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wolverine [178]
2 years ago
7

A person walks 5.0kilometers north, then 5.0 kilometers east. His displacement is closest to ? A. 10 kilometers northwest B. 7.1

kilometers northwest C. 10 kilometers northeast D. 7.1 kilometers northeast
Physics
1 answer:
Rainbow [258]2 years ago
4 0
Use vector analysis and calculate resultant vector using Pythagoras theorem:
5^2 + 5^2 = 50
Square root of 50 = approx 7.1 km NE
Therefore the answer is D
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Assume your eye has an aperture diameter of 3.00 mm at night when bright headlights are pointed at it. 1) At what distance can y
krek1111 [17]

Answer:

The distance is 6259.31 meters.

Explanation:

We shall use the Reyligh criterion to solve the problem

For diffraction due to circular aperture we have

Assuming that human eye is circular we have

\frac{x}{D}=\frac{1.22\lambda }{d}

\therefore D=\frac{xd}{1.22\lambda }

Applying the given values we have

D=\frac{1.40\times 3\times 10^{-3}}{1.22\times 550\times 10^{-9}}\\\\\therefore D=6259.31m\\D=6.26km

3 0
2 years ago
Read 2 more answers
The diagram below shows a person using a device called a jetpack. Water is forced downwards from the jetpack and produces an upw
klasskru [66]

Answer:

The condition necessary for the person to be able to remain stationary in mid-air is that there must be an upward force also referred to as THRUST

Explanation:

Thrust in Physics is defined as the force that propels, forces or pushes the mass of a certain object in a specified or particular direction.

Thrust can defined in the terms of a Jet pack can be defined as the force that is required to propel the mass of a person in an upward direction.

Based on the diagram that we have been shown the question, we can see that water is the fuel used to proper the user or person upwards hence, the jet pack used in the question is an HydroJet pack

The condition necessary for the person to be able to remain stationary in mid-air is that the propelling force called the thrust must be upwards such that it can overcome these 4 things:

a) Gravitational force

b) Mass of the jet pack user

c) Mass of the jet pack itself

d) The water which serves as fuel for the jet pack.

5 0
2 years ago
What would the speed of each particle be if it had the same wavelength as a photon of yellow light (????=575.0 nm)? Proton (mass
PilotLPTM [1.2K]

Answer:

Proton: v=0.689 m/s

Neutron: v=0.688 m/s

Electron: v=1265.078 m/s

Alpha particle: v=0.173 m/s

Explanation:

De Broglie equation allows you to calculate the “wavelength” of an electron or any other particle or object of mass m that moves with velocity v:

λ=\frac{h}{mv}

h is the Planck constant: 6.626×10⁻³⁴\frac{kg.m^2}{s}

We know that the wavelength of the particle is 575 nm (575×10⁻⁹m), so we find the velocity v for each particle:

λ=\frac{h}{mv}

v=h÷(mλ)

<u>Proton:</u>

m=1.673×10⁻²⁴ g · \frac{1kg}{1000g}=1.673×10⁻²⁷ kg

v=h÷(mλ)

v=6.626×10⁻³⁴\frac{kg.m^2}{s}÷(1.673×10⁻²⁷ kg×575×10⁻⁹m)

v=0.689 m/s

<u>Neutron:</u>

m=1.675×10⁻²⁴ g · \frac{1kg}{1000g}=1.675×10⁻²⁷ kg

v=h÷(mλ)

v=6.626×10⁻³⁴\frac{kg.m^2}{s}÷(1.675×10⁻²⁷ kg×575×10⁻⁹m)

v=0.688 m/s

<u>Electron:</u>

m= 9.109×10⁻²⁸ g · \frac{1kg}{1000g}=9.109×10⁻³¹ kg

v=h÷(mλ)

v=6.626×10⁻³⁴\frac{kg.m^2}{s}÷(9.109×10⁻³¹ kg×575×10⁻⁹m)

v=1265.078 m/s

<u>Alpha particle:</u>

m=6.645×10⁻²⁴ g · \frac{1kg}{1000g}=6.645×10⁻²⁷ kg

v=h÷(mλ)

v=6.626×10⁻³⁴\frac{kg.m^2}{s}÷(6.645×10⁻²⁷ kg×575×10⁻⁹m)

v=0.173 m/s

3 0
2 years ago
Read 2 more answers
What is the porosity of the sand sample?(The sediment volume for each sample is 400ml.) a. 90.25% b. 72.00% c. 25.50% d. 16.75%
Katyanochek1 [597]
C.25.50% Hope this helps.
7 0
2 years ago
Read 2 more answers
Write the equivalent formulas for velocity, acceleration, and force using the relationships covered for UCM, Newton’s Laws, and
yKpoI14uk [10]

Answer:

The newton’s second law is F=ma

The Gravitational force is F=\dfrac{Gm_{1}m_{2}}{r^2}

Explanation:

Given that,

The equivalent formulas for velocity, acceleration, and force using the relationships covered for UCM, Newton’s Laws, and Gravitation.

We know that,

Velocity :

The velocity is equal to the rate of position of the object.

v=\dfrac{dx}{dt}....(I)

Acceleration :

The acceleration is equal to the rate of velocity of the object.

a=\dfrac{dv}{dt}....(II)

Newton’s second Laws

The force is equal to the change in momentum.

In mathematically,

F=\dfrac{d(p)}{dt}

Put the value of p

F=\dfrac{d(mv)}{dt}

F=m\dfrac{dv}{dt}

Put the value from equation (II)

F=ma

This is newton’s second laws.

Gravitational force :

The force is equal to the product of mass of objects and divided by square of distance.

In mathematically,

F=\dfrac{Gm_{1}m_{2}}{r^2}

Where, m₁₂ = mass of first object

m= mass of second object

r = distance between both objects

Hence, The newton’s second law is F=ma

The Gravitational force is F=\dfrac{Gm_{1}m_{2}}{r^2}

3 0
2 years ago
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