1. A train is moving north at 5 m/s on a straight track. The engine is causing it to accelerate northward at 2 m/s^2.
How far will it go before it is moving at 20 m/s?
A) 83
B) 43
C) 39
D) 94
E) 20
Answer:
It will go up to 93.75 m before it is moving at 20 m/s
Explanation:
As we know that
[tex]v^2 - u^2 = 2aS[/tex]
here v is the final speed i.e 20 m/s
u is the initial speed i.e 5 m/s
a is the acceleration due to gravity i.e 2 m/s^2
Substituting the given values in above equation, we get -
[tex]20^2 - 5^2 = 2*2*S\\S = 93.75[/tex]meters
If you weigh 690 N on the earth, what would be your weight on the surface of a neutron star that has the same mass as our sun and a diameter of 15.0 km ? Take the mass of the sun to be ms = 1.99×1030 kg , the gravitational constant to be G = 6.67×10−11 N⋅m2/kg2 , and the free-fall acceleration at the earth's surface to be g = 9.8 m/s2 . Express your weight wstar in newtons.
Answer:
W' = 1.66 x 10¹⁴ N
Explanation:
First, we will calculate the mass:
[tex]W = mg[/tex]
where,
W = weight on earth = 690 N
m = mass = ?
g = acceleration due to gravity on earth = 9.8 m/s²
Therefore,
[tex]m = \frac{W}{g} = \frac{690\ N}{9.8\ m/s^2}\\\\m = 70.4\ kg[/tex]
Now, we will calculate the value of g on the neutron star:
[tex]g' = \frac{GM}{R^2}[/tex]
where,
g' = acceleration due to gravity on the surface of the neutron star = ?
G = Gravitational Constant = 6.67 x 10⁻¹¹ N.m²/kg²
M = Mass of the Neutron Star = 1.99 x 10³⁰ kg
R = Radius of the Neutron Star = 15 km/2 = 7.5 km = 7500 m
Therefore,
[tex]g' = \frac{(6.67\ x\ 10^{-11}\ N.m^2/kg^2)(1.99\ x\ 10^{30}\ kg)}{(7500\ m)^2}\\\\g' = 2.36\ x\ 10^{12}\ m/s^2[/tex]
Therefore, the weight on the surface of the neutron star will be:
[tex]W' = mg'\\W' = (70.4\ kg)(2.36\ x\ 10^{12}\ m/s^2)[/tex]
W' = 1.66 x 10¹⁴ N
A block of weight 1200N is on an incline plane of 30° with the horizontal, a force P is applied to the body parallel to the plane, if the coefficient of the static friction is 0.20 and kinetic friction is 0.15 (1) find the value of P to cause motion up the plane (2) find P to prevent motion down the plane. (3) Find P to cause continuous motion up the plane.
Answer:
a) P = 807.85 N, b) P = 392.15 N, c) P = 444.12 N
Explanation:
For this exercise, let's use Newton's second law, let's set a reference frame with the x-axis parallel to the plane and the direction rising as positive, and the y-axis perpendicular to the plane.
Let's use trigonometry to break down the weight
sin θ = Wₓ / W
cos θ = W_y / W
Wₓ = W sin θ
W_y = W cos θ
Wₓ = 1200 sin 30 = 600 N
W_y = 1200 cos 30 = 1039.23 N
Y axis
N- W_y = 0
N = W_y = 1039.23 N
Remember that the friction force always opposes the movement
a) in this case, the system will begin to move upwards, which is why friction is static
P -Wₓ -fr = 0
P = Wₓ + fr
as the system is moving the friction coefficient is dynamic
fr = μ N
fr = 0.20 1039.23
fr = 207.85 N
we substitute
P = 600+ 207.85
P = 807.85 N
b) to avoid downward movement implies that the system is stopped, therefore the friction coefficient is static
P + fr -Wx = 0
fr = μ N
fr = 0.20 1039.23
fr = 207.85 N
we substitute
P = Wₓ -fr
P = 600 - 207,846
P = 392.15 N
c) as the movement is continuous, the friction coefficient is dynamic
P - Wₓ + fr = 0
P = Wₓ - fr
fr = 0.15 1039.23
fr = 155.88 N
P = 600 - 155.88
P = 444.12 N
Plutonium-238 has a half life of 87.7 years. What percentage of a 5 kilogram (kg) sample remains after 50 years?
Answer:
i dont know but i should know try g o o g l e
Explanation:
Can someone take there time and answer this :)
Answer: I think B.)
Explanation:
Unpolarized light with intensity 370 W/m2 passes first through a polarizing filter with its axis vertical, then through a second polarizing filter. It emerges from the second filter with intensity 132 W/m2 . You may want to review (Pages 897 - 898) . Part A What is the angle from vertical of the axis of the second polarizing filter
Answer:
θ = 32.4º
Explanation:
For this exercise let's use Malus's law
I = Io cos² θ
in this case it indicates that the incident intensity is 370 W/m², when the first polarization passes, only the radiation with the same polarization of the polarizer emerges, that is, vertical
I₀ = 370/2 = 185 W / m²
this is the radiation that affects the second polarizer, let's apply the expression of Maluz
θ = cos⁻¹ ([tex]\sqrt{\frac{I}{I_o} }[/tex])
θ = cos⁻¹ ([tex]\sqrt{132/185}[/tex])
θ = cos⁻¹ (0.844697)
θ = 32.4º
What happens when Earth rotates on its axis and how long does it take
Answer:
You get Day and Night
It takes 24 hour
Answer:
Explanation:
The Earth's orbit makes a circle around the sun. At the same time the Earth orbits around the sun, it also spins.Since the Earth orbits the sun and rotates on its axis at the same time we experience seasons, day and night, and changing shadows throughout the day.It only takes 23 hours, 56 minutes and 4.0916 seconds for the Earth to turn once on its axis.
An ideal horizontal spring-mass system has a mass of 1.0 kg and a spring with constant 78 N/m. It oscillates with a period of 0.71 seconds. When this same spring-mass system oscillates vertically instead, the period is _______ seconds. Enter 2 significant figures (a total of three digits) and use g = 10.0 m/s2 if necessary.
Answer:
T = 0.71 seconds
Explanation:
Given data:
mass m = 1Kg, spring constant K = 78 N/m, time period of oscillation T = 0.71 seconds.
We have to calculate time period when this same spring-mass system oscillates vertically.
As we know
[tex]T = 2\pi \sqrt{\frac{m}{K} }[/tex]
This relation of time period is true under every orientation of the spring-mass system, whether horizontal, vertical, angled or inclined. Therefore, time period of the same spring-mass system oscillating vertically too remains the same.
Therefore, T = 0.71 seconds
If the girl skater has a mass of 30 kg and moves backward at 5 m/s, what is the velocity or the boy skater
his mass is 50 kg?
Question: Two people stand facing each other at a roller-skating rink then push off each other. If the girl skater has a mass of 30 kg and moves backward at 5 m/s, what is the velocity of the boy skater if his mass is 50 kg?
Answer:
3 m/s
Explanation:
Applying,
The Law of conservation of momentum
Momentum of the girl skater = momentum of the boy skater
MV = mv...................... Equation 1
Where M = mass of the girl skater, V = velocity of the girl skater, m = mass of the boy skater, v = velocity of the boy skater
From the question, we were asked to calculate v
v = MV/m.................. Equation 1
Given: M = 30 kg, V = 5 m/s, m = 50 kg
Substitute these values into equation 1
v = (30×5)/50
v = 3 m/s
Hence the velocity of the the boy skater is 3m/s
Transverse thrusters are used to make large ships fully maneuverable at low speeds without tugboat assistance. A transverse thruster consists of a propeller mounted in a duct; the unit is then mounted below the waterline in the bow or stern of the ship. The duct runs completely across the ship. Calculate the thrust developed by a 1900 kW unit supplied to the propeller if the duct is 2.6 m in diameter and the ship is stationary.
Answer:
Thrust developed = 212.3373 kN
Explanation:
Assuming the ship is stationary
Determine the Thrust developed
power supplied to the propeller ( Punit ) = 1900 KW
Duct distance ( diameter ; D ) = 2.6 m
first step : calculate the area of the duct
A = π/4 * D^2
= π/4 * ( 2.6)^2 = 5.3092 m^2
next : calculate the velocity of propeller
Punit = (A*v*β ) / 2 * V^2 ( assuming β = 999 kg/m^3 ) also given V1 = 0
∴V^3 = Punit * 2 / A*β
= ( 1900 * 10^3 * 2 ) / ( 5.3092 * 999 )
hence V2 = 8.9480 m/s
Finally determine the thrust developed
F = Punit / V2
= (1900 * 10^3) / ( 8.9480)
= 212.3373 kN
15 points!
a. Calculate the electric potential energy stored in a 1.4 x 10-7 F capacitor
that stores 3.40 x 10-6 C of charge at 24.0 V.
Answer:
[tex]4.12\times 10^{-5}\ J[/tex].
Explanation:
Given that,
Capacitance, [tex]C=1.4\times 10^{-7}\ F[/tex]
Charge stored in the capacitor, [tex]Q=3.4\times 10^{-6}\ C[/tex]
We need to find the electric potential energy stored in the capacitor. The formula for the electric potential energy stored in the capacitor is given by :
[tex]E=\dfrac{Q^2}{2C}[/tex]
Put all the values,
[tex]E=\dfrac{(3.4\times 10^{-6})^2}{2\times 1.4\times 10^{-7}}\\\\=4.12\times 10^{-5}\ J[/tex]
So, the required electric potential eenergy is equal to [tex]4.12\times 10^{-5}\ J[/tex].
How much work will a 500 watt motor do in 10 seconds?
Answer:
50j
Explanation:
Watts are units used to measure power. power can be defined as rate of energy transfer
500 watts means - 500 J of energy per second
in 1 second - 500 J of work is done
therefore within 10 seconds - 500 J/s x 10 s = 5000 J
work of 5000 J is carried out in 10 seconds
Answer:
Watts are units used to measure power. power can be defined as rate of energy transfer
500 watts means - 500 J of energy per second
in 1 second - 500 J of work is done
therefore within 10 seconds - 500 J/s x 10 s = 5000 J
work of 5000 J is carried out in 10 seconds
Explanation:
1.What is the Kinetic energy of a 3 kg object moving at 4 m/s?
Plz help I’ll give points
Answer:
24 J
Explanation:
[tex]K = \frac{1}{2} mv^{2} = \frac{1}{2} (3kg)(4m/s)^{2} = 24 J[/tex]
Scientists are constantly exploring the universe, looking for new planets that support life similar to the life on
Earth. A new planet that supports life would have all of the following characteristics except -
A. a gaseous atmosphere.
B. an orbiting moon.
C. liquid water.
D. protection from radiation.
A new planet that supports life would have all the following characteristics except an orbiting moon. Hence, option B is correct.
What is a Planet?An enormous, spherical celestial object known as a planet is neither a star nor its remains. The nebular hypothesis, which states how an interstellar cloud falls out of a nebula to produce a young protostar encircled by a protoplanetary disk, is now the best explanation for planet formation.
By gradually accumulating material under the influence of gravity, or accretion, planets develop in this disk.
The rocky planets Mercury, Venus, Earth, and Mars, as well as the giant planets Jupiter, Saturn, Uranus, and Neptune, make up the Solar System's minimum number of eight planets. These planets all revolve around axes that are inclined relative to their respective polar axes.
To know more about Planet:
https://brainly.com/question/14581221
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A truck is traveling on a level road. The driver suddenly applies the brakes, causing the truck to decelerate by an amount g/2. This causes a box in the rear of the truck to slide forward. If the coefficient of sliding friction between the box and the truckbed is 2/5, find the acceleration of the box relative to the truck and relative to the road.
Answer:
Truck [tex]\dfrac{g}{10}[/tex]
Road [tex]-\dfrac{g}{10}[/tex]
Explanation:
[tex]a_1[/tex] = Acceleration of truck = [tex]-\dfrac{g}{2}[/tex]
[tex]\mu[/tex] = Coefficient of friction = [tex]\dfrac{2}{5}[/tex]
Frictional force is given by
[tex]f=-\mu mg\\\Rightarrow f=-\dfrac{2}{5}mg\\\Rightarrow ma_2=-\dfrac{2}{5}mg\\\Rightarrow a_2=-\dfrac{2}{5}g[/tex]
Net acceleration is given by
[tex]a=a_2-a_1\\\Rightarrow a=-\dfrac{2}{5}g+\dfrac{g}{2}\\\Rightarrow a=\dfrac{g}{10}[/tex]
The acceleration of the box relative to the truck is [tex]\dfrac{g}{10}[/tex] and [tex]-\dfrac{g}{10}[/tex] relative to the road.
