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-BARSIC- [3]
3 years ago
15

A bomb is dropped from a bomber traveling at the speed of 120 km / h, destroying a military objective located at a distance of 2

000 m. How high was the plane traveling?
Physics
1 answer:
schepotkina [342]3 years ago
7 0

Answer:

18 km

Explanation:

Convert km/h to m/s:

120 km/h × (1000 m/km) × (1 h / 3600 s) = 33.3 m/s

The time it takes the bomb to travel the 2000 meters is:

2000 m / (33.3 m/s) = 60 s

So it takes 60 seconds for the bomb to fall.  The distance it fell is therefore:

Δy = v₀ᵧ t + ½ aᵧ t²

Δy = (0 m/s) (60 s) + ½ (10 m/s²) (60 s)²

Δy = 18,000 m

Δy = 18 km

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Lúc 7g bạn an đi từ nhà đến trường với tóc độ trung bình là 20km/h . Bạn đến trường lúc 7g20. Tính khoảng cách từ nhà tới trường
Len [333]

Answer:

Distance = 6.667 kilometres

Explanation:

Given the following data;

Speed = 20 km/h

Departure time = 7:00

Arrival time = 7:20

Time taken = 20 minutes

To calculate the distance travelled from home to school;

First of all, we would have to convert the value of time in minutes to hours.

Conversion:

60 minutes = 1 hour

20 minutes = X hours

Cross-multiplying, we have;

X = 20/60 = 1/3 hours

Mathematically, the distance travelled by an object is calculated by using the formula;

Distance = speed * time

Distance = 20 * 1/3

Distance = 20/3 =

Distance = 6.667 kilometres

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2 years ago
What would be some benefits of living on a planet with less surface gravity
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You could jump high!
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3 years ago
Read 2 more answers
What is the density of a piece of quartz with a mass of 30g and a volume of 6cm^3
FromTheMoon [43]

Answer:

\boxed {\boxed {\sf d= 5 \ g/cm^3}}

Explanation:

Density can be found by dividing the mass by the volume.

d=\frac{m}{v}

The mass of the quartz is 30 grams and the volume is 6 cubic centimeters.

m=30 \ g \\v= 6 \ cm^3

Substitute the values into the formula.

d=\frac{30 \ g}{6 \ cm^3}

Divide.

d= 5 \ g/cm^3

The density of this piece of quartz is 5 grams per cubic centimeter.

7 0
3 years ago
1, 2, & 3.........................
oksian1 [2.3K]

Answer:

1 is correct

2 is -150  

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3 years ago
The box leaves position x=0x=0 with speed v0v0. The box is slowed by a constant frictional force until it comes to rest at posit
const2013 [10]

Answer:

fr = ½ m v₀²/x

Explanation:

This exercise the body must be on a ramp so that a component of the weight is counteracted by the friction force.

The best way to solve this exercise is to use the energy work theorem

            W = ΔK

Where work is defined as the product of force by distance

           W = fr x cos 180

The angle is because the friction force opposes the movement

          Δk =K_{f} –K₀

          ΔK = 0 - ½ m v₀²

We substitute

         - fr x = - ½ m v₀²      

           fr = ½ m v₀²/x

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