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Black_prince [1.1K]
3 years ago
15

One interesting fact bout golf

Physics
2 answers:
Art [367]3 years ago
5 0

Answer:

You hit the ball and it flies

Explanation:

You hit the ball and it flies

GaryK [48]3 years ago
4 0

Answer:

you hit a golf ball to the moon

Explanation:

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Three cars (car F, car G, and car H) are moving with the same speed and slam on their brakes. The most massive car is car F, and
Crazy boy [7]

To solve this problem it is necessary to apply the concepts related to Normal Force, frictional force, kinematic equations of motion and Newton's second law.

From the kinematic equations of motion we know that the relationship of acceleration, velocity and distance is given by

v_f^2=v_i^2+2ax

Where,

v_f = Final velocity

v_i = Initial Velocity

a = Acceleration

x = Displacement

Acceleration can be expressed in terms of the drag coefficient by means of

F_f = \mu_k (mg)  \rightarrowFrictional Force

F = ma \rightarrow Force by Newton's second Law

Where,

m = mass

a= acceleration

\mu_k = Kinetic frictional coefficient

g = Gravity

Equating both equation we have that

F_f = F

\mu_k mg=ma

a = \mu_k g

Therefore,

v_f^2=v_i^2+2ax

0=v_i^2+2(\mu_k g)x

Re-arrange to find x,

x = \frac{v_i^2}{2(-\mu_k g)}

The distance traveled by the car depends on the coefficient of kinetic friction, acceleration due to gravity and initial velocity, therefore the three cars will stop at the same distance.

3 0
3 years ago
at a certain instant an object is moving to the right with speed 1.0 m/s and has a constant acceleration to the left of 1.0 m/s^
chubhunter [2.5K]
For this problem, you would use the equation v=u+at. In this case, u=1 v=0 (when the object is at rest) a=-1
v=u+at
0=1+(-1)t
t=1 second
3 0
4 years ago
This is a net gain or loss of electrons. *
krok68 [10]
A net gain of electrons.
8 0
3 years ago
Read 2 more answers
A car traveling 28 mi/h accelerates uniformly for 8.9 s, covering 599 ft in this time. what was its acceleration? round your ans
almond37 [142]

Answer:

5.90 ft/s^2

Explanation:

There are mixed units in this question....convert everything to miles or feet

    and hr  to s

28 mi / hr = 41.066 ft/s

Displacement = vo t + 1/2 at^2

         599       =  41.066 (8.9)  + 1/2 a (8.9^2)

                      solve for a = ~ 5.90 ft/s^2

5 0
2 years ago
We investigated a jet landing on an aircraft carrier. In a later maneuver, the jet comes in for a landing on solid ground with a
AlexFokin [52]

Answer:

Time  is 14.8 s and cannot landing

Explanation:

This is a kinematic exercise with constant acceleration, we assume that the acceleration of the jet to take off and landing  are the same

Calculate the time to stop, where it has zero speed

       Vf² = Vo² + a t

       t = - Vo² / a

       t = - 110²/(-7.42)

       t = 14.8 s

This is the time it takes to stop the jet

Let's analyze the case of the landing at the small airport, let's look for the distance traveled to land, where the speed is zero

Vf² = Vo² + 2 to X

X = -Vo² / 2 a

X = -110² / 2 (-7.42)

X = 815.4 m

Since this distance is greater than the length of the runway, the jet cannot stop

Let's calculate the speed you should have to stop on a track of this size

Vo² = 2 a X

Vo = √ (2 7.42 800)

Vo = 109 m / s

It is conclusion the jet must lose some speed to land on this track

4 0
3 years ago
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