ m v²

    Em=   Em_{f}

    ½ m v² = m g h

    v = √ 2gh

Now we can use the moment to find the speed with which objects collide

The large object has a mass M = 5.41 kg a velocity starts v₁₀, the small object has a mass m = 1.68 kg an initial velocity of zero v₂₀ = 0 and  final velocity v

Initial before the crash

    p₀ = M v₁₀ + 0

Final after the crash

      p_{f} = M v1f + m v

   p₀ =   p_{f}

   M v₁₀ = M v_{1f}+ m v

As the shock is elastic the kinetic energy is conserved

     K₀ = K_{f}

    ½ M v₁₀² = ½ M v_{1f}² + ½ m v²

Let's write the system of equations

    M v₁₀ = M  v_{1f} + m v

    M v1₁₀² = M v_{1f}² + m v²

We cleared v1f in the first we replaced in the second

   v_{1f} = (M v₁₀ - mv) / M

    M v₁₀² = M (M v₁₀ - mv)² / M² + m v²

    M v₁₀² = 1 / M (M² v₁₀² - 2mM v v₁₀ + m² v²) +m v²

     v₁₀² (M - M) + 2 m v v₁₀ - v² (m2 + m) / M = 0

     2 m v₁₀ - v (m + 1) m/ M = 0

     v₁₀ = v (m +1) / (2M)

Let's substitute the value of v

     v1₁₀= √ (2gh) (m +1) / (2M)

Let's calculate

    v₁₀ = √ (2 9.8 3) (1+ 1.68) / (2  5.41)

    V₁₀ = 7.668 (2.68) / 10.82

   v₁₀ = 1.90 m / s

5 0
3 years ago
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