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monitta
3 years ago
13

A football player is preparing to punt the ball down the field. He drops the ball from rest and it falls vertically 1.0 m down o

nto his foot. After he kicks it, the ball leaves the foot with a speed of 18.5 m/s at an angle 57° above the horizontal.
What is the magnitude of the impulse delivered to the ball in kilogram meters per second after he kicks it? An American football has a mass of 425 g.

Physics
1 answer:
Doss [256]3 years ago
4 0

Answer: Impulse = 6.37kgm/s

Explanation: please find the attached file for the solution.

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A scientist wants to measure the relationship between humidity in the
ICE Princess25 [194]

Answer:

D

Explanation:

<em>The most suitable testable question. in this case, would be that 'are there more home runs during the more humid months of the  summer?'</em>

Since the aim of the investigation is to find the relationship between humidity and the number of home runs, measuring the number of home runs during the more humid months in the summer and comparing the data to the number of home runs during the less humid months in the same summer would provide the answer.

<u>Only option D raises a valid question that is relevant to the aim of the investigation.</u>

7 0
3 years ago
Cavendish used the attraction between large and small lead balls to measure an experimental value for the gravitational constant
irina1246 [14]

The gravitational constant was experimentally measured by W Cavendish using the attraction between big and small lead balls. is true

The correct answer is true

<h3>How do you define gravitational constant?</h3>

the strength of gravity. a factor in use in Newton's gravity law to relate the strength of the gravitational pull between two bodies with their masses and distance from one another. 6.67259 X 10-11 newtons per square kilogram is roughly the gravitational constant. G is its identifier.

<h3> where is the strongest gravity is?</h3>

The gravitational pull of the earth is greatest near sea level, normally, and weakens as you get further from the center, such as to the summit of Mt. Everest. Because the obloid earth was slightly wider, but only by a minor ratio, the gravity just at poles is stronger than that at the equator.

To know more about gravitational constant visit:

brainly.com/question/858421
#SPJ9

8 0
9 months ago
A solid circular disk has a mass of 1.2 kg and a radius of 0.16m. Each of three identical thin rods has a mass of 0.16kg. The ro
Juli2301 [7.4K]

Answer:

0.027648 kgm²

Explanation:

M = Mass of disc = 1.2 kg

r = Radius of disc = 0.16 m

m = Mass of rod = 0.16 kg

R = Rod distance = 0.16 m

Moment of inertia of disk is given by

I_1=\dfrac{1}{2}Mr^2\\\Rightarrow I_1=\dfrac{1}{2}1.2\times 0.16^2\\\Rightarrow I_1=0.01536\ kgm^2

Moment of inertia of the three rods

I_2=3mr^2\\\Rightarrow I_2=3\times 0.16\times 0.16^2\\\Rightarrow I_2=0.012288\ kgm^2

The total moment of inertia is given by

I=I_1+I_2=0.01536+0.012288\\\Rightarrow I=0.027648\ kgm^2

The moment of inertia of the stool with respect to an axis that is perpendicular to the plane of the disk at its center is 0.027648 kgm²

8 0
3 years ago
At takeoff, a commercial jet has a speed of 72 m/s. Its tires have a diameter of 0.89 m. Part (a) At how many rev/min are the ti
34kurt

Answer:

a) Revolutions per minute = 2.33

b) Centripetal acceleration = 11649.44 m/s²

Explanation:

a) Angular velocity is the ratio of linear velocity and radius.

Here linear velocity = 72 m/s

Radius, r  = 0.89 x 0. 5 = 0.445 m

Angular velocity

         \omega =\frac{72}{0.445}=161.8rad/s

Frequency

         f=\frac{2\pi}{\omega}=\frac{2\times \pi}{161.8}=0.0388rev/s=2.33rev/min

Revolutions per minute = 2.33

b) Centripetal acceleration

               a=\frac{v^2}{r}

  Here linear velocity = 72 m/s

  Radius, r  = 0.445 m

Substituting

   a=\frac{72^2}{0.445}=11649.44m/s^2

Centripetal acceleration = 11649.44m/s²

6 0
2 years ago
E asteroid belt circles the sun between the orbits of mars and jupiter. one asteroid has a period of 6.0 earth years. part a wha
anzhelika [568]

To solve the problem, use Kepler's 3rd law : 

T² = 4π²r³ / GM 

Solved for r : 

r = [GMT² / 4π²]⅓ 

but first covert 6.00 years to seconds : 

6.00years = 6.00years(365days/year)(24.0hours/day)(6... 
= 1.89 x 10^8s 

The radius of the orbit then is : 

r = [(6.67 x 10^-11N∙m²/kg²)(1.99 x 10^30kg)(1.89 x 10^8s)² / 4π²]⅓ 
= 6.23 x 10^11m

6 0
3 years ago
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