How is the gravitational force acting on a falling object calculated

An 80- quarterback jumps straight up in the air right before throwing a 0.43- football horizontally at 15 . How fast will he be

lord [1]

Answer:

a)

the quarterback will be moving back at speed of 0.080625 m/s

b)

the distance moved horizontally by the quarterback is 0.0241875 m or 2.41875 cm

Explanation:

Given the data in the question;

a)

How fast will he be moving backward just after releasing the ball?

using conservation of momentum;

m₁v₁ = m₂v₂

v₂ = m₁v₁ / m₂

where m₁ is initial mass ( 0.43 kg )

m₂ is the final mass ( 80 kg )

v₁ is the initial velocity ( 15 m/s )

v₂ is the final velocity

so we substitute

v₂ = ( 0.43 × 15 ) / 80

v₂ = 6.45 / 80

v₂ = 0.080625 m/s

Therefore, the quarterback will be moving back at speed of 0.080625 m/s

b) Suppose that the quarterback takes 0.30 to return to the ground after throwing the ball. How far d will he move horizontally, assuming his speed is constant?

we make use of the relation between time, distance and speed;

s = d/t

d = st

where s is the speed ( 0.080625 m/s )

t is time ( 0.30 s )

so we substitute

d = 0.080625 × 0.30

d = 0.0241875 m or 2.41875 cm

Therefore, the distance moved horizontally by the quarterback is 0.0241875 m or 2.41875 cm

5 0

3 years ago

Hey people! I sorta need help. :))

il63 [147K]

Answer:

1. Attract

2. Repel

3. Attract

4 Repel

always remember when dealing with magnets, Opposites attract

4 0

3 years ago

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The components of vector A are:

Korvikt [17]

here as it is given that x component of the vector is positive while y component of the vector is negative so we can say the vector must inclined in Fourth quadrant.

So angle must be more than 270 degree and less than 360 degree

Now in order to find the value we can say that

$tan\theta = \frac{opposite\: side}{adjacent\: side}$

$tan\theta = \frac{8.6}{6.1}$

$\theta = tan^{-1}1.41$

$\theta = 54.65^0$

so it is inclined at above angle with X axis in fourth quadrant

Now if angle is to be measured counterclockwise then its magnitude will be

$\theta = 360 - 54.65 = 305.3^0$

so the correct answer will be 305 degree

3 0

3 years ago

Janelle wants to buy some strings of decorative lights for her home. She is trying to decide between two strings of lights that

inysia [295]

If she has a choice and the wiring details are stated on the packaging,
then Janelle should look for lights that are wired in parallel within the
string, and she should avoid lights that are wired in series within the string.

If a single light in a parallel string fails, then only that one goes out.
The rest of the lights in the string continue to shimmer and glimmer.

If a single light in a series string fails, then ALL of the lights in that string
go out, and it's a substantial engineering challenge to determine which light
actually failed.

6 0

3 years ago

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A stone is dropped into a river from a bridge at a height h above the water. Another stone is thrown vertically down at a time t

Mumz [18]

Answer:

$v_{y_0} = \frac{\frac{g}{2}t(t - 2\sqrt{\frac{2h}{g}})}{\sqrt{\frac{2h}{g}} - t}$

Explanation:

We will apply the equations of kinematics to both stones separately.

First stone:

Let us denote the time spent after the second stone is thrown as 'T'.

$y - y_0 = v_{y_0}(t+T) + \frac{1}{2}a(t+T)^2\\0 - h = 0 + \frac{1}{2}(-g)(t+T)^2\\(t+T)^2 = \frac{2h}{g}\\T = \sqrt{\frac{2h}{g}}-t$

Second stone:

$y - y_0 = v_{y_0}T + \frac{1}{2}aT^2\\0 - h = v_{y_0}T -\frac{1}{2}gT^2\\-h = v_{y_0}(\sqrt{\frac{2h}{g}} - t) - \frac{g}{2}(\sqrt{\frac{2h}{g}} - t)^2\\-h = v_{y_0}(\sqrt{\frac{2h}{g}} - t) - \frac{g}{2}(\frac{2h}{g} + t^2 - 2t\sqrt{\frac{2h}{g}})\\-h = v_{y_0}\sqrt{\frac{2h}{g}} - v_{y_0}t - h -\frac{g}{2}t^2 + gt\sqrt{\frac{2h}{g}}\\v_{y_0}(\sqrt{\frac{2h}{g}} - t) = \frac{g}{2}t^2 - gt\sqrt{\frac{2h}{g}}\\v_{y_0} = \frac{\frac{g}{2}t(t - 2\sqrt{\frac{2h}{g}})}{\sqrt{\frac{2h}{g}} - t}$

6 0

2 years ago

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