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2 years ago

A Force has a size and a(n)?

2 answers:

6 0

A force is a vector quantity. As learned in an earlier unit, a vector quantity is a quantity that has both magnitude and direction. To fully describe the force acting upon an object, you must describe both the magnitude (size or numerical value) and the direction.

Gekata [30.6K]2 years ago

4 0

Answer:

Force have both magnitude and direction since its considered to be a scalar quantity

Explanation:

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What is the difference in KE between a 52.5 kg person running 3.50 m/s and a 0.0200 kg bullet flying 450 m/s?

rusak2 [61]

Answer:

Ek = 1705.28 [J]

Explanation:

In order to solve this problem, we must remember that kinetic energy can be calculated by means of the following equation.

$E_{k}=\frac{1}{2} *m*v^{2}$

where:

m = mass [kg]

v = velocity [m/s]

Ek = kinetic energy [J] (Units of Joules)

For the person running

$E_{k} =\frac{1}{2}*52.2*(3.5)^{2} \\ E_{k} =319.72[J]$

For the bullet

$E_{k} =\frac{1}{2} *m*v^{2}$

$E_{k} =\frac{1}{2} *0.02*(450)^{2} \\E_{k}=2025 [J]$

The difference in Kinetic energy is equal to:

Ek = 2025 - 319.72

Ek = 1705.28 [J]

8 0

2 years ago

An airplane pilot wishes to fly due west. A wind of 80.0 km/h (about 50 mi/h) is blowing toward the south. (a) If the airspeed o

Vesnalui [34]

Answer:

a) 75.5 degree relative to the North in north-west direction

b) 309.84 km/h

Explanation:

a)If the pilot wants to fly due west while there's wind of 80km/h due south. The north-component of the airplane velocity relative to the air must be equal to the wind speed to the south, 80km/h in order to counter balance it

So the pilot should head to the West-North direction at an angle of

$cos(\alpha) = 80/320 = 0.25$

$\alpha = cos^{-1}(0.25) = 1.32 rad = 180\frac{1.32}{\pi} = 75.5^0$ relative to the North-bound.

b) As the North component of the airplane velocity cancel out the wind south-bound speed. The speed of the plane over the ground would be the West component of the airplane velocity, which is

$320sin(\alpha) = 320sin(75.5^0) = 309.84 km/h$

7 0

3 years ago

Read 2 more answers

How people get along with one another is called

Orlov [11]

I am not sure what kind of answer you are looking for but the start of friendship.

5 0

3 years ago

According to the text, there is no energy shortage now, nor will there ever be. what reason (s) is given to support this stateme

Bas_tet [7]

The reason why there is no energy shortage nor will there ever be is because energy is being preserved and conserved and only changes form. It never gets lost or increased.

8 0

2 years ago

A ball is thrown from a rooftop with an initial downward velocity of magnitude vo = 2.9 m/s. The rooftop is a distance above the

Step2247 [10]

Answer:

a) The velocity of the ball when it hits the ground is -20.5 m/s.

b) To acquire a final velocity of 27.3 m/s, the ball must be thrown from a height of 38 m.

Explanation:

I´ve found the complete question on the web:

A ball is thrown from a rooftop with an initial downward velocity of magnitude v0=2.9 m/s. The rooftop is a distance above the ground, h= 21 m. In this problem use a coordinate system in which upwards is positive.

(a) Find the vertical component of the velocity with which the ball hits the ground.

(b) If we wanted the ball's final speed to be exactly 27, 3 m/s from what height, h (in meters), would we need to throw it with the same initial velocity?

The equation of the height and velocity of the ball at any time "t" are the following:

h = h0 + v0 · t + 1/2 · g · t²

v = v0 + g · t

Where:

h = height of the ball at time t.

h0 = initial height.

v0 = initial velocity.

t = time.

g = acceleration due to gravity (-9.8 m/s² considering the upward direction as positive).

v = velocity of the ball at a time "t".

First, let´s find the time it takes the ball to reach the ground (the time at which h = 0)

h = h0 + v0 · t + 1/2 · g · t²

0 = 21 m - 2.9 m/s · t - 1/2 · 9.8 m/s² · t²

Solving the quadratic equation using the quadratic formula:

t = 1.8 s ( the other solution of the quadratic equation is rejected because it is negative).

Now, using the equation of velocity, let´s find the velocity of the ball at

t = 1.8 s:

v = v0 + g · t

v = -2.9 m/s - 9.8 m/s² · 1.8 s

v = -20.5 m/s

The velocity of the ball when it hits the ground is -20.5 m/s.

b) Now we have the final velocity and have to find the initial height. Using the equation of velocity we can obtain the time it takes the ball to acquire that velocity:

v = v0 + g · t

-27.3 m/s = -2.9 m/s - 9.8 m/s² · t

(-27.3 m/s + 2.9 m/s) / (-9.8 m/s²) = t

t = 2.5 s

The ball has to reach the ground in 2.5 s to acquire a velocity of 27.3 m/s.

Using the equation of height, we can obtain the initial height:

h = h0 + v0 · t + 1/2 · g · t²

0 = h0 -2.9 m/s · 2.5 s - 1/2 · 9.8 m/s² · (2.5 s)²

-h0 = -2.9 m/s · 2.5 s - 1/2 · 9.8 m/s² · (2.5 s)²

h0 = 38 m

To acquire a final velocity of 27.3 m/s, the ball must be thrown from a height of 38 m.

6 0

3 years ago