Answer:
the force exerted by the umbrella against the drops is the same in both cases
Explanation:
In this exercise the total mass of the raindrops and the hail are equal, therefore the momentum that these drops create is
raindrops I₁ = ∫ F dt
hail I₂ = ∫ F dt
since the mass and the velocity is the same the force is the weight of the drops
F = mg
whereby the impulse is
I₁ = mg t
I₂ = mg t
we can see that the impulses generated by hail and raindrops are equal, therefore the force exerted by the umbrella against the drops is the same in both cases
Electron A is fired horizontally with speed 1.00 Mm/s into a region where a vertical magnetic field exists. Electron B is fired along the same path with speed 2.00 Mm/s. (i) Which electron has a larger magnetic force exerted on it
B will have the greater force
Fc=MV2 /R=Fm
The A particle has less centipetal force and larger radius so larger curve
The Earth’s orbit around the Sun is slightly elliptical. At Earth's closest approach to the Sun (perihelion) the orbital radius is 1.471×10^11m, and at its farthest distance (aphelion) the orbital radius is 1.521×10^11m.
a. Find the difference in gravitational potential energy between when the Earth is at its aphelion and perihelion radii.
b. If the orbital speed of the Earth is 29,290 m/s at aphelion, what is its orbital speed at perihelion?
Answer:
1.25
Explanation:
Assume that a friend hands you a 10-newton box to hold for her. If you hold the box without moving it at a height of 10 meters above the ground, how much work do you do
Answer:
100 Joules
Explanation:
Applying,
W = mgh................... Equation 1
Where W = workdone to hold the box above the ground, mg = weight of the box, h = height of the box.
From the question,
Given: mg = 10 newtons, h = 10 meters.
Substitute these values into equation 1
W = 10×10
W = 100 Joules.
Hence the amount of workdone is 100 Joules
A force of 200 N, acting at 60° to the horizontal, accelerates a block of mass 50 kg along a horizontal plane. Calculate the component of the 200N force that accelerates the block horizontally
Answer:
Explanation:
a) Fx = F cos (θ)
= (200) cos(60)
= 100 N
b) FR = ma
Fx + Ff = ma
100 + Ff = (50)(1,5)
Ff = 75 - 100
= -25 N
c) Fy = F sin θ
= (200) sin(60)
= 173,2 N
A fan is turned off, and its angular speed decreases from 10.0 rad/s to 6.3 rad/s in 5.0 s. What is the magnitude of the angular acceleration of the fan?
A) 0.37 rad/s2
B) 11.6 rad/s2
C) 0.74 rad/s2
D) 0.86 rad/s2
E) 1.16 rad/s2
Answer:
chk photo
Explanation:
Consider a 200-ft-high, 1200-ft-wide dam filled to capacity. Determine (a) the hydrostatic force on the dam and (b) the force per unit area of the dam near the top and near the bottom. Note: we will see that the resultant hydrostatic force will be
Answer:
a) [tex]F_g=1.5*10^9Ibf[/tex]
b) [tex]F_t=12490Ibf/ft^2[/tex]
[tex]F_b=0[/tex]
Explanation:
From the question we are told that:
Height [tex]h=200ft[/tex]
Width [tex]w=1200ft[/tex]
a)
Generally the equation for Dam's Hydro static force is mathematically given by
[tex]F_g=\rho*g*\frac{h}{2}(w*h)[/tex]
Where
[tex]\rho=Density\ of\ water[/tex]
[tex]\rho=62.4Ibm/ft^3[/tex]
Therefore
[tex]F_g=62.4*32.2*\frac{200}{2}(1200*200)[/tex]
[tex]F_g=1.5*10^9Ibf[/tex]
b)
Generally the equation for Dam's Force per unit area is mathematically given by
[tex]F=\rho*g*h[/tex]
For Top
[tex]F_t=\rho*g*h[/tex]
[tex]F_t=62.4*32.2*200[/tex]
[tex]F_t=12490Ibf/ft^2[/tex]
For bottom
[tex]Here \\H=0 zero[/tex]
Therefore
[tex]F_b=0[/tex]
The hydrostatic force on the dam is [tex]2.995 \times 10^9 \ lbF[/tex].
The force per unit area near the top is 86.74 psi.
The force per unit area near the bottom is zero.
Hydrostatic force
The hydrostatic force on the dam is the force exerted on the dam by the column of the water.
[tex]F = PA\\\\F = (\rho gh) \times (wh)\\\\F = (62.4 \times 32.17 \times 200) \times (1200 \times 200)\\\\F = 9.636 \times 10^{10} \ lb-ft/s^2\\\\1 \ lbF = 32.17\ lb-ft/s^2\\\\F = 2.995 \times 10^9 \ lbF[/tex]
Force per unit area near the topThe force per unit area is the pressure exerted near the top of the dam.
[tex]P = \rho gh\\\\P = 0.052 \times \rho h[/tex]
where;
P is pressure in PSI
ρ is density of water in lb/gal
h is the vertical height in ft
[tex]P = 0.052 \times 8.34 \times 200\\\\P = 86.74 \ Psi[/tex]
The pressure near the bottom is zero, become the vertical height is zero.
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A proton traveling at 17.6° with respect to the direction of a magnetic field of strength 3.28 mT experiences a magnetic force of 9.14 × 10-17 N. Calculate (a) the proton's speed and (b) its kinetic energy in electron-volts.
Answer:
a) The proton's speed is 5.75x10⁵ m/s.
b) The kinetic energy of the proton is 1723 eV.
Explanation:
a) The proton's speed can be calculated with the Lorentz force equation:
[tex] F = qv \times B = qvBsin(\theta) [/tex] (1)
Where:
F: is the force = 9.14x10⁻¹⁷ N
q: is the charge of the particle (proton) = 1.602x10⁻¹⁹ C
v: is the proton's speed =?
B: is the magnetic field = 3.28 mT
θ: is the angle between the proton's speed and the magnetic field = 17.6°
By solving equation (1) for v we have:
[tex]v = \frac{F}{qBsin(\theta)} = \frac{9.14 \cdot 10^{-17} N}{1.602\cdot 10^{-19} C*3.28 \cdot 10^{-3} T*sin(17.6)} = 5.75 \cdot 10^{5} m/s[/tex]
Hence, the proton's speed is 5.75x10⁵ m/s.
b) Its kinetic energy (K) is given by:
[tex] K = \frac{1}{2}mv^{2} [/tex]
Where:
m: is the mass of the proton = 1.67x10⁻²⁷ kg
[tex] K = \frac{1}{2}mv^{2} = \frac{1}{2}1.67 \cdot 10^{-27} kg*(5.75 \cdot 10^{5} m/s)^{2} = 2.76 \cdot 10^{-16} J*\frac{1 eV}{1.602 \cdot 10^{-19} J} = 1723 eV [/tex]
Therefore, the kinetic energy of the proton is 1723 eV.
I hope it helps you!
Two horizontal pipes have the same diameter, but pipe B is twice as long as pipe A. Water undergoes viscous flow in both pipes, subject to the same pressure difference across the lengths of the pipes. If the flow rate in pipe B is Q=ΔV/Δt what is the flow rate in pipe A? Viscosity: Two horizontal pipes have the same diameter, but pipe B is twice as long as pipe A. Water undergoes viscous flow in both pipes, subject to the same pressure difference across the lengths of the pipes. If the flow rate in pipe B is what is the flow rate in pipe A?
a) Q√2
b) 16Q
c) 2Q
d) 4Q
e) 8Q
Answer:
c) 2Q
Explanation:
From the given information:
The pressure inside a pipe can be expressed by using the formula:
[tex]\Delta P = \dfrac{128 \mu L Q}{\pi D^4}[/tex]
Since the diameter in both pipes is the same, we can say:
[tex]D = D_A = D_B[/tex]
where;
length of the first pipe A [tex]L_A = L[/tex] and the length of the second pipe B [tex]L_B = 2L[/tex]
Since the difference in pressure is equivalent in both pipes:
Then:
[tex]\dfrac{128 \mu L_1Q_1}{\pi D_1^4} = \dfrac{128 \mu L_2Q_2}{\pi D_2^4}[/tex]
[tex]\dfrac{ L_1Q_1}{D_1^4} = \dfrac{ L_2Q_2}{D_2^4}[/tex]
[tex]\dfrac{ LQ_1}{D^4} = \dfrac{ 2LQ}{D^4}[/tex]
[tex]\mathbf{Q_1 = 2Q}[/tex]
The flow rate in pipe B is 2Q of the flow rate of the pipe A
What is flow rate?
The flow rate is defined as the flow of the fluid across the cross section in per unit time.
From the given information:
The pressure inside a pipe can be expressed by using the formula:
[tex]\Delta p=\dfrac{128\mu LQ}{\pi D^4}[/tex]
Since the diameter in both pipes is the same, we can say:
[tex]D=D_A=D_B[/tex]
where;
length of the first pipe A [tex]L_A=L[/tex] and the length of the second pipe B
[tex]L_B=2L[/tex]
Since the difference in pressure is equivalent in both pipes:
Then:
[tex]\dfrac{128\mu L_1Q_1}{\pi D_1^4}=\dfrac{128\mu L_2Q_2}{\pi D_2^4}[/tex]
[tex]\dfrac{L_1Q_1}{D_1^4}=\dfrac{L_2Q_2}{D_2^4}[/tex]
[tex]\dfrac{LQ_1}{D_1^4}=\dfrac{2LQ}{D_2^4}[/tex]
[tex]Q_1=2Q[/tex]
Hence the flow rate in pipe B is 2Q of the flow rate of the pipe A
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2. A parallel-plate capacitor has a capacitance of C. If the area of the plates is doubled and
the distance between the plates is doubled, what is the new capacitance?
A) C/4
B) C/2
C)C
D) 4C
(C)
Explanation:
The capacitance C of a parallel plate capacitor is given by
[tex]C = \epsilon_0 \dfrac{A}{d}[/tex]
Let C' be the new capacitance where the area and the plate separation distance are doubled. This gives us
[tex]C' = \epsilon_0\dfrac{A'}{d'} = \epsilon_0\left(\dfrac{2A}{2d}\right) = \epsilon_0 \dfrac{A}{d} = C[/tex]
A 25g rock is rolling at a speed of 5 m/s. What is the kinetic energy of the rock?
Answer:
The answer is 312.5j
Explanation:
The kinetic energy (KE):
KE=1/2*m*v^2
M= mass of the object
v= velocity of the object
We have;
m=25g
v=5m/s
KE=1/2*25g*5^2m/s
KE =312.5j
Two long straight wires are suspended vertically. The wires are connected in series, and a current from a battery is maintained in them. What happens to the wires? What happens if the battery is replaced by an a-c source?
Answer:
(i) When a battery is connected inseries to two long parallel wires, the currents in the two wires will be in opposite directions. Due to which a force of repulsion will be acting between them and they are moving further apart.
(ii) When a battery is connected in parallel to two long parallel wires, the currents in the two wires will be in same direction. Due to it, a force of attraction will be acting between them and they are coming closer to each other.
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Explanation:
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Assume that the car on the left makes a quick turn to the left. According to inertia, your body will resist a change and still want to go in the original direction. In which direction with the passenger slide?
Answer:
to the right
Explanation:
if the car turns to the lift, the body forces energy to the left side, so according to the first law of Newton, the body will move to the right side to resist the sudden motion.
A large metal sphere has three times the diameter of a smaller sphere and carries three times the charge. Both spheres are isolated, so their surface charge densities are uniform. Compare (a) the potentials (relative to infinity) and (b) the electric field strengths at their surfaces.
Answer:
A. Equals to that of the smaller sphere
B. 3 times less than that of the smaller sphere
Explanation:
(a) Equals to that of the smaller sphere
The potential of an isolated metal sphere, with charge Q and radius R, is kQ=R, so a sphere with charge 3Q and radius 3R has the same potential
b) 3 times less than that of the smaller sphere
However, the electric field at the surface of the smaller sphere is ?=? 0 = kQ=R2 , so tripling Q and R reduces the surface field by a factor of 1/3
A long string is moved up and down with simple harmonic motion with a frequency of 46 Hz. The string is 579 m long and has a total mass of 46.3 kg. The string is under a tension of 3423 and is fixed at both ends. Determine the velocity of the wave on the string. What length of the string, fixed at both ends, would create a third harmonic standing wave
Answer:
a) [tex]v=206.896m/s[/tex]
b) [tex]L=6.749m[/tex]
Explanation:
From the question we are told that:
Frequency [tex]F=46Hz[/tex]
Length [tex]l=579m[/tex]
Total Mass [tex]T=4.3kg[/tex]
Tension [tex]T=3423[/tex]
a)
Generally the equation for velocity is mathematically given by
[tex]v=\sqrt{\frac{T}{\rho}}[/tex]
Where
[tex]\pho=m*l\\\\\pho=46*579\\\\\pho=0.0799kg/m[/tex]
Therefore
[tex]v=\sqrt{\frac{3423}{0.0799}}[/tex]
[tex]v=206.896m/s[/tex]
b)
Generally the equation for length of string is mathematically given by
[tex]L=\frac{3\lambda}{2}[/tex]
Where
[tex]\lambda=\frac{v}{f}[/tex]
[tex]\lambda=\frac{206.89}{46}[/tex]
[tex]\lambda=4.498[/tex]
Therefore
[tex]L=\frac{3*4.498}{2}[/tex]
[tex]L=6.749m[/tex]
The cart travels the track again and now experiences a constant tangential acceleration from point A to point C. The speeds of the cart are 11.0 ft/s at point A and 18.0 ft/s at point C. The cart takes 5.00 s to go from point A to point C, and the cart takes 1.30 s to go from point B to point C. What is the cart's speed at point B
Answer:
The speed at B is 16.18 ft/s .
Explanation:
Speed at A, u = 11 ft/s
Speed at C, v' = 18 ft/s
Time from A to C = 5 s
Time from B to C = 1.3 s
Let the speed of car at B is v.
Let the acceleration is a.
From A to B
Use first equation of motion
v = u + a t
18 = 11 + a x 5
a = 1.4 ft/s^2
Let the time from A to B is t' .
t' = 5 - 1.3 = 3.7 s
Use first equation of motion from A to B
v = 11 + 1.4 x 3.7 = 16.18 ft/s
In the diagram, the crest of the wave is show by:
A
B
C
D
Answer:
D.
Explanation:
The crest of a wave refers to the highest point of a wave. This is illustrated by D.
A surveyor measures the distance across a straight river by the following method: Starting directly across from a tree on the opposite bank, he walks x = 106 m along the riverbank to establish a baseline. Then he sights across to the tree. The angle from his baseline to the tree is = 32.8°. How wide is the river?
Answer:
x = 68.3 m
Explanation:
tan 32.8 = x / 106
Ayudaaa :(
Calcula la resistencia total del siguiente circuito eléctrico.
A mountain biker takes a jump in a race and goes airborne. The mountain bike is travelling at 10.0 m/s before it goes airborne. If the mass of the front wheel on the bike is 750 g and has radius 35 cm, what is the angular momentum of the spinning wheel in the air the moment the bike leaves the ground?
Answer:
Explanation:
The formula for angular momentum is
L = mvr where L is the angular momentum, m is the mass of the object, v is the velocity of the object, and r is the radius of the object. The problem we have that prevents us from just throwing those numbers in there is that mass has to be in kg and it's not, and radius has to be in meters and it's not.
Changing the mass to kg:
750 g = .750 kg
Changing the radius to m:
35 cm = .35 m
Now we can fill in the variables with their respective values:
L = .750(10.0)(.35) gives us
[tex]L=2.625\frac{kg*m^2}{s}[/tex]
A proposed communication satellite would revolve around the earth in a circular orbit in the equatorial plane at a height of 35880Km above the earth surface. Find the period of revolution of the satellite. (Take the mass of earth =5.98×10²⁴kg, the radius of the earth 6370km and G=6.6×10–¹¹Nm²/kg2)
Answer:
Period is 86811.5 seconds.
Explanation:
[tex]{ \boxed{ \bf{T {}^{2} = (\frac{4 {\pi}^{2} }{GM}) {r}^{3} }}}[/tex]
[tex]{ \tt{T {}^{2} = \frac{4 {(3.14)}^{2} }{(6.6 \times {10}^{ - 11} ) \times (5.98 \times {10}^{24} )} \times {((35880\times {10}^{3}) } + (6370 \times {10}^{3} )) {}^{3} }} \\ \\ { \tt{T {}^{2} = 7.54 \times {10}^{9} }} \\ { \tt{T = \sqrt{7.54 \times {10}^{9} } }} \\ { \tt{T = 86811.5 \: seconds}}[/tex]
One charge is fixed q1 = 5 µC at the origin in a coordinate system, a second charge q2 = -3.2 µC the other is at a distance of x = 90 m from the origin.
What is the potential energy of this pair of charges?
Answer:
5.4uC
Explanation:
which one is more powerful hydrogen bomb or atom bomb and why?
Hydrogen bomb is more powerful than atom bomb
Hydrogen has a calorie value of 150000KJ .It is very much than nuclear bomb or atom bombScientists also told that Hydrogen bomb is more powerful.But both bombs are destructive.A Geiger counter registers a count rate of 8,000 counts per minute from a sample of a radioisotope. The count rate 24 minutes later is 1,000 counts per minute. What is the half-life of the radioisotope?
11.54 minutes
Explanation:
The decay rate equation is given by
[tex]N = N_0e^{-\frac{t}{\lambda}}[/tex]
where [tex]\lambda[/tex] is the half-life. We can rewrite this as
[tex]\dfrac{N}{N_0} = e^{-\frac{t}{\lambda}}[/tex]
Taking the natural logarithm of both sides, we get
[tex]\ln \left(\dfrac{N}{N_0}\right) = -\left(\dfrac{t}{\lambda}\right)[/tex]
Solving for [tex]\lambda[/tex],
[tex]\lambda = -\dfrac{t}{\ln \left(\frac{N}{N_0}\right)}[/tex]
[tex]\:\:\:\:= -\dfrac{(24\:\text{minutes})}{\ln \left(\frac{1000\:\text{counts/min}}{8000\:\text{counts/min}}\right)}[/tex]
[tex]\:\:\:\:=11.54\:\text{minutes}[/tex]
Andrea's near point is 20.0 cm and her far point is 2.0 m. Her contact lenses are designed so that she can see objects that are infinitely far away. What is the closest distance that she can see an object clearly when she wears her contacts?
Answer:
the closest distance that she can see an object clearly when she wears her contacts is 22.2 cm
Explanation:
Given the data in the question,
near point = 20 cm
far point = 2 m = 200 cm
Now, for an object that is infinitely far away, the image is at is its far point.
so using the following expression, we can determine the focal length
1/f = 1/i + 1/o
where f is the focal length, i is the image distance and o is the object distance.
here, far point i = 2 m = 200 cm and v is ∞
so we substitute
1/f = 1/(-200 cm) + 1/∞
f = -200 cm
Also, for object at its closest point, the image appear at near point,
so
1/f = 1/i + 1/o
we make o the subject of formula
o = ( i × f ) / ( i - f )
given that near point i = 20 cm
we substitute
o = ( -20 × -200 ) / ( -20 - (-200) )
o = 4000 / 180
o = 22.2 cm
Therefore, the closest distance that she can see an object clearly when she wears her contacts is 22.2 cm
In these formulas, it is useful to understand which variables are parameters that specify the nature of the wave. The variables E0E0E_0 and B0B0B_0 are the __________ of the electric and magnetic fields. Choose the best answer to fill in the blank.
Derive the dimension of coefficient of linear expansivity
Answer:
The SI unit of coefficient of linear expansion can be expressed as °C-1 or °K-1. ... The dimension of coefficient of linear expansion will be M0L0T0K−1.
A rod of length L and electrical resistance R moves through a constant uniform magnetic field ; both the magnetic field and the direction of motion are parallel to the rod. The force that must be applied by a person to keep the rod moving with constant velocity is:
Answer:
don't know what class are you you are using which mobile or laptop
Find out other examples of bodies showing more than one type of motion Tabulate your findings.
Answer:
down below
Explanation:
Image 1- wheels of train showing both translatory motion as well as rotatory motion.
Image 2- rotation of ball shows both rotatory motion as well as translatory motion.
Image 3- the earth rotates about its axis, same time it revolves around the sun thus showing both rotatory motion and curvilinear motion in a fixed time. (perodic motion)
Image 4- while cutting wood, the
carpenter's saw has both
translatory motion and oscillatory
motion, as it moves down while
oscillating.
What is science?Give two examples of living beings?
Answer:
the study of the past
Explanation:
dogs and cats
A roller coaster has a total track length of 500 yards. A complete ride on the roller coaster is considered two times around the track. The start and stop places for the ride are virtually the same. What are the distance and displacement for a ride on the roller coaster? Explain your answers.
Answer:
Explanation:
its 1000 yards if its going around the track 2 times and that if one whole around the track is 500 its 500 x 2
