Answer:
a) The initial total mechanical energy of the projectile is 498556.296 joules.
b) The work done on the projectile by air friction is 125960.4 joules.
c) The speed of the projectile immediately before it hits the ground is approximately 82.475 meters per second.
Explanation:
a) The system Earth-projectile is represented by the Principle of Energy Conservation, the initial total mechanical energy ([tex]E[/tex]) of the project is equal to the sum of gravitational potential energy ([tex]U_{g}[/tex]) and translational kinetic energy ([tex]K[/tex]), all measured in joules:
[tex]E = U_{g} + K[/tex] (Eq. 1)
We expand this expression by using the definitions of gravitational potential energy and translational kinetic energy:
[tex]E = m\cdot g\cdot y + \frac{1}{2}\cdot m\cdot v^{2}[/tex] (Eq. 1b)
Where:
[tex]m[/tex] - Mass of the projectile, measured in kilograms.
[tex]g[/tex] - Gravitational acceleration, measured in meters per square second.
[tex]y[/tex] - Initial height of the projectile above ground, measured in meters.
[tex]v[/tex] - Initial speed of the projectile, measured in meters per second.
If we know that [tex]m = 54\,kg[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], [tex]y = 132\,m[/tex] and [tex]v = 126\,\frac{m}{s}[/tex], the initial mechanical energy of the earth-projectile system is:
[tex]E = (54\,kg)\cdot \left(9.807\,\frac{m}{s^{2}}\right)\cdot (132\,m)+\frac{1}{2}\cdot (54\,kg)\cdot \left(126\,\frac{m}{s} \right)^{2}[/tex]
[tex]E = 498556.296\,J[/tex]
The initial total mechanical energy of the projectile is 498556.296 joules.
b) According to this statement, air friction diminishes the total mechanical energy of the projectile by the Work-Energy Theorem. That is:
[tex]W_{loss} = E_{o}-E_{1}[/tex] (Eq. 2)
Where:
[tex]E_{o}[/tex] - Initial total mechanical energy, measured in joules.
[tex]E_{1}[/tex] - FInal total mechanical energy, measured in joules.
[tex]W_{loss}[/tex] - Work losses due to air friction, measured in joules.
We expand this expression by using the definitions of gravitational potential energy and translational kinetic energy:
[tex]W_{loss} = E_{o}-K_{1}-U_{g,1}[/tex]
[tex]W_{loss} = E_{o} -\frac{1}{2}\cdot m\cdot v_{1}^{2}-m\cdot g\cdot y_{1}[/tex] (Eq. 2b)
Where:
[tex]m[/tex] - Mass of the projectile, measured in kilograms.
[tex]g[/tex] - Gravitational acceleration, measured in meters per square second.
[tex]y_{1}[/tex] - Maximum height of the projectile above ground, measured in meters.
[tex]v_{1}[/tex] - Current speed of the projectile, measured in meters per second.
If we know that [tex]E_{o} = 498556.296\,J[/tex], [tex]m = 54\,kg[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], [tex]y_{1} = 297\,m[/tex] and [tex]v_{1} = 89.3\,\frac{m}{s}[/tex], the work losses due to air friction are:
[tex]W_{loss} = 498556.296\,J -\frac{1}{2}\cdot (54\,kg)\cdot \left(89.3\,\frac{m}{s} \right)^{2} -(54\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (297\,m)[/tex]
[tex]W_{loss} = 125960.4\,J[/tex]
The work done on the projectile by air friction is 125960.4 joules.
c) From the Principle of Energy Conservation and Work-Energy Theorem, we construct the following model to calculate speed of the projectile before it hits the ground:
[tex]E_{1} = U_{g,2}+K_{2}+1.5\cdot W_{loss}[/tex] (Eq. 3)
[tex]K_{2} = E_{1}-U_{g,2}-1.5\cdot W_{loss}[/tex]
Where:
[tex]E_{1}[/tex] - Total mechanical energy of the projectile at maximum height, measured in joules.
[tex]U_{g,2}[/tex] - Potential gravitational energy of the projectile, measured in joules.
[tex]K_{2}[/tex] - Kinetic energy of the projectile, measured in joules.
[tex]W_{loss}[/tex] - Work losses due to air friction during the upward movement, measured in joules.
We expand this expression by using the definitions of gravitational potential energy and translational kinetic energy:
[tex]\frac{1}{2}\cdot m \cdot v_{2}^{2} = E_{1}-m\cdot g\cdot y_{2}-1.5\cdot W_{loss}[/tex] (Eq. 3b)
[tex]m\cdot v_{2}^{2} = 2\cdot E_{1}-2\cdot m \cdot g \cdot y_{2}-3\cdot W_{loss}[/tex]
[tex]v_{2}^{2} = 2\cdot \frac{E_{1}}{m}-2\cdot g\cdot y_{2}-3\cdot \frac{W_{loss}}{m}[/tex]
[tex]v_{2} = \sqrt{2\cdot \frac{E_{1}}{m}-2\cdot g\cdot y_{2}-3\cdot \frac{W_{loss}}{m} }[/tex]
If we know that [tex]E_{1} = 372595.896\,J[/tex], [tex]m = 54\,kg[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], [tex]y_{2} =0\,m[/tex] and [tex]W_{loss} = 125960.4\,J[/tex], the final speed of the projectile is:
[tex]v_{2} =\sqrt{2\cdot \left(\frac{372595.896\,J}{54\,kg}\right)-2\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (0\,m)-3\cdot \left(\frac{125960.4\,J}{54\,kg}\right) }[/tex]
[tex]v_{2} \approx 82.475\,\frac{m}{s}[/tex]
The speed of the projectile immediately before it hits the ground is approximately 82.475 meters per second.
What is the difference between a rule and a law?
Electric charges are either positive or ____
Answer:
Negative
Explanation:
duh
Answer:
:)
Explanation:
negative.
go add the snap carmel.bratz
Julie takes a cold glass of dear, doness houd out the register and pusi on the babe hathewwad warme up to room temperature, Julie sees buthles of gas form in the und and me to the face Which conclusion is best supported by Jules onervations? The liquid is a pure sutrance because it is closers O The liquid is a mixture because its temperature res The liquid is a pure substance because it remains lowd when The liquid is a mixture because it has a gas dissolved in a loud
Answer:
D. The liquid is a mixture because it has a gas dissolved in a liquid.
Cooling causes a material to
Answer:
whats the question?
Explanation:
Answer:
2 Key Concepts Heating and cooling can cause materials to change characteristics, such as state, color, and texture. Heating causes ice to become liquid water and cooling causes condensation to form on a window, mirror, or on the outside of a glass of water.
Explanation:
The human circulatory system is closed-that is, the blood pumped out of the left ventricle of the heart into the arteries is constrained to a series of continuous, branching vessels as it passes through the capillaries and then into the veins as it returns to the heart. The blood in each of the heart’s four chambers comes briefly to rest before it is ejected by contraction of the heart muscle. If the aorta (diameter dad a) branches into two equal-sized arteries with a combined area equal to that of the aorta, what is the diameter of one of the branches? (a) da−−√ d a ; (b) da/2–√d a / 2 ; (c) 2da2d a ; (d) da/2d a /2.
The question is missing parts. The complete question is here.
The human circulatory system is closed, that is, the blood pumped out of the left ventricle of the heart into the arteries is constrained to a series of continuoous, branching vessels as it passes through the capillaries and then into the veins as it returns to the heart. The blood in each of the heart's four chambers comes briefly to rest before it is ejected by cotraction of the heart muscle. If the aorta (diameter [tex]d_{a}[/tex]) branches into two equal-sized arteries with a combined area equal to that of the aorta, what is the diameter of one of the branches?
(a) [tex]\sqrt{d_{a}}[/tex]
(b) [tex]\frac{d_{a}}{\sqrt{2} }[/tex]
(c) [tex]2d_{a}[/tex]
(d) [tex]\frac{d_{a}}{2}[/tex]
Answer: (b) [tex]\frac{d_{a}}{\sqrt{2} }[/tex]
Explanation: The cross-sectional area of a vessel is a circle. So area is:
[tex]A=\pi.r^{2}[/tex]
Radius is half of a diameter, i.e.:
[tex]r=\frac{d}{2}[/tex]
Suppose [tex]d_{b}[/tex] is diameter of the branches, radius of one the branches is:
[tex]r_{b}=\frac{d_{b}}{2}[/tex]
Both branches are equal sized, which means, both have the same radius. So, combined area of both branches is:
[tex]A_{b}=\pi\frac{d_{b}^{2}}{4} +\pi\frac{d_{b}^{2}}{4}[/tex]
[tex]A_{b}=2\frac{d_{b}^{2}}{4}[/tex]
Area of aorta is
[tex]A_{a}=\pi.(\frac{d_{a}}{2})^{2}[/tex]
[tex]A_{a}=\pi.\frac{d_{a}^{2}}{4}[/tex]
Area of aorta is equal the combined area of the branches, then:
[tex]2\pi\frac{d_{b}^{2}}{4}= \pi\frac{d_{a}^{2}}{4}[/tex]
Rearraging:
[tex]d_{b}^{2}=\frac{d_{a}^{2}}{2}[/tex]
[tex]d_{b}=\frac{d_{a}}{\sqrt{2} }[/tex]
The diameter of one of the branches is [tex]\frac{d_{a}}{\sqrt{2}}[/tex].
blank refers to the method of spreading fertilizer evenly over the entire field by hand it is done at the blank stage
Answer:
Broadcasting is the method, not sure about the stage it is done in
Explanation:
Answer:
Broadcasting is the first (blank) second (blank) is Cultivation.
Explanation:
I took the test & got this answer correct.
how can you observe the law of conservation of energy in action at the skatepark?
Explanation:
When the skater is dropped onto the ramp from above, the potential energy decreases and the kinetic energy increases.
Every time the skater bounces from the impact, thermal energy is gained, and both potential and kinetic energy are lost.
Long flights at midlatitudes in the Northern Hemisphere encounter the jet stream, an eastward airflow that can affect a plane's speed relative to Earth's surface. If a pilot maintains a certain speed relative to the air (the plane's airspeed), the speed relative to the surface (the plane's ground speed) is more when the flight is in the direction of the jet stream and less when the flight is opposite the jet stream. Suppose a round-trip flight is scheduled between two cities separated by 4300 km, with the outgoing flight in the direction of the jet stream and the return flight opposite it. The airline computer advises an airspeed of 930 km/h, for which the difference in flight times for the outgoing and return flights is 61 min. What jet-stream speed is the computer using
Answer:
103.52 km/h
Explanation:
We are given;
Distance between two cities; d = 4300 km
airspeed; v_as = 930 km/h
Difference in flight time; Δt = 61 min = 1.0167 h
Now, the equation of motion to find the distance is given as;
d = vt
Where v = v_as + v_js
v_as is the airspeed
v_js is the jet speed
Thus;
d = (v_as + v_js)t
Thus, time(t1) for outgoing flight is;
t1 = d/(v_as + v_js)
Meanwhile, time(t2) for the return flight, the jet stream velocity will be negative and time is;
t2 = d/(v_as - v_js)
Recall that Difference in flight time; Δt = 61 min.
Thus;
Δt = t2 - T1 = [d/(v_as - v_js)] - [d/(v_as + v_js)]
Factorizing out, we have;
Δt = d[1/(v_as - v_js)] - [1/(v_as + v_js)]
Furthermore, it gives;
Δt = d[(v_as + v_js - v_as + v_js)]/((v_as - v_js) × (v_as + v_js))
Δt = d(2v_js)/((v_as)² - (v_js)²)
Cross multiply to get;
(2dv_js)/Δt = ((v_as)² - (v_js)²)
(v_js)² + ((2dv_js)/Δt) - (v_as)² = 0
Plugging in values for d,v_as and Δt gives;
(v_js)² - ((2 × 4300 × v_js)/1.0167) - (930)² = 0
(v_js)² - (8458.7329v_js) - 864900 = 0
Using quadratic formula, we have;
v_js = 103.52 km/h
The airline computer advises an airspeed of 930 km/h, for which the difference in flight times for the outgoing and return flights is 61 min. The jet stream speed obtained using the quadratic formula is 103.52 km/h.
Given:
Distance between two cities; d = 4300 km
airspeed; v(as) = 930 km/h
Difference in flight time; Δt = 61 min = 1.0167 h
Now, the equation of motion to find the distance is given as;
d = vt
Where v = v(as) + v(js)
d = v(as) + v(js)t
At time t₁, t₁ = d/v(as) + v(js)
At time t₂, t₂ = d/v(as) + v(js)
Recall that Difference in flight time; Δt = 61 min.
Δt = d/v(as) + v(js) - d/v(as) + v(js)
Δt = d[(v(as) + v(js) - v(as) + v(js)]/((v(as) - v(js)) × (v(as) + v(js)))
Δt = d(2v(js))/((v(as))² - (v(js))²)
(2dv(js))/Δt = ((v(as))² - (v(js))²)
(v(js))² + ((2dv(js))/Δt) - (v(as))² = 0
(v(js))² - ((2 × 4300 × v(js))/1.0167) - (930)² = 0
(v(js))² - (8458.7329v(js)) - 864900 = 0
Using the quadratic formula, we have;
v(js) = 103.52 km/h
Therefore, The jet stream speed obtained using the quadratic formula is 103.52 km/h.
To know more about the quadratic formula:
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What is reduction division mean ?
Answer:
Reduction division: The first cell division in meiosis, the process by which germ cells are formed. In reduction division, the chromosome number is reduced from diploid (46 chromosomes) to haploid (23 chromosomes). Also known as first meiotic division and first meiosis.
Explanation:
Please mark as brainliest
100 POINTS. PLEASE PROVIDE EXPLANATION
Answer:
60 kg
80 kg
Explanation:
Work is equal to the change in energy.
W = ΔE = E − E₀
Let's start with block B. The work done by the tension force is equal to the change in energy. Initially, the block has potential energy. Finally, the block has kinetic energy.
W = ΔE
FΔy = ½ mv² − mgh
T (-2.0 m) = ½ m (6.00 m/s)² − m (10 m/s²) (2.0 m)
T (-2.0 m) = m (-2 m²/s²)
T = m (1 m/s²)
Now let's look at block A. The work done by tension and against friction is equal to the change in energy. Initially, the block has no energy. Finally, it has both kinetic and potential energy.
W = ΔE
Fd = ½ mv² + mgh − 0
(T − Nμ) (2.0 m) = ½ (4.00 kg) (6.00 m/s)² + (4.00 kg) (10 m/s²) (⅗ × 2.0 m)
(T − Nμ) (2.0 m) = 120 J
T − Nμ = 60 N
Draw a free body diagram of block A and sum the forces in the perpendicular direction to find the normal force N.
N = mg cos θ
N = (4.00 kg) (10 m/s²) (⅘)
N = 32 N
Substitute:
T − 32μ = 60 N
If μ = 0, then T = 60 N and m = 60 kg.
If μ = ⅝, then T = 80 N and m = 80 kg.
Answer:did you find you answer also i answered so you can give him brainiest
Explanation:
What feature does not require a planet to have any particular characteristics?
Stream Beds
Dunes
Impact Craters
Volcanic Lava Flows
What is the correct answer?
Answer:
Impact Craters.
Explanation:
An impact crater can be defined as a circular depression that is caused by impact on any planet or asteroids or any other celestial body's surface. When smaller body in galaxy impacts these larger bodies, they form a circular depression on it's surface.
This is a major feature found in solid object such as the Moon, Mercury, etc.
Therefore, the feature that a planet does not require is an impact crater. As other features are important to define a planet. Thus correct option is C.
what is power?
a- the magnitude of a force needed to move an object
b- how much work can be done in a given time
c- the distance over time that an object moves
d- the energy needed to create work
Answer:
b- how much work can be done in a given time
how much work can be done in a given time
what is the principal of moment
Answer:
hope it helps...
Explanation:
The Principle of Moments states that when a body is balanced, the total clockwise moment about a point equals the total anticlockwise moment about the same point.
Answer:
The Principle of Moments states that when a body is balanced, the total clockwise moment about a point equals the total anticlockwise moment about the same point.
A man climbs on to a wall that is 3.6 m high and gains 2222.64 J of potential
energy. What is his mass?
63 kg
17.5 kg
405 kg
617.4 kg
How can models help us understand energy?
Which statement describes the interaction between the north and south poles of two magents?
Answer:
A south pole attracts a north pole :)
Explanation:
A car drives horizontally off a cliff that is 30 meters high killing the driver on impact. The grieving widow claims her husband would never have driven faster than the posted speed limit of 35 mph and has hired a lawyer to sue the city for negligence. You are sent to investigate the crash site. You measure that the car landed 75 m from the base of the cliff [1 mph = 0.447 m/s].
A. Draw the detailed sketch for this motion with all of the relevant information, variables, and numbers contained in the sketch.
B. Write the three projectile equations with the known values and unknown variables placed into the equation.
C. Determine the time of fall. Then determine vix of the car as it flies off the cliff.
D. Based on what you were able to determine, assess whether the city is held liable for the accident
Answer:
b) 0 = y₀ + 0 - ½ g t² , v₀ₓ = x / t
c) v₀ₓ = 30.31 m / s
Explanation:
This is a projectile launching exercise, in your statement you have units of several systems, we are going to reduce everything to the SI system
v = 35 mph (0.447 m / s / 1 mph) = 15.645 m / s
y₀ = 75 m
A) In the attachment we can see a diagram of the movement of the vehicle in its fall
B) let's find the time for the fall to the base of the cliff (y = 0)
y = y₀ + [tex]v_{oy}[/tex] t - ½ g t²
when the vehicle leaves the cliff it goes horizontally, so its vertical speed is zero (v_{oy} = 0)
0 = y₀ + 0 - ½ g t²
t = √2y₀ / g
with this time we can use the equation of motion on the x axis
x = v₀ₓ t
v₀ₓ = x / t
C) we perform the calculations
t = √ (2 30 / 9.8)
t = 2.474 s
v₀ₓ = 75 / 2.474
v₀ₓ = 30.31 m / s
D) as we can see, the vehicle speed when leaving the cliff is almost twice the allowed speed (15.6 m / s)
therefore the city is not responsible for the accident
5 13. If a train going 60 m/s hits the brakes, and it takes the train 2 minute 25 seconds to stop, what is the train's acceleration?
Answer:
Acceleration = -12/17 m/s ^2
Explanation:
Name the variables
V = final velocity( velocity after the train stops) u= intial velocity(velocity when the train was moving at constant speed, before brakes are applied)
A = acceleration( in this case, it is deceleration)
T = time ( time taken to fully stop)
Second Step: What equation should we use?
The 3 Big Motion in One Dimension Formulas are:
V = u+at
v^2 = u^2 + 2as
S = ut+ (1/2) at^2
For this question, we will be using the first one.
Third Step: Solve
V = 0 (the final velocity is 0, because the train stopped)
U = 60 (the train was going at 60 mps)
T = 85 seconds( 1 min and 25 sec)
0 = 60+ 85a
A package is dropped from an air balloon two times. In the first trial the distance between the balloon and the surface is Hand in the second trial 4H. Compare the FINAL VELOCITY'S for the packages right as they hit the
ground (or right before)?
Answer:
The final speed of the second package is twice as much as the final speed of the first package.
Explanation:
Free Fall Motion
If an object is dropped in the air, it starts a vertical movement with an acceleration equal to g=9.8 m/s^2. The speed of the object after a time t is:
[tex]v=gt[/tex]
And the distance traveled downwards is:
[tex]\displaystyle y=\frac{gt^2}{2}[/tex]
If we know the height at which the object was dropped, we can calculate the time it takes to reach the ground by solving the last equation for t:
[tex]\displaystyle t=\sqrt{\frac{2y}{g}}[/tex]
Replacing into the first equation:
[tex]\displaystyle v=g\sqrt{\frac{2y}{g}}[/tex]
Rationalizing:
[tex]\displaystyle v=\sqrt{2gy}[/tex]
Let's call v1 the final speed of the package dropped from a height H. Thus:
[tex]\displaystyle v_1=\sqrt{2gH}[/tex]
Let v2 be the final speed of the package dropped from a height 4H. Thus:
[tex]\displaystyle v_2=\sqrt{2g(4H)}[/tex]
Taking out the square root of 4:
[tex]\displaystyle v_2=2\sqrt{2gH}[/tex]
Dividing v2/v1 we can compare the final speeds:
[tex]\displaystyle v_2/v_1=\frac{2\sqrt{2gH}}{\sqrt{2gH}}[/tex]
Simplifying:
[tex]\displaystyle v_2/v_1=2[/tex]
The final speed of the second package is twice as much as the final speed of the first package.
How long does it take a cheetah that runs with a velocity of 34m/s to run 750m?
Answer:
It would take 5 seconds
Explanation:
I can't find the 'delta' sign nor the vector sign so just pretend that displacement, time and velocity has them.
V = d / t
34 = 170 / t
34 x t = 170
34t = 170
t = 170 / 34
t = 5
A car rounds a banked curve as we will discuss in class on Tuesday. The radius of curvature of the road is R and the banking angle is θ. (a) In the absence of friction, what is the safe speed for the car to take this curve? (b) Now assume the coefficient of friction between the car’s tires and the road is µs. Determine the range of speeds the car can have without slipping up or down the road. (c) What is the minimum value of µs that makes the minimum speed zero? (d) If θ = 25.0 ◦ , for what values of µs can the curve be taken at any speed? Note: The upper limit of µs you will find is practically impossible to achieve for the car’s tires and the road.
Answer:
A) v = √[(rg(tan θ - µ_s))/(1 + (µ_s•tan θ))]
B)√[(rg(tan θ - µ_s))/(1 + (µ_s•tan θ))] ≤ v ≤ v = √[(rg(tan θ + µ_s))/(1 - (µ_s•tan θ))]
C) µ_s = tan θ
D) µ_s = 0.4663
Explanation:
A) The forces acting on the car will be;
Force due to friction; F_f
Force due to Gravity; F_g
Normal Force; F_n
Now, let us take the vertical direction to be j^ and the direction approaching the centre to be i^ downwards and parallel to the road surface by k^.
Also, we will assume that initially, F_n is in the negative k^ direction and that it will have a maximum possible value of; F_f = µ_s × F_n
Thus, sum of forces about the vertical j^ direction gives;
ΣF_j^ = F_n•cos θ − mg + F_f•sin θ = 0
Since F_f = µ_s × F_n ;
F_n•cos θ − mg + (µ_s × F_n × sin θ) =0
F_n = mg/[cos θ + (µ_s•sin θ)]
Also, sum of forces about the centre i^ direction gives;
ΣF_i^ = F_n(sin θ + (µ_s•cos θ)) = mv²/r
Plugging in formula for F_n gives;
ΣF_i^ = [mg/[cos θ + (µ_s•sin θ)]] × (sin θ + (µ_s•cos θ)) = mv²/r
Making v the subject gives;
v = √[(rg(tan θ - µ_s))/(1 + (µ_s•tan θ))]
B) What we got in a above is the minimum speed the car can have while going round the turn.
The maximum speed will be gotten by making the frictional force(F_f) to point in the positive k^ direction. This means that F_f will be negative.
Now, if we change the sign in front of F_f in the equation in part a that led to the minimum velocity, we will have the maximum as;
v = √[(rg(tan θ + µ_s))/(1 - (µ_s•tan θ))]
Thus the range is;
√[(rg(tan θ - µ_s))/(1 + (µ_s•tan θ))] ≤ v ≤ v = √[(rg(tan θ + µ_s))/(1 - (µ_s•tan θ))]
C) For the minimum speed to be 0, it implies that F_f will be in the negative k^ direction. Thus, Sum of the forces in the k^ direction gives;
ΣF_k^ = mg(sin θ - µ_s•cos θ) = 0
Thus;
mg(sin θ - µ_s•cos θ) = 0
Making µ_s the subject gives;
µ_s = sin θ/cos θ
µ_s = tan θ
D) If θ = 25.0°;
Thus;
µ_s = tan 25
µ_s = 0.4663
PLEASEEEEE!!!!
A cube has a mass of 100 g and a volume of 50 mL. What is the density of the cube?
Answer:
The answer is 2.0 g/mLExplanation:
The density of a substance can be found by using the formula
[tex]density = \frac{mass}{volume} \\[/tex]
From the question
mass = 100 g
volume = 50 mL
We have
[tex] density = \frac{100}{50} = \frac{10}{5} \\ [/tex]
We have the final answer as
2.0 g/mLHope this helps you
9. A plane starts at rest & accelerates along the ground before takeoff. It
moves 1600m in 18s. Calculate the acceleration rate of the plane. *
Answer:
9.877 m/s^2
Explanation:
The acceleration can be computed from ...
d = (1/2)at^2
(1600 m) = (1/2)a(18 s)^2
a = (1600/162) m/s^2 ≈ 9.877 m/s^2
Alcohol wiped across a tabletop rapidly disappears. What happened to the temperature of the tabletop
Answer:
it gets colder
Explanation:
because if you put anything on a table top it will become colder
A 55-kg packing crate is pulled with constant speed across a rough floor with a rope that is at an angle of 40.0o above the horizontal. If the tension in the rope is 125 N, how much work is done on the crate to move it 5.0 m?
Answer:
478.75 J
Explanation:
W=force* displacement
constant speed= (a=0) net F=0
Horizontal component of tension
Tcosx
125Ncos40= 95.76 N
W= (95.76 N)(5 m)
=478.75 J
The work done in moving the crate across the given distance is 478.75 J.
The given parameters;
Mass of the packing create, m = 55 kgAngle of inclination of the rope, Ф = 40°Tension on the rope, T = 125 NDistance through which the crate is the moved, d = 5 mThe work done in moving the crate is the product of the horizontal component of the tension and the distance through which the crate is moved.
The work-done in moving the crate is calculated as;
W = Tcos(Ф) x d
W = 125cos(40) x 5
W = 478.75 J.
Thus, the work done in moving the crate across the given distance is 478.75 J.
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If a person has the values for an object's density and volume, what value can be calculated?
o the object's size
o the object's mass
the shape the object forms in a container
the amount of space the object takes up
Answer:
Answer:
The object's mass.
Explanation:
The formula d=m/v.
d --> density
m--> mass
v --> volume
With density and volume given, you can calculate the mass of the object.
Explanation:
What do virtually all daily task require ?
Answer:muscle strength
Explanation: bc well you would need it
How much distance did this object travel in meters between 0 and 10 seconds
Answer:
5m
Explanation:
Answer:
5
Explanation:
because that's the average of how far it when the most
distinguish between current and current density.
Answer:
1- the rate of flow of charge through a conductor is called current .whereas ,current density is the current per unit area of conductor
43. The particle consists of fast moving electrons is.
Answer:
Explanation: someone help me please you want free points right