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Umnica [9.8K]
3 years ago
14

[K1 A2 T1 C1] A golfer strikes a golf ball on level ground.

Physics
1 answer:
gregori [183]3 years ago
6 0

Answer:

Explanation:

See the file attached .

b ) Range of projectile

= u²sin2θ / g

= 42² sin32 x 2 / g

= 42² sin64 / 9.8

= 161.8 m

c )

Max height = u² sin²32 / 2 g

= 42² sin²32 / 2x 9.8

= 25.27  m .

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6 0
3 years ago
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Refer to the figure for this question: Four particles form a square. the charges are q1=q4=Q and q2=q3=q. (Part A) What is Q/q i
nikitadnepr [17]
Draw a vector diagram. The net force on particle 1 = F12 + F13 + F14 These forces have to be added as vectors. 
We will resolve our forces along the direction 1-4 F12 (tot) = -kQq / a^2 in the direction of particle 4 F12 = -kQq *sin (45) / a^2 F12 = -kQq /( a^2 * sqrt(2) ) 
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F14 = -kQQ / (Sqrt(2)*a) ^ 2 
For net force on particle 1 : 
F12+F13+F14 = 0 -2kQq /( a^2 * sqrt(2) ) + -kQQ / (Sqrt(2)*a) ^ 2 = 0 
Some simple manipulation should give you : 
Q/q = -2 sqrt(2) 
4 0
3 years ago
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Difference beteween phycial change and chemical change​
pychu [463]

Answer:

Physical changes only change the appearance of a substance, not its chemical composition. Chemical changes cause a substance to change into an entirely substance with a new chemical formula. Chemical changes are also known as chemical reactions.

4 0
3 years ago
A point charge q1 is at the center of a sphere of radius 20 cm. Another point charge q2 = 5 nC is located at a distance r = 30 c
valina [46]

Answer:

q_1 =7.08*10^{-9}C.

Explanation:

Gauss's Law says that the electric flux \Phi_E through a closed surface is directly proportional to the charge Q_{enc} inside it. More precisely,

$\Phi_E=\oint_S E\cdot dA = \dfrac{Q_{enc}}{\epsilon_0}. $

This means what is outside this closed surface S does not contribute to the flux through it because field lines that go in must come out, <em>resulting a zero flux from an external charge. </em>

In our context, this means the charge q_2 which is outside the sphere will have zero flux through the surface; therefore, Gauss's law will only be concerned with charge q_1 which is inside the sphere; Hence,

$\Phi_E=\oint_S E\cdot dA = \dfrac{q_1}{\epsilon_0} = 800 N\cdot m^2/C. $

Solving for q_1 gives

$ q_1= (800 N\cdot m^2/C)\epsilon_0, $

$ q_1= (800 N\cdot m^2/C)*(8.85*10^{-12}C^2/N\cdot m^2) $

\boxed{q_1 =7.08*10^{-9}C. }

which is the charge inside the sphere.

5 0
3 years ago
The solubility of a gas is .55 g/l at 8.0 atm pressure. what will be the solubility of the gas at 5.0 atm partial pressure?
garik1379 [7]
Henry's law states that:

P1/C1 = P2/C2

Where P = Pressure, and C = Solubility
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Therefore,
C2 = (P2*C1)/P1 = (5*0.55)/8 = 0.344 g/l
8 0
3 years ago
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