Answer:
The average velocity and average speed of the dog are 2.262 meters per second and 6.787 meters per second, respectively.
Explanation:
From Physics we must remember the definitions of average speed and average velocity, both measured in meters per second. Velocity is a vectorial quantity, that is, it has both magnitude and direction, whereas speed is an scalar quantity, which is a quantity that is represented solely by its magnitude. We assume that dog moves at constant speed.
For the case of the dog, we get that average speed and average velocity of the animal are, respectively:
Average velocity:
[tex]\vec v_{avg} = \frac{1}{\Delta t}\cdot (\vec r_{B}-\vec r_{A})[/tex] (Eq. 1)
Where:
[tex]\Delta t[/tex] - Travelling time of the dog, measured in seconds.
[tex]\vec r_{A}[/tex] - Initial vector position of the dog, measured in meters.
[tex]\vec r_{B}[/tex] - Final vector position of the dog, measured in meters.
Average speed:
[tex]v_{avg} = \frac{1}{\Delta t} \cdot (s_{A}+s_{B})[/tex] (Eq. 2)
Where [tex]s_{A}[/tex] and [tex]s_{B}[/tex] are the travelled distances of each stage, measured in meters.
If we know that [tex]\Delta t = 11.05\,s[/tex], [tex]\vec r_{A} = 0\,\hat{i}\,\,\,[m][/tex] and [tex]\vec r_{B} = 25\,\hat{i}\,\,\,[m][/tex], [tex]s_{A} = 50\,m[/tex] and [tex]s_{B} = 25\,m[/tex], average velocity and average speed are, respectively:
[tex]\vec v_{avg} = \frac{1}{11.05\,s}\cdot (25\,\hat{i})\,\,\,[m][/tex]
[tex]\vec v_{avg} = 2.262\,\hat{i}\,\,\,\left[\frac{m}{s} \right][/tex]
[tex]v_{avg} = \frac{75\,m}{11.05\,s}[/tex]
[tex]v_{avg} = 6.787\,\frac{m}{s}[/tex]
The average velocity and average speed of the dog are 2.262 meters per second and 6.787 meters per second, respectively.
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:___________.
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).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?
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.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.
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]
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
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.
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?
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]
what is a proper way and a safe way to dispose batteries
Answer:
throw them away............
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
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
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?
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]
A cat chases a mouse for a distance of 9.0 m over 16 s before getting tired. What is the cats average speed?
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
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).
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]
How is the average American diet affected by our current food system?"
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
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
Forces are expressed in ________. (newtons or mass)
Suppose the angle of incidence of a light ray is 42°.What is the angle of reflection?
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
Wind eroding the rocks on a moutain is an example of the atmosphere interacting with the cryosphere
true or false
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.A man is passing barrels of water to another person from a height of 7 meters. The barrel is attached to a roof that is 10 meters above the ground. The rope holding the barrel is 8 meters long and will break if the tension exceeds 638 N. How many liters of water can the man put in the barrel without the rope breaking (assuming that the barrel is massless)? (The density of water = 1000 kg/m3 and 1 m3 = 1000 L. Density = mass/volume.)
The name for this type of energy is
A.potential energy
B.motion
C.position
D.kinetic energy
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♨
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?
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.
Question 1 of 15
All digits shown on the measuring device, plus one estimated digit, are
considered
Answer here
SUBMIT
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).
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?
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
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
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
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)
1. Which of these is not a natural fibre?
a. leather
b. jute
C.Wool
d. cotton
Answer:
leather
Explanation:
plz mark as brainliest......hope it helps
Answer:
a. leather
Explanation:
Hello friend!!!!
a. leather is the correct option because jute, wool and cotton are all natural fibres whereas leather is a synthetic fibre.
Hope this helps
plz mark as brainliest!!!!!!
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.
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
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
Answer:
4m/s^2 ( A)
Explanation:
The solution is in the attached file
What are the three different types of muscle tissue?
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???
Which of the following is a form of kinetic energy?
A. Electrical energy
B. Potential energy
C. Sound energy
D. Nuclear energy
HELP URGENT
Answer:
Electrical energy ⚡ Hope this helps you Stay happy and safe Do mark as brainliest ✌️Answer:
Potential Energy. It's wrong but it was worth a try
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?
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
can humans be considered carbon sinks? If so,for how long
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