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3 years ago

A passenger jet flies from one airport to another 1293 miles away in 2.1hours . Find average speed

Physics

1 answer:

JulsSmile [24]3 years ago

6 0

The average speed is 615.7 mph (275.2 m/s)

Explanation:

The average speed of an object gives a measure of how fast an object is moving over a certain time interval. It is a scalar quantity, and it is calculated as follows

$v=\frac{d}{t}$

where

v is the average speed

d is the distance covered

t is the time interval

For the jet in this problem, we have:

d = 1293 mi is the distance covered

t = 2.1 h is the time interval

Therefore, the average speed is:

$v=\frac{1293}{2.1}=615.7 mph$

We can also convert into SI units (m/s), keeping in mind that:

$1 mi = 1609 m\\1 h = 3600 s$

And so

$v=615.7 \frac{mi}{h}\cdot \frac{1609 m/mi}{3600 s/h}=275.2 m/s$

Learn more about average speed:

brainly.com/question/8893949

brainly.com/question/5063905

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Andrews [41]

Answer:

6.26 m/s

Explanation:

Pretty slow.... the PE (Potential Energy) at 2m will be converted to KE (Kinetic Energy) at the bottom of the track (neglecting friction)

PE = KE

mgh = 1/2 mv^2 divide both sides of the equation by 'm'

gh = 1/2 v^2 multiply both sides by 2

2 gh = v^2 take sqrt of both sides

v = sqrt ( 2gh) = sqrt ( 2*9.81*2) = 6.26 m/s

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2 years ago

The sound from a bolt of lightning travelled 4.08 km in 12.0 s. What was the speed

LenaWriter [7]

Answer:

83.67 m/s

Explanation:

Set up a calculation to convert units of measure to what you need.

You have km/s and you need m/s.

4.08km 1000 m 83.67m

----------- X ---------- = --------------- the km will cancel out and you are left

12.0 s 1 km s with m/s

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3 years ago

a 42.3 kg girl and a 7.93 kg sled are on the surface of a frozen lake, 15.0m apart and linked by a rope, but not moving yet. the

ycow [4]

Answer:

they meet from the girl's original position at: 2.37 (meters)

Explanation:

We need to use the Newton's law, exactly the second law that relate force, mass and acceleration as: $F=m*a$ with this we can get both accelerations; solving for acceleration $a=\frac{F}{m}$ . Now $a_{girl}=\frac{5.76}{42.3}=0.14 (m/s^{2})$ and $a_{sled}=\frac{5.76}{7.93}=0.73(m/s^{2})$ . Then knowing that they both travel at the same time and assuming that the distance among the girl and the sled is: 15.0-x, so, $x=\frac{1}{2}*a_{girl}*t^{2}$ and $15.0-x=\frac{1}{2}*a_{sled}*t^{2}$ , solving for the time we get: $t=\sqrt{\frac{2x}{a_{girl} } }$ and $t=\sqrt{\frac{2*(15.0-x)}{a_{sled} } }$ with this equations we solving for the x that is the distance between the girl and the sled after the apply the force and we get: $\sqrt{\frac{2x}{a_{girl}}} = \sqrt{\frac{2*(15.0-x)}{a_{sled} }$ . Finally we get: $\frac{x}{a_{girl} }=\frac{(15.0-x)}{a_{sled} }$ and replacing the values we have got: $\frac{x}{0.14} =\frac{(15.0-x)}{0.73}$ so $5.33*x=15-x$ so x=2.37 (meters).

5 0

3 years ago

In boxing, the use of 16-ounce gloves rather than 12-ounce gloves reduces the chance of injury because the force is distributed

mr Goodwill [35]

Answer:

true

Explanation:

Here we have assumed that increasing the mass of a glove will increase the surface area.

Injury is caused by the application of pressure at a point on the body. The application of pressure takes place via the area of the gloves. Pressure is given by

$P=\dfrac{F}{A}$

where

F = Force

A = Area to which the force is applied

So, a bigger glove will increase the surface area and reduce the pressure resulting in a lower chance of injury.

Hence, the statement is true.

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2 years ago

A 3,000 kg truck moving at +10 m/s hits a 1,000 kg parked car which moves off at +15 m/s What is the velocity of

Rina8888 [55]

Answer:

v₃ = 5 [m/s]

Explanation:

To solve this problem we must use the definition of linear momentum, which tells us that momentum is equal to the product of mass by Velocity.

P = m*v

where:

P = linear momentum [kg*m/s]

m = mass [kg]

v = velocity [m/s]

We must also clarify that the momentum is preserved i.e. it is equal before the collision and after the collision

Pbeforecollision = Paftercollision

(m₁*v₁) + (m₂*v₂) = (m₁*v₃) + (m₂*v₄)

where:

m₁ = mass of the truck = 3000 [kg]

v₁ = velocity of the truck = 10 [m/s]

m₂ = mass of the car = 1000 [kg]

v₂ = velocity of the car before the collision = 0 (the car is parked)

v₃ = velocity of the truck after the collision [m/s]

v₄ = velocity of the car after the collision = 15 [m/s]

(3000*10) + (1000*0) = (3000*v₃) + (1000*15)

30000 = 3000*v₃ + 15000

3000*v₃ = 30000 - 15000

3000*v₃ = 15000

v₃ = 5 [m/s]

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2 years ago