Answer:
The Aluminum
Explanation:
With a larger specific heat, water requires more heat to raise its temperature by a temperature degree.
In this system, with equal masses of water and aluminum, the heat moving from the aluminum lowering its temperature by one degree is not sufficient to raise the water temperature by one degree.
A train slows its speed from 52 kilometers per hour to 46 kilometers per hour in 0.04 hour. What is the acceleration o the train during this time?
Answer: here you go i have to put 20 letters in so just ignore this and look at the link.
I need the answer fast for c pleaseeee
Answer:
F = - K x for spring (note that that F here is given in grams, F = m g is correct)
K here is 100 g / cm for the spring constant
x = -420 g / 100 g/cm = -4.2 cm
The spring would compress 4.2 cm for a total length of 20 - 4.2 = 15.8
d) to compress the spring 6.8 cm one can see that the load would be 680 g
Disk A, with a mass of 2.0 kg and a radius of 40 cm , rotates clockwise about a frictionless vertical axle at 50 rev/s . Disk B, also 2.0 kg but with a radius of 20 cm , rotates counterclockwise about that same axle, but at a greater height than disk A, at 50 rev/s . Disk B slides down the axle until it lands on top of disk A, after which they rotate together.
After the collision, what is magnitude of their common angular velocity (in rev/s)?
Hi there!
For this problem, we must use the conservation of angular momentum. This is an example of an inelastic "collision", so:
I₁w₁ + I₂w₂ = (I₁ + I₂)wf
We know that the moment of inertia of a disk is 1/2mR², so we can calculate the moments of inertia for both disks:
Disk 1: 1/2(2)(0.40²) = .16 kgm²/s
Disk 2: 1/2(2)(0.20²) = .04 kgm²/s
Plug in the values. Let counterclockwise be positive.
.16(-50) + .04(50) = (.16 + .04)wf
Solve:
wf = -30 rev/s
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Answer:
jsbdhdndmlsusgsbkaksudgnslsosufhbf ffb
g What is the CD's moment of inertia for rotation about a perpendicular axis through the edge of the disk
Answer:
Explanation:
A CD has an OD of 120 mm and an ID of 15 mm and has a mass between 14 and 33 grams. Let's call it m
Lets call the outer and inner radii R and r respectively
Find the moment of inertia about a line perpendicular to the surface of the disc through its center. We can integrate or look up the result from standard tables
I = ½m(R² + r²)
then use the parallel axis theorem to shift the position of the axis
I = ½m(R² + r²) + md²
where d is the distance of the shift. In this case d = R
I = ½m(R² + r²) + mR²
I = m(1.5R² + 0.5r²)
If we select a mass of say 20 grams
I = 0.020(1.5(0.060²) + 0.5(0.0075²))
I = 0.0001085625 kg•m²
The elevation at the base of a ski hill is 350 m above sea level. A ski lift raises a skier (total mass=72 kg, including equipment) to the top of the hill. If the skier's gravitational potential energy relative to the base of the hill is now 9.2 x 105 J, what is the elevation at the top of the hill?
The elevation at the top of the hill is 1,653.85 m.
The given parameters;
initial height of the skier, h₁ = 350 mlet the final height of the skier at the hill top, = h₂total mass, m = 72 kggravitational potential energy of the skier, P.E = 9.2 x 10⁵ JThe elevation at the top of the hill is calculated as follows;
[tex]P.E = mg\Delta h\\\\P.E = mg(h_2 -h_1)\\\\h_2 -h_1 = \frac{P.E}{mg} \\\\h_2 = \frac{P.E}{mg} + h_1\\\\h_2 = \frac{9.2 \times 10^5 }{72 \times 9.8} \ + \ 350 \ m\\\\h_2 = 1,653.85 \ m[/tex]
Thus, the elevation at the top of the hill is 1,653.85 m.
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Elevations on the tongue are called
sulci
taste buds
papillae
gyri
Answer:
Papillae is correct
Explanation:
hope it helps you
Answer:
Papillae is the correct answer of this question
Seven friends equally split a restaurant bill that
comes to $93.17. How much does each person pay?
Answer:
$13.31
Explanation:
We know that the bill comes to $93.17 and that 7 people will split the bill equally
We can just use the equation
bill = $93.17/7
bill = $13.31
A 1-kg mass at the Earth's surface weighs how much
Answer:
the answer is weight=10N
Answer:
[tex]\boxed {\boxed {\sf 9.8 \ Newtons}}[/tex]
Explanation:
Weight is also called the force of gravity. This force acts on all objects at all times, pulling them down toward the center of the Earth.
It is calculated by multiplying the mass by the acceleration due to gravity.
[tex]F_g=mg[/tex]
The mass of the object is 1 kilogram. This scenario is occurring on Earth, so the acceleration due to gravity is 9.8 meters per second squared.
m= 1 kg g= 9.8 m/s²Substitute the values into the formula.
[tex]F_g= 1 \ kg *9.8 \ m/s^2[/tex]
Multiply.
[tex]F_g= 9.8 \ kg*m/s^2[/tex]
Convert the units. 1 kilogram meter per second squared is equal to 1 Newton, so our answer of 9.8 kilogram meters per second squared is equal to 9.8 Newtons.
[tex]F_g= 9.8 \ N[/tex]
A 1 kilogram mass at Earth's surface weighs 9.8 Newtons.
F (N)
4
* 0
3
A
2
FIGURE 2
t(s)
5
0
1
2
3
4
3) A force of magnitude Fx acting in the x-direction on a 2.00 kg particle varies in time as shown
in FIGURE 2. Find
a) The impulse of the force
b) The final velocity of the particle if it is initially at rest
c) The final velocity of the particle if it is initially moving along the x-axis with velocity
of -2.00 ms -1
Answer:
Mark me as brainlist please.
6. How are the temperature of the universe and Cosmic Microwave Background (CMB) related?
A. Astronomers use the temperature of CMB as the warmest temperature in the universe
B. Astronomers calculate the temperature of the universe based on the coldest part of the CMB
C. Astronomers consider the temperature of the universe to be the temperature of CMB
D. Astronomers never consider the temperature of CMB when looking at the temperature of the universe
Answer:
I think the answer is (A)...
Hope this helps!
Pendulum makes 12 complete swings in 8 seconds, what are its frequency and period on earth
Hi there!
We can begin by finding the period of the pendulum.
[tex]T = \text{ # of complete swings / seconds} = 12 / 8 = \boxed{\text{1.5 sec}}[/tex]
The frequency is simply the reciprocal of the period, so:
[tex]f = \frac{1}{T} = \frac{1}{1.5} = \frac{2}{3}Hz \text{ or } \boxed{0.67 Hz}[/tex]
The equation for a progressive wave is y=6 cos(20t-4x) What is the equation of another progressive wave which has twice the amplitude and frequency, and moving in the same direction?
The equation of the progressive wave is y = 12 cos(40t - 4x)
The general wave equation is given by:
y = A sin(ωt - kx)
Where A is the amplitude, ω is the angular frequency = 2πf, f is the frequency, k is the wave number and y, x is the displacement.
Given the equation for a progressive wave is y=6 cos(20t-4x). Hence:
The amplitude A = 6,
ω = 20 = 2πf
f = 20/2π = 3.183 Hz
Twice the amplitude = 2 * 6 = 12, twice the frequency = 2 * 3.183.
ω = 2π(3.183*2) = 40
Therefore the other progressive wave has an equation of:
y = 12 cos(40t - 4x)
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Rachel drops a ball from a hot–air balloon while her friend Lisa is watching her from the ground. Which statement about the ball's motion is true from Lisa's point of view?
Assume that there is no air resistance and the hot–air balloon is moving horizontally.
A. The ball drops to the ground along a straight–line path.
B.When the ball lands, the hot–air balloon will be ahead of it.
C. When the ball lands, the hot–air balloon will be behind it.
D. When the ball lands, the hot–air balloon will be directly above it.
Answer:
According Lisa, both the ball and the balloon have the same forward velocity of Vx.
(D) is correct
Convert :
36°C = ... °F
373 K = ... °C
Question easy
Answer:
36 C= 96.8 F
373 K= 99.85
Explanation:
C to F: (36 x 1.8) + 32
= 64.8 +32
= 96.8 F
K to C: C= K- 273.15
C= 373-273.15
C= 99.85
____
= 36°C
=( 36 × 9/5 ) + 32
=(36 ÷ 5 × 9) + 32
=(7,2 × 9) + 32
= 64,8 + 32
= 96,8°F______
______
= 373 K
= 373 - 273
= 100°C[tex] \boxed { \sf semoga \: membantu \: :v }[/tex]
A 0.50-kg mass is attached to a spring of spring constant 20 N/m along a horizontal, frictionless surface. The object oscillates in simple harmonic motion and has a speed of 1.5 m/s at the equilibrium position. What is the total energy of the system?
Hi there!
With the work-energy theorem for oscillating springs:
ME = KE + PE
[tex]ME = \frac{1}{2}mv^2 + \frac{1}{2}kx^2[/tex]
Where:
m = mass (kg)
v = velocity (m/s)
k = Spring Constant (N/m)
x = displacement from equilibrium (m)
If the object is at the equilibrium position, there is NO potential energy since:
[tex]\frac{1}{2}k(0^2) = 0 J[/tex]
Thus:
[tex]ME = \frac{1}{2}mv^2[/tex]
Plug in the given values:
[tex]ME = \frac{1}{2}(0.50)(1.5^2) = \boxed{0.5625 \text{ J}}[/tex]
The symbol or variable to used find initial velocity is
Answer:
v down exponenet 1 brainlest
Explanation:
Answer:
v0 [vee nought] is the initial velocity when time=0
a convex mirror and a plane mirror both give virtual and erect images still a convex mirror is used in vehicles. why?
Pls answer thiss
Answer:Convex mirrors are used because these mirrors provide a wider viewing angle than a plane mirror. This wide angle will help you getting more information/overview than what is happening at a narrow spot right behind the car if you use a plane mirror.
With a convex mirror you are for example able to detect an overtake (by the car behind you) early, if you for some reason wanted to turn left into another lane at the same moment the overtake took place - so you then can prevent a collision. Convex mirrors are simply covering a much larger area behind the car than plane mirrors do. And in the US, on the mirrors there is a text explaining that the vehicle behind you is closer than it appears - some kind of an idiot explanation in case some driver took the mirror image literally….because in the mirror image of a convex mirror, everything looks smaller and further away than they actually are.
Explanation: mark me as brainliest this is my best answer till now
A skater is spinning with his arms outstretched. He has a 2 lb weight in each hand. In an attempt to change his angular velocity he lets go of both weights (by just opening his grip). Does he succeed in changing his angular velocity
Answer:
No
Explanation:
Changing momentum of any kind requires work. Work is a force acting over a distance. While holding the weights at arms length and spinning will create a force (centripetal), there is no radial distance change incurred. Releasing the weights will reduce the force to zero, still no work done and no change in angular momentum.
If he was holding the weights at arms length while spinning and he pull his hands to his chest, there now exists both the centripetal force and a distance in the direction of that force (inward radial) this work will result in an increase in angular velocity as moment of inertia has decreased with the work done.
No, the skater doesn't succeed in changing his angular velocity.
Conservation of angular momentumThe final angular velocity of the skater is determined by applying the principle of conservation of angular momentum as shown below;
Li = Lf
[tex]Ii\omega _i = I_f \omega _f[/tex]
where;
Ii is the initial moment of inertia of the skaterIf is the final moment of inertia of the skaterωi is the initial angular speed of the skaterωf is the final angular speed of the skaterWhen the skater holds the weight, the momnet of inertia of both arms is the same. Also when the skater drops the weight, the moment of inertia of both arms is still the same. Thus, at any instant, the moment of inertia of the two arms is the same.
To change the angular speed, the initial and final moment of inertia of the two arms must be different. Thus, the skater doesn't succeed in changing his angular velocity.
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Objects 1 and 2 attract each other with a gravitational force of 178 units. If the mass of object 1 is one-fourth the original value AND the mass of object 2 is tripled AND the distance separating objects 1 and 2 is halved, then the new gravitational force will be _____ units.
Explanation:
Fgravity = G*(mass1*mass2)/D²
G is the gravitational constant, which has the same value throughout our universe.
D is the distance between the objects.
now, several numbers change.
Fgravitynew = G*((1/4)*mass1*3*mass2)/(1/2 * D)² =
= G*((3/4)*mass1*mass2)/(D²/4) =
= (3/4)* (G*(mass1*mass2)/D²) *4 =
= 4*(3/4)* (G*(mass1*mass2)/D²) =
= 3* (G*(mass1*mass2)/D²) = 3* Fgravity
the new gravitational force will be 3×178 = 534 units.
A man is whirling a 0.25 kg ball on a 1.5 m long string at 3 m/s. Find the centripetal acceleration of this ball.
Question 2 options:
0.5 m/s2
13.5 m/s2
6 m/s2
2 m/s2
The centripetal acceleration of this ball is equal to 12 [tex]m/s^2[/tex]
Given the following data:
Diameter = 1.5 mSpeed, V = 3 m/s.Mass = 0.25 kgRadius = [tex]\frac{Diameter}{2} = \frac{1.5}{2} = 0.75 \;meters[/tex]
To find the centripetal acceleration of this ball:
The acceleration of an object along a circular track is referred to as centripetal acceleration.
Mathematically, the centripetal acceleration of an object is given by the formula:
[tex]A_c = \frac{V^2}{r}[/tex]
Where:
Ac is the centripetal acceleration.r is the radius of the circular track.V is the velocity of an object.
Substituting the given parameters into the formula, we have;
[tex]A_c = \frac{3^2}{0.75}\\\\A_c = \frac{9}{0.75}\\\\A_c = \frac{9}{0.75}[/tex]
Centripetal acceleration = 12 [tex]m/s^2[/tex]
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A small 1240-kg SUV has a wheelbase of 3.2 m. If 67% of its weight rests on the front wheels, how far behind the front wheels is the wagon's center of mass
Answer:
Explanation:
Let d be the distance to the center of mass from the front wheels
Sum moments about the front wheel contact point to zero
1240(9.8)[d] - 1240(9.8)(1 - 0.67)[3.2] = 0
1240(9.8)[d] = 1240(9.8)(1 - 0.67)[3.2]
d = (1 - 0.67)[3.2]
d = 1.056 m
If m1 is 24 kilograms, m2 is 12 kilograms, and mbar is 10 kilograms, what is the direction of rotation and the sign of the angular acceleration
From the diagram, the angular speed will increase clockwise, the sign of the angular acceleration will be negative and the direction of rotation will be clockwise direction and the sign of the angular acceleration is negative. The correct answer is option B
Given that two objects of masses m1 and m2 are attached to a seesaw. The seesaw is made of a bar that has length l and is pivoted so that it is free to rotate in the vertical plane without friction. Counterclockwise is considered the positive rotational direction.
If m1 is 24 kilograms, m2 is 12 kilograms, and mbar is 10 kilograms, The moment of object m1 will be equal to the moment of object m2 without the Mbar
Let assume that the length L of the seesaw is 9 cm.
Anticlockwise moment = 24 x 9/3 = 72Nm
Clockwise moment = 12 x 2(9/3) = 72 Nm
With the consideration of mass of the bar Mbar, this will add to clockwise moment of the seesaw.
Therefore, the direction of rotation will be clockwise direction.
Angular acceleration is positive when object is speeding up and negative when slowing down. Also, angular acceleration is positive when speed increases in an anticlockwise direction and negative when speed increases in the clockwise direction.
From the diagram, since the angular speed increase clockwise, the sign of the angular acceleration will be negative.
We can conclude that the direction of rotation will be clockwise direction and the sign of the angular acceleration is negative.
The correct answer is option B
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Can anyone help me with question 10 a.
Answer:
it's ahfdfhhh hhgfdjjjjuyggffdddcff
two billiard balls moving along the same line hit each other head-on. each has a mass of 0.220 kg; one has an initial velocity of 1.84 m/s, the other an initial velocity of 0.530 m/s. if the collision is elastic, what are their final velocities? ignore friction.
Hi there!
Since the collision is elastic, we must also satisfy the following condition:
Ei = Ef, or:
KEi = KEf
Begin by writing an expression for momentum. (p = mv) Remember that one ball's direction is negative; in this instance, we can let the second ball be moving LEFT.
mv1 + mv2 = mvf1 + mvf2
0.220(1.84) + 0.220(-.530) = 0.220(vf1 + vf2)
0.2882/0.220 = vf1 + vf2
1.31 = vf1 + vf2
Now, we can express this as a conservation of energy:
1/2mv1² + 1/2mv2² = 1/2mvf1² + 1/2mvf2²
Plug in values and simplify:
0.403315 = 1/2m(vf1² + vf2²)
Simplify further:
3.6665 = vf1² + vf2²
Use the equation derived from momentum above and solve for one variable:
vf2 = 1.31 - vf1
Plug in this expression for vf2:
3.6665 = vf1² + (1.31 - vf1)²
Expand:
3.6665 = vf1² + 1.7161 - 2.62vf1 + vf1²
Simplify:
1.9504 = -2.62vf1 + 2vf1²
Solve for vf1 using a graphing calculator:
vf1 = -0.53 m/s or 1.84 m/s; we must figure out which one is correct.
Since v1 is heading to the right initially with a velocity of 1.84 m/s, we know that the ball's velocity could not have stayed the same in both magnitude and direction, so the final velocity must be -0.53 m/s.
Now, we can solve for the velocity of the other ball (initial of 0.53 m/s):
vf2 = 1.31 - (-0.53) = 1.84 m/s.
Now, you could have also made the connection that when two balls of the SAME MASS experience an ELASTIC collision, the velocities are simply "exchanged" from one to another. I just used this more "extensive" method to prove this.
The volume of a toy car was calculated by displacing water. The water
rose by 20ml when the object was placed into the graduated cylinder. The balance showed the toy car had a
mass of 500grams. Calculate the density of the toy car
Answer:
D = 25g/cm³
Explanation:
1ml = 1cm³
D = m/V
D = 500g/20cm³
D = 25g/cm³
How many states of matter are there?
Answer:
3
Explanation:
state of matter are solid
liquid and
gases
The current in a resistor is 3.0 A, and its power is 60 W. What is the voltage?
Answer:
20 volts
Explanation:
Use the equation [tex]P=VI[/tex]
[tex]60=V(3)[/tex]
[tex]V=20[/tex]
PLEASE HELP!
A 2457 kg car moves with initial speed of 18 ms-?. It is stopped in 62 m by its brakes.
How much work is done by the brakes?
Answer:
Explanation:
The work of the brakes will equal the initial kinetic energy of the car
Fd = ½mv²
F = mv²/2d
F = 2457(18²) / (2(62))
F = 6,419.903...
F = 6.4 kN
2) A rolling disk, mass m and radius R, approaches a step of height R/2 with velocity v. (i) Taking the corner of the step as the pivot point, what is the initial angular momentum of the disk
The rolling disk's initial angular momentum is mR√[2(gR + v²)]/2
Using the law of conservation of energy, the initial mechanical energy E of the disk equals its final mechanical energy E' as it climbs the step.
So, E = E'
1/2Iω + 1/2mv² + mgh = 1/2Iω' + 1/2mv'² + mgh'
where I = rotational inertia of disk = 1/2mR² where m = mass of disk and R = radius of disk, ω = initial angular speed of disk, v = initial velocity of disk, h = initial height of disk = 0 m, ω' = final angular speed of disk = 0 rad/s (assumung it stops at the top of the step), v' = final velocity of disk = 0 m/s (assumung it stops at the top of the step), and h' = final height of disk = R/2.
Substituting the values of the variables into the equation, we have
1/2Iω² + 1/2mv² + mgh = 1/2Iω'² + 1/2mv'² + mgh'
1/2(1/2mR² )ω² + 1/2mv² + mg(0) = 1/2I(0)² + 1/2m(0)² + mgR/2
mR²ω²/4 + 1/2mv² + 0 = 0 + 0 + mgR/2
mR²ω²/4 + 1/2mv² = mgR/2
R²ω²/4 = gR/2 + 1/2v²
R²ω²/4 = (gR + v²)/2
ω² = 2(gR + v²)/R²
ω² = √[2(gR + v²)/R²]
ω = √[2(gR + v²)]/R
Since angular momentum L = Iω, the rolling disk's initial angular momentum is
L = 1/2mR² ×√[2(gR + v²)]/R
L = mR√[2(gR + v²)]/2
the rolling disk's initial angular momentum is mR√[2(gR + v²)]/2
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