Answer: hello some of your values are wrongly written hence I will resolve your question using the right values
answer:
stiffness = 1.09 * 10^-6 N/m
Explanation:
Given data:
Length ( l ) = 16 m
radius of wire ( r ) = 3.5 m
mass ( m ) = 5kg
Distance stretched ( Δl ) = 4 * 10^-3 m ( right value )
average bond length ( between atoms ) = 2.3 * 10^-10 m ( right value)
first step : calculate the area
area ( A ) = πr^2 = π * ( 3.5)^2 = 38.48 m^2
γ = MgL / A Δl
= [ (5 * 9.81 * 16 ) / ( 38.48 * (4.3*10^-3) ) ]
= 784.8 / 0.165 = 4756.36 N/m^2
hence : stiffness = γ * bond length
= 4756.36 * 2.3 * 10^-10 = 1.09 * 10^-6 N/m
An electron is moving at speed of 6.3 x 10^4 m/s in a circular path of radius of 1.7 cm inside a solenoid the magnetic field of the solenoid is perpendicular to the plane of the electron's path. Find its relevatn motion.
Answer:
Here, m=9×10
−31
kg,
q=1.6×10
−19
C,v=3×10
7
ms
−1
,
b=6×10
−4
T
r=
qB
mv
=
(1.6×10
−19
)(6×10
−4
)
(9×10
−31
)×(3×10
7
)
=0.28m
v=
2πr
v
=
2πm
Bq
=
2×(22/7)×9×10
−31
(6×10
−4
)×(1.6×10
−19
)
=1.7×10
7
Hz
Ek=
2
1
mv
2
=
2
1
×(9×10
−31
)×(3×10
7
)
2
J
=40.5×10
−17
J=
1.6×10
−16
40.5×10
−17
keV
=2.53keV
the plane of a 5.0 cm by 8.0 cm rectangular loop wire is parallel to a 0.19 t magnetic field. if the loop carries a current of 6.2 amps, what is the magnitude of the torque on the loop
S.I unit for distance =______
(A) m (B)cm
(c) km (d) mm
Answer:
opinion a
Explanation:
the si units of distance is metre (m)
Answer:
A
Explanation:
Rachel has good distant vision but has a touch of presbyopia. Her near point is 0.60 m. Part A When she wears 2.0 D reading glasses, what is her near point
Answer:
The right answer is "0.273 m".
Explanation:
Given:
Power (P),
[tex]\frac{1}{f} = 2D[/tex]
Near point,
u = 0.6 m
As we know,
⇒ [tex]\frac{1}{v} -\frac{1}{u}=\frac{1}{f} = 2[/tex]
By substituting the values, we get
⇒ [tex]\frac{1}{v} -\frac{1}{0.6} =2[/tex]
[tex]\frac{1}{v}=2+\frac{1}{0.6}[/tex]
[tex]\frac{1}{v} =\frac{1.2+1}{0.6}[/tex]
[tex]\frac{1}{v}=\frac{2.2}{0.6}[/tex]
By applying cross-multiplication, we get
[tex]0.6=2.2 \ v[/tex]
[tex]v = \frac{0.6}{2.2}[/tex]
[tex]S_{near} = 0.273 \ m[/tex]
. A ball of mass 0.50 kg is rolling across a table top with a speed of 5.0 m/s. When the ball reaches the edge of the table, it rolls down an incline onto the floor 1.0 meter below (without bouncing). What is the speed of the ball when it reaches the floor?
Answer:
4
Explanation:
A friend lends you the eyepiece of his microscope to use on your own microscope. He claims that since his eyepiece has the same diameter as yours but twice the focal length, the resolving power of your microscope will be doubled. Is his claim valid? Explain.
Answer:
The resolving power remains same.
Explanation:
The resolving power of the lens is directly proportional to the diameter of the lens not on the focal length.
As the diameter is same but the focal length is doubled so the resolving power remains same.
What is the volume of a metal block 3cm long by 2cm wide by 4cm high? What would be the volume of a block twice as long, wide, and high?
Answer:
Volume of a metal block = 24 cm^3
Volume of a block twice as long, wide and high = 192 cm^3
Explanation:
Volume of a block is measured in l*w*h and in the first block, the sides are 3, 2 and 4 and 3*2*4 = 24
Second block, just double each of the lengths to get 6*4*8 = 192
Good evening everyone Help me i n my hw ,The wall of cinema hall are covered with sound absorbing materials. Why?Answer it ASAP.Good day
what do you mean about it
A 1.5kg block slides along a frictionless surface at 1.3m/s . A second block, sliding at a faster 4.3m/s , collides with the first from behind and sticks to it. The final velocity of the combined blocks is 2.0m/s . What was the mass of the second block?
Answer:
The mass of the second block=0.457 kg
Explanation:
We are given that
m1=1.5 kg
v1=1.3m/s
v2=4.3 m/s
V=2.0 m/s
We have to find the mass of the second block.
[tex]m_1v_1+m_2v_2=(m_1+m_2)V[/tex]
Let m2=m
Substitute the values
[tex]1.5(1.3)+m(4.3)=(1.5+m)(2)[/tex]
[tex]1.95+4.3m=3+2m[/tex]
[tex]4.3m-2m=3-1.95[/tex]
[tex]2.3m=1.05[/tex]
[tex]m=\frac{1.05}{2.3}[/tex]
[tex]m=0.457 kg[/tex]
Hence, the mass of the second block=0.457 kg
You are to connect resistors R1 andR2, with R1 >R2, to a battery, first individually, then inseries, and then in parallel. Rank those arrangements according tothe amount of current through the battery, greatest first. (Useonly the symbols > or =, for exampleseries>R1=R2>parallel.)
Answer:
The current is more in the parallel combination than in the series combination.
Explanation:
two resistances, R1 and R2 are connected to a battery of voltage V.
When they are in series,
R = R1 + R2
In series combination, the current is same in both the resistors, and it is given by Ohm's law.
V = I (R1 + R2)
[tex]I = \frac{V}{R_1 + R_2}[/tex]..... (1)
When they are connected in parallel.
the voltage is same in each resistor.
The effective resistance is R.
[tex]R = \frac{R_1R_2}{R_1 + R_2}[/tex]
So, the current is
[tex]I = \frac{V(R_1+R_2)}{R_1 R_2}[/tex]..... (2)
So, the current is more is the parallel combination.
If a jet travels 350 m/s, how far will it travel each second?
Answer:
It will travel 350 meters each second.
Explanation:
The unit rate, 350 m/s, tells us that the jet will travel 350 meters per every second elapsed.
Answer:
5.83 seconds
Explanation:
60 seconds in 1 minute
350 meters per second
350/60
=5.83
Determine the acceleration of a pendulum bob as it passes through an angle of 15 degrees to the right of the equilibrium point.
Answer:
Explanation:
Since energy is conserved:
2
mu
2
=
2
mv
2
+mgh
⇒u
2
=v
2
+2gh
⇒(3)
2
=v
2
+2(9.8)(0.5−0.5cos60)
⇒v=2m/s
Acceleration of the simple pendulum is 2.62 m/s².
What is meant by a simple pendulum ?When a point mass is suspended from a fixed support by a light, non-extensible string, the instrument is said to be a simple pendulum.
Here,
Let the mass of the bob be m. The simple pendulum is attached to the fixed support with a string having length l. The pendulum makes an angle of 15° with the vertical from the equilibrium point.
Let T be the tension acting on the string.
As, the bob passes through the angle,
The weight of the bob becomes equal to the vertical component of the tension.
mg = T cos15°
Also, the horizontal component of the tension,
T sin15° = ma
By solving these two equations, we get that,
Acceleration of the simple pendulum,
a = g tan15°
a = 9.8 x 0.267
a = 2.62 m/s²
Hence,
Acceleration of the simple pendulum is 2.62 m/s².
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the rate of cooling determines ....... and ......
Answer:
freezing point and melting point
A man standing in an elevator holds a spring scale with a load of 5 kg suspended from it. What would be the reading of the scale, if the elevator is accelerating downward with an acceleration 3.8 m/s?.
Answer:
3.1 kg
Explanation:
Applying,
R = m(g-a)..................... Equation 1
Where R = weight of the scale when the elevator is coming down, a = acceleration of the elevator, g = acceleration due to gravith.
From the question,
Given: m = 5 kg, a = 3.8 m/s²
Constant: g = 9.8 m/s²
Substitute these values into equation 1
R = 5(9.8-3.8)
R = 5(6)
R = 30 N
Hence the spring scale is
m' = R/g
m' = 30/9.8
m' = 3.1 kg
Why we use semiconductor instead of metal in thermopile.
Answer:
Semiconductors are not normal materials. They have special properties which conductors/metals cannot exhibit. The main reason for the behavior of semiconductors is that they have paired charge carriers-the electron-hole pair. This is not available in metals.
The cation that is reabsorbed from the urine in response to aldosterone
Answer:
If decreased blood pressure is detected, the adrenal gland is stimulated by these stretch receptors to release aldosterone, which increases sodium reabsorption from the urine, sweat, and the gut. This causes increased osmolarity in the extracellular fluid, which will eventually return blood pressure toward normal.
calculate the length of wire.
Answer:
L = 169.5 m
Explanation:
Using Ohm's Law:
V = IR
where,
V = Voltage = 1.5 V
I = Current = 10 mA = 0.01 A
R = Resistance = ?
Therefore,
1.5 V = (0.01 A)R
R = 150 Ω
But the resistance of a wire is given by the following formula:
[tex]R = \frac{\rho L}{A}[/tex]
where,
ρ = resistivity = 1 x 10⁻⁶ Ω.m
L = length of wire = ?
A = cross-sectional area of wire = πr² = π(0.6 mm)² = π(0.6 x 10⁻³ m)²
A = 1.13 x 10⁻⁶ m²
Therefore,
[tex]150\ \Omega = \frac{(1\ x\ 10^{-6}\ \Omega .m)L}{1.13\ x\ 10^{-6}\ m^2}\\\\L = \frac{150\ \Omega(1.13\ x\ 10^{-6}\ m^2)}{1\ x\ 10^{-6}\ \Omega .m}\\\\[/tex]
L = 169.5 m
A 0.160 kg glider is moving to the right on a frictionless, horizontal air track with a speed of 0.710 m/s. It has a head-on collision with a 0.296 kg glider that is moving to the left with a speed of 2.23 m/s. Suppose the collision is elastic.
Required:
a. Find the magnitude of the final velocity of the 0.157kg glider.
b. Find the magnitude of the final velocity of the 0.306kg glider.
The masses of the gliders provided in the question differ from the masses mentioned in the "Required" section. I'll use the first masses throughout.
Momentum is conserved, so the total momentum of the system is the same before and after the collision:
m₁ v₁ + m₂ v₂ = m₁ v₁' + m₂ v₂'
==>
(0.160 kg) (0.710 m/s) + (0.296 kg) (-2.23 m/s) = (0.160 kg) v₁' + (0.296 kg) v₂'
==>
-0.546 kg•m/s ≈ (0.160 kg) v₁' + (0.296 kg) v₂'
where v₁' and v₂' are the gliders' respective final velocities. Notice that we take rightward to be positive and leftward to be negative.
Kinetic energy is also conserved, so that
1/2 m₁ v₁² + 1/2 m₂ v₂² = 1/2 m₁ (v₁' )² + 1/2 m₂ (v₂' )²
or
m₁ v₁² + m₂ v₂² = m₁ (v₁' )² + m₂ (v₂' )²
==>
(0.160 kg) (0.710 m/s)² + (0.296 kg) (-2.23 m/s)² = (0.160 kg) (v₁' )² + (0.296 kg) (v₂' )²
==>
1.55 kg•m²/s² ≈ (0.160 kg) (v₁' )² + (0.296 kg) (v₂' )²
Solve for v₁' and v₂'. Using a calculator, you would find two solutions, one of which we throw out because it corresponds exactly to the initial velocities. The desired solution is
v₁' ≈ -3.11 m/s
v₂' ≈ -0.167 m/s
and take the absolute values to get the magnitudes.
If you want to instead use the masses from the "Required" section, you would end up with
v₁' ≈ -3.18 m/s
v₂' ≈ -0.236 m/s
Cuando el pistón tiene un volumen de 2x10^-4 m^3, el gas en el pistón está a una presión de 150 kPa. El área del pistón es 0.00133 m^2. Calcular la fuerza que el gas ejerce sobre el embolo del pistón.
Answer:
F = 1.128 10⁸ Pa
Explanation:
Pressure is defined by
P = F / A
If the gas is ideal for equal force eds on all the walls, so on the piston area we have
F = P A
We reduce the pressure to the SI system
P = 150 kpa (1000 Pa / 1kPa = 150 103 Pa
we calculate
F = 150 10³ / 0.00133
F = 1.128 10⁸ Pa
A load of 25 kg is applied to the lower end and of a steal wire of length 25 m and thickness 3.0mm .The other end of wire is suspeded from a rigid support calculate strain and stress produced in the wire
Answer:
the weight of the wire + 25kg
Explanation:
Suppose a 60-turn coil lies in the plane of the page in a uniform magnetic field that is directed out of the page. The coil originally has an area of 0.325 m2. It is stretched to have no area in 0.100 s. What is the magnitude (in V) and direction (as seen from above) of the average induced emf if the uniform magnetic field has a strength of 1.60 T
Answer:
emf = 312 V
Explanation:
In this exercise the electromotive force is asked, for which we must use Faraday's law
emf = [tex]- N \frac{d \Phi }{dt}[/tex]- N dfi / dt
Ф = B. A = B A cos θ
bold type indicates vectors.
They indicate that the magnetic field is constant, the angle between the normal to the area and the magnetic field is parallel by local cosine values 1
It also indicates that the area is reduced from a₀ = 0.325 me² to a_f = 0 in a time interval of ΔT = 0.100 s, suppose that this reduction is linear
emf = -N B [tex]\frac{dA}{dT}[/tex]
emf = - N B (A_f - A₀) / Dt
we calculate
emf = - 60 1.60 (0 - 0.325) /0.100
emf = 312 V
The direction of this voltage is exiting the page
A 700N marine in basic training climbs a 10m vertical rope at constant speed in 8sec. what is power put
Answer:
875 Watts
Explanation:
P = W/t = mgh/t = 700(10)/8 = 875 Watts
A block with a mass of 0.26 kg is attached to a horizontal spring. The block is pulled back from its equilibrium position until the spring exerts a force of 1.2 N on the block. When the block is released, it oscillates with a frequency of 1.4 Hz. How far was the block pulled back before being released?
Answer:
2
Explanation:
pulling force because of it force
Answer:
5.9 cm
Explanation:
f: frequency of oscillation
frequency of oscillationk: spring constant
frequency of oscillationk: spring constantm: the mass
[tex]f = \frac{1}{2\pi} \sqrt{ \frac{k}{m} } [/tex]
in this problem we know,
F= 1.4 Hz
m= 0.26 kg
By re-arranging the formula we get
[tex]k = {(2\pi \: f )}^{2} m = {(2\pi(1.4hz))}^{2} 0.26kg = 20.1 \frac{n}{m} [/tex]
The restoring force of the spring is:
F= kx
where
F= 1.2 N
k= 20.1 N/m
x: the displacement of the block
[tex]x = \frac{f}{k} = \frac{1.2 \: n}{20.1 \frac{n}{m} } = 0.059m \: = 5.9 \: cm[/tex]
Q)what are convex mirrors?
Answer:
A curved mirror is a mirror with a curved reflecting surface. The surface may be either convex or concave. Most curved mirrors have surfaces that are shaped like part of a sphere, but other shapes are sometimes used in optical devices.
A convex mirror (or lens) is one constructed so that it is thicker in the middle than it is at the edge.
A space ship has four thrusters positioned on the top and bottom, and left and right as shown below. The thrusters can be operated independently or together to help the ship navigate in all directions.
Initially, the Space Probe is floating towards the East, as shown below, with a velocity, v. The pilot then turns on thruster #2.
Select one:
a.
Space ship will have a velocity to the West and will be speeding up.
b.
Space ship will have a velocity to the East and will be speeding up.
c.
Space ship will have a velocity to the East and will be slowing down.
d.
Space ship will have a velocity to the West and will be slowing down.
e.
Ship experiences no change in motion.
Answer:
The correct answer is - c. Spaceship will have a velocity to the East and will be slowing down.
Explanation:
In this case, if turned on thruster #2 then it will exert force on the west side as thruster 2 is on the east side and it can be understood by Newton's third law that says each action has the same but opposite reaction.
As the spaceship engine applies force on the east side then according to the law the exhauster gas applies on towards west direction. It will try to decrease the velocity of the spaceship however, the direction of floating still be east side initally.
i don't understand this, can someone help please??
Explanation:
N2 + H2 --> NH3
balance them:
N2 + 3 H2 --> 2 NH3
so if 6 moles of N2 react, 12 moles of NH3 will form.
(you have to look at the big number in front, in this case its N2 and 2 NH3, therefore the amount of N2 will produce double the amount of NH3 )
5. Steve is driving in his car to take care of some errands. The first errand has him driving to a location 2 km East and 6 km North of his starting location. Once he completes that errand, he drives to the second one which is 4 km East and 2 km South of the first errand. What is the magnitude of the vector that describes how far the car has traveled from its starting point, rounded to the nearest km?
Answer:
gshshs
Explanation:
hshsksksksbsbbshd
Two forces are acting on a body. One acts east, the other at 35° north of east. If the
two forces are equal in magnitude of 50 N, find the resultant using the Law of Sines
and the Law of Cosines. Please answer with full solution. Thanks
A=B=50NAngle=theta=35°
We know
[tex]\boxed{\sf R=\sqrt{A^2+B^2+2ABcos\Theta}}[/tex]
[tex]\\ \sf\longmapsto R=\sqrt{50^2+50^2+2(50)(50)cos35}[/tex]
[tex]\\ \sf\longmapsto R=\sqrt{2500+2500+2(2500)\times (-0.9)}[/tex]
[tex]\\ \sf\longmapsto R=\sqrt{5000+5000(-0.9)}[/tex]
[tex]\\ \sf\longmapsto R=\sqrt{5000+(-4500)}[/tex]
[tex]\\ \sf\longmapsto R=\sqrt{5000-4500}[/tex]
[tex]\\ \sf\longmapsto R=\sqrt{-500}[/tex]
[tex]\\ \sf\longmapsto R=22.4i[/tex]
Resultant using the Law of Sines and the Law of Cosines will be R=95 N
What is force?Force is an external agent applied on any object to displace it from its position. Force is a vector quantity, so with magnitude it also requires direction. Direction is necessary to examine the effect of the force and to find the equilibrium of the force.
The Magnitude of two forces =50 N
Angle between the forces = 35
By using the resultant formula
[tex]\rm R=\sqrt{A^2+B^2+2ABCos\theta}[/tex]
[tex]\rm R=\sqrt{50^2+50^2+2(50)(50)Cos35}[/tex]
[tex]\rm R=\sqrt{5000+5000(0.81)}[/tex]
[tex]\rm R=\sqrt{5000+4500}[/tex]
[tex]\rm R=95\ N[/tex]
Hence the Resultant using the Law of Sines and the Law of Cosines will be R=95 N
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A object of mass 3.00 kg is subject to a force Fx that varies with position as in the figure below. A coordinate plane has a horizontal axis labeled x (m) and a vertical axis labeled Fx (N). There are three line segments. The first segment runs from the origin to (4,3). The second segment runs from (4,3) to (11,3). The third segment runs from (11,3) to (17,0). (a) Find the work done by the force on the object as it moves from x = 0 to x = 4.00 m. J (b) Find the work done by the force on the object as it moves from x = 4.00 m to x = 11.0 m. J (c) Find the work done by the force on the object as it moves from x = 11.0 m to x = 17.0 m. J (d) If the object has a speed of 0.450 m/s at x = 0, find its speed at x = 4.00 m and its speed at x = 17.0 m.
Answer:
Explanation:
An impulse results in a change of momentum.
The impulse is the product of a force and a distance. This will be represented by the area under the curve
a) W = ½(4.00)(3.00) = 6.00 J
b) W = (11.0 - 4.00)(3.00) = 21.0 J
c) W = ½(17.0 - 11.0)(3.00) = 9.00 J
d) ASSUMING the speed at x = 0 is in the direction of applied force
½(3.00)(v₄²) = ½(3.00)(0.450²) + 6.00
v₄ = 2.05 m/s
½(3.00)(v₁₇²) = ½(3.00)(0.450²) + 6.00 + 21.0 + 9.00
v₁₇ = 4.92 m/s
If the initial speed is NOT in the direction of applied force, the final speed will be slightly less in both cases.
Chameleons catch insects with their tongues, which they can rapidly extend to great lengths. In a typical strike, the chameleon's tongue accelerates at a remarkable 220 m/s^2 for 20 msms, then travels at constant speed for another 30 ms.
Required:
During this total time of 50 ms, 1/20 of a second, how far does the tongue reach?
Solution :
We know,
Distance,
[tex]$S=ut+\frac{1}{2}at^2$[/tex]
[tex]$S=ut+0.5(a)(t)^2$[/tex]
For the first 20 ms,
[tex]$S=0+0.5(220)(0.020)^2$[/tex]
S = 0.044 m
In the remaining 30 ms, it has constant velocity.
[tex]$v=u+at$[/tex]
[tex]$v=0+(220)(0.020)[/tex]
v = 4.4 m/s
Therefore,
[tex]$S=ut+0.5(a)(t)^2$[/tex]
[tex]$S'=4.4 \times 0.030[/tex]
S' = 0.132 m
So, the required distance is = S + S'
= 0.044 + 0.132
= 0.176 m
Therefore, the tongue can reach = 0.176 m or 17.6 cm
Answer:
The total distance is 0.176 m.
Explanation:
For t = 0 s to t = 20 ms
initial velocity, u = 0
acceleration, a = 220 m/s^2
time, t = 20 ms
Let the final speed is v.
Use first equation of motion
v = u + at
v = 0 + 220 x 0.02 = 4.4 m/s
Let the distance is s.
Use second equation of motion
[tex]s = u t + 0.5 at^2\\\\s = 0 + 0.5 \times 220 \times 0.02\times 0.02\\\\s = 0.044 m[/tex]
Now the distance is
s' = v x t
s' = 4.4 x 0.03 = 0.132 m
The total distance is
S = s + s' = 0.044 + 0.132 = 0.176 m
