All categories

3 years ago

As a 5.0 × 10^2-newton basketball player jumps

1 answer:

4 0

This player is initially at rest, then the Force Weight (W) and the Normal Force (N) have a same module. When applies a Extra Force (E) on the floor, this reacts with this increase, causing the player to be released upwards.
Using the Newton's Secound Law, we have:

$F_{R}=ma \\ W+E-N=ma \\ N =W+E \\ N=5*10^2+10*10^2 \\ \boxed {N=1.5*10^3N}$

Number 3

If you notice any mistake in my english, please let me know, because i am not native.

You might be interested in

A 2.0 moles of a monatomic ideal gas expands isothermally from state a to state b, Pa = 600 Pa, Va = 3.0 m3, and Vb = 9.0 m3.

krek1111 [17]

Answer:

a) Pb= 200 PA

b).work done= -3600 joules

c).3600joules

D).the system works under isothermal condition so no heat was transferred

Explanation:

2.0 moles of a monatomic ideal gas expands isothermally from state a to state b, Pa = 600 Pa, Va = 3.0 m3, and Vb = 9.0 m3.

a). PbVb= PaVa

Pb= (PaVa)/VB

Pb= (600*3)/9

Pb= 1800/9

Pb= 200 PA

b). work done= n(Pb-Pa)(Vb-Va)

Work done= 2*(200-600)(9-3)

Work done= -600(6)

Work done=- 3600 Pam³

work done= -3600 joules

C). Change in internal energy I the work done on the system

= 3600joules

D).the system works under isothermal condition so no heat was transferred

4 0

3 years ago

Consider a block of mass m attached to two springs, one on the left with spring constant k1 and one on the right with spring con

Andrew [12]

Answer:

$T= 2\pi\times sqrt(m/(k1+k2))$

Explanation:

When the block is displaced by x units

F= spring force

two springs are connected parallel

$F =-k_1x - k_2x$

Writing Newtons second law, F = ma

$-k_1x - k_2x =ma$

$-k_1x - k_2x = mx''$

a= x" ( differentiating x w.r.t time twice)

$x''+(k_1/m + k_2/m) x=0$

this the standard form of equation of oscillation spring mass system

This is the differential equation, x'' means that double differentiation of x , i.e, x'' is acceleration

since, Period $T=2\pi\sqrt{\frac{m}{K_{eq.}} }$

therefore,

$T= 2\pi\times sqrt(m/(k1+k2))$

3 0

3 years ago

17) The charge of an anti-charm quark isapproximately equal to

Gnom [1K]

The charge of an anti-charm quark is approximately equal to +5.33 ×10⁻²⁰ C.

<h3>What are electrons?</h3>

The electrons are the spinning objects around the nucleus of the atom of the element in an orbit.

The charge on electron is equal to e = -1.6 x 10⁻¹⁹C

The charge of an anti- charm quark is +23 times the magnitude of charge on electron.

So, charge on anti- charm quark is

q = +1/3 x ( 1.6 x 10⁻¹⁹C)

q = +5.33 ×10⁻²⁰ C

Thus, the charge of an anti-charm quark is approximately equal to +5.33 ×10⁻²⁰ C.

Learn more about electrons.

brainly.com/question/1255220

#SPJ1

7 0

2 years ago

Use the Divergence Theorem to calculate the surface integral

MrRa [10]

Divergense theorem says flux across the closed surface = ∫∫∫ div F dV

div F = 3y^2+ 3 z^2 
convert to cylindrical 
y = r cos t 
z = r sin t 
x = x 
dx dy dz = r dx dr dt 

∫ [0, 2pi]∫[0,3]∫ [-1,3] 3r^3 dx dr dt 

2pi(3)(81/4)(4) 
486pi

Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions.

3 0

3 years ago

A wire with a circular cross section has radius 0.16 mm and is lying along the x axis from x=0 to x=0.20 m. The wire is made of

kipiarov [429]

Answer:

Resistance = 68.23 Ω

Explanation:

Let's start off by remembering the fact that when resistors are connected in series, their resistances is added up to find the total resistance.

The equation for resistance using resistivity is given below:

Resistance = Resistivity * Length / Area

where Length = x

Resistivity = 3x^5