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shusha [124]
3 years ago
12

A 0.046 kg golf ball hit by a driver can accelerate from rest to 67 m/s in 1 ms while the driver is in contact with the ball. Ho

w much average force does the golf ball experience
Physics
1 answer:
Flauer [41]3 years ago
3 0

Answer:

Average force = 67 mn

Explanation:

Given:

Initial velocity u = 0 m/s

Final velocity v = 67 m/s

Time t = 1 ms = 0.001 sec.

Computation:

Using Momentum theory

Change in momentum  = F × Δt

 (v-u)/t =  F × Δt

F × 0.001 = (67 - 0)/0.001

F= 67,000,000

Average force = 67 mn

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The Earth's axis of rotation is tilted. Because of this, the amount of sunlight an area on the planet receives each day varies
Sergio039 [100]

Answer: The Northern Hemisphere experiences the start of <u>summer</u>

Explanation:

Due to this tilt in Earth's axis of rotation, some regions receive different amounts of sunlight according to the seasons of the year. These variations are more evident near the poles and softer or imperceptible near the equator. This explains the fact that while in the northern hemisphere it is summer and there are high temperatures (because the Sunlight incides more perpendicularly in this area during this season), in the southern hemisphere it is winter and very low temperatures are recorded, and vice versa.  

In this sense, according to the image the Earth is at the point of its orbit in which, due to the inclination of its axis, the North Pole is closer to the Sun.  

In other words, the northern hemisphere is tilted toward the Sun, and this astronomical phenomenon brings the summer season for the Northern Hemisphere.

4 0
3 years ago
Read 2 more answers
The water flowing through a 2.0 cm (inside diameter) pipe flows out through three 1.3 cm pipes. (a) If the flow rates in the thr
Verdich [7]

Answer:

a)54L/min

b)0.845

Explanation:

a) A x V=A_1V_1+ A_2V_2+A_3V_3

where suffix 1,2,3 refers to the three pipes.

            =27L/min+16L/min+11 L/min

            =54L/min

b) A x V=54L/min => \frac{\pi }{4} d^2 x v

   d= 2 cm

\frac{\pi }{4} d^2 x v = 54

v= \frac{4}{\pi } x \frac{54}{2^2}

-> A_1 x V_1=27L/min => \frac{\pi }{4} d_1^2 x v_1

d_1= 1.3cm

\frac{\pi }{4} d^2 x v_1 = 27

v_1= \frac{4}{\pi } x \frac{27}{1.3^2}

Next is to find the ratio of speed i.e \frac{v}{v_1}

\frac{4}{\pi } x \frac{54}{2^2} / \frac{4}{\pi } x \frac{27}{1.3^2} => \frac{54}{27} \frac{1.3^2}{2^2}

\frac{v}{v_1}= 0.845

8 0
3 years ago
A ball is projected horizontally from the top of a bertical building 25.0m above the ground level with an initial velocity of 8.
kirill115 [55]

Answer:

Solution given:

height [H]=25m

initial velocity [u]=8.25m/s

g=9.8m/s

now;

a. How long is the ball in flight before striking the ground?

Time of flight =?

Now

Time of flight=\sqrt{\frac{2H}{g}}

substituting value

  • =\sqrt{\frac{2*25}{9.8}}
  • =2.26seconds

<h3><u>the ball is in flight before striking the ground for 2.26seconds</u>.</h3>

b. How far from the building does the ball strike the ground?

<u>H</u><u>o</u><u>r</u><u>i</u><u>z</u><u>o</u><u>n</u><u>t</u><u>a</u><u>l</u><u> </u>range=?

we have

Horizontal range=u*\sqrt{\frac{2H}{g}}

  • =8.25*2.26
  • =18.63m

<h3><u>The ball strikes 18.63m far from building</u>. </h3>
7 0
3 years ago
Find the energy u of the capacitor in terms of c and q by using the definition of capacitance and the formula for the energy in
gtnhenbr [62]

The formula for the energy in a capacitor , u in terms of q and c is q²/2c

<h3>What is the energy of a capacitor?</h3>

The energy of a capacitor u = 1/2qv where

  • q = charge on capacitor and
  • v = voltage across capacitor.

<h3>What is the capacitance of a capacitor?</h3>

Also, the capacitance of a capacitor c = q/v where

  • q = charge on capacitor and
  • v = voltage across capacitor.

So, v = q/c

<h3>The formula for energy of the capacitor in terms of q and c</h3>

Substituting v into u, we have

u = 1/2qv

= 1/2q(q/c)

= q²/2c

So, the formula for the energy in a capacitor , u in terms of q and c is q²/2c

Learn more about energy in a capacitor here:

brainly.com/question/10705986

#SPJ12

3 0
2 years ago
Travels 11,000 feet along a dark desert highway if the car averages 84 mph find the amount of time to cover this distance
laila [671]

The time taken by traveler to cover the distance is,

t=\frac{d}{v}

Substitute the known values,

\begin{gathered} t=\frac{(11000\text{ ft)}}{(84\text{ mph)(}\frac{1.46667\text{ ft/s}}{1\text{ mph}})_{}} \\ \approx89.3\text{ s} \end{gathered}

Therefore, the time taken by traveler to cover the distance is 89.3 s.

5 0
1 year ago
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