Answer: [tex]240\ rad/s^2[/tex]
Explanation:
Given
Length of beam [tex]l=2\ m[/tex]
mass of beam [tex]m=5\ kg[/tex]
Two forces of equal intensity acted in the opposite direction, therefore, they create a torque of magnitude
[tex]\tau =F\times l=200\times 2=400\ N.m[/tex]
Also, the beam starts rotating about its center
So, the moment of inertia of the beam is
[tex]I=\dfrac{ml^2}{12}=\dfrac{5\times 2^2}{12}\\\\I=\dfrac{5}{3}\ kg.m^2[/tex]
Torque is the product of moment of inertia and angular acceleration
[tex]\Rightarrow \tau=I\alpha\\\\\Rightarrow 400=\dfrac{5}{3}\times \alpha\\\\\Rightarrow \alpha =240\ rad/s^2[/tex]
A baby carriage is sitting at the top of a hill that is 21 m high. The carriage with the baby weighs 20
kg. The carriage has
energy. Calculate it
Answer:
Energy in carriage (Potential energy) = 4,116 J
Explanation:
Given:
Mass of baby = 20 kg
Height = 21 m
Find:
Energy in carriage (Potential energy)
Computation:
The energy accumulated in an object as a result of its location relative to a neutral level is known as potential energy.
In carriage accumulated energy is potential energy.
Energy in carriage (Potential energy) = mgh
Energy in carriage (Potential energy) = (20)(9.8)(21)
Energy in carriage (Potential energy) = 4,116 J
The elastic energy stored in your tendons can contribute up to 35 % of your energy needs when running. Sports scientists have studied the change in length of the knee extensor tendon in sprinters and nonathletes. They find (on average) that the sprinters' tendons stretch 41 mm , while nonathletes' stretch only 33 mm .
Hello. Your question is incomplete. However, I managed to find it completely on the internet and I realized that you forgot to mention that the question asks you for the maximum energy difference between velovistas and non-athletes, considering that the spring constant for the tendon of the two groups is equal to 33n/mm.
To make this calculation you will need to use Hooke's law, using the formula: ¹/2*K*x², where "K" will be the value of the spring constant for the tendon and "X" will be the value of the sprinter and non-athlete terms.
So for the sprinter we will have the calculation:
¹/2*33*41² -------> 0,5*33*1681 = 27736. 5 Nmm
(To facilitate the calculation, first solve the division of ¹/2 and then multiply 41 by 41, lastly, just multiply all the results.)
For the non-athlete we will have the calculation:
¹/2*33*33² -------> 0,5*33*1089 = 17968. 5 Nmm
(To facilitate the calculation, first solve the division of ¹/2 and then multiply 41 by 41, lastly, just multiply all the results.)
Now, to reach the final result, you only need to subtract the two values presented by the sprinter and the non-athlete.
27736.5 - 17968.5 = 9768 Nmm
When a 20 kg explosive detonates and sends a 5 kilogram piece traveling to the right at 105 m/s
what is the speed and direction of the other 15 kilogram piece of the explosive!
Answer:
speed: 35m/s
direction: left
Explanation:
Assuming the right side is the positive direction:
before explosion:
P = mv = 0
after explosion:
P' = 15P + 5P
(Set the velocity of the 15kg piece after explosion as v1' and the velocity of the 5kg piece after explosion as v2')
P' = 0.75mv1' + 0.25mv2'
P' = (15kg)v' + (5kg)(105m/s)
P' = 525kg/m/s + (15kg)v1'
P = P'
525kg/m/s + (15kg)v1' = 0
(15kg)v1' = -525kg/m/s
v1' = -35m/s
speed = |-35| = 35m/s
direction is to the left since the right side is the positive direction.
A 2.0 kg breadbox on a fric-
tionless incline of angle u 40 is
connected, by a cord that runs over a
pulley, to a light spring of spring con-
stant k 120 N/m, as shown in
Fig. 8-43. The box is released from rest when the spring is unstretched. Assume that the pulley is massless and frictionless. (a) What is the speed of the box when it has moved 10 cm down the in- cline? (b) How far down the incline from its point of release does the box slide before momentarily stopping, and what are the (c) magnitude and (d) direction (up or down the incline) of the box’s acceleration at the instant the box momentarily stops?
The moment of inertia of the club head is a design consideration for a driver in golf. A larger moment of inertia about the vertical axis parallel to the club face provides more resistance to twisting of the club face for off-center hits. The mass of one club head is 200 g and its moment of inertia is 5000 g cm2 . What is the radius of gyration of this club head
Answer:
Explanation:
Moment of inertia I = M k² , where M is mass and k is radius of gyration .
Putting the given values in the equation
5000 = 200 x k²
k² = 25
k = 5 cm .
Radius of gyration is 5 cm .
Calculate the terminal velocity of
the following nain drops faning
through air (a) one with a diameter
of 0.3cm 6 one with a a diameter
of o. Olm. Take the density of
water to be looo Kym3 and the
eis cosity of air to be ixlos pas.
The buoyancy effect of the air
may be ignored)
Two students are sitting 1.50 m apart. One student has a mass of 70.0 kg and
the other has a mass of 52.0 kg. What is the gravitational force between them?
A. 8.01 x 10-9
B. 1.08 x 10-2
C. 2.28 x 10-8
Answer:
B
Explanation:
A proton is moved so that its electric potential energy increases from 4.0 × 10-14 J to 9.0 × 10-14 J. The magnitude of the charge on a proton is 1.602 × 10-19 C.
What is the electric potential difference through which the proton moved?
2.5 × 105 V
3.1 × 105 V
5.6 × 105 V
8.1 × 105 V
Answer:
B. 3.1 × 10^5 V
Explanation:
Answer:
B
Explanation:
e2021
What is the unit of measurement of mass and weight?
Answer:
kilogram
In the International System of Units (SI), the kilogram is the basic unit of mass, and the newton is the basic unit of force. The non-SI kilogram-force is also a unit of force typically used in the measure of weight.
Why is it harder to breathe on a
mountain?
A. The air pressure is so high the lungs can't expand.
B. The air is denser and oxygen can't flow easily into the
lungs.
C. The denser oxygen molecules sink below the
surrounding air.
D. The air is less dense so there are fewer oxygen
molecules.
HELP URGENT PLEASE!!!!!!!
Answer:
I think c I dont know sorry if I'm wrong
What causes the Coriolis effect?
A
The sun's position relative to Earth
B.
Earth's orbit around the sun
с
Moon phases
D
Earth's rotation
Which of the following actions will increase the current induced in a wire by a
magnetic field?
Answer:
The induced current can be increased in the coil in the following ways: By increasing the strength of the magnet. By increasing the speed of the magnet through the coil.
Explanation:
Fairly easy question I’ll give extra points help.
1. third law
2. first law
3. third law
4. second law
A wave has a wavelength of 1.5 meters and period of 0.083s. What is the waves speed?
The augue
1) What will be number of image if the angle
between two mirroro is
a) 45
b:36
A scientist notices that an oil slick floating on water when viewed from above has many different colors reflecting off the surface, making it look rainbow-like (an effect known as iridescence). She aims a spectrometer at a particular spot and measures the wavelength to be 750 nm (in air). The index of refraction of water is 1.33.
Part A: The index of refraction of the oil is 1.20. What is the minimum thickness of the oil slick at that spot? t= 313nm
Part B: Suppose the oil had an index of refraction of 1.50. What would the minimum thickness be now? t=125nm
Answer:
a) The minimum thickness of the oil slick at the spot is 313 nm
b) the minimum thickness be now will be 125 nm
Explanation:
Given the data in the question;
a) The index of refraction of the oil is 1.20. What is the minimum thickness of the oil slick at that spot?
t[tex]_{min[/tex] = λ/2n
given that; wavelength λ = 750 nm and index of refraction of the oil n = 1.20
we substitute
t[tex]_{min[/tex] = 750 / 2(1.20)
t[tex]_{min[/tex] = 750 / 2.4
t[tex]_{min[/tex] = 312.5 ≈ 313 nm
Therefore, The minimum thickness of the oil slick at the spot is 313 nm
b)
Suppose the oil had an index of refraction of 1.50. What would the minimum thickness be now?
minimum thickness of the oil slick at the spot will be;
t[tex]_{min[/tex] = λ/4n
given that; wavelength λ = 750 nm and index of refraction of the oil n = 1.50
we substitute
t[tex]_{min[/tex] = 750 / 4(1.50)
t[tex]_{min[/tex] = 750 / 6
t[tex]_{min[/tex] = 125 nm
Therefore, the minimum thickness be now will be 125 nm
Which device converts electric energy into mechanical energy?
O A. An electromagnet
O B. A motor
O C. A transformer
O D. A generator
Answer:
B motor
Explanation:
How would increasing the pressure of this reaction affect the equilibrium
Explanation:
c because there is element
Answer:
C. H2 and N2 would react to produce more NH3
Explanation:
A.P.E.X
would it be m/s or kg?
Answer:
m.s
Explanation:
The low pressure area near Earth's equator is filled by cool air moving in from
А
Europe and South America
B
the North and South Pole
с
the Prime Meridian
D
the Atlantic and Pacific Ocean
Basic Science!! Helppp
4) Which statement about teamwork is not true?
A) Team members should not have to make personal sacrifices for the success of the team.
B) To be successful, all team members need to agree about how to achieve the goal.
C) To achieve agreement, teams must be able to communicate and negotiate.
D) Team members need to be ready to resolve conflicts in an open and honest way
Answer: A) Team should not have to make personal sacrifices for the success of the team.
Explanation:
A bullet of mass 4.00 g is fired horizontally into a wooden block of mass 1.30 kg resting on a horizontal surface. The coefficient of kinetic friction between block and surface is 0.170. The bullet remains embedded in the block, which is observed to slide a distance 0.240 m along the surface before stopping. Part A What was the initial speed of the bullet
Answer:
[tex]291.67\ \text{m/s}[/tex]
Explanation:
[tex]m_1[/tex] = Mass of bullet = 4 g
[tex]m_2[/tex] = Mass of block = 1.3 kg
[tex]\mu[/tex] = Coefficient of friction = 0.17
[tex]s[/tex] = Displacement of block = 0.24 m
[tex]v_1[/tex] = Velocity of bullet
[tex]v[/tex] = Velocity of combined mass
[tex]g[/tex] = Acceleration due to gravity = [tex]9.81\ \text{m/s}^2[/tex]
The energy balance of the system is given by
[tex]\dfrac{1}{2}(m_1+m_2)v^2=\mu(m_1+m_2)gs\\\Rightarrow v=\sqrt{2\mu gs}[/tex]
As the momentum is conserved in the system we have
[tex]m_1v_1=(m_1+m_2)v\\\Rightarrow m_1v_1=(m_1+m_2)\sqrt{2\mu gs}\\\Rightarrow v_1=\dfrac{(m_1+m_2)\sqrt{2\mu gs}}{m_1}\\\Rightarrow v_1=\dfrac{(4\times 10^{-3}+1.3)\times \sqrt{2\times 0.17\times 9.81\times 0.24}}{4\times 10^{-3}}\\\Rightarrow v_1=291.67\ \text{m/s}[/tex]
The initial speed of the bullet is [tex]291.67\ \text{m/s}[/tex].
What is the order of the events for the water cycle on a typical warm day?
А
rain, snow, sleet
B
precipitation, evaporation, rain
с
evaporation, condensation, precipitation
D
condensation, evaporation, precipitation
Which of the following happens to
density as air pressure decreases?
С C
A. Density increases.
B. Density stays the same.
C. Density decreases.
D. There is no correlation between air pressure and
density.
Explanation:
As pressure increases, with temperature constant, density increases. Conversely when temperature increases, with pressure constant, density decreases. Air density will decrease by about 1% for a decrease of 10 hPa in pressure or 3 °C increase in temperature.
Can someone tell me anything useful about energy management in the human body?
Answer:
The human body carries out its main functions by consuming food and turning it into usable energy. Immediate energy is supplied to the body in the form of adenosine triphosphate (ATP). Since ATP is the primary source of energy for every body function, other stored
Explanation:
this what teacher explain to us
Please help me with this review question.
Answer:
28.7%
Explanation:
efficiency = work output /work input × 100
Convert (a) 50 oF, (b) 80 oF, (c) 95 oF to Celsius
Predicted height and total energy
Answer:
The predicted height is 2.809 meters, writing this in centimeters we get (1m = 100cm):
h = 2.809 m = (2.809)*(100cm) = 280.9 cm
And the total energy is:
E = 6.696 J
Explanation:
First let's see the problem.
We have an object of mass m = 274g which is thrown upwards with an initial velocity v0 = 6.991 m/s, in a place with a gravitational acceleration of g = 8.7 m/s^2
When the object is on the air, the only force acting on it will be the gravitational force, then the acceleration of the object will be equal to the gravitational acceleration, then we can write:
a(t) = -8.7 m/s^2
Where the negative sign is because this acceleration points down.
Now to get the velocity of the object we can integrate over time to get:
v(t) = (-8.7 m/s^2)*t + v0
Where v0 is a constant of integration, which is the initial velocity, then we can write this as:
v(t) = (-8.7 m/s^2)*t + 6.991 m/s
Now we can integrate again over the time to get the position equation.
p(t) = (1/2)*(-8.7 m/s^2)*t^2 + (6.991 m/s)*t + p0
Where p0 is the initial position, because the ball is being thrown from the ground, the initial position is 0.
Then the position equation is:
p(t) = (1/2)*(-8.7 m/s^2)*t^2 + (6.991 m/s)*t
Ok, now we know all the movement equations for the object.
The first thing we want to know is the maximum height of the object.
We know that the object reaches its maximum height when the velocity is zero (this is, the velocity stops being positive, meaning that the object stops going up, then in that time we have the maximum height)
We need to solve:
v(t) = 0m/s = (-8.7 m/s^2)*t + 6.991 m/s
(8.7 m/s^2)*t = 6.991 m/s
t = 6.991 m/s/( (8.7 m/s^2) = 0.804 seconds
The maximum height of the object is given by:
p(0.804s) = (1/2)*(-8.7 m/s^2)*(0.804)^2 + (6.991 m/s)*(0.804) = 2.809 m
The maximum height of the object is 2.809 meters.
Now let's find the maximum energy.
Remember that the energy of an object can be written as the sum of the potential energy U and the kinetic energy K.
E = K + U
Such that for an object of mass m and velocity v, the kinetic energy is:
K = (1/2)*m*v^2
And for an object of mass m, at a height h from the ground and with gravitational acceleration g, the potential energy is:
U = m*g*h
Now, when the object is at its maximum height, the velocity is zero.
Then K = 0
And for conservation of energy, the total energy of the object becomes potential energy.
E = 0 + U
E = U
So if we find the potential energy at the maximum height of the object's path, we can find the total energy of the object.
We know that:
mass = m = 274g = 0.274 kg (here i used that 1kg = 1000g)
height = h = 2.809 meters.
gravitational acceleration = g = 8.7 m/s^2
Then the potential energy at this point is:
U = 0.274 kg*(2.809 meters)*(8.7 m/s^2) = 6.696 J
This means that the total energy of the object is:
E = 6.696 J