PHYSICS COLLEGE

The best rebounders in basketball have a vertical leap (that is, the vertical movement of a fixed point on their body) of about 100 cm . a) What is their initial "launch" speed off the ground?b)How long are they in the air?

Answers

Answer 1

Answer:

Answer:

a) 4.45 m/s

b) 0.9 seconds

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.81 m/s²

$v^2-u^2=2as\n\Rightarrow -u^2=2as-v^2\n\Rightarrow u=√(v^2-2as)\n\Rightarrow u=√(0^2-2* -9.81* 1)\n\Rightarrow u=4.45\ m/s$

a) The vertical speed when the player leaves the ground is 4.45 m/s

$v=u+at\n\Rightarrow t=(v-u)/(a)\n\Rightarrow t=(0-4.45)/(-9.81)\n\Rightarrow t=0.45\ s$

Time taken to reach the maximum height is 0.45 seconds

$s=ut+(1)/(2)at^2\n\Rightarrow 1=0t+(1)/(2)* 9.81* t^2\n\Rightarrow t=\sqrt{(1* 2)/(9.81)}\n\Rightarrow t=0.45\ s$

Time taken to reach the ground from the maximum height is 0.45 seconds

b) Time the player stayed in the air is 0.45+0.45 = 0.9 seconds

Related Questions

When some stars use up their fuel, they undergo a catastrophic explosion called a supernova. This explosion blows much or all of a star's mass outward, in the form of a rapidly expanding spherical shell. As a simple model of the supernova process, assume that the star is a solid sphere of radius R that is initially rotating at 2.4 revolutions per day. After the star explodes, find the angular velocity, in revolutions per day, of the expanding supernova shell when its radius is 4.3R. Assume that all of the star's original mass is contained in the shell.
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Which law does the following statement express? "In all cases of electromagnetic induction, the induced voltages have adirection such that the currents they produce oppose the effect that produces them."
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Which describes one feature of the image formed by a convex mirror?????

Determine whether or not each of the following statement is true. If a statement is true, prove it. If the statement is false, provide a counterexample and explain how it constitutes a counterexample. Diagrams can be useful in explaining such things. If the electric potential in a certain region of space is constant, then the charge enclosed by any closed surface completely contained within that region is zero.

Answers

Answer:

True

Explanation:

This is a representation of Gauss law.

Gauss’s law does hold for moving charges, and in this respect Gauss’s law is more general than Coulomb’s law. In words, Gauss’s law states that: The net outward normal electric flux through any closed surface is proportional to the total electric charge enclosed within that closed surface. The law can be expressed mathematically using vector calculus in integral form and differential form, both are equivalent since they are related by the divergence theorem, also called Gauss’s theorem.

An internal explosion breaks an object, initially at rest,intotwo pieces, one of which has 1.5 times the mass of the other.If
7500 J were released in the explosion, how much kinetic energydid
each piece acquire?

Answers

Answer:

4500 J and 3000 J

Explanation:

According to conservation of momentum

$0 = m_1 V_1 + m_2 V_2$

Given that m_2 = 1.5 m_1 , so

$V_1 = -1.5 V_2$

the kinetic energy of each piece is

$K_2= (1)/(2) m_2v_2^2$

$K_1= (1)/(2) m_1v_1^2$

substituting the value of V1 in the above equation

$K_1 = (1/2)( m_2 / 1.5 )( -1.5 V_2)^2 = 1.5 (1/2)m_2 V_2^2 = 1.5 K_2$

Given that

K_1 + k_2 = 7500 J

1.5 K_2 + K_2 = 7500

K_2 = 7500 / 2.5

= 3000 J

this is the KE of heavier mass

K_1 = 7500 - 3000 = 4500 J

this is the KE of lighter mass

Final answer:

The question is about finding the kinetic energy acquired by each of two pieces of an object following an internal explosion, using principles of conservation of energy and momentum in physics.

Explanation:

The student has asked about an internal explosion that breaks an object into two pieces with different masses, releasing a certain amount of kinetic energy in the process. This question involves applying the principle of conservation of energy and momentum to find the kinetic energy acquired by each piece post-explosion.

Assuming piece 1 has a mass of m and piece 2 has a mass of 1.5m, the total mass of the system is 2.5m. Since 7500 J of energy was released in the explosion, to find the kinetic energy of each piece, we can use the fact that the total kinetic energy is equal to the energy released during the explosion. Let the kinetic energy of the smaller piece be K1 and of the larger piece be K2. Because the object was initially at rest and momentum must be conserved, the momenta of the two pieces must be equal and opposite. This relationship allows us to derive the ratio of the kinetic energies. We can solve for K1 and K2 proportionally. Finally, because the kinetic energy is a scalar quantity, adding the kinetic energies of the two pieces will equal the total energy released.

Learn more about Kinetic Energy Acquisition here:

brainly.com/question/28545352

#SPJ3

Which sling can the crane use to lift the 1000kg pipe?A.
800kg rated sling
B. 1000kg rated sling
C. 2000kg rated sling
D. Band C

Answers

Answer:

C. 2000kg rated sling

Explanation:

ensures better safety and can carry twice more mass than current mass.

A battery with an emf of 12.0 V shows a terminal voltage of 11.7 V when operating in a circuit with two lightbulbs, each rated at 4.0 W (at 12.0 V), which are connected in parallel.What is the battery's internal resistance?

Answers

Answer:

0.46Ω

Explanation:

The electromotive force (E) in the circuit is related to the terminal voltage(V), of the circuit and the internal resistance (r) of the battery as follows;

E = V + Ir --------------------(a)

Where;

I = current flowing through the circuit

But;

V = I x Rₓ ---------------------(b)

Where;

Rₓ = effective or total resistance in the circuit.

First, let's calculate the effective resistance in the circuit:

The effective resistance (Rₓ) in the circuit is the one due to the resistances in the two lightbulbs.

Let;

R₁ = resistance in the first bulb

R₂ = resistance in the second bulb

Since the two bulbs are both rated at 4.0W ( at 12.0V), their resistance values (R₁ and R₂) are the same and will be given by the power formula;

P = $(V^(2) )/(R)$

=> R = $(V^(2) )/(P)$ -------------------(ii)

Where;

P = Power of the bulb

V = voltage across the bulb

R = resistance of the bulb

To get R₁, equation (ii) can be written as;

R₁ = $(V^(2) )/(P)$ --------------------------------(iii)

Where;

V = 12.0V

P = 4.0W

Substitute these values into equation (iii) as follows;

R₁ = $(12.0^(2) )/(4)$

R₁ = $(144)/(4)$

R₁ = 36Ω

Following the same approach, to get R₂, equation (ii) can be written as;

R₂ = $(V^(2) )/(P)$ --------------------------------(iv)

Where;

V = 12.0V

P = 4.0W

Substitute these values into equation (iv) as follows;

R₂ = $(12.0^(2) )/(4)$

R₂ = $(144)/(4)$

R₂ = 36Ω

Now, since the bulbs are connected in parallel, the effective resistance (Rₓ) is given by;

$(1)/(R_(X) )$ = $(1)/(R_1)$ + $(1)/(R_2)$ -----------------(v)

Substitute the values of R₁ and R₂ into equation (v) as follows;

$(1)/(R_X)$ = $(1)/(36)$ + $(1)/(36)$

$(1)/(R_X)$ = $(2)/(36)$

Rₓ = $(36)/(2)$

Rₓ = 18Ω

The effective resistance (Rₓ) is therefore, 18Ω

Now calculate the current I, flowing in the circuit:

Substitute the values of V = 11.7V and Rₓ = 18Ω into equation (b) as follows;

11.7 = I x 18

I = $(11.7)/(18)$

I = 0.65A

Now calculate the battery's internal resistance:

Substitute the values of E = 12.0, V = 11.7V and I = 0.65A into equation (a) as follows;

12.0 = 11.7 + 0.65r

0.65r = 12.0 - 11.7

0.65r = 0.3

r = $(0.3)/(0.65)$

r = 0.46Ω

Therefore, the internal resistance of the battery is 0.46Ω

Answer:

$R_i_n_t=0.45 \Omega$

Explanation:

Internal resistance is a concept that helps model the electrical consequences of the complex chemical reactions that occur within a battery. When a charge is applied to a battery, the internal resistance can be calculated using the following equation:

$R_i_n_t=((V_N_L)/(V_F_L) -1)R_L$

Where:

$V_F_L=Load\hspace{3}voltage=11.7V\nV_N_L= O pen\hspace{3}circuit\hspace{3}voltage=12V\nR_L=Load\hspace{3}resistance$

As you can see, we don't know the exactly value of the $R_L$ . However we can calculated that value using the next simple operations:

The problem tell us that the power of each lightbulb is 4.0 W at 12.0 V, hence let's calculated the power at 11.7V using Cross-multiplication:

$(12)/(11.7) =(4)/(P)$

Solving for $P$ :

$P=(11.7*4)/(12) =3.9W$

Now, the electric power is given by:

$P=(V^2)/(R_b)$

Where:

$R_b=Resistance\hspace{3}of\hspace{3}each\hspace{3}lightbulb$

So:

$R_b=(V^2)/(P) =(11.7^2)/(3.9) =35.1\Omega$

Now, because of the lightbulbs are connected in parallel the equivalent resistance is given by:

$(1)/(R_L) =(1)/(R_b) +(1)/(R_b) =(2)/(R_b) \n\n R_L= (R_b)/(2) =(35.1)/(2)=17.55\Omega$

Finally, now we have all the data, let's replace it into the internal resistance equation:

$R_i_n_t=((12)/(11.7) -1)17.55=0.45\Omega$

John and Linda are arguing about the definition of density. John says the density of an object is proportional to itsmass. Linda says the object's mass is proportional to its density and to its volume. Which one, if either, is correct?A. They are both wrong
B. John is correct, but Linda is wrong
C. John is wrong, but Linda is correct
D. They are both correct.
E. John must be wrong, because Linda always wins these arguments.

Answers

Answer:

They are both correct.

Explanation:

The density of an object is defined as the ratio of its mass to its volume. This implies that the density of the object is both proportional to the mass and also to the volume of the object. John only mentioned mass which is correct. Linda mentioned the second variable on which density depends which is the volume of the object.

Hence considering the both statements objectively, one can say that they are both correct.

Guitar string has an overall length of 1.22 m and a total mass of 3.5 g before being strung on a guitar. Once it is used on the guitar, there is a distance of 70 cm between fixed end points. The guitar string is tightened to a tension of 255 N.What is the frequency of the fundamental wave on the guitar string?