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John, who has a mass of 65kg stands at rest on the ice. He catches a 10kg ball that is thrown to him at 5m/s.

Answers

Answer 1

Answer:

The momentum of John after catching the ball is 50 kg.m/s.

"Your question is not complete, it seems to be missing the following information";

find John's momentum

The given parameters;

mass of John, m = 65 kg
mass of the ball caught by John, m' = 10 kg
initial velocity of John, u = 0
initial velocity of the ball, v = 5 m/s

Apply the principles of conservation of linear momentum to determine the momentum of John.

The momentum of John is calculated as follows;

P = mu + mv

P = (65 x 0) + (10 x 5)

P = 0 + 50

P = 50 kg.m/s

Thus, the momentum of John after catching the ball is 50 kg.m/s.

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A nearsighted person has a far point of 40cm. What power spectacle lens is needed if the lens is 2cm from the eye

Answers

Answer:

The value is $p = - 2.63 \ Diopters$

Explanation:

From the question we are told that

The value of the far point is $a = 40 \ cm = 0.4 \ m$

The distance of the lens to the eye is $b = 2 \ cm = 0.02$

Generally

$1 Diopter = > 1 m^(-1)$

Generally the power spectacle lens needed is mathematically represented as

$p = (1)/(d_o ) + (1)/(d_i)$

Here $d_o$ is the object distance which for a near sighted person is $d_o = \infty$

And $d_i$ is the image distance which is evaluated as

$d_i = b - a$

=> $d_i = 0.02 - 0.4$

=> $d_i = -0.38 \ m$

$p = (1)/(\infty ) + (1)/(-0.38)$

=> $p = 0 - 2.63$

=> $p = - 2.63 \ Diopters$

Plzzz helppppp!!! I need answers A, B, C & D

Answers

Answer: the answer i for c is yes 0& 10

Explanation:

Three children want to play on a see saw that is 6 meters long and has a fulcrum in the middle. Two of the children are twins and weigh 40 kg each and sit on the same side at a distance 2m and 3m away from the fulcrum. The other child weighs 80 kg. How far away should he sit from the fulcrum so that the see saw is balanced

Answers

Given Information:

mass of child 1 = m₁ = 40 kg

distance from fulcrum of child 1 = d₁ = 2 m

mass of child 2 = m₂ = 40 kg

distance from fulcrum of child 2 = d₂ = 3 m

mass of child 3 = m₃ = 80 kg

Required Information:

distance from fulcrum of child 3 = d₃ = ?

Answer:

distance from fulcrum of child 3 = 2.5 m

Explanation:

In order to balance the see-saw, the moment of force should be same on both sides of the fulcrum.

Since 2 children are sitting on one side and only 1 on the other side

F₁d₁ + F₂d₂ = F₃d₃

Where Force is given by

F = mg

m₁gd₁ + m₂gd₂ = m₃gd₃

m₁d₁ + m₂d₂ = m₃d₃

Re-arrange the equation for d₃

m₃d₃ = m₁d₁ + m₂d₂

d₃ = (m₁d₁ + m₂d₂)/m₃

d₃ = (40*2 + 40*3)/80

d₃ = 2.5 m

Therefore, the child on the other side should sit 2.5 m from the fulcrum so that the see-saw remains balanced.

Dogs can hear higher-pitched whistles that humans do. How do you
think the sound frequencies that dogs can
hear compare to the frequencies that humans
can hear?

Answers

Dogs can hear sounds at higher frequencies than humans. The range of sound frequencies that dogs can hear is approximately 40 Hz to 60,000 Hz, while the range for humans is 20 Hz to 20,000 Hz. This means that dogs can hear ultrasonic sounds that are beyond the range of human hearing.

What is sound about?

In terms of physics, sound is a vibration that travels through a transmission medium like a gas, liquid, or solid as an acoustic wave.

Sound is the reception of these waves and the brain's perception of them in terms of human physiology and psychology. Dogs have the ability to hear ultrasonic sounds that are audible only to them.

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At what partial pressure are argon atoms expected to have a free travel of approximately 5 µm, if the gas is at a temperature of 400 K? The cross section of collision, σ, or Argon is 0.28 nm2Ar molar mass is 39.9 g/mole

Answers

Answer:

2790 Pa

Explanation:

Given wavelength λ= 5μm

temperature T= 400 K

cross section of collision σ= 0.28 nm^2

molar mass = 39.9 g/mole

pressure = $P= (RT)/(√(2)N_A\sigma\lambda )$

putting values we get

= $(8.314*400)/(√(2)*6.022*10^(23)*0.28*10^(-18)*5*10^(-6) )$

⇒P = 2790 J/m^3

the partial pressure are argon atoms expected= 2790 Pa

5. (Serway 9th ed., 7-3) In 1990, Walter Arfeuille of Belgium lifted a 281.5-kg object through a distance of 17.1 cm using only his teeth. (a) How much work was done on the object by Arfeuille in this lift, assuming the object was lifted at constant speed? (b) What total force was exerted on Arfeuille’s teeth during the lift? (Ans. (a) 472 J; (b) 2.76 kN)

Answers

Para resolver este problema es necesario aplicar los conceptos de Fuerza, dados en la segunda Ley de Newton y el concepto de Trabajo, como expresión de la fuerza necesaria para realizar una actividad en una distancia determinada.

El trabajo se define como

W = F*d

Where,

F = Force

d = Distance

At the same time we have that the Force by second's Newton law is equal to

F = mg

Where,

m = mass

g = Gravitational acceleration

PART A) Using our values and replacing we have that

$W = F*d\nW = mg*d\nW=281.5*9.8(17.1*10^(-2)\nW = 471.738 J\approx 472J$