Answers
Answer:
The car appears to have a constant, positive acceleration for most of the video clip.
Related Questions
The drawing shows two identical airplanes at an air show. The airplanes are flying at the same speed. Airplane W is flying 50 m higher than airplane X. Which statement best describes the energy of the two airplanes?
Answers
Answer:
Airplane X has more gravitational potential energy than Airplane W
Explanation:
Gravitational potential energy is defined as "the energy acquired by an object due to its positional change in presence of gravitational force."
That being said, gravitational potential energy depends on the height of an object above the ground. It also depends on the mass of the object and even further, the amount of gravitational force that is applied.
And if we take a look at the question again, we'd agree that the two airplanes are flying at different heights, this means their gravitational potential energy will be different. And as such, Airplane X has more gravitational potential energy than Airplane W
In the figure below, a block of 1.67 slides on a track with different levels, which has friction only at the highest point where the kinetic coefficient of friction is uk = 0.35. If the block has an initial speed V0 = 7.5m/s and the highest point of the track is at ℎ = 2.1 above the initial position of the block, calculate the distance where the friction force for the block is.
Answers
Answer:
2.0 m
Explanation:
Energy is conserved.
Initial KE = Final PE + Work done by friction
½ mv² = mgh + Fd
½ mv² = mgh + mgμd
½ v² = gh + gμd
½ v² − gh = gμd
d = (½ v² − gh) / (gμ)
d = (½ (7.5 m/s)² − (10 m/s²) (2.1 m)) / ((10 m/s²) (0.35))
d = 2.0 m
1. A 100-kg crate is pulled across a warehouse floor using a rope with a force of 250 N at an angle of 45o from the horizontal. The coefficient of friction between the crate and the floor is 0.12.
a) Calculate the net force and acceleration of the crate.
b) If the crate was moving at 1.0 m/s when it was pulled, what would be its velocity after pulling it for 5.0 s?
Answers
Answer:
(a) The net force is 80.394 N
The acceleration of the crate is 0.804 m/s²
(b) the final velocity of the crate is 5.02 m/s
Explanation:
Given;
mass of the crate, m = 100 kg
applied force, F = 250 N
angle of inclination, θ = 45°
coefficient of friction, μ = 0.12
Applied force in y-direction, [tex]F_y = Fsin \theta = 250sin45 = 176.78 \ N[/tex]
Applied force in x-direction, [tex]F_x = Fcos \theta = 250cos45 = 176.78 \ N[/tex]
The normal force is calculated as;
N + Fy -W = 0
N = W - Fy
N = (100 x 9.8) - 176.78
N = 980 - 176.78 = 803.22 N
The frictional force is given by;
Fk = μN
Fk = 0.12 x 803.22
Fk = 96.386 N
(a) The net force is given by;
[tex]F_{net} = F_x - F_k\\\\F_{net} = 176.78-96.386\\\\F_{net} = 80.394 \ N[/tex]
Apply Newton's second law of motion;
F = ma
[tex]a = \frac{F_{net}}{m}\\\\ a = \frac{80.394}{100}\\\\ a = 0.804 \ m/s^2[/tex]
(b) the velocity of the crate after 5.0 s
[tex]F = ma= \frac{m(v-u)}{t} \\\\Ft =m(v-u)\\\\v-u = \frac{Ft}{m}\\\\ v = \frac{Ft}{m} + u\\\\v = \frac{F_{net}*t}{m} + u\\\\v = \frac{80.394*5}{100} + 1\\\\v = 5.02 \ m/s[/tex]
In an RL series circuit, an inductor of 4.74 H and a resistor of 9.33 Ω are connected to a 26.4 V battery. The switch of the circuit is initially open. Next close the switch and wait for a long time. Eventually the current reaches its equilibrium value. At this time, what is the corresponding energy stored in the inductor? Answer in units of J.
Answers
Answer:
The energy is [tex]U = 18.98 \ J [/tex]
Explanation:
From the question we are told that
The inductor is [tex]L = 4.74 \ H[/tex]
The resistance of the resistor is [tex]R = 9.33 \ \Omega[/tex]
The voltage of the battery is [tex]V = 26.4 \ V[/tex]
Generally the current flowing in the circuit is mathematically represented as
[tex]I = \frac{V}{R}[/tex]
=> [tex]I = \frac{26.4}{9.33 }[/tex]
=> [tex]I = 2.83 \ A[/tex]
Generally the corresponding energy stored in the circuit is
[tex]U = \frac{1}{2} * L * I^2[/tex]
[tex]U = \frac{1}{2} * 4.74 * 2.83 ^2[/tex]
[tex]U = 18.98 \ J [/tex]
4. Tires on the road show how friction produces
a. lubrication
b. heat
c. gravity
d. force
help :(
Answers
I think it is D. Force
Answer:
Explanation:
Do you mean tire tracks?
If you do then the tracks show that there must have been a lot of heat involved. When the tires skid, the friction produced must be awfully hot to stop and 800 kg vehical (which is pretty small). The tire tracks that you see are bits of melted rubber taken from the tire.
Force is involved, but heat is the better answer.
gravity is necessary for friction to take place. If you were out in space, you could not get enough friction to leave tire tracks. (Force is awfully small).
Lubrication is there to give you a 4th choice. The exact opposite is what takes place. Lubrication reduces friction.
Water flows from one reservoir to another a height, 41 m below. A turbine (η=0.77) generates power from this flow. 1 m3/s passes through the turbine. If 12 m head loss occurs between the two reservoirs, determine the actual (i.e. useable) power generated by the turbine (in kW).
Answers
Complete Question
A diagram representing this question is shown on the first uploaded image
Answer:
The value is [tex]P = 294594.3 \ W[/tex]
Explanation:
From the question we are told that
The height is [tex]h = 41 \ m[/tex]
The efficiency of the turbine is [tex]\eta = 0.77[/tex]
The flow rate is [tex]\r V = 1 m^3 / s[/tex]
The head loss is g = 2 m
Generally the head gain by the turbine is mathematically represented as
[tex]H = h - d[/tex]
=> [tex]H = 41 - 2[/tex]
=> [tex]H = 39 \ m [/tex]
Generally the actual power generated by the turbine is mathematically represented as
[tex]P = \eta *\gamma * \r V * H[/tex]
Here [tex]\gamma[/tex] is the specific density of water with value
[tex]\gamma = 9810 N/m^3 [/tex]
So
[tex]P =0.77 *9810 * 1 * 39[/tex]
[tex]P = 294594.3 \ W[/tex]
6. A cat walks 1.5km South and then 2.4km East. What is the total displacement of the cat?
7. A boy scout troop hikes 10.0 km East and then hikes an additional 7km North. What is the total displacement of the boy scout troop?
8. A go-kart is moving 23 m/s on the track. The track is 152m long. How long will take for the go-kart to reach the end of the track?
9. If a snail moves at a pace of .01m/s and it travels for 1 hr. How far does the snail get?
10. It takes 3 hr and 10 min to ride the tram to the top of Pike’s Peak in Colorado. The tram will travel a total distance of 14.32 km. What is the speed of the tram?
Answers
Answer:
6. 3.9 km SE
7. 17 km NE
8. 3496 seconds
9. 36 m
10. 13.27 km/m
Explanation:
6. and 7. added both numbers
8. multiplied both numbers
9. 60 x 60 = 3600
3600 x 0.01 = 36
10. 60 x 3 = 180
180 + 10 =190
190 / 14.32 = 13.27
What is the ratio of the displacement amplitudes of two sound waves given that they are both5.0 kHz but have a 3.0 dB intensity level difference?
Answers
Answer:
The ratio of the displacement amplitudes of two sound waves is 1.16.
Explanation:
Given that,
Frequency = 5.0 kHz
Intensity level difference = 3.0 dB
We know that,
The sound intensity is inversely proportional to the square of distance.
[tex]I\propto\dfrac{1}{r^2}[/tex]
The sound intensity for first wave is
[tex]\beta_{1}=10\log\dfrac{I_{1}}{I_{0}}[/tex]...(I)
The sound intensity for second wave is
[tex]\beta_{2}=10\log\dfrac{I_{2}}{I_{0}}[/tex]...(II)
We need to calculate the ratio of intensity
From equation (I) and (II)
[tex]\beta_{2}-\beta_{1}=10\log\dfrac{I_{2}}{I_{0}}-10\log\dfrac{I_{1}}{I_{0}}[/tex]
[tex]\Delta \beta=10\log(\dfrac{I_{2}}{I_{1}})[/tex]
Put the value into the formula
[tex]3.0=10\log(\dfrac{I_{2}}{I_{1}})[/tex]
[tex]\dfrac{I_{2}}{I_{1}}=e^{\dfrac{3.0}{10}}[/tex]
[tex]\dfrac{I_{2}}{I_{1}}=1.34[/tex]
We need to calculate the ratio of the displacement
Using formula of displacement
[tex]\dfrac{r_{1}}{r_{2}}=\sqrt{\dfrac{I_{2}}{I_{1}}}[/tex]
Put the value into the formula
[tex]\dfrac{r_{1}}{r_{2}}=\sqrt{1.34}[/tex]
[tex]\dfrac{r_{1}}{r_{2}}=1.16[/tex]
Hence, The ratio of the displacement amplitudes of two sound waves is 1.16.
Wind eroding the rocks on a moutain is an example of the atmosphere interacting with the cryosphere
true or false
Answers
Answer:
False
Explanation:
The cryosphere is all the part of the earth where water is in solid form. Where wind interacts with rocks, it is an example of atmosphere - geosphere interaction.
Rocks are part of the geosphere The geosphere is the part of the earth made up of solid rocks. Wind erosion occurs when wind wears down part of the geosphere.Question 1 of 15
All digits shown on the measuring device, plus one estimated digit, are
considered
Answer here
SUBMIT
Answers
Answer:
significant
Explanation:
The digits in a measurement that are considered significant are all of those digits that represent marked calibrations on the measuring device plus one additional digit to represent the estimated digit (tenths of the smallest calibration).
A stone is released from rest from the edge of a building roof 190 m above the ground. Neglecting air resistance, the speed of the stone, just before striking the ground, is:___________.
Answers
Answer:
61 m/s
Explanation:
If the stone is realeased from rest, this means that its initial velocity is 0.As tha stone is only influenced by gravity, and the acceleration due to it is constant (near the surface of the Earth), we can apply the following kinematic equation:[tex]v_{f}^{2} - v_{o}^ {2} = 2* g* h (1)[/tex]
Replacing by the values of g=9.8 m/s², and h=190 m, rearranging and solving for vf, we get:vf = √2*g*h =√2*9.8 m/s²*190 m = 61 m/s (assuming that the downward direction is the positive one).Sally is on a large sailboat that comes to a stop a small distance from the dock. Since it is such a small distance, Sally decides to jump to the dock. She makes the jump, but the large sailboat moves away from her as she jumps. Since Sally is interested to see what happens on other boats, she makes the same jump from a rowboat that is much smaller than the large sailboat. Which boat will move away from Sally more slowly
Answers
Answer:
The rowboat will move away from sally more quickly because the rowboat because the sailboat is larger in mass
Explanation:
Gnerally the row boat will move away from her quicker than the sailboat this is because the mass of the sail boat is larger than the row boat , hence the frictional force that opposes motion will be greater in the sailboat than in the row boat.
What are the three different types of muscle tissue?
Answers
Answer:
skeletal,cardiac,and smooth.
Explanation:
Answer:
Skeletal, smooth, and cardiac.
Explanation:
Skeletal Muscles:
Skeletal muscles are the most familiar type of muscles; they make up most of the muscle mass in the body. Flexible bands of connective tissue called tendons attach these muscles to the bones in the body. Skeletal muscles control voluntary movement in the body.
Smooth Muscles:
Smooth muscles are involuntary muscles that we don’t consciously control. They are found within the walls of many organs and control the movement of these organs. For example, they enable the movement of food through the digestive system.
Cardiac Muscles:
Cardiac muscles are a special type of involuntary muscle. Located in the heart, these muscles control the contractions of the heart.
can u give me brainliest???
2 forces are acting on a body. I forgot the body doesn't it's position or shape, the forces.......a)must be of equal magnitude b)must be parallel and opposite c)must add upto zero d)must be in a single line
Answers
Answer:
C
Explanation:
only if there is a net force of zero, the body will not move
some people may say B but that is wrong because maybe one force is greater than the other so the object would still move even though the forces are in opposite directions and parallel
Suppose the angle of incidence of a light ray is 42°.What is the angle of reflection?
Answers
Answer:
angle of reflection will be also 42°Explanation:
we know that ------------- angle of incidence=angle of reflectionProtons and neutrons grouped in a specific pattern
Ob
Protons and electrons spread around randomly
Ос
А group of protons and neutrons that are surrounded by electrons
Od
A ball of electrons and neutrons surrounded by protons
Answers
A cat chases a mouse for a distance of 9.0 m over 16 s before getting tired. What is the cats average speed?
Answers
Answer:
The answer is 0.56 m/sExplanation:
The speed of the cat can be found by using the formula
[tex]v = \frac{d}{t} \\ [/tex]
d is the distance
t is the time taken
From the question we have
[tex]v = \frac{9}{16} = \\ = 0.5625[/tex]
We have the final answer as
0.56 m/sHope this helps you
A force is applied to a block sliding along a surface (Figure 2). The magnitude of the force is 15 N, and the horizontal component of the force is 4.5 N. At what angle (in degrees) above the horizontal is the force directed?
Answers
Answer:
Fy = 14.3 [N]
Explanation:
To be able to solve this problem we must know that the force is a vector and has magnitude and direction, therefore it can be decomposed into the force in the X & y components:
When we have the components on the horizontal and vertical axes we must use the Pythagorean theorem.
[tex]F = \sqrt{F_{x}^{2} +F_{y}^{2} }[/tex]
where:
F = 15 [N]
Fx = horizontal component = 4.5 [N]
Fy = vertical component [N]
[tex]15=\sqrt{4.5^{2}+F_{y}^{2}}\\ 15^{2}= (\sqrt{4.5^{2}+F_{y}^{2}})^{2} \\225 = 4.5^{2}+F_{y} ^{2}\\ F_{y}^{2} =225 -4.5^{2}\\ F_{y}^{2}=204.75\\F_{y}=\sqrt{204.75}\\ F_{y}=14.3 [N][/tex]
The name for this type of energy is
A.potential energy
B.motion
C.position
D.kinetic energy
Answers
Explanation:
[tex]kinetic \: energy \: is \: the \: energy \: of \: a \: body \: in \: motion. \\ that \: is \: energy \: of \: a \: body \: that \: is \: moving.\\ while \\ potential \: energy \: is \: the \: energy \: of \: a \: body \: by \: the \: virtue \: of \: its \: position \: \\ that \: is \: energy \: of \: a \: body \: that \: is \: not \: moving.[/tex]
♨Rage♨
Watching the World Series (only as an example of physics in action), you wonder about the ability of the catcher to throw out a base runner trying to steal second. Suppose a catcher is crouched down behind the plate when he observes the runner breaking for second. After he gets the ball from the pitcher, he throws as hard as necessary to second base without standing up. If the catcher throws the ball at an angle of 30 degrees from the horizontal so that it is caught at second base at about the same height as that catcher threw it, how much time does it take for the ball to travel the 120 feet from the catcher to second base
Answers
Answer:
The time is [tex]t_t = 3.7583 \ s [/tex]
Explanation:
From the question we are told that
The angle is [tex]\theta = 30^o[/tex]
The horizontal distance is [tex]d = 120 \ ft[/tex]
Generally when the ball is at maximum height before descending the velocity is zero and this velocity can be mathematically represented as
[tex]v = u + at[/tex]
here a = -g the negative sign is because the direction of motion is against gravity
So
[tex]v = v_i + at[/tex]
Here[tex] v_i [/tex] is the vertical component of the initial velocity of the ball which is mathematically
represented as
[tex]v_i = usin(\theta )[/tex]
So
=> [tex]0 = usin(\theta ) -9.8t[/tex]
Generally the total time taken to travel the 120 ft is mathematically represented as
[tex]t_t = \frac{120}{v_h}[/tex]
Here [tex]v_h[/tex] is the horizontal component of the initial velocity which is mathematically represented as
[tex]v_h = u cos(\theta )[/tex]
So
[tex]t_t = \frac{120}{ u cos(\theta )}[/tex]
Generally the time taken to reach the maximum height is
[tex]t = \frac{t_t}{2}[/tex]
=> [tex]t = \frac{120}{ u cos(\theta )} * \frac{1}{2} [/tex]
=> [tex]t = \frac{60}{ u cos(\theta )} [/tex]
So
[tex]0 = usin(\theta ) -9.8* [\frac{60}{ u cos(\theta )}][/tex]
[tex] usin(\theta ) = 9.8* [\frac{60}{ u cos(\theta )}][/tex]
[tex] usin(\theta ) * u cos(\theta) =60* 9.8 [/tex]
[tex] u^2 sin(\theta ) cos(\theta) =60* 9.8 [/tex]
[tex] u^2 sin(30 ) cos(30) =60* 9.8 [/tex]
[tex] u^2 * \frac{1}{2}* \frac{\sqrt{3}}{2} =588.6 [/tex]
[tex] u^2 *\sqrt{3} =2354.4 [/tex]
[tex] u^2 = 1359.31 [/tex]
[tex] u = 36.87 \ ft/s [/tex]
Substituting this value into the equation for total time
[tex]t_t = \frac{120}{36.87 cos(30 )}[/tex]
[tex]t_t = 3.7583 \ s [/tex]
Forces are expressed in ________. (newtons or mass)
Answers
Mass isn’t even a unit of measurement
How is the average American diet affected by our current food system?"
Answers
Determine the acceleration that results when a 12 N net force is applied to a 3 kg object.
a. 4 m/s 2
b. 6 m/s 2
c. 12 m/s 2
d. 36 m/s 2
Answers
Answer:
4m/s^2 ( A)
Explanation:
The solution is in the attached file
A cannon ball is shot horizontally off a 37.0 m cliff and lands a distance of 18.5 m
from the base of the cliff. Whall was the initial horizontal velocity of the cannon ball?
Answers
Answer:
vₓ = 6.73 m/s
Explanation:
Assuming no other external influences than gravity, in the horizontal direction (which we make to coincide with the x- axis) , speed is constant, so, applying the definition of average velocity, we can write the following equation:[tex]v_{x} = \frac{\Delta x}{\Delta t} (1)[/tex]
Now, in the vertical direction (coincident with the y- axis) , as both movements are independent each other, initial velocity is zero, so we can write the following equation for the vertical displacement:[tex]\Delta h = \frac{1}{2} * g * t^{2} (2)[/tex]
where Δh = -37.0 m , g = -9.8 m/s2Solving (2) for t, we get:[tex]t = \sqrt{\frac{2*\Delta h}{g} } =\sqrt{\frac{2*37.0m}{9.8m/s2}} = 2.75 s (3)[/tex]
Taking t₀ = 0, ⇒ Δt = tReplacing (3) in (1), we get:[tex]v_{x} = \frac{\Delta x}{\Delta t} = \frac{x}{t} = \frac{18.5m}{2.75s} = 6.73 m/s[/tex]
As the horizontal velocity is constant, the initial horizontal velocity is just the average one, i.e., 6.73 m/s.Which of the following is not true about taxes? A. Mandatory sum of money by government so that it can operate B. Due on April 15th C. largely collected to support private businesses D. Collected by the Internal Revenue Service (IRS)
Answers
can humans be considered carbon sinks? If so,for how long
Answers
Answer:
Humans be considered carbon sinks. Not only do humans
have a lot of carbon in them, they also use a lot of carbon.
hope this helps
QUESTION 10
An archer fires an arrow towards a tree with initial speed 65 m/s and angle 25 degrees above the horizontal. If the arrow takes 0.85
seconds to hit the tree, calculate the horizontal distance between the archer and the tree.
QUESTION 11
A monkey throws a banana from a tree into a nearby river. The banana has initial speed 7.6 m/s, is angled 40 degrees below the
horizontal, and takes 0.75 seconds to land in the river. Calculate the speed of the banana when it hits the water.
Answers
Answer:
10) The distance between the archer and the tree is 50.074 meters.
11) The speed of the banana when it hits the water is approximately 13.554 meters per second.
Explanation:
10) The arrow experiments a parabolic motion, which is the combination of horizontal motion at constant velocity and vertical uniform accelerated motion. In this case we need to find the horizontal distance between the archer and the tree, calculated by the following kinematic equation:
[tex]x = x_{o} +v_{o}\cdot t \cdot \cos \theta[/tex] (Eq. 1)
Where:
[tex]x_{o}[/tex] - Initial position of the arrow, measured in meters.
[tex]x[/tex] - Final position of the arrow, measured in meters.
[tex]v_{o}[/tex] - Initial speed of the arrow, measured in meters per second.
[tex]t[/tex] - Time, measured in seconds.
[tex]\theta[/tex] - Launch angle, measured in sexagesimal degrees.
If we know that [tex]x_{o} = 0\,m[/tex], [tex]v_{o} = 65\,\frac{m}{s}[/tex], [tex]t = 0.85\,s[/tex] and [tex]\theta = 25^{\circ}[/tex], the horizontal distance between the archer and the tree is:
[tex]x = 0\,m + \left(65\,\frac{m}{s}\right)\cdot (0.85\,s)\cdot \cos 25^{\circ}[/tex]
[tex]x = 50.074\,m[/tex]
The distance between the archer and the tree is 50.074 meters.
11) The final speed of the banana ([tex]v[/tex]), measured in meters per second, just before hitting the water is determined by the Pythagorean Theorem:
[tex]v = \sqrt{v_{x}^{2}+v_{y}^{2}}[/tex] (Eq. 2)
Where:
[tex]v_{x}[/tex] - Horizontal speed of the banana, measured in meters per second.
[tex]v_{y}[/tex] - Vertical speed of the banana, measured in meters per second.
Each component of the speed are obtained by using these kinematic equations:
[tex]v_{x} = v_{o}\cdot \cos \theta[/tex] (Eq. 3)
[tex]v_{y} = v_{o}\cdot \sin \theta +g\cdot t[/tex] (Eq. 4)
Where [tex]g[/tex] is the gravitational acceleration, measured in meters per square second.
If we know that [tex]v_{o} = 7.6\,\frac{m}{s}[/tex], [tex]\theta = -40^{\circ}[/tex], [tex]g = -9.807\,\frac{m}{s^{2}}[/tex] and [tex]t = 0.75\,s[/tex], the components of final speed are, respectively:
[tex]v_{x} = \left(7.6\,\frac{m}{s} \right)\cdot \cos (-40^{\circ})[/tex]
[tex]v_{x} = 5.822\,\frac{m}{s}[/tex]
[tex]v_{y} = \left(7.6\,\frac{m}{s}\right)\cdot \sin (-40^{\circ})+\left(-9.807\,\frac{m}{s^{2}} \right) \cdot (0.75\,s)[/tex]
[tex]v_{y} = -12.240\,\frac{m}{s}[/tex]
And the speed of the banana right before hitting the water is:
[tex]v = \sqrt{\left(5.822\,\frac{m}{s} \right)^{2}+\left(-12.240\,\frac{m}{s} \right)^{2}}[/tex]
[tex]v \approx 13.554\,\frac{m}{s}[/tex]
The speed of the banana when it hits the water is approximately 13.554 meters per second.
what is a proper way and a safe way to dispose batteries
Answers
Answer:
throw them away............
A large semi-truck, with mass 31x crashes into a small sedan with mass x . If the semi-truck exerts a force F on the sedan, what force will the sedan exert on the semi-truck
Answers
Answer:
Force Exerted by Sedan on Truck = - F
Explanation:
This question can easily be solved by using Newton's Third Law of Motion. Newton's Third Law of Motion clearly states that for every action force there exists an equal in magnitude reaction force. However, the direction of the reaction force is opposite to the direction of the action force. Mathematically,
Action Force = - Reaction Force
Hence, we have here:
Force Exerted by Semi Truck on Sedan = F
So, from Newton's Third Law of motion:
Force Exerted by Sedan on Truck = - Force Exerted by Semi Truck on Sedan
Force Exerted by Sedan on Truck = - F
What is the magnitude of the force exerted by the biceps FbicepsFbicepsF_biceps? What is the magnitude of the force exerted by the elbow FelbowFelbowF_elbow
Answers
Answer:
The force of forearm is 239.9 N
The force of elbow is 215.89 N
Explanation:
Suppose, When you lift an object by moving only your forearm, the main lifting muscle in your arm is the biceps. Suppose the mass of a forearm is 1.50 kg. If the biceps is connected to the forearm a distance [tex]d_{b}[/tex] 2.50 cm from the elbow, how much force [tex]F_{b}[/tex] must the biceps exert to hold a 950 g ball at the end of the forearm at distance dball J 36.0 cm from the elbow, with the forearm parallel to the floor? How much force [tex]F_{l}[/tex] must the elbow exert,
Given that,
Mass of forearm = 1.50 kg
Distance of forearm = 2.50 cm
Mass of ball = 950 g
Distance of ball = 36.0 cm
We need to calculate the force of forearm
Using balancing torque about elbow
[tex]F_{b}\times d_{b}=w_{f}\times\dfrac{d_{f}}{2}+w_{ball}\times d_{ball}[/tex]
Put the value into the formula
[tex]F_{b}\times 0.025= 1.50\times9.8\times\dfrac{0.36}{2}+0.95\times9.8\times0.36[/tex]
[tex]F_{b}=\dfrac{1.50\times9.8\times\dfrac{0.36}{2}+0.95\times9.8\times0.36}{0.025}[/tex]
[tex]F_{b}=239.9\ N[/tex]
We need to calculate the force of elbow
Using balancing force
[tex]F_{b}=F_{l}+w_{f}+w_{b}[/tex]
[tex]F_{l}=F_{b}-w_{f}-w_{b}[/tex]
Put the value into the formula
[tex]F_{l}=239.9-(1.50\times9.8)-(0.95\times9.8)[/tex]
[tex]F_{l}=215.89\ N[/tex]
Hence, The force of forearm is 239.9 N
The force of elbow is 215.89 N