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pickupchik [31]

3 years ago

8

In a circus act, a uniform board (length 3.00 m, mass 25.0 kg ) is suspended from a bungie-type rope at one end, and the other e

nd rests on a concrete pillar. When a clown (mass 79.0 kg ) steps out halfway onto the board, the board tilts so the rope end is 30∘ from the horizontal and the rope stays vertical. Calculate the force exerted by the rope on the board with the clown on it.

1 answer:

tester [92]3 years ago

4 0

Answer:

Force of Rope = 122.5 N

Force of Rope = 480.2N

Explanation:

given data

length = 3.00 m

mass = 25.0 kg

clown mass = 79.0 kg

angle = 30°

solution

we get here Force of Rope on with and without Clown that is

case (1) Without Clown

pivot would be on the concrete pillar so Force of Rope will be

Force of Rope × 3m = (25kg)×(9.8ms²)×(1.5m)

solve it and we get

Force of Rope = 122.5 N

and

case (2) With Clown

so here pivot is still on concrete pillar and clown is standing on the board middle and above the centre of mass so Force of Rope will be

Force of Rope × 3m = (25kg+73kg)×(9.8ms²)×(1.5m)

solve it and we get

Force of Rope = 480.2N

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Air flows through a nozzle at a steady rate. At the inlet the density is 2.21 kg/m3 and the velocity is 20 m/s. At the exit, the

Vitek1552 [10]

To solve the problem, it is necessary to apply the concepts related to the change of mass flow for both entry and exit.

The general formula is defined by

$\dot{m}=\rho A V$

Where,

$\dot{m} =$ mass flow rate

$\rho =$ Density

V = Velocity

Our values are divided by inlet(1) and outlet(2) by

$\rho_1 = 2.21kg/m^3$

$V_1 = 20m/s$

$A_1 = 60*10^{-4}m^2$

$\rho_2 = 0.762kg/m^3$

$V_2 = 160m/s$

PART A) Applying the flow equation we have to

$\dot{m} = \rho_1 A_1 V_1$

$\dot{m} = (2.21)(60*10^{-4})(20)$

$\dot{m} = 0.2652kg/s$

PART B) For the exit area we need to arrange the equation in function of Area, that is

$A_2 = \frac{\dot{m}}{\rho_2 V_2}$

$A_2 = \frac{0.2652}{(0.762)(160)}$

$A_2 = 2.175*10^{-3}m^2$

Therefore the Area at the end is $21.75cm^2$

3 0

3 years ago

How does a parallel circuit differ from a series circuit?

Levart [38]

A series circuit has only one path for current,
while a parallel circuit has more than one.

4 0

3 years ago

A 0.311 kg tennis racket moving 30.3 m/s east makes an elastic collision with a 0.0570 kg ball moving 19.2 m/s west. Find the ve

amm1812

<u>Answer</u>:

The velocity of the tennis racket after the collision 14.966 m/s.

<u>Step-by-step explanation:</u>

let the following:

m₁ = mass of tennis racket = 0.311 kg

m₂ = mass of the ball = 0.057 kg

u₁ = velocity of tennis racket before collision = 30.3 m/s

u₂ = velocity of the ball before collision = -19.2 m/s

v₁ = velocity of tennis racket after collision

v₂ = velocity of the ball after collision

Right (+) , Left (-)

An elastic collision is an encounter between two bodies in which the total kinetic energy of the two bodies remains the same.

So, the total kinetic energy before collision = the total kinetic energy after collision.

So, 0.5 m₁ u₁² + 0.5 m₂ u₂² = 0.5 m₁ v₁² + 0.5 m₂ v₂² ⇒ (1)

Also, the total momentum before collision = the total momentum after collision.

So, m₁ u₁ + m₂ u₂ = m₁ v₁ + m₂ v₂ ⇒ (2)

Solving (1) and (2):

∴ v₁ = [ u₁ * (m₁ - m₂) + u₂ * 2m₂ ]/ (m₁ + m₂)

= ( 30.3 * (0.311 - 0.057) - 19.2 * 2 * 0.057 ) / ( 0.311 + 0.057)

= 14.966 m/s.

So, the velocity of the tennis racket after the collision 14.966 m/s.

7 0

3 years ago

Use the dimensional analysis and check the correctness of given equation:-<br> PV= nRT

Harman [31]

PV=nRT

Here

P=Pressure

V=Volume

n=Molarity

R=universal gas constant

T=Temperature.

LHS

$\\ \tt\bull\leadsto PV$

$\\ \tt\bull\leadsto [ML^2T^{-2}][M^0L^3T^0]$

$\\ \tt\bull\leadsto [ML^5T^{-2}]$

RHS

$\\ \tt\bull\leadsto nRT$

$\\ \tt\bull\leadsto [M^0L^{3}T^0][M^1 L^2 T^{-2}K^{-1}][M^0L^0T^0K^1]$

$\\ \tt\bull\leadsto [ML^5T^{-2}]$

LHS=RHS

hence verified

6 0

3 years ago

An object is placed perpendicular to the the principal axis of convex lens of focal length 8,the distance of the object from the

Digiron [165]

Answer:

v = 24 cm and inverted image

Explanation:

Given that,

The focal length of the object, f = +8 cm

Object distance, u = -12 cm

We need to find the position &nature of the image. Let v be the image distance. Using lens formula to find it :

$\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}$

Put all the values,

$\dfrac{1}{v}=\dfrac{1}{f}+\dfrac{1}{u}\\\\\dfrac{1}{v}=\dfrac{1}{8}+\dfrac{1}{(-12)}\\\\v=24\ cm$

So, the image distance from the lens is 24 cm.

Magnification,

$m=\dfrac{v}{u}\\\\m=\dfrac{24}{-12}\\\\m=-2$

The negative sign of magnification shows that the formed image is inverted.

7 0

2 years ago