Ag,Au and Cu are called coinage metals why plzzzz hurry its urgent plzzz

A smooth circular hoop with a radius of 0.400 m is placed flat on the floor. A 0.325-kg particle slides around the inside edge o

aleksandrvk [35]

Answer:

a) W = - 6.825 J, b) θ = 1.72 revolution

Explanation:

a) In this exercise the work of the friction force is negative and is equal to the variation of the kinetic energy of the particle

W = ΔK

W = K_f - K₀

W = ½ m v_f² - ½ m v₀²

W = ½ 0.325 (5.5² - 8.5²)

W = - 6.825 J

b) find us the coefficient of friction

Let's use Newton's second law

fr = μ N

y-axis (vertical) N-W = 0

fr = μ W

work is defined by

W = F d

the distance traveled in a revolution is

d₀ = 2π r

W = μ mg d₀ = -6.825

μ = $\frac{ -6.825}{d_o \ mg}$

The total work as the object stops the final velocity is zero v_f = 0

W = 0 - ½ m v₀²

W = - ½ 0.325 8.5²

W = - 11.74 J

μ mg d = -11.74

we subtitle the friction coefficient value

( $\frac{-6.8525 }{d_o mg}$ ) m g d = -11.74

6.825 $\frac{d}{d_o}$ = 11.74

d = 11.74/6.825 d₀

d = 1.7201 2π 0.400

d = 4.32 m

this is the total distance traveled, the distance and the angle are related

θ = d / r

θ = 4.32 / 0.40

θ = 10.808 rad

we reduce to revolutions

θ = 10.808 rad (1rev / 2π rad)

θ = 1.72 revolution

3 0

3 years ago

25% part (c) assume that d is the distance the cheetah is away from the gazelle when it reaches full speed. Derive an expression

levacccp [35]

maximum speed of cheetah is

$v_1 = v_{max}$

speed of gazelle is given as

$v_2 = v_{g}$

Now the relative speed of Cheetah with respect to Gazelle

$v_{12} = v_1 - v_2$

$v_{12} = v_{max} - v_g$

now the relative distance between Cheetah and Gazelle is given initially as "d"

now the time taken by Cheetah to catch the Gazelle is given as

$d = v_{12}* t$

so by rearranging the terms we can say

$t = \frac{d}{v_{12}}$

$t = \frac{d}{v_{max} - v_g}$

so above is the relation between all given variable

6 0

4 years ago

Water flows through a water hose at a rate of Q1 = 860 cm3/s, the diameter of the hose is d1 = 1.85 cm. A nozzle is attached to

Snowcat [4.5K]

Answer:

a) A1 = $\frac{\pi (d1)^{2} }{4}$

b) A1 = 2.688 cm $^{2}$

c) Q1 = A1 x v1

d) v1 = 3.1994 m/s

e) A2 = $\frac{A1 X v1}{v2}$

f) A2 = 0.7963cm $^{2}$

Explanation:

a) Area = $\pi r^{2}$

r = $\frac{d}{2}$

thus,

area = $\pi (\frac{d}{2})^{2}$

A1 = $\frac{\pi (d1)^{2} }{4}[/tex]b) d1 = 1.85 cmsubstituting in the above equation,A1 = [tex]\frac{\pi (d1)^{2} }{4}$

A1 = $\frac{\pi (1.85)^{2} }{4}$

A1 = 2.688 cm $^{2}$

c) Flow rate = Area x velocity ( refer brainly.com/question/13997998)

Q1 = A1 x v1

d) From the above equation,

v1 = $\frac{Q1}{A1}$ = $\frac{860}{2.688}$ = 319.94 cm/s = 3.1994 m/s

e) Since the flow rate Q1 is constant throughout the hose, Av is a constant.

i.e. A1 x v1 = A2 x v2

thus,

A2 = $\frac{A1 X v1}{v2}$

f) v2 = 10.8 m/s.

substituting the values in the above equation,

A2 = $\frac{2.688 X 3.1994}{10.8}$ = 0.7963cm $^{2}$

3 0

4 years ago

The amount of force required to deform a spring is _____.

Mama L [17]

The first one, directly proportional to the length that the spring is stretched.

8 0

4 years ago

Read 2 more answers

Imagine a third particle, which we will call a cyberon. It has three times the mass of an electron (3m). It has a positive charg

GalinKa [24]

Answer:

Explanation:

Let the potential difference between plate be V .

Electrical energy lost for cyberon = V x 3e

Kinetic energy lost by cyberon = kinetic energy gain by it

V x 3e = 1/2 x 3m x vc²

For electron , the above equation becomes

V x e = 1/2 x m x ve²

Dividing the two equations

1 = vc² / ve²

vc = ve .

vc / ve = 1 .

6 0

3 years ago

Ag,Au and Cu are called coinage metals why plzzzz hurry its urgent plzzz​

Ag,Au and Cu are called coinage metals why plzzzz hurry its urgent plzzz