W: work
F: force required to lift load; F = 800 N
d: distance lifted: d = 40 m
W = Fd = 800*40
W = 32000 J
If 100000J were exerted on a load for which only 32000J was necessary, the other 100000-32000 = 68000J was not transferred to the load.
68000/100000 = 0.68 = 68%
URGENT!! ILL GIVE
BRAINLIEST!!!! AND 100 POINTS!!!!!!
What type of energy transfer is shown in this image?
A. electrical to mechanical
B. chemical to electrical
C. electrical to chemical
D. chemical to mechanical
Answer:
The Correct Answer Is b.
The magnetic field at point P is zero. Findthe distance, r2, from P to the second wire,1₂.I₁ = 4.32 A P------0.831 m1₂ = 7.50 Ar2r₂ = [?] m
The formula for calculating the magnetic field is
B = μo[(I1/2πR1) - (I2/2πR2)]
B = μo/2π[(I1/R1) - (I2/R2)]
where
B is the magnetic field
μo is the permeability of free space
From the information given,
B = 0
I1 = 4.32
R1 = 0.831
I2 = 7.5
R2 = ?
μo = 4π x 10^-7 Tm/A
Thus,
0 = 4π x 10^-7/2π[(4.32/0.831) - (7.5/R2)]
0 = 4 x 10^-7/2[(4.32/0.831) - (7.5/R2)]
0 = 4 x 10^-7/2(4.32/0.831) - 4 x 10^-7/2(7.5/R2)]
4 x 10^-7/2(7.5/R2)] = 4 x 10^-7/2(4.32/0.831)
7.5/R2 = 4.32/0.831
By cross multiplying,
4.32R2 = 7.5 x 0.831
R2 = (7.5 x 0.831)/4.32
R2 = 1.443 m
A girl holds an arrow in her hands, ready to shoot the arrow at a dragon who is just outside her arrow’s range of 45 meters. The girl then slides down a cliffside, giving her a velocity of 8.5 m/s at an angle of 45 below the horizontal. She points her bow 30 degrees above the horizontal and shoots, the arrow leaves the bow with a velocity of 45 m/s. The girl is 55 meters away from the dragon horizontally. How far above the girl is the dragon? (Assuming the arrow hits.)I already figured out how long it takes for the arrow to reach the dragon (4.72 seconds) but I don’t know how to calculate the distance between the girl and the dragon. Please help!
First, assume that the arrow leaves the bow with a velocity of 45 m/s above the horizontal with respect to the bow.
Since the bow is moving at 8.5 m/s 45º below the horizontal, find the initial velocity of the arrow with respect to the ground:
[tex]\vec{v}_{AG}=\vec{v}_{AB}+\vec{v}_{BG}[/tex]This equation reads:
The velocity of the arrow with respect to the ground is equal to the velocity of the arrow with respect to the bow plus the velocity of the bow with respect to the gound.
Notice that this is a vector equation. Then, the vertical and horizontal components of the velocities must be added separately:
[tex]\begin{gathered} v_{AG-x}=v_{AB-x}+v_{BG-x} \\ v_{AG-y}=v_{AB-y}+v_{BG-y} \end{gathered}[/tex]Find the vertical and horizontal components of the velocity of the arrow with respect to the bow and the velocity of the bow with respect to the ground:
[tex]\begin{gathered} v_{AB-x}=v_{AB}\cos (\theta) \\ =45\frac{m}{s}\cdot\cos (30º) \\ =38.97\frac{m}{s} \end{gathered}[/tex][tex]\begin{gathered} v_{AB-y}=v_{AB}\sin (\theta) \\ =45\frac{m}{s}\sin (30º) \\ =22.5\frac{m}{s} \end{gathered}[/tex]Similarly, for the velocity of the bow with respect to the ground:
[tex]\begin{gathered} v_{BG-x}=6.01\frac{m}{s} \\ v_{BG-y}=-6.01\frac{m}{s} \end{gathered}[/tex]Then, the vertical and horizontal components of the initial velocity of the arrow with respect to the ground, are:
[tex]\begin{gathered} v_{AG-x}=38.97\frac{m}{s}+6.01\frac{m}{s}=44.98\frac{m}{s} \\ \\ v_{AG-y}=22.5\frac{m}{s}-6.01\frac{m}{s}=16.49\frac{m}{s} \end{gathered}[/tex]Use the horizontal component of the velocity to find how long it takes for the arrow to travel a horizontal distance x of 55 meters. Then, use that time to find the vertical position of the arrow.
Since the horizontal movement of the arrow is uniform, then:
[tex]v_{AG-x}=\frac{x}{t}_{}[/tex]Isolate t and substitute x=55m, v_{AG-x}=44.98 m/s:
[tex]\begin{gathered} t=\frac{x}{v_{AG-x}} \\ =\frac{55m}{44.98\frac{m}{s}} \\ =1.2227s \end{gathered}[/tex]The vertical motion of the arrow is a uniformly accelerated motion. Then, the vertical position is given by:
[tex]y=v_{AG-y}t-\frac{1}{2}gt^2[/tex]Replace v_{AG-y}=16.49 m/s, t=1.2227s and g=9.81 m/s^2 to find the vertical position of the arrow when the horizontal position is 55 meters. This matches the elevation of the dragon with respect to the girl when the girl shoots:
[tex]\begin{gathered} y=(16.49\frac{m}{s})(1.2227s)-\frac{1}{2}(9.81\frac{m}{s^2})(1.2227s)^2 \\ =12.829\ldots m \\ \approx12.8m \end{gathered}[/tex]Therefore, the dragon is 12.8 meters above the girl when the arrow is shoot.
I need full explanation on how to solve both these questions I don't understand haha
Given that weight of 1 kg is equivalent to the weight of 2.2 lb.
The weight of one kg or 2.2 lbF is 9.8 N
(a)
The weight of 2.2 lbF is 9.8 N
Thus the force of 1.0 lbF is,
[tex]\begin{gathered} 1.0\text{ lbF}=\frac{1.0\text{ lbF}\times9.8\text{ N}}{2.2\text{ lbF}} \\ =4.45\text{ N} \end{gathered}[/tex]Thus the force of the weight of 1.0 lbF is 9.8 N
(b)
If the thrusters are meant to use the value in N but used it in lbF, then the trusters would have used, say, 9.8 lbF in place of 9.8 N.
The force of 9.8 lbF is greater than the force of 9.8 N. Thus the force applied to slow the craft is higher than intended.
Thus they would slow the craft more.
(c)
Yes, the failure is correctly explained by the unit mix-up. As it is explained in part b, the force applied will be higher than intended. Which is exactly the case.
Thus the failure is correctly explained by the mix-up.
A ship's wheel has a moment of inertia of 0.930 kilogram·meters squared. The inner radius of the ring is 26 centimeters, and the outer radius of the ring is 32 centimeters. Disagreeing over which way to go, the captain and the helmsman try to turn the wheel in opposite directions. The captain applies a force of 314 newtons at the inner radius, while the helmsman applies a force of 290 newtons at the outer radius. What is the magnitude of the angular acceleration of the wheel?
We can use the formula of the moment of inertia given by:
[tex]r\cdot F=I\alpha[/tex]Where:
r = Distance from the point about which the torque is being measured to the point where the force is applied
F = Force
I = Moment of inertia
α = Angular acceleration
So:
[tex]\begin{gathered} r\cdot F=(-0.26\times314+290\times0.32)=92.8-81.64=11.16 \\ I=0.930 \\ so,_{\text{ }}solve_{\text{ }}for_{\text{ }}\alpha: \\ \alpha=\frac{r\cdot F}{I} \\ \alpha=\frac{11.16}{0.930} \\ \alpha=\frac{12rad}{s^2} \end{gathered}[/tex]Answer:
12 rad/s²
The diagram below shows a block on a horizontal, frictionless surface. A 100.-newton force acts on the block at an angle of 30.0° above the horizontal. 100. N F 30.0° Block 77 Frictionless Surface What is the magnitude of force F, to the nearest tenth of a newton, if it establishes equilibrium?
ANSWER:
STEP-BY-STEP EXPLANATION:
You have a light spring which obeys Hooke's law. This spring stretches 3.02 cm vertically when a 2.50 kg object is suspended from it. Determine the following.(a) the force constant of the spring (in N/m)N/m(b) the distance (in cm) the spring stretches if you replace the 2.50 kg object with a 1.25 kg objectcm(c) the amount of work (in J) an external agent must do to stretch the spring 8.90 cm from its from unstretched positionJ
We are given that a spring stretches 3.02 cm vertically when a 2.5 kg object is suspended.
Part (a) To determine the constant of the spring we need first to determine the weight of the object, to do that we will use the following formula:
[tex]W=mg[/tex]Where:
[tex]\begin{gathered} m=\text{ mass} \\ g=\text{ acceleration of gravity} \end{gathered}[/tex]Now we plug in the values:
[tex]W=(2.5\operatorname{kg})(9.8\frac{m}{s^2})[/tex]Solving the operations we get:
[tex]W=24.5N[/tex]Now we use Hooke's law:
[tex]F=kx[/tex]Where:
[tex]\begin{gathered} F=\text{ force} \\ k=\text{ spring contant} \\ x=\text{ distance stretched} \end{gathered}[/tex]Now we solve for "k" by dividing both sides by "x":
[tex]\frac{F}{x}=k[/tex]Now, since the object is placed vertically this means that the only force acting on the spring in the weight of the object, therefore:
[tex]F=W[/tex]Now we plug in the known values:
[tex]\frac{24.5N}{3.02\operatorname{cm}}=k[/tex]Solving the operations we get:
[tex]8.11\frac{N}{\operatorname{cm}}=k[/tex]Since we are required to express the constant in N/m, we need to convert the centimeters into meters. To do that we will use the following conversion factor:
[tex]100\operatorname{cm}=1m[/tex]Now we multiply by the conversion factor in decimal form, placing the centimeters as numerator:
[tex]8.11\frac{N}{\operatorname{cm}}\times\frac{100\operatorname{cm}}{1m}=811.26\frac{N}{m}[/tex]Therefore, the constant of the spring is 811.26 N/m.
part (b) Now we are asked to determine the distance is an object of 1.25 kg is place. First, we determine the weight of the new object:
[tex]W=mg[/tex]Now we plug in the values:
[tex]W=(1.25\operatorname{kg})(9.8\frac{m}{s^2})=12.25N[/tex]Now we use Hooke's law, but we solve for the distance "x" by dividing both sides by the constant "k", we get:
[tex]\frac{F}{k}=x[/tex]Just as before, the only force acting is the weight, therefore, we plug in the values we got:
[tex]\frac{12.25N}{811.26\frac{N}{m}}=x[/tex]Solving the operations:
[tex]0.015m=x[/tex]Now we convert the meters into centimeters using the same conversion factor:
[tex]0.015m\times\frac{100\operatorname{cm}}{1m}=1.5\operatorname{cm}[/tex]Therefore, the new mass stretches the spring 1.5 centimeters.
Part (c) Now we are asked to determine the work that has to be done on the spring to stretch it 8.9 centimeters. To determine that we will use the following formula for the work done on a spring:
[tex]w=\frac{1}{2}kx^2[/tex]Now we replace the values:
[tex]w=\frac{1}{2}(811.26\frac{N}{m})(8.9\operatorname{cm})^2[/tex]We will first convert the 8.9 centimeters into meters:
[tex]8.9\operatorname{cm}\times\frac{1m}{100\operatorname{cm}}=0.089m[/tex]Now we replace this in the formula for the work:
[tex]w=\frac{1}{2}(811.26\frac{N}{m})(0.089m)^2[/tex]Solving the operations we get:
[tex]w=3.21J[/tex]Therefore, the work is 3.21 Joules.
The melting point of ethyl alcohol is -129 F. What is the Celsius reading?
We are asked to convert -129 F to Celsius. To do that we will use the following formula:
[tex]C=\frac{5}{9}(F-32)[/tex]Where:
[tex]\begin{gathered} C=\text{ reading in Celsius} \\ F=\text{ reading in Fahrenheit} \end{gathered}[/tex]Now we plug in the reading in Fahrenheit:
[tex]C=\frac{5}{9}(-129-32)[/tex]Solving the operations:
[tex]C=-89.44\text{Celsius}[/tex]Therefore, the Celsius reading is -89.44°C.
A cord is wrapped around the rim of a wheel 0.230 m in radius, and a steady pull of 37.0 N is exerted on the cord. The wheel is mounted on frictionless bearings on a horizontal shaft through its center. The moment of inertia of the wheel about this shaft is 4.60 kg⋅m2.
The angular acceleration of the wheel is 1.85m/s²
Acceleration is the rate at which speed and direction of velocity vary over time. Something is said to be accelerating when it starts to move faster or slower.
We are given that,
The pull force of the wheel = τ = 37.0N
The radius of the wheel = r = 0.230m
The moment of inertia = I = 4.60kg-m²
Here the angular acceleration = α = ?
The formula for angular acceleration can be given as,
α = τ /I
α = [(37.0N)×( 0.230m)]/4.60kg-m²
α = 1.85 m/s²
Hence , the angular acceleration of the wheel would be 1.85m/s²
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A ray of light is traveling through a mineral sample is submerged inwater. The ray refracts as it enters the water, as shown in the diagrambelow.NormalWater41°149°63°27°MineralCalculate the absolute index of refraction of the mineral.
We are asked to determine the absolute index of refraction of a mineral submerged in water. To do that we will use Snell's law:
[tex]n_1\sin \theta_1=n_2\sin \theta_2[/tex]Where n1 and n2 are the refraction indices of water and mineral respectively and the angles "theta 1" and "theta 2" are the incidence and refraction angles. We will solve for n1:
[tex]n_1=\frac{n_2\sin \theta_2}{\sin \theta_1}[/tex]Replacing the values:
[tex]n_1=\frac{(589.29nm)\sin 41}{\sin 27}[/tex]Solving the operations:
[tex]n_1=851.58nm[/tex]Therefore, the index of refraction of the mineral is 851.58 nm.
Which statement describes the sign of gravitational force?(1 point)
A Gravitational force is negative because it is attractive.
B Gravitational force is negative because it is repulsive.
C Gravitational force is positive because it is attractive.
D Gravitational force is positive because it is repulsive
The statement that describes the sign of gravitational force is as follows: Gravitational force is negative because it is attractive (option A).
What is gravitational force?Gravitational force is a very long-range, but relatively weak fundamental force of attraction that acts between all particles that have mass. It is believed to be mediated by gravitons.
According to the Newton’s law of gravitation, which explains the gravitational force, every particle in the universe attracts every other particle with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.
The negative sign in the gravitational force equation indicates that the gravitational force interaction is always attractive.
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At what speed must a satellite be travelling so that it shall remain in a circular orbit 1683049 m above the surface of the Earth. Take the mass of the Earth as 6.0 × 1024 kg
Given data
*The given mass of the Earth is M_e = 6.0 × 10^24 kg
*The distance above the surface of the Earth is r = 1683049 m
The formula for the speed of the satellite must be traveling so that it shall remain in a circular orbit is given as
[tex]v=\sqrt[]{\frac{GM_e}{r}}[/tex]Substitute the known values in the above expression as
[tex]\begin{gathered} v=\sqrt[]{\frac{(6.67\times10^{-11})(6.0\times10^{24})}{(1683049)}} \\ =1.54\times10^4\text{ m/s} \end{gathered}[/tex]Hence, the speed of the satellite must be traveling so that it shall remain in a circular orbit is v = 1.54 × 10^4 m/s
If it takes 0.8 s for your voice to be heard at a distance of 272 m, what is the temperature of the air?
V = 331 + 0.59 Tc
v= d/t = 272 / 0.8 = 340 m/s
340 = 331 + 0.59 Tc
340 - 331 = 0.59 TC
9 = 0.59 Tc
9/0.59 = Tc
Tc = 15.25°C
Equations of linear motion
The Newton's equations of motion are stated below
v = u + at
s = ut 1/2at^2
v^2 = u^2 + 2as
s = 1/2(u + v)t
where
u represents initial velocity
v represents final velocity
t represents time
a represents acceleration
which Vector goes from 5, 5 to 4, 0
The required vector is î + 5ĵ
Those physical quantities that have magnitude, as well as direction, are represented by a Vector. Some examples of Vector quantities are Force, Acceleration, Velocity, and Torque.
According to the given information, we represent the points in vector form,
The 1st vector is [tex]R_{1}[/tex] = 5î + 5ĵ
And the Second Vector is [tex]R_{2}[/tex] = 4î
Now, it is given that the required vector goes from 5,5 to 4,0.
So, the required vector is R,
Thus, R = (5-4)î + (5-0)ĵ
R = î + 5ĵ
Hence, the required vector is î + 5ĵ
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Answer:
Vector B
Explanation:
Tsunami waves flood coastal and inland areas and affect coastal life. Which of these properties of tsunami waves most contributes to the flooding? The low friction of waves at the shoreline The high friction of waves at the shoreline The high amplitude of the waves at origination The huge volume of water that surges across shore
Take into account that the flooding produced by Tsunami waves has a great level of energy.
Moreover, consider that the amplitude of the waves determines the energy of the waves and then, the flooding is produced primarily by the high amplitude of the waves at origination.
A force of 585 N is exerted on a 407 kg mass a distance of 13660 km above the surface of a planet having a mass of 7.9E24 kg. Determine the average density of the planet in kg/cubic meter. Derive and express algebraic solution in terms of givens: F, m, mp, alt and G.
The average density of a planet is given by:
[tex]\rho=\frac{m}{V}[/tex]where m is the mass of the planet and V is its volume. We know the mass of the planet but we don't know its volume, to find it we will need find its radius.
To find the radius of the planet we can use Newton's Law of gravitation:
[tex]F=G\frac{mM}{d^2}[/tex]where G is the gravitational constant, m is the mass of the object, M is the mass of the planet and d is the distance between the planet and the object. Let r be the radius of the planet, and x be the distance from the surface of the planet to the object (x=13660 in this case); then we have:
[tex]\begin{gathered} F=G\frac{mM}{(r+x)^2} \\ (r+x)^2=\frac{GmM}{F} \\ r+x=\pm\sqrt[]{\frac{GmM}{F}} \\ r=-x\pm\sqrt[]{\frac{GmM}{F}} \end{gathered}[/tex]Plugging the values given we have:
[tex]\begin{gathered} r=-13660\times10^3\pm\sqrt[]{\frac{(6.67\times10^{-11})(407)(7.9\times10^{24})}{585}} \\ \text{ Using the positive root we have:} \\ r=5.49\times10^6 \\ \text{ Using the negative root we have:} \\ r=-3.28\times10^7 \end{gathered}[/tex]Since the radius of the planet has to be positive we choose the positive solution.
Now, that we know the radius of the planet we can calculate its volume; assuming the planet is spherical we have that:
[tex]V=\frac{4}{3}\pi r^3[/tex]then we have:
[tex]\begin{gathered} V=\frac{4}{3}\pi(5.49\times10^6)^3 \\ V=6.92\times10^{20} \end{gathered}[/tex]Finally we can calculate the density:
[tex]\begin{gathered} \rho=\frac{7.9\times10^{24}}{6.92\times10^{20}} \\ \rho=11398 \end{gathered}[/tex]Therefore, the average density is 11398 kg/m^3
A crate is at rest on an inclined plane. As the slope increases the crate remains at rest until the incline reaches an angle of 32.7° from the horizontal. At this angle the crate begins to slidedown the ramp.Apply Newton's 1st or 2nd law to each axis of the Free body diagram.
The free body diagram of the crate can be shown as,
According to free body diagram, the net force acting on the crate is,
[tex]F=mg\sin \theta-f\ldots\ldots\text{ (1)}[/tex]The frictional force acting on the crate can be given as,
[tex]f=\mu N[/tex]According to free body diagram, the normal force acting on the crate is,
[tex]N=mg\cos \theta[/tex]Therefore, the frictional force becomes,
[tex]f=\mu mg\cos \theta[/tex]According to Newton's second law, the net force acting on the crate is,
[tex]F=ma[/tex]Therefore, equation (1) becomes,
[tex]\begin{gathered} ma=mg\sin \theta-\mu mg\cos \theta \\ a=g\sin \theta-\mu g\cos \theta \end{gathered}[/tex]Thus, the net acceleration of the crate can be expressed using Newton's second law as,
[tex]a=g\sin \theta-\mu\cos \theta[/tex]A standing wave is produced by reflecting a wave off a wall (which acts like a fixed end). If the standing wave consists of 6 anti-nodes and 7 nodes and the wall is 12m away from the source determine the wavelength.
Answer:
4 m
Explanation:
We can represent the standing wave as follows
So, we can divide the total distance into 6 parts as follows
12 m/6 = 2 m
Then, 2 m is the distance from node to node. It means that the wavelength will be twice this distance, so
wavelength = 2 x 2m = 4m
Therefore, the wavelength is 4m
What is physics and explain elements of physics
Physics is the branch of science that deals with mass , time , space, energy and their inter-relations .
The application of
How would you go about answering this question?
The new angular velocity becomes 0.286 rev/s.
Given parameters:
Mass of the marry go round: M = 120 kg.
Radius of the marry go round: r = 1.80 m.
Mass of the boy: m = 27 kg.
Initial angular velocity of the marry go round: ω₁ = 0.350 rev/s.
Final angular velocity of the marry go round: ω₂ = ?
From the principle of conservation of angular momentum;
Initial angular momentum = final angular momentum
⇒ 1/2 Mr²ω₁ = 1/2( M+m)r²ω₂
⇒ ω₂ = Mω₁/(m+M)
= 120×0.35/(120+27) rev/s
= 0.286 rev/s.
So, final angular velocity of the marry go round: ω₂ = 0.286 rev/s.
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1)A weight lifter raises a 180 kg barbell to a height of 1.95 m. What is the increase in the potential energy of the barbell? 2)A 20.0 kg rock is on the edge of a 100 m cliff. What is the velocity of the rock just before it hits the grounds when it is pushed off the cliff?
Question 1.
Given:
Mass of barbell = 180 kg
Height = 1.95 m
Let's find the increase in the potential energy of the barbell.
To find the potential energy, apply the formula:
PE = mgh
Where:
m is the mass = 180 kg
g is acceleration due to gravity = 9.8 m/s²
h is the height = 1.95 m
Input values into the formula and solve:
PE = 180 x 9.8 x 1.95 = 3439.8 J
Therefore, the increase in the potential energy of the barbell is 3439.8 J
ANSWER:
3439.8 J
_7. What happens to the total resistance of the resistors if they are connected inparallel? A. increases C. decreases B. remains the same D. increases then decreases
In order to find what happens to the total resistance when resistors are connected in parallel, let's analyze the formula below for the total resistance between two resistors in parallel:
[tex]\begin{gathered} \frac{1}{RT}=\frac{1}{R_1}+\frac{1}{R_2}\\ \\ RT=\frac{R_1R_2}{R_1+R_2} \end{gathered}[/tex]When two resistors are in parallel, they have the same voltage.
That means more current will flow, because there will be current flowing from each resistor and the voltage is the same.
Calculating the total resistance, we will find a smaller resistance than the two resistors used, because with the same voltage, the current is greater.
Therefore the correct option is C.
An object is placed 3.00 cm in front of a concave lens with a focal length of 5.00 cm. Find the image distance.
ANSWER:
-1.88 cm
STEP-BY-STEP EXPLANATION:
The situation would be the following:
Given:
Object distance (u): -3 cm
Focal distance (f): -5 cm
Using the lens formula, we calculate the distance of the image:
[tex]\begin{gathered} \frac{1}{v}-\frac{1}{u}=\frac{1}{f} \\ \\ \text{We replacing} \\ \\ \frac{1}{v}-\frac{1}{-3}=\frac{1}{-5} \\ \\ \frac{1}{v}\cdot \:15v-\frac{1}{-3}\cdot \:15v=\frac{1}{-5}\cdot \:15v \\ \\ 15+5v=-3v \\ \\ 5v=-3v-15 \\ \\ 8v=-15 \\ \\ v=-\frac{15}{8} \\ \\ v=-1.88\text{ cm} \end{gathered}[/tex]The image distance is -1.88 cm
Answer:
Explanation:
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In a tug-of-war two teams are pulling in opposite directions, but neither team ismoving. What do the net forces equal in this example?
so, the net forces equal zero
Explanation
Step 1
Draw the situation
Newton's first law says that if the net force on an object is zero, then that object will have zero acceleration
in this case due to neither team is moving, we can say the center point is at rest,so its acceleration is zero
According to Newton's law, as the object is at rest, the sum of the forces acting on it equals zero
so
sum of forces = 0
the forces are due to the teams pulling so,
[tex]\begin{gathered} \text{Force team 1 -force team 2= 0} \\ \text{if we move the second term to the rigth} \\ \text{Force team 1=Force team }2 \end{gathered}[/tex]so, the net forces equal zero
Yea I think and this the other day and
Given:
The radius of the tank is: R = 7.5 feet.
Volume rate is: dV/dt = 60 cubic feet per hour.
h = 2.8 feet.
To find:
V, R, h, dV/dt, dR/dt, dh/dt
Explanation:
The given expression for volume is:
[tex]V=\pi h^2(R-\frac{1}{3}h)[/tex]Substituting the values in the above equation, we get:
[tex]V=3.14\times2.8^2\times(7.5-\frac{1}{3}\times2.8)=161.74\text{ cubic feet}[/tex]The volume is 161.74 cubic feet.
The radius of the tank is given as: R = 7.5 feet
The height of the water in the tank is given as: h = 2.8 feet
The volume rate of water is given as: dV/dt = 60 cubic feet per hour
As the radius of the tank is constant, its derivative will be zero. Thus, dR/dt = 0.
dh/dt can be calculated by differentiating the given volume equation with respect to time as:
[tex]\begin{gathered} \frac{dV}{dt}=\pi^2\times\frac{dh^2}{dt}\times(R-\frac{1}{3}\frac{dh}{dt}) \\ \\ \frac{dV}{dt}=\pi^2\times\frac{2dh}{dt}\times(R-\frac{1}{3}\frac{dh}{dt}) \\ \\ \frac{dV}{dt}=2\pi^2R\times\frac{dh}{dt}-\frac{2\pi^2}{3}\times(\frac{dh}{dt})^2 \end{gathered}[/tex]Substituting the values in the above equation, we get:
[tex]\begin{gathered} 60=2\pi^2\times7.5\times\frac{dh}{dt}-\frac{2\pi^2}{3}\times(\frac{dh}{dt})^2 \\ \\ 60=148.044\frac{dh}{dt}-6.580(\frac{dh}{dt})^2 \\ \\ 6.580(\frac{dh}{dt})^2-148.044\frac{dh}{dt}+60=0 \end{gathered}[/tex]Solving the above quadratic equation, we get:
dh/dt = 0.412 ft/hr or dh/dt = 22.086 ft/hr.
Final answer:
V = 161.74 cubic feet
R = 7.5 feet
h = 2.5 feet
dV/dt = 60 cubic feet per hour
dR/dt = 0
dh/dt = 22.086 ft/hr or 0.412 ft/hr
A 6 kg sign is suspended by two strings making angles with the ceiling as shown in the diagram. Determine the magnitudes of the tensions in each string. Remember that 1kg exerts a force of 9.8 N. please use sine/cosine law for this question
The triangle representing the given scenario is shown below
T1 and dT2 represents the tension in each spring
We would apply the sine rule which is expressed as
a/SinA = b/SinB = c/SinC
where
A, B and C are the angles of the triangle
a, b and c are the sides opposite the respective angles
Thus,
A = 42
a = T1
B = 23
b = T2
C = 115
c = 58.8
T1/sin42 = 58.8/sin115
By cross multiplying, it becomes
T1sin115 = 58.8sin42
T1 = 58.8sin42/sin115
T1 = 43.41
T2/sin23 = 58.8/sin115
By cross multiplying, it becomes
T2sin115 = 58.8sin23
T2 = 58.8sin23/sin 115
T2 = 25.35
The magnitude of the tension in each string are 43.41 N and 25.35 N
what is this math problem Solve for x.
0.5x = 4.5
A ball is moving at 4 m/s and has momentum 48 kg.m/s. What is the ball's mass (in kilograms)?
Given:
The speed of the ball is v = 4 m/s
The momentum of the ball is p = 48 kg m/s
Required: The mass of the ball
Explanation:
The mass of the ball can be calculated as
[tex]\begin{gathered} p=mv \\ m=\frac{p}{v} \\ m=\frac{48\text{ kg m/s}}{4\text{ m/s}} \\ =12\text{ kg} \end{gathered}[/tex]Final Answer: The mass of the ball is 12 kg.
Two billiard balls collide elastically ball A has a mass of 1.22 kg and an initial velocity of 0.8 M/S Ball B has a mass of 1.37 KG and an initial velocity of 0 M/S assuming that all of ball A’s is momentum is transferred to ball B after the collision what is the final velocity of Ball B ? *a perfect momentum transfer means that what was moving before the collision will not move after the collision and vice versa.A) 0.377 m/sB) 0.976 m/sC) 0.712 m/sD) 0 m/s
We will have the following:
[tex]\begin{gathered} (1.22kg)(0.8m/s)+(1.37kg)(0m/s)=(1.22kg)(0m/s)+(1.37kg)v_f \\ \\ \Rightarrow0.976kg\ast m/s=1.37kg\ast v_f\Rightarrow v_f=\frac{0.976kg\ast m/s}{1.37kg} \\ \\ \Rightarrow v_f=\frac{488}{685}m/s\Rightarrow v_f\approx0.712m/s \end{gathered}[/tex]So, the final velocity will be approximately 0.712 m/s.