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Rama09 [41]
3 years ago
13

A jet takes 7 hours to travel a distance of 5586mi against the wind. the return trip takes 6 hours with the wind. what is the ra

te of the jet in still air and what is the rate of the wind?
Physics
1 answer:
MakcuM [25]3 years ago
8 0
Lets say speed of jet is u
& speed of wind is v
5586=(u-v)7--------1
5586=(u+v)6--------2
by 1,2 eqations,
we can get u&v
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i throw a tennis ball straight down from the roof of a building 400 meters tall.id it leaves my hand at 10 m/s, how fast will it
Sidana [21]

consider the motion of the tennis ball in downward direction

Y = vertical displacement = 400 m

a = acceleration = acceleration due to gravity = 9.8 m/s²

v₀ = initial velocity of the ball at the top of building = 10 m/s

v = final velocity of the ball when it hits the ground = ?

using the kinematics equation

v² = v²₀ + 2 a Y

inserting the values

v² = 10² + 2 (9.8) (400)

v = 89.11 m/s

6 0
3 years ago
What is the kinetic energy of a baseball moving at a speed of 40 m/s if the baseball has mass of 0.15 kg?
bonufazy [111]

0.5mv^2 > 0.5*0.15*40*40=120J

4 0
3 years ago
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How do I find the cosine of theta.
NNADVOKAT [17]

The value of cos θ in the given figure is 0.98.

<h3>What is cosine of an angle?</h3>

The cosine of an angle is defined as the sine of the complementary angle.

The complementary angle equals the given angle subtracted from a right angle, 90.

cos θ = sin(90 - θ)

For example, if the angle is 30°, then its complement is 60°

cos 30 = sin(90 - 30)

cos 30 = sin 60

0.866 = 0.866

<h3>Cosine of an angle with respect to sides of a right triangle</h3>

cos θ = adjacent side / hypotenuse side

adjacent side of the given right triangle is calculated as follows;

adj² = 10² - 2²

adj² = 100 - 4

adj² = 96

adj = √96

adj = 9.8

cos θ = 9.8/10

cos θ = 0.98

Thus, the value of cos θ in the given figure is 0.98.

Learn more about cosine of angles here: brainly.com/question/23720007

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5 0
1 year ago
Please help me with this for science
finlep [7]

Answer:Force is to the right

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2 years ago
What is the magnitude of the momentum of a 3400 kg airplane traveling at a speed of 450 miles per hour?
Vitek1552 [10]

Answer:

The magnitude of momentum of the airplane is 6.83\times 10^5\ kg-m/s.

Explanation:

Given that,

Mass of the airplane, m = 3400 kg

Speed of the airplane, v = 450 miles per hour

Since, 1 mile per hour = 0.44704 m/s

v = 201.16 m/s

We need to find the magnitude of momentum of the airplane. It is given by the product of mas and velocity such that,

p=mv

p=3400\ kg\times 201.16 \ m/s

p=683944\ kg-m/s

or

p=6.83\times 10^5\ kg-m/s

So, the magnitude of momentum of the airplane is 6.83\times 10^5\ kg-m/s. Hence, this is the required solution.

7 0
3 years ago
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