Answer:
(d) is your answer for your question
A block slides down a smooth ramp, starting from rest at a height h. When it reaches the bottom it's moving at speed v. It then continues to slide up a second smooth ramp. At what height is its speed equal to v/2
Answer:
3h/4
Explanation:
At speed v/2 height will be 3/4 h
What is equation of motion in kinematics?Equation that describes the motion of point , bodies , and system of bodies without considering the force that cause them to move is called equation of motion in kinematics
When block is at top of first ramp
u=0 ( block was at rest )
a = g ( acceleration due to gravity
using equation of motion
2as = v^2 - u^2
2gh = v^2
Then the block continued and reached a speed of v1 = v/2 on second ramp
now , final velocity = v= v1 =[tex]\sqrt{2gh}[/tex] / 2
u= [tex]\sqrt{2gh\\}[/tex]
s= h1
using equation of motion , we get
2as = v^2 - u^2
2(-g)h1 =( [tex]\sqrt{2gh}[/tex]/2)^2 - [tex]\sqrt{2gh}[/tex]
2(-g)h1 = (g h - 4 g h) / 2
h1 = 3/4 h
At speed v/2 height will be 3/4 h
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Work is required to lift a barbell. How many times more work is required to lift the barbell 2 times as high?
Answer:
2 times more work
Explanation:
Work is a force times distance
W = Fd
The barbell does not change mass so its weight (force) is constant.
That means work is directly proportional to the distance traveled.
twice the distance means twice the work.
A Vector that starts from
Origin is called what?
Find the work done by the force field F in moving an object from A to B. F(x, y) = 6y3/2i + 9x y j A(1, 1), B(3, 4)
Answer:
138
Explanation:
(since there's a blank next to "9x y j", I'm assuming the y of F(x, y) is 9x[tex]\sqrt{y}[/tex] j)
1) find the partial derivative of each:
[tex](6y^{\frac{3}{2} })i + (9x\sqrt{y} )j[/tex]
[tex]f_{x} =(6y^{\frac{3}{2} })_{x} = \int\limits{6y^{\frac{3}{2} }} \, dx = 6xy^{\frac{3}{2} } +c \\\\f_{y} = (9x\sqrt{y} )_{y} = \int\limits{9xy^{\frac{1}{2} } } \, dy = 9x(\frac{2}{3} )y^{\frac{3}{2} } +c = 6xy^{\frac{3}{2} } +c[/tex]
2) use partial integrals to make gradient of f:
take whatever you got from partial integral and add them together (if they repeat, just use it once)
[tex]F =[/tex] Vf (V = gradient of)
[tex]F(x, y) = 6xy^{\frac{3}{2} }[/tex]
3) Evaluate the integrals with given points:
Integral of F dotted with dr = F(point B) - F(point A) = F(3, 4) - F(1, 1)
[tex]F(point B) = 6(3)(4)^{\frac{3}{2} }\\F(Point A) = 6(1)(1)^{\frac{3}{2}}\\F(point B) - F(pointA) = 6(3)(4)^{\frac{3}{2} }-(6(1)(1)^{\frac{3}{2}})[/tex]
= 144 - 6 = 138 units of work
Work done by the force field F in moving an object from A to B = 138 J
Given data :
Force field F(x,y) = [tex]6y^{\frac{3}{2} }i + (9x\sqrt{y} ) j[/tex]
Step 1 : determine the partial derivatives of the vector quantity
Fx = ∫ [tex]6y^{\frac{3}{2} }i = 6xy^{\frac{3}{2} } + c[/tex]
Fy = ∫ [tex](9x \sqrt{y})_{y} = 9x(\frac{2}{3})y^{\frac{3}{2} } + c[/tex]
Equating the partial derivatives :
[tex]9x(\frac{2}{3})y^{\frac{3}{2} } + c[/tex] = [tex]6xy^{\frac{3}{2} } + c[/tex]
therefore the gradient of F i.e. F = vF = F( x,y ) = [tex]6xy^{\frac{3}{2} }[/tex]
Next step : Determine the work done
Work done ( F.dr ) = [ F(point b ) = F( 3,4 ) ] - [ F(point A) = F( 1,1 ) ]
F(3,4 ) = 6(3)(4)[tex]^{\frac{3}{2} }[/tex] = 144
F( 1,1 ) = 6(1)(1)[tex]^{\frac{3}{2} }[/tex] = 6
Therefore the work done by the force field = 144 - 6 = 138 J
Hence we can conclude that the work done by the force field F is = 138 J
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electric field due to on the surface of the charge sphere
Explanation:
static electricity draws energy
What are the two factors that affect the frictional force between objects
What happened when a lobster release a claw? a. The lobster attacks with the claw. b. The lobster let the claw go. C. The lobster bites off the claw. D. The lobster eats with the claw.
Answer:b
Explanation:
Answer:
The lobster let the claw go. so, B.
Explanation:
i hope this helps.
Two point particles, with masses m and 2 m, are separated by some distance Lin an uniform gravitational field. The center of gravity of this two particle system is:________
А. located L/2 distance away from either mass and on the line joining the two particles.
B. located 1/3 distance away from 2 m and on the line joining the two particles.
C. located L/4 distance away from 2 m and on the line joining the two particles.
D. located L/3 distance away from mand on the line joining the two particles.
E. located L/4 distance away from m and on the line joining the two particles.
F. located where the mass 2 m is located.
The center of gravity of this two particle system is located 1/3 distance away from 2 m and on the line joining the two particles.
The given parameters:
Mass of the first particle, = mMass of the second particle, = 2mDistance between the two particles, = LThe center of gravity of the two particles when first particle is fixed is calculated as;
[tex]C_G = \frac{m(0 ) \ +2m(L) }{m+ 2m} \\\\C_G = \frac{2mL}{3m} \\\\C_G = \frac{2L}{3}[/tex]
The center of gravity of the two particles when second particle is fixed is calculated as;
[tex]C_G = \frac{m(L) \ + \ 2m(0)}{m + 2m} \\\\C_G = \frac{mL}{3m} \\\\C_G = \frac{L}{3}[/tex]
Thus, the center of gravity of this two particle system is located 1/3 distance away from 2 m and on the line joining the two particles.
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What’s 1 + 1 many half a window
Answer:
1
+
1
is
2
Explanation:
because if you add one to a one it would end up being 2 <3
Answer: 2
Explanation: 1+1=2
This was THE hardest question I've ever decided to answer
2. Match the term to the definition
1. Reversibility
a. A break between activities during a workout.
2. Overload
3. Progression
4 Duration
5. Rest
b. Length of an individual workout.
c. The Improvement seen when you are training will be lost if you stop
training.
d. Moving forward through a sequence to gradually build up the body.
e Number of training sessions per week.
f. Level of difficulty of your workout.
9. Improvements will only be seen in the areas that you work on.
h The time between workouts that allows your body to heal and adapt to
the stress from your workout.
6. Recovery
7. Frequency
8. Specificity
1. Increase the amount of stress you put on your body.
9. Intensity
Over half the global population faces a shortage of clean water. Which example is a point source water pollutant? pesticide run-off from farmers’ fields a sewer pipe draining directly into a river road oil washing into sewer drains, and eventually reaching waterways erosion from a denuded slope after a heavy rainfall
An example of a point source water pollutant is a sewer pipe draining directly into a river.
According to the United States Environmental Protection Agency, a point source pollution is defined as; “any single identifiable source of pollution from which pollutants are discharged, such as a pipe, ditch, ship or factory smokestack.”
Hence, an example of a point source water pollutant is a sewer pipe draining directly into a river. The pollution is coming from an easily identifiable source which is a sewer pipe connected to a river directly.
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Throwing all over the place will___________the environment.
Answer:
pollute the environment
PLEASE HELP ME GET THIS RIGHT
Explanation:
I'm not sure, but I would go for the more than A since its orbital speed is at its fastest and the sweep occurs in about the same period of days.
A patient comes to an outpatient laboratory for a physician-ordered fasting test. The patient indicates that he forgot that the test was to be fasting and ate a candy bar 2 hours ago. The patient insists that you should draw the test because he cannot come back at another time. What should you do?
The patient eating a candy bar instead of fasting for the test should be told
that the test results will be wrong and he may receive a wrong diagnosis.
The medical practitioners told him to fast when coming for a reason and he
forgetting and eating something means the objective for the test may have
been defeated.
This is why it's best to explain to him about the consequences of his actions
which is likely getting a wrong result and diagnosis.
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A solid sphere starts from rest and rolls down a slope that is 6.4 m long. If its speed at the bottom of the slope is 5.3 m/s, what is the angle of the slope
From the relationship between acceleration a and g on an inclined plane, the angle of the slope is 13 degrees
Given that a solid sphere starts from rest and rolls down a slope that is 6.4 m long. The speed at the bottom of the slope is 5.3 m/s, the distance travelled is 6.4 m. That is,
Initial velocity U = 0 ( since it starts from rest)
Final velocity V = 5.3 m/s
distance S = 6.4 m
Let us first calculate its acceleration by using third equation of motion.
[tex]V^{2}[/tex] = [tex]U^{2}[/tex] + 2aS
[tex]5.3^{2}[/tex] = 0 + 2 x 6.4a
28.09 = 12.8a
a = 28.09 / 12.8
a = 2.2 m / [tex]s^{2}[/tex]
To calculate the angle of the slope, let us use the relationship between acceleration a and g on an inclined plane.
acceleration a = gsin∅
substitute all the relevant parameters
2.2 = 9.8 sin∅
sin∅ = 2.2/9.8
sin∅ = 0.224
∅ = [tex]Sin^{-1}[/tex](0.224)
∅ = 12.97 degrees
∅ = 13 degrees (approximately)
Therefore, the angle of the slope is 13 degrees
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PLEASE HELP DUE TODAY
When same vectors are in the same direction they add. When same vectors are in opposite directions, they
a, also add
b, subtract
c, multiply
d, divide
Answer:
The answer is a (also add)
Two identical vertical springs S1 and S2 have masses m1 = 400 g and m2 = 800 g attached to them. If m1 causes spring S1 to stretch by 4 cm, what is the ratio of the potential energy of S1 and S2? Use g = 10 m/s^2
Select one:
a. 1:2
b. 4:1
c. 1:4
d. 1:3
e. 2:1
Answer:
potential energy = mgh
= 400÷1000 × 10× 4÷100
= 0.4 × 10 × 0.04
=4/10 ×10×4/100
= 4/10 × 4/10
=16/100
= 0.16 joules
m1 (400) stretches 4cm
m1 (100g) stretches 1cm
so, m2(800g) stretches 8 cm
potential energy of m2 = mgh
= 800/1000 ×10×8/100
= 0.8 × 0.8
=8/10 ×8/10
= 64/100
=0.64 joules
Ratio of s1 to s2
16/100 ÷ 64/100
= 1:4 ( answer)
)Suppose you have two insulated buckets containing the same amount of water at room temperature. You also happen to have two blocks of metal of the same mass, both at the same temperature, warmer than the water in the buckets. One block is made of aluminum and one is made of copper. You put the aluminum block into one bucket of water, and the copper block into the other. After waiting a while you measure the temperature of the water in both buckets. Which is warmer
The water in the bucket containing the aluminium block is warmer than the bucket containing the copper block.
The specific heat is the amount of heat needed or required to elevate the temperature of 1 gram of a substance by 1° C.
At standard conditions;
the specific heat of aluminium = 0.215 cal/gm/° Cthe specific heat of copper = 0.0923 cal/gm/° CWe know that:
Heat loss by the metal block = Heat gained by the water bucketTherefore, since the specific heat of aluminium is higher than that of copper, the water bucket containing aluminium block will be warmer than the bucket containing the copper block.
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. If the frequency of a system undergoing simple harmonic motion doubles, by what factor does the maximum value of acceleration change
Answer:
a1 = -w^2 A sin w t where w is the angular frequency
a2 = - (2 w)^2 A sin 2 w t where w2 = 2 w1
a2 / a1 = 4 sin 2 w t / sin w t
Since sin 2 w t / sin w t = 2 sin w t will change twice as fast
a2 / a1 = 8 the max acceleration must change 8 tims as fast
what is democratic means in science
Answer:
relating to or supporting democracy or its principles.
Explanation:
A merry-go-round rotates from rest with an angular acceleration of 1.08 rad/s2. How long does it take to rotate through (a) the first 3.74 rev and (b) the next 3.74 rev
Answer:
Let ω1 be the initial angular speed and ω2 the final angular speed:
α = (ω2- α1) / t
corresponding to a = (v2 - v1) / t
S (distance corresponds to theta)
1 rev = 2 pi and 3.74 rev = 7.48 pi = 23.5 radians
S = 1/2 a t^2 linear or S = 1/2 α t^2 angular acceleration
23.5 = 1/2 * 1.08 t^2 and t = 6.60 sec for first 3.84 rev
b) ω1 = 1.08 * 6.6 = 7.13 rad/sec initial speed for second 3.74
23.5 = 7.13 t + .54 t^2 compare to S = v1 t + 1/2 a t^2
.54 t^2 + 7.13 t - 23.5 = 0
t^2 + 13.2 t - 43.5 = 0
t = 2.7 sec for next 3.74
Check:
7.13 * 2.7 + .54 * 2.7^2 = 23.2 rad = 3.7 rad
define moment of a force about a point and give the si unit of moment?
Answer:
Moment of a force is the product of the force and the perpendicular distance of force from axis of rotation.
The SI unit of force is newton (N).
Explanation:
Hope it helps you again!
A 30.0-g object moving to the right at 20.5 cm/s overtakes and collides elastically with a 13.0-g object moving in the same direction at 15.0 cm/s. Find the velocity of each object after the collision. (Take the positive direction to be to the right. Indicate the direction with the sign of your answer.)
Let m₁ and m₂ be the masses of the two objects, and v₁ and v₂ their initial velocities. So
m₁ = 30.0 g = 0.0300 kg
m₂ = 13.0 g = 0.0130 kg
v₁ = + 20.5 cm/s = 0.205 m/s
v₂ = + 15.0 cm/s = 0.150 m/s
and we want to find v₁' and v₂', the final velocities of either object after their collision.
Momentum is conserved throughout the objects' collision, so that
m₁v₁ + m₂v₂ = m₁v₁' + m₂v₂'
where v₁' and v₂' are the first and second object's velocities after the collision.
Kinetic energy is also conserved, so that
1/2 m₁v₁² + 1/2 m₂v₂² = 1/2 m₁(v₁')² + 1/2 m₂(v₂')²
or
m₁v₁² + m₂v₂² = m₁(v₁')² + m₂(v₂')²
From the first equation (omitting units), we have
0.0300 • 0.205 + 0.0130 • 0.150 = 0.0300 v₁' + 0.0130 v₂'
0.0810 = 0.0300 v₁' + 0.0130 v₂'
81 = 30 v₁' + 13 v₂'
From the second equation,
0.0300 • 0.205² + 0.0130 • 0.150² = 0.0300 (v₁')² + 0.0130 (v₂')²
0.00155 ≈ 0.0300 (v₁')² + 0.0130 (v₂')²
1.55 ≈ 30 (v₁')² + 13 (v₂')²
Solving both equations simultaneously gives two solutions, one of which corresponds to the initial conditions. The other yields
v₁' ≈ + 0.172 m/s
and
v₂' ≈ + 0.227 m/s
The force of earth’s gravity is 10N downward. What us the acceleration of a 15kg backpack if lifted with a a 15N force?
Answer:
F-F(gr) = ma
a= {F-F(gr)}/m =
=(15-10)/15=0.33 m/s² (upward)
which of the following statements describes a perfectly inelastic collision
Answer: An ice hockey player picks up a trophy as he slides past it.
Two pool balls colliding precisely inelastically and rebounding off one another is an example. So, the correct option is A.
What is an Inelastic collision?Inelastic collisions are those in which the total kinetic energy is lower after the impact than it was before. The stick is travelling quickly in the direction of the ball before to the contact. The stick comes to a stop following the accident. It transfers some of its kinetic energy to the cue ball, which rolls forward.
The type of collision mentioned in the given example is known as an inelastic collision is one in which the system's momentum is preserved but its kinetic energy is not. In a ballistic pendulum, an example of an inelastic collision. Dropped ball of clay is another illustration of an inelastic collision.
Thus, the correct option is A.
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Your question is incomplete, most probably the complete question is:
Which of the following statements describes a perfectly inelastic collision?
A. Two billiard balls bounce off of each other,
B. A car crashes into a tree and rebounds in the other direction,
C. A wad of chewing gum is thrown and sticks to a moving truck,
D. Two ice skaters hit each other and fall over in opposite directions,
A tow truck pulls a car 5.00 km along a horizontal roadway using a cable having a tension of 850 N. (a.) How much work does the cable do on the car if it pulls horizontally? If it pulls at 35.0° above the horizontal? (b.) How much work does the cable do on the tow truck in both cases of part (a)? (c.) How much work does gravity do on the car in part (a)?
I think 1980is the answer because you add???
Identify:
In each case the forces are constant and the displacement is along a straight line, so
[tex]$$W=F s \cos \phi \text {. }$$[/tex]
Set-Up:
In part (a), when the cable pulls horizontally [tex]$\phi=0^{\circ}$[/tex] and when it pulls [tex]$35.0^{\circ}$[/tex] above the horizontal [tex]$\phi=35.0^{\circ}$[/tex]. In part (b), if the cable pulls horizontally[tex]$\phi=180^{\circ}$[/tex]. If the cable pulls on the car [tex]$35.0^{\circ}$[/tex] above the horizontal it pulls on the truck at below the horizontal and [tex]$\phi=145.0^{\circ}$[/tex]. For the gravity force [tex]$\phi=90^{\circ}$[/tex], since the force is vertical and the displacement is horizontal.
Execute:
(a) When the cable is horizontal, [tex]$W=(850 \mathrm{~N})\left(5.00 \times 10^{3} \mathrm{~m}\right) \cos 0^{\circ}=4.25 \times 10^{6} \mathrm{~J}$[/tex].
When the cable is[tex]$35.0^{\circ}$[/tex] above the horizontal,[tex]$W=(850 \mathrm{~N})\left(5.00 \times 10^{3} \mathrm{~m}\right) \cos 35.0^{\circ}=3.48 \times 10^{6} \mathrm{~J}$[/tex].
(b)[tex]$\cos 180^{\circ}=-\cos 0^{\circ}$[/tex] and [tex]$\cos 145.0^{\circ}=-\cos 35.0^{\circ}$[/tex],
So the answers are [tex]$-4.26 \times 10^{6} \mathrm{~J}$[/tex] and [tex]$-3.48 \times 10^{6} \mathrm{~J}$[/tex].
(c) Since [tex]$\cos \phi=\cos 90^{\circ}=0, W=0$[/tex] in both cases.
Evaluate: If the car and truck are taken together as the system, the tension in the cable does no net wnetwork
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A magnet is located above circular current. What is the direction of the magnetic force on the magnet
The magnet is attracted to the ring since the north pole of the current loop is above the ring and the south pole is below the ring.
Question 3 of 20 :
Select the best answer for the question.
3. Which of the following accurately describes the behavior of water when subjected to
temperature change?
A. The volume of water will decrease if heated from 6°C to 7°C.
B. The volume of water will increase if cooled from 3°C to 2°C.
O C. A mass of water will contract if cooled from 1°C to 0°C.
D. A mass of water will expand if heated from 0°C to 2°C.
As water cools the volume expands.
Answer: B. The volume of water will increase if cooled from 3°C to 2°C.
who the football games yesterday 2021
what are the greenhouse gasses in the earth that are primarily responsible for the greenhouse effect on Earth
Answer:
Water Vapour, Carbon dioxide, methane,nitrous oxide,ozone.
