A basketball player jumps 76cm to get a rebound. How much time does he spend in the top 15cm of the jump (ascent and descent)?

Answers

Answer 1

Answer:

Answer:

The time for final 15 cm of the jump equals 0.1423 seconds.

Explanation:

The initial velocity required by the basketball player to be able to jump 76 cm can be found using the third equation of kinematics as

$v^2=u^2+2as$

where

'v' is the final velocity of the player

'u' is the initial velocity of the player

'a' is acceleration due to gravity

's' is the height the player jumps

Since the final velocity at the maximum height should be 0 thus applying the values in the above equation we get

$0^2=u^2-2* 9.81* 0.76\n\n\therefore u=√(2* 9.81* 0.76)=3.86m/s$

Now the veocity of the palyer after he cover'sthe initial 61 cm of his journey can be similarly found as

$v^(2)=3.86^2-2* 9.81* 0.66\n\n\therefore v=√(3.86^2-2* 9.81* 0.66)=1.3966m/s$

Thus the time for the final 15 cm of the jump can be found by the first equation of kinematics as

$v=u+at$

where symbols have the usual meaning

Applying the given values we get

$t=(v-u)/(g)\n\nt=(0-1.3966)/(-9.81)=0.1423seconds$

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20 examples of scalar quantity

Answers

Answer:

Length

Time

Mass

Temperature

Energy

Direct Current (DC)

Frequency

Volume

Speed

Amount of substance

Luminous Intensity

Density

Concentration

Refractive Index

Work

Pressure

Power

Charge

Electric Potential

Entropy

The 160-lblb crate is supported by cables ABAB, ACAC, and ADAD. Determine the tension in these wire

Answers

Answer:

$F_(AB) = 172.1356\nF_(AC) = 258.2033\nF_(AD) = 368.8004$

Explanation:

Using the diagram (see attachment) we extract the following position vectors:

$Vector (OA) = 6i + 0j + 0k \nVector (OB) = 0i + 3j + 2k \nVector (OC) = 0i - 2j + 3k$

Next step is to find unit vectors $u_(AB) ,u_(AC), u_(AD), u_(AE)$ as follows:

$u_(AB) = (vector(AB))/(magnitude(AB)) \n= \frac{OB - OA}{magnitude({vector(OB - OA))} }\n=(-6i +3j+2k)/(√(6^2 + 3^2+2^2) ) \n\n=-0.857 i +0.429j+0.286k\n\nu_(AC) = (vector(AC))/(magnitude(AC)) \n= \frac{OC - OA}{magnitude({vector(OC - OA))} }\n=(-6i -2j-3k)/(√(6^2 + 2^2+3^2) ) \n\n=-0.857 i -0.286j+0.429k\n\nu_(AD) = +1i\nu_(AC) = -1k$

Using the diagram we find the corresponding vectors Forces:

$F_(AB) = F_(AB) i + F_(AB)j +F_(AB)k\nF_(AC) = F_(AC) i + F_(AC)j +F_(AC)k\nF_(AD) = F_(AD) i + F_(AD)j +F_(AD)k\nW = -160 k$

Equation of Equilibrium:

$Sum of forces = 0\nF_(AB). u_(AB) + F_(AC).u_(AC) + F_(AD).u_(AD) + W = 0\n(-0.857F_(AB)i + 0.429F_(AB)j +0.286F_(AB)k) + (-0.857F_(AC)i - 0.286F_(AC)j +0.429F_(AC)k) + (+1F_(AD) i) + (-160k) = 0$

Comparing i , j and k components as follows:

$-0.857F_(AB) -0.857F_(AC) +1F_(AD) = 0\n+ 0.429F_(AB) - 0.286F_(AC) = 0\n+0.286F_(AB) +0.429F_(AC) = 160$

Solving Above equation simultaneously we get:

$F_(AB) = 172.1356\nF_(AC) = 258.2033\nF_(AD) = 368.8004$

A capacitor consists of two closely spaced metal conductors of large area, separated by a thin insulating foil. It has an electrical capacity of 3000.0 μF and is charged to a potential difference of 60.0 V. Calculate the amount of energy stored in the capacitor. Tries 0/20 Calculate the charge on this capacitor when the electrical energy stored in the capacitor is 6.53 J. Tries 0/20 If the two plates of the capacitor have their separation increased by a factor of 5 while the charge on the plates remains constant, by what factor is the energy stored in the capacitor increased?

Answers

Answer:

1 = 5.4 J

2 = 0.1979 C

3 = 5

Explanation:

Energy in a capacitor, E is

E = 1/2 * C * V²

E = 1/2 * 3000*10^-6 * 60²

E = 1/2 * 3000*10^-6 * 3600

E = 1/2 * 10.8

E = 5.4 J

E = Q²/2C = 6.53 J

E * 2C = Q²

Q² = 6.53 * 2 * 3000*10^-6

Q² = 13.06 * 3000*10^-6

Q² = 0.03918

Q = √0.03918

Q = 0.1979 C

The Capacitor, C is inversely proportional to the distance of separation, D. Thus, if D is increased by 5 to be 5D, then C would be C/5. And therefore, our energy stored in the capacitor is increased by a factor of 5.

An electromagnetic wave is traveling the +y direction. The maximum magnitude of the electric field associated with the wave is Em, and the maximum magnitude of the magnetic field associated with the wave is Bm. At one instant, the electric field has a magnitude of 0.25 Em and points in the +x direction. At this same instant, the wave's magnetic field has a magnitude which is ... and it is in the ... direction.

Answers

Answer:

0.83x10^-9 T

Direction is towards +z axis.

Explanation:

E = cB

E = magnitude of electrical 0.25 Em

c = speed of light in a vacuum 3x10^8 m/s

Therefore,

B = E/c = 0.25 ÷ 3x10^8

B = 0.83x10^-9 T

Magnetic fueld of a EM wave acts perpendicularly to its electric field, therefore it's direction is towards the +Z axis

What force causes oppositely charged particles to attract each other? A. Magnetic force B. Compression
C. Electrical Force D. Gravity

Answers

Answer:

It is electrical force

Explanation:

i got it wrong on A P E X with magnetic hope this helps!

Answer:

The electromagnetic force.

A Michelson interferometer operating at a 400 nm wavelength has a 3.95-cm-long glass cell in one arm. To begin, the air is pumped out of the cell and mirror M2 is adjusted to produce a bright spot at the center of the interference pattern. Then a valve is opened and air is slowly admitted into the cell. The index of refraction of air at 1.00 atmatm pressure is 1.00028.Required:
How many bright-dark-bright fringe shifts are observed as the cell fills with air?