Answer:
K = 0.076 J
Explanation:
The height of the target, h = 0.860 m
The mass of the steel ball, m = 0.0120 kg
Distance moved, d = 1.50 m
We need to find the kinetic energy (in joules) of the target ball just after it is struck. Let t is the time taken by the ball to reach the ground.
[tex]h=ut+\dfrac{1}{2}at^2\\\\t=\sqrt{\dfrac{2h}{g}}[/tex]
Put all the values,
[tex]t=\sqrt{\dfrac{2\times 0.860 }{9.8}} \\\\=0.418\ s[/tex]
The velocity of the ball is :
[tex]v=\dfrac{1.5}{0.418}\\\\= $$3.58\ m/s[/tex]
The kinetic energy of the ball is :
[tex]K=\dfrac{1}{2}mv^2\\\\K=\dfrac{1}{2}\times 0.0120\times 3.58^2\\\\=0.076\ J[/tex]
So, the required kinetic energy is 0.076 J.
The only variable we found that affects the speed of a wave on a string was the tension of the string. How does this relate to how a musician tunes a stringed instrument?
Answer:
We know that in a string with two fixed points (like the one you will find in a guitar, where the string is "fixed" at the bridge and at the nut) the only thing that defines the speed (the frequency) at which the string vibrates is the tension (the length is also important, but for the 6 strings in the guitar all of them have the same length, but, as you know, when you press in a given fret the note changes, this happens because you are changing the length of the string, finally, the mass of the string is also important, but all the strings have almost the same mass, so we can ignore this).
Then for the open notes (the notes that you play when you don't fret any note) the only thing that defines the note that will sound is the tension of the string.
And a musician can tune the stringed instrument by changing the tension of each string using a tuner, which is the mechanism in the headstock of the instrument). As more tense is the string, higher will be the open note.
Which method of powering a vehiclewill help to reduce air pollution
using oil
using biofuels
using gasoline
using diesel fuel
Answer:
Using biofuels
Explanation:
The emissions of NOx and total VOCs lead to the formation of ozone in the troposphere, the main component of smog. ... Biofuels has a number of health and environmental benefits including improvement in air quality by reducing pollutant gas emissions relative to fossil fuels
Define universal gravitational constant.
Three packing crates of masses, M1 = 6 kg, M2 = 2 kg and M3 = 8 kg are connected by a light string of negligible mass that passes over the pulley as shown. Masses M1 and M3 lies on a 30o incline plane which slides down the plane. The coefficient of kinetic friction on the incline plane is 0.28. A. Draw a free body diagram of all the forces acting in the masses M1 and M2. B. Determine the tension in the string that connects M2 and M3.
Answer:
39.81 N
Explanation:
I attached an image of the free body diagrams I drew of crate #1 and #2.
Using these diagram, we can set up a system of equations for the sum of forces in the x and y direction.
∑Fₓ = maₓ
∑Fᵧ = maᵧ
Let's start with the free body diagram for crate #2. Let's set the positive direction on top and the negative direction on the bottom. We can see that the forces acting on crate #2 are in the y-direction, so let's use Newton's 2nd Law to write this equation:
∑Fᵧ = maᵧ T₁ - m₂g = m₂aᵧNote that the tension and acceleration are constant throughout the system since the string has a negligible mass. Therefore, we don't really need to write the subscripts under T and a, but I am doing so just so there is no confusion.
Let's solve for T in the equation...
T₁ = m₂aᵧ + m₂gT₁ = m₂(a + g)We'll come back to this equation later. Now let's go to the free body diagram for crate #1.
We want to solve for the forces in the x-direction now. Let's set the leftwards direction to be positive and the rightwards direction to be negative.
∑Fₓ = maₓ F_f - F_g sinΘ = maₓThe normal force is equal to the x-component of the force of gravity.
(F_n · μ_k) - m₁g sinΘ = m₁aₓ (F_g cosΘ · μ_k) - m₁g sinΘ = m₁aₓ [m₁g cos(30) · 0.28] - [m₁g sin(30)] = m₁aₓ [(6)(9.8)cos(30) · 0.28] - [(6)(9.8)sin(30)] = (6)aₓ [2.539595871] - [-58.0962595] = 6aₓ 60.63585537 = 6aₓ aₓ = 10.1059759 m/s²Now let's go back to this equation:
T₁ = m₂(a + g)We have 3 known variables and we can solve for the tension force.
T = 2(10.1059759 + 9.8)T = 2(19.9059759)T = 39.8119518 NThe tension force is the same throughout the string, therefore, the tension in the string connecting M2 and M3 is 39.81 N.
The unit of distance used in astronomy is the light year, defined as the distance travelled by light in one calender year. How far away from earth (in km) is a star if its distance is quoted as 10 light years?
Answer:
9.7 trillion kilometers
Explanation:
A student must use an object attached to a string to graphically determine the gravitational field strength near Earth's surface. The student attaches the free end of the string to the ceiling and pulls the object-string system so that the string makes an angle of 5 degrees from the object's vertical hanging position. The student then releases the object from rest and uses a stopwatch to measure the time it takes for the object to make one complete oscillation. Which of the following is the next step that will allow the student to determine the gravitational field strength?
А) Repeat the experiment by adding additional mass to the object for multiple trials
B) Repeat the experiment by changing the length of the string for multiple trials
C) Repeat the experiment by changing the angle that the string makes with the object's vertical hanging position
D) Repeat the experiment by measuring the time it takes to make two oscillations, three oscillations, and additional oscillations for multiple trials
Answer:
B) True. By changing the length get a different period and with a graph you can find the best value of the gravity pull
Explanation:
The student is reacting a simple pendulum experiment where he can determine the value of the relationship of gravity with the expression
T = 2π [tex]\sqrt{\frac{L}{g} }[/tex]
let's analyze each statement
A) False. The mass is not a paramer of the period expression
B) True. By changing the length get a different period and with a graph you can find the best value of the gravity pull
C) False. The angle while it is small does not influence the period
D) True. By changing the number of oscillations the period does not change, so you can get the value of the pull of gravity.
We can see that the expressions B and d are true, the most exact value is obtained using procedure B since the graphs allow to reduce the errors
How fast much an 816kg Volkswagen travel to have the same momentum as (a) a 2650kg Cadillac going 16.0 km/h? (b) a 9080-kg truck also going 16.0 km/hr?
Answer:
(a) v₁ = 51.96 km/h
(b) v₁ = 178 km/h
Explanation:
(a)
For having the same momentum:
m₁v₁ = m₂v₂
where,
m₁ = mass of Volkswagen = 816 kg
v₁ = speed of Volkswagen = ?
m₂ = mass of Cadillac = 2650 kg
v₂ = speed of Cadillac = 16 km/h
Therefore, using these values in the equation, we get:
[tex](816\ kg)v_1 = (2650\ kg)(16\ km/h)\\\\v_1 = (16\ km/h)(\frac{2650\ kg}{816\ kg})[/tex]
v₁ = 51.96 km/h
(b)
For having the same momentum:
m₁v₁ = m₂v₂
where,
m₁ = mass of Volkswagen = 816 kg
v₁ = speed of Volkswagen = ?
m₂ = mass of Truck = 9080 kg
v₂ = speed of Truck = 16 km/h
Therefore, using these values in the equation, we get:
[tex](816\ kg)v_1 = (9080\ kg)(16\ km/h)\\\\v_1 = (16\ km/h)(\frac{9080\ kg}{816\ kg})[/tex]
v₁ = 178 km/h
What is the efficiency of a ramp that is 5.5 m long when used to move a 66 kg object to a height of 110 cm when the object is pushed by a 150 N force .
Answer and I will give you brainiliest
Explanation:
Energy input = F×d = (150 N)(5.5 m) = 825 J
Energy output = mgh = (66 kg)(9.8 m/s^2)(1.10 m) = 711 J
efficiency = [tex]\dfrac{\text{output}}{\text{input}}[/tex]×100% = 86.2%
what is the main limitation of debye huckel theory
Answer:
Explanation:
For very low values of the ionic strength the value of the denominator in the expression above becomes nearly equal to one. In this situation the mean activity coefficient is proportional to the square root of the ionic strength. This is known as the Debye–Hückel limiting law.
What is the internal resistance of a current source with an EMF of 12 V if, when a resistor with an unknown resistance is connected to it, a current of 2 A flows through the circuit? A voltmeter connected to the source terminals shows 8 V.
A 3-kg projectile is launched at an angle of 45o above the horizontal. The projectile explodes at the peak of its flight into two pieces. A 2-kg piece falls directly down and lands exactly 50 m from the launch point. Determine the horizontal distance from the launch point where the 1-kg piece lands.
1517.4 m
Step-by-step explanation:
Since the projectile broke up at the peak of its flight, it already traveled half its initial range so we can find its initial launch velocity [tex]v_0[/tex] from the equation
[tex]\frac{1}{2}R= \dfrac{1}{2} \left(\dfrac{v_0^2}{g}\sin 2\theta_0 \right)[/tex]
where [tex]\theta_0 = 45°[/tex] and [tex]\frac{1}{2}R = 50\:\text{m}[/tex] so we will get [tex]v_0=31.3\:\text{m/s}[/tex]. Next, we can use the equation
[tex]v_y = v_0y - gt = v_0 \sin 45 - gt[/tex]
and since [tex]v_y=0[/tex] at its peak, we get t = 22.1 s. Let's set this aside for a moment and we'll use it later.
At the top of its peak, we can use the conservation law of linear momentum. Let M be the mass if of the original projectile, [tex]m_1[/tex] be the mass of the larger fragment (2 kg) and [tex]m_2[/tex] be the mass of the smaller fragment (1 kg). We can write the conservation law as
[tex]Mv_0x = m_1V_1 + m_2V_2[/tex]
where [tex]V_1\:\text{and}\:V_2[/tex] are the velocities of the fragments immediately after the break up. But we also know that [tex]V_1=0[/tex] so the velocity of [tex]m_2[/tex] can be calculated from the conservation law as
[tex]Mv_0 \cos 45° = m_2V_2[/tex]
or
[tex]V_2 = \dfrac{M}{m_2}v_0 \cos 45° = 66.4\:\text{m/s}[/tex]
Now we can calculate the horizontal distance the smaller fragment traveled after the break up. Recall that the amount of time for it to go up is also the amount of time to get down so the horizontal distance x is
[tex]x = V_2 t = (66.4\:\text{m/s})(22.1\:\text{s})= 1467.4\:\text{m}[/tex]
Therefore, the total distance traveled from the launch point is
[tex]D = 50\:\text{m} + 1467.4\:\text{m}=1517.4\:\text{m}[/tex]
How are hypotheses tested?
Answer:
by making observation hope it's helpful
THE ANSWER!!! Please
Answer:
I think -7 N. Netforce is 3N-10N= -7N
Explanation:
How do solar panels work with conduction, convection and radiation?
Answer:
In the case of a solar thermal panel we are trying to heat above the ambient temperature so conduction and convection will work against us by taking heat from the panel to the out- side world. ... The sun (at 6000 C surface temperature) is hotter than the solar panel so the panel will get hot due to the solar radiation.
Explanation:
Một vật chuyển động tròn đều có chu kì T = 0,25 s. Tính tần số chuyển động f của vật?
Answer:8pi
Explanat:Omega =2pi/T
state what is meant by graviration potential at a point in an orbit 6.5×10^7
Explanation:
The gravitational potential at a point in a gravitational field is the work done per unit mass that would have to be done by some externally applied force to bring a massive object to that point from some defined position of zero potential,
If you pull with your lower leg such that you exert a 90 N force on the cord attached to your ankle, determine the magnitude of the tension force of your hamstring on your leg and the compression force at the knee joint.
This question is incomplete, the missing diagram is uploaded along this answer below.
Answer:
- the magnitude of the tension force exerted by the hamstring muscles on the leg is 990 N
- the magnitude of compression force at the knee joint is 900 N
Explanation:
Given the data in the question and diagram below;
Net torque = 0
Torque = force × lever arm
so
F[tex]_{ConF[/tex] × ( 15.0 in + 1.5 in ) = T[tex]_{HonL[/tex] × 1.5 in
given that F[tex]_{ConF[/tex] = 90 N
90 × ( 15.0 in + 1.5 in ) = T[tex]_{HonL[/tex] × 1.5 in
90 N × 16.5 in = T[tex]_{HonL[/tex] × 1.5 in
T[tex]_{HonL[/tex] = ( 90 N × 16.5 in ) / 1.5 in
T[tex]_{HonL[/tex] = 990 N
Therefore, the magnitude of the tension force exerted by the hamstring muscles on the leg is 990 N
b) magnitude of compression force at the knee joint;
In equilibrium, net force = 0
along horizontal
F[tex]_{FonB[/tex] - T[tex]_{HonL[/tex] + F[tex]_{ConF[/tex] = 0
we substitute
F[tex]_{FonB[/tex] - 990 + 90 = 0
F[tex]_{FonB[/tex] - 900 = 0
F[tex]_{FonB[/tex] = 900 N
Therefore, the magnitude of compression force at the knee joint is 900 N
How fast much an 816kg Volkswagen travel to have the same momentum as (a) a 2650kg Cadillac going 16.0 km/h? (b) a 9080-kg truck also going 16.0 km/hr?
Answer:
Explanation:
From the given information:
the car's momentum = momentum of the truck
∴
(a) 816 kg × v = 2650 kg × 16.0 km/h
v = (2650 kg × 16.0 km/h) / 816 kg
v = 51.96 km/hr
(b) 816 kg × v = 9080 kg × 16.0 km/h
v = (9080 kg × 16.0 km/h) / 816 kg
v = 178.04 km/hr
How does the density of water change when: (a) it is heated from 0o
C to
4o
C; (b) it is heated from 4o
C to 10o
C ?
Answer:
[b] it id heated from 4o
Explanation:
Rank the six combinations of electric charges on the basis of the electric force acting on q1.
a.
q1 = -1nC
q2= +1nC
q3= +1nC
b.
q1 = -1nC
q2= -1nC
q3= -1nC
c.
q1 = +1nC
q2= +1nC
q3= -1nC
d.
q1 = +1nC
q2= -1nC
q3= +1nC
(a) From an atomic point of view, why do you have to heat a solid to melt it? (b) If you have a solid and a liquid at room temperature, what conclusion can you draw about the relative strengths of their inter-atomic forces?
Answer:
A. & B. Heat energy is needed to convert solid into a liquid because heat energy increases the kinetic energy of the particles. The heat energy that it used to change 1 kg of solid into liquid at atmospheric pressure and at its melting point is called the latent heat of fusion.
A meter stick has a mass of 0.30 kg and balances at its center. When a small chain is suspended from one end, the balance point moves 28.0 cm toward the end with the chain. Determine the mass of the chain.
Answer:
M L1 = m L2 torques must be zero around the fulcrum
M = m L2 / L1 = .3 kg * 28 cm / 22 cm = .382 kg
the minimum charge on any object cannot be less than
Answer:
1.6 x 10^{-19} Coulombs
Explanation:
In Physics, the standard unit of measurement of a charge is Coulombs and it's denoted by C. Also, the symbol for denoting a charge is Q.
In Chemistry, electrons can be defined as subatomic particles that are negatively charged and as such has a magnitude of -1.
The minimum charge on any object such as an electron cannot be less than 1.6 x 10^{-19} Coulombs and it's usually referred to as the fundamental unit of charge.
Scientists studying an anomalous magnetic field find that it is inducing a circular electric field in a plane perpendicular to the magnetic field. The electric field strength 1.5 m from the center of the circle is 7 mV/m.
Required:
At what rate is the magnetic field changing?
This question is incomplete, the complete question is;
Scientists studying an anomalous magnetic field find that it is inducing a circular electric field in a plane perpendicular to the magnetic field. The electric field strength 1.5 m from the center of the circle is 7 mV/m.
At what rate is the magnetic field changing?
Answer:
the magnetic field changing at the rate of 9.33 m T/s
Explanation:
Given the data in the question;
Electric field E = 7 mV/m
radius r = 1.5 m
Now, from Faraday law of induction;
∫E.dl = d∅/dt
E∫dl = A( dB/dt )
E( 2πr ) = πr² ( dB/dt )
( 0.007 ) = (r/2) ( dB/dt )
( 0.007 ) = 0.75 ( dB/dt )
dB/dt = 0.007 / 0.75
dB/dt = 0.00933 T/s
dB/dt = ( 0.00933 × 1000) m T/s
dB/dt = 9.33 m T/s
Therefore, the magnetic field changing at the rate of 9.33 m T/s
Question 9 of 10
What causes the different seasons on Earth?
A. The angles at which the suns rays strike the Earth
Ο Ο Ο
B. The distance between Earth and the sun
C. The speed at which the Earth rotates on its axis
O
D. Increasing levels of carbon dioxide in the atmosphere.
SUBMIT
Answer:
B
Explanation:
The seasons are measured in how far or close the earth is to the sun.
Atoms of which two elements could combine with atoms of chromium (Cr) to
form ionic bonds?
O A. F.
B. Au
C. Se
D. Ti
E. Mg
Answer:
D and E
Ionic bonds are formed between metals and non-metals and both of those are metals
Kinematics equations tells us the position of an object under constant acceleration increases linearly with time.
A. True
B. False
Answer:
False.
Explanation:
Suppose that we have an object that moves with constant acceleration A.
Then the acceleration of the object is defined by the equation:
a(t) = A
The acceleration is the rate of change of the velocity, then the velocity equation is given by the integration of the acceleration equation, we will get:
v(t) = A*t + V₀
Where V₀ is the velocity of the object at the time t = 0s.
Now, if we integrate it again, we will get the position equation:
p(t) = (1/2)*A*t^2 + V₀*t + P₀
Where P₀ is the initial position equation.
Here, we can see that the position equation is a quadratic equation (not a linear equation), then the statement is false.
To a man running east at the rate of 3m/s vain appears to fall vertically with a speed of 4m/s. Find the actual speed and direction of rain...
Answer:
The actual speed of the rain is 5 m/s and its direction is -53.13°
Explanation:
The actual speed of the rain V = speed of man, v + speed of rain relative to man, v'.
V = v + v'
We add these vectorially.
Since the man's speed is 3 m/s east, in the positive x - direction, we have v = 3i and the rain's speed is falling vertically at 4 m/s, in the negative y- direction, we have v' = -4j
So, V = v + v'
V = 3i + (-4j)
V = 3i - 4j
So, the magnitude of V which is its speed is V = √(3² + (-4)²) = √(9 + 16) = √25 = 5 m/s
The direction of V, Ф = tan⁻¹(vertical component/horizontal component) = tan⁻¹(-4/3) = tan⁻¹(-4/3) = tan⁻¹(-1.333) = -53.13°
So, the actual speed of the rain is 5 m/s and its direction is -53.13°
occurs when an air mass and its clouds encounter a mountain. This forces the air mass to move from a low elevation to a high elevation as it crosses over the mountain.
Frontal wedging
Orographic lifting
Localized convective lifting
Convergence
Jet streams
Answer:
Localized convective lifting
A ball is thrown into the air with a velocity of 39 ft/s. Its height, in feet, after t seconds is given by s(t) = 39t − 16t2. Find the velocity (in ft/s) of the ball at time t = 1 second.
Answer:
7 ft/s
Explanation:
Applying,
V(t) = ds(t)/dt
Where V(t) = velocity of the ball at a given time
From the question,
Given: s(t) = 39t-16t²
Therefore,
V(t) = ds(t)/dt = 39-32t............. Equation 1
at t = 1 seconds,
Substitute the value of t into equation 1
V(t) = 39-32(1)
V(t) = 39-32
V(t) = 7 ft/s