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valentina_108 [34]
4 years ago
12

a golf club is swung and strikes a golf ball with a force of 12.2 N. The golf ball accelerates at 265.2 m/s2. Calculate the mass

of the golf ball. Show your work and include the correct unit.
Physics
1 answer:
MaRussiya [10]4 years ago
5 0

Answer:

0.046\ kg

Explanation:

We\ are\ given\ that,\\Acceleration\ produced\ on\ the\ golf\ ball\ due\ to\ Force\ exerted=265.2\ m/s^2\\Magnitude\ of\ the\ Force\ exerted\ over\ the\ Golf-Ball=12.2\ N\\We\ know\ that,\\'The\ force\ acting\ on\ an\ object\ is\ equivalent\ to\ the\ product\ of\ it's\ mass\\ and\ the\ acceleration\ produced'.\\Or,\\F=ma\\Let\ the\ mass\ of\ the\ golf\ ball\ be\ m,\\Hence,\\12.2=265.2\ m\\m=\frac{12.2}{265.2}=0.046\ kg

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If a rock has a speed of 12 m/s as it hits the ground, from what height did it
grin007 [14]

Answer:

To find the height the following formula should be used:  

v 2 = u 2 + 2aH

Explanation:

Assuming this occurs on earth,  a= 9.8 ms -2

 -2        2

12=0+2 x9.8 x H

144

_______ =H

2 x 9.8

H= 7.35m

6 0
3 years ago
A ball which is dropped from the top of a building strikes the ground with a speed of 30m/s. Assume air resistance can be ignore
Vika [28.1K]

Answer:

               h= 45.87 m.

Explanation:

Data given:

Vf= 30m/s ,     and we know that g = 9.8 m/s²    

The ball has an initial velocity of zero      Vi = 0 m/s²    

To Find:

Height of the building = ?

Solution:

       According to 3rd law of the motion;

                                      2aS= Vf² - Vi²                      ( S= h , a=g)

                                     2*9.81*h = (30)² - (0)²

                                      h= 45.87 m.

3 0
3 years ago
If it takes a planet 2.8 × 108 s to orbit a star with a mass of 6.2 × 1030 kg, what is the average distance between the planet
Shtirlitz [24]

The average distance between the planet and the star is:

R=9.36*10^11 m

Orbital velocity  v=√{(G*M)/R},

G = gravitational constant =6.67*10^-11 m³ kg⁻¹ s⁻²,

M = mass of the star

R =distance from the planet to the star.

v=ωR, with ω as the angular velocity and R the radius

ωR=√{(G*M)/R},

ω=2π/T,

T = orbital period of the planet

To get R we write the formula by making R the subject of the equation

(2π/T)*R=√{(G*M)/R}

{(2π/T)*R}²=[√{(G*M)/R}]²,

(4π²/T²)*R²=(G*M)/R,

(4π²/T²)*R³=G*M,

R³=(G*M*T²)/4π²,

R=∛{(G*M*T²)/4π²},

Substitute values

R=9.36*10^11 m

As was already said, Earth is located roughly 150 million kilometres (93 million miles) from the Sun on average. It is 1 AU. Mars is on our fictitious football field's three-yard line. On average, the distance between the Sun and the red planet is around 142 million miles (228 million kilometres).

Learn more about average distance:

brainly.com/question/18366547

#SPJ4

The complete question is ''If it takes a planet 2.8 × 108 s to orbit a star with a mass of 6.2 × 10^30 kg, what is the average distance between the planet and the star? 1.43 × 10^9 m 9.36 × 10^11 m 5.42 × 10^13 m 9.06 × 10^17 m''.

4 0
1 year ago
A rocket, initially at rest on the ground, accelerates straight upward from rest with constant acceleration 34.3 m/s^2 . The acc
Eddi Din [679]

Answer:

The maximum height reached by the rocket is 1.94 × 10³ m.

Explanation:

The height of the rocket can be calculated using the following equations:

y = y0 + v0 · t + 1/2 · a · t²    (when the rocket is accelerated upward).

y = y0 +  v0 · t + 1/2 · g · t² (after the rocket runs out of fuel).

Where:

y = height at time t.

y0 = initial height.

v0 = initial velocity.

t = time.

a = acceleration due to engines of the rocket.

g = acceleration due to gravity.

In the same way, the velocity of the rocket can be calculated as follows:

v = v0 + a · t  (when the rocket has fuel)

v = v0 + g · t   (when the rocket runs out of fuel)

Where "v" is the velocity at time "t"

First, let´s find the height reached until the rocket runs out of fuel.

y = y0 + v0 · t + 1/2 · a · t²

y = 0 m + 0 m/s · t + 1/2 · 34.3 m/s² · (5.00 s)²

y = 429 m

And now, let´s find the velocity reached in that time of upward acceleration:

v = v0 + a · t

v = 0 m/s + 34.3 m/s² · 5.00 s

v = 172 m/s

When the rocket runs out of fuel, it is accelerated downward due to gravity. But, since the rocket has initially an upward velocity (172 m/s), it will not fall immediately and will continue to go up until the velocity becomes 0. In that instant, the rocket is at its maximum height and thereafter it will start to fall with negative velocity.

Then, using the equation for velocity, we can calculate the time it takes the rocket to reach its maximum height:

v = v0 + g · t

0 = 172 m/s - 9.80 m/s² · t

-172 m/s / -9.80 m/s² = t

t = 17.6 s

With this time, we can now calcualte the maximum height. Notice that the initial velocity and height are the ones reached during the upward acceleration phase:

y = y0 +  v0 · t + 1/2 · g · t²

ymax = 429 m + 172 m/s · 17.6 s - 1/2 · 9.80 m/s² · (17.6 s)²

ymax = 1.94 × 10³ m

4 0
3 years ago
In a string of lights, if one light goes out and the others stay lit, which type of circuit do you have?
zepelin [54]
You have a parallel circuit since it allows the electrons to travel in 2 paths, allowing the other light to stay lit.
3 0
4 years ago
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