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quester [9]
2 years ago
5

A golfer starts with the club over her head and swings it to reach maximum speed as it contacts the ball. Halfway through her sw

ing, when the golf club is parallel to the ground, does the acceleration vector of the club head point straight down, parallel to the ground, approximately toward the golfer's shoulders, approximately toward the golfer's feet, or toward a point above the golfer's head?
Physics
1 answer:
lara [203]2 years ago
3 0

Answer:

a) parallel to the ground True

c) parallel to the ground towards man True

Explanation:

To examine the possibilities, we propose the solution of the problem.

Let's use Newton's second law

      F = m a

The force is exerted by the arm and the centripetal acceleration of the golf club, which in this case varies with height.

In our case, the stick is horizontal in the middle of the swing, for this point the centripetal acceleration is directed to the center of the circle or is parallel to the arm that is also parallel to the ground;

Ask the acceleration vector

a) parallel to the ground True

b) down. False

c) parallel to the ground towards True men

d) False feet

e) the head. False

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The energy from 0.015 moles of octane was used to heat 250 grams of water. The temperature of the water rose from 293.0 K to 371
arsen [322]

Answer : The correct option is, (B) -5448 kJ/mol

Explanation :

First we have to calculate the heat required by water.

q=m\times c\times (T_2-T_1)

where,

q = heat required by water = ?

m = mass of water = 250 g

c = specific heat capacity of water = 4.18J/g.K

T_1 = initial temperature of water = 293.0 K

T_2 = final temperature of water = 371.2 K

Now put all the given values in the above formula, we get:

q=250g\times 4.18J/g.K\times (371.2-293.0)K

q=81719J

Now we have to calculate the enthalpy of combustion of octane.

\Delta H=\frac{q}{n}

where,

\Delta H = enthalpy of combustion of octane = ?

q = heat released = -81719 J

n = moles of octane = 0.015 moles

Now put all the given values in the above formula, we get:

\Delta H=\frac{-81719J}{0.015mole}

\Delta H=-5447933.333J/mol=-5447.9kJ/mol\approx -5448kJ/mol

Therefore, the enthalpy of combustion of octane is -5448 kJ/mol.

5 0
2 years ago
In a scientific test conducted in Arizona, a special cannon called HARP (High Altitude Research Project) shot a projectile strai
Alexus [3.1K]

Answer:

It took the projectile 120 s to reach the maximum height.

Explanation:

Given;

maximum height of the projectile, s = 180 km = 180,000 m

initial speed of the projectile, u = 3 km/s = 3000 m/s

final velocity at maximum height, v = 0

Apply the following kinematic equation for average velocity of the projectile;

s = (\frac{v+u}{2} )t\\\\(v+u)t = 2s\\\\t = \frac{2s}{v+u} \\\\t = \frac{2*180,000}{0+3,000}\\\\ t = 120 \ s

Therefore, it took the projectile 120 s to reach the maximum height.

5 0
2 years ago
A block moves at 5 m/s in the positive x direction and hits an identical block, initially at rest. A small amount of gunpowder h
Anestetic [448]

Answer:

Speed of 1.83 m/s and 6.83 m/s

Explanation:

From the principle of conservation of momentum

mv_o=m(v_1 + v_2) where m is the mass, v_o is the initial speed before impact, v_1 and v_2 are velocity of the impacting object after collision and velocity after impact of the originally constant object

5m=m(v_1 +v_2)

Therefore v_1+v_2=5

After collision, kinetic energy doubles hence

2m*(0.5mv_o)=0.5m(v_1^{2}+v_2^{2})

2v_o^{2}=v_1^{2} + v_2^{2}

Substituting 5 m/s for v_o then

2*(5^{2})= v_1^{2} + v_2^{2}

50= v_1^{2} + v_2^{2}

Also, it’s known that v_1+v_2=5 hence v_1=5-v_2

50=(5-v_2)^{2}+ v_2^{2}

50=25+v_2^{2}-10v_2+v_2^{2}

2v_2^{2}-10v_2-25=0

Solving the equation using quadratic formula where a=2, b=-10 and c=-25 then v_2=6.83 m/s

Substituting, v_1=-1.83 m/s

Therefore, the blocks move at a speed of 1.83 m/s and 6.83 m/s

6 0
2 years ago
To a stationary observer, a man jogs east at 2.5 m/s and a woman jogs west at 1.5 m/s. from the woman's frame of reference, what
neonofarm [45]
T o a stationary observer, a man jogs east at 2.5 m/s and a woman jogs west at 1.5 m/s. from the woman's frame of reference, what is the man's velocity? it is 4m/s east
5 0
2 years ago
Read 2 more answers
A) The current theory of the structure of the Earth, called plate tectonics, tells us that the continents are in constant motion
suter [353]

A) The mass of the continent is 2.5\cdot 10^{21} kg

B) The kinetic energy is 2016 J

C) The speed of the jogger should be 7.1 m/s

Explanation:

A)

The mass of the continent can be calculated as

m = \rho V

where

\rho = 2800 kg/m^3 is its density

V is its volume

We have to calculate its volume. We know that the continent is represented as a slab of side 5900 km (so its surface is 5900 x 5900, assuming it is a square) and depth of 26 km, so its volume is:

V=(5900 km)^2 (26 km)=9.05\cdot 10^8 km^3 =9.05 \cdot 10^8 \cdot (10^9 m^3/k^3)=9.05\cdot 10^7 m^3

So, the mass of the continent is

m=\rho V = (2800)(9.05\cdot 10^{17})=2.5\cdot 10^{21} kg

B)

The kinetic energy of a body is given by

K=\frac{1}{2}mv^2

where

m is the mass of the body

v is its speed

For the continent, we have:

m=2.5\cdot 10^{21} kg is the mass

v=4 cm/year is the speed

We have to convert the speed into SI units. we have:

1 cm = 0.01 m

1 year = (365)(24)(60)(60) s = 3.15\cdot 10^7 s

So, the speed is

v=4 cm/year = 0.04 m/year \cdot \frac{1}{3.15\cdot 10^7}=1.27\cdot 10^{-9} m/s

Therefore, the kinetic energy is

K=\frac{1}{2}(2.5\cdot 10^{21} kg)(1.27\cdot 10^{-9} m/s)^2=2016 J

C)

Again, the kinetic energy of an object is

K=\frac{1}{2}mv^2

For the jogger in this problem, his mass is

m = 80 kg

And we want its kinetic energy to be equal to that of the continent, so

K = 2016 J

Re-arranging the equation for v, we find what speed the jogger needs to have this kinetic energy:

v=\sqrt{\frac{2K}{m}}=\sqrt{\frac{2(2016)}{80}}=7.1 m/s

Learn more about kinetic energy here:

brainly.com/question/6536722

#LearnwithBrainly

8 0
2 years ago
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