Answer:
Explanation:
Your body weighs more at the pole for two important reasons. Both have to do to the spin of the earth on its axis.
Because of its spin the earth is thicker around the equator than it is through the poles. This means that when you stand on the equator, you are farther away from the center of earth than you would be at the poles. As gravity decreases with the inverse of the square of distance, gravity will be weaker at the equator.
As you are also spinning with the earth, you will have a required centripetal acceleration and force to keep you attached to the ground, This force decreases the effect of gravity so again, you would weigh less at the equator.
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
A spring whose stiffness is 3500 N/m is used to launch a 4 kg block straight up in the classroom. The spring is initially compressed 0.2 m, and the block is initially at rest when it is released. When the block is 1.3 m above its starting position, what is its speed
Answer:
the speed of the block at the given position is 21.33 m/s.
Explanation:
Given;
spring constant, k = 3500 N/m
mass of the block, m = 4 kg
extension of the spring, x = 0.2 m
initial velocity of the block, u = 0
displacement of the block, d =1.3 m
The force applied to the block by the spring is calculated as;
F = ma = kx
where;
a is the acceleration of the block
[tex]a = \frac{kx}{m} \\\\a = \frac{(3500) \times (0.2)}{4} \\\\a = 175 \ m/s^2[/tex]
The final velocity of the block at 1.3 m is calculated as;
v² = u² + 2ad
v² = 0 + 2ad
v² = 2ad
v = √2ad
v = √(2 x 175 x 1.3)
v = 21.33 m/s
Therefore, the speed of the block at the given position is 21.33 m/s.
The speed of the block at a height of 1.3 m above the starting position is 21.33 m/s
To solve this question, we'll begin by calculating the acceleration of the block.
How to determine the acceleration Spring constant (K) = 3500 N/m Mass (m) = 4 KgCompression (e) = 0.2 mAcceleration (a) =?F = Ke
Also,
F = ma
Thus,
ma = Ke
Divide both side by m
a = Ke / m
a = (3500 × 0.2) / 4
a = 175 m/s²
How to determine the speed Initial velocity (u) = 0 m/sAcceleration (a) = 175 m/s²Distance (s) = 1.3 mFinal velocity (v) =?v² = u² + 2as
v² = 0² + (2 × 175 × 1.3)
v² = 455
Take the square root of both side
v = √455
v = 21.33 m/s
Learn more about spring constant:
https://brainly.com/question/9199238
19 point please please answer right need help
Block on an incline
A block of mass m1 = 3.9 kg on a smooth inclined plane of angle 38is connected by a cord over a small frictionless
pulley to a second block of mass m2 = 2.6 kg hanging vertically. Take the positive direction up the incline and use 9.81
m/s2 for g.
What is the tension in the cord to the nearest whole number?
Explanation:
We can write Newton's 2nd law as applied to the sliding mass [tex]m_1[/tex] as
[tex]T - m_1g\sin38 = m_1a\:\:\:\:\:\:\:(1)[/tex]
For the hanging mass [tex]m_2,[/tex] we can write NSL as
[tex]T - m_2g = -m_2a\:\:\:\:\:\:\:(2)[/tex]
We need to solve for a first before we can solve the tension T. So combining Eqns(1) & (2), we get
[tex](m_1 + m_2)a = m_2g - m_1g\sin38[/tex]
or
[tex]a = \left(\dfrac{m_2 - m_1\sin38}{m_1 + m_2}\right)g[/tex]
[tex]\:\:\:\:= 0.30\:\text{m/s}^2[/tex]
Using this value for the acceleration on Eqn(2), we find that the tension T is
[tex]T = m_2(g - a) = (2.6\:\text{kg})(9.51\:\text{m/s}^2)[/tex]
[tex]\:\:\:\:=24.7\:\text{N}[/tex]
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
Each rarefraction on a longitudinal wave correspond to what point on a transverse wave?
1. Why do only some people get addicted to
drugs?
Answer:
When drugs are taken in are body are brain release dopamine: which make us feel so pleasure and good, and for this some people are addicted to drugs which makes them feel good. on other hand damaging their health.
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
Match the following properties to the type of wave.
Answer:
hi there
Explanation:
1 - III
2- 1
3-1
HOPE IT HELP YOU
PLz mark me as a BRAINLIST
Explanation:
1 . 3
2. 1
3. 2
I hope it is helpful to you.
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:
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 car is running with the velocity of 72 km per hour what will be its velocity after 5 seconds if its acceleration is -2 metre per second square
Answer:
initial velocity (u)=72×1000/60×60
=72000/3600
=20m/s
final velocity(v)=v
Time(t)=5s
acceleration(a)=-2m/s
now,
acceleration(a)=v-u/t
-2=v-20/5
-2×5=v-20
-10=v-20
-10+20=v
v=10m/s
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.
1:
Forces and Motion:Question 2
A car is travelling east, when suddenly a more massive car travelling
north hits it with a greater force. What is likely to happen to the car
that was originally travelling east?
Explanation:
the car will be brought back
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
Open the sash half way up, take the beaker containing the dry ice / water out of the hood, and slowly move it from right in front of the hood all the way down to the floor. At what point do the fumes stop getting sucked up by the fume hood?
Answer:
The fumes stop getting sucked up by the fume hood once the beaker is pulled out of the hood.
a vector starts at the point (0.0) and ends at (2,-7) what is the magnitude of the displacement
Answer:
|x| = √53
Explanation:
We are told that the vector starts at the point (0.0) and ends at (2,-7) .
Thus, magnitude of displacement is;
|x| = √(((-7) - 0)² + (2 - 0)²)
|x| = √(49 + 4)
|x| = √53
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.
A block of mass 2 kg is launched by compressing a spring of force constant 1200 N/m. The block slides on a frictionless surface, up a 1 meter tall ramp, then it enters a region of rough surface. It comes to a stop after traveling 3 meters over the rough surface. The coefficient of kinetic friction between the block and the rough surface is 0.40.
Required:
a. How many forces end up doing work on the block from release to stop?
b. What is the total non-conservative work done on the block?
c. What is the change in the spring potential energy of the block?
Answer:
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Explanation:
a point object is 10 cm away from a plane mirror while the eye of an observer(pupil diameter is 5.0 mm) is 28 cm a way assuming both eye and the point to be on the same line perpendicular to the surface find the area of the mirror used in observing the reflection of the point
Answer:
1.37 mm²
Explanation:
From the image attached below:
Let's take a look at the two rays r and r' hitting the same mirror from two different positions.
Let x be the distance between these rays.
[tex]d_o =[/tex] distance between object as well as the mirror
[tex]d_{eye}[/tex] = distance between mirror as well as the eye
Thus, the formula for determining the distance between these rays can be expressed as:
[tex]x = 2d_o tan \theta[/tex]
where; the distance between the eye of the observer and the image is:
[tex]s = d_o + d_{eye}[/tex]
Then, the tangent of the angle θ is:
[tex]tan \theta = \dfrac{R}{d_o+d_{eye}}[/tex]
replacing [tex]tan \theta = \dfrac{R}{d_o+d_{eye}}[/tex] into [tex]x = 2d_o tan \theta[/tex], we have:
[tex]x = 2d_o \Big( \dfrac{R}{d_o+d_{eye}}\Big)[/tex]
[tex]x = 2(10) \Big( \dfrac{0.25}{10+28}\Big)[/tex]
[tex]x = 20\Big( \dfrac{0.25}{38}\Big) cm[/tex]
x = (0.13157 × 10) mm
x = 1.32 mm
Finally, the area A = π r²
[tex]A = \pi(\frac{x}{2})^2[/tex]
[tex]A = \pi(\frac{1.32}{2})^2[/tex]
A = 1.37 mm²
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
What type of wave is a microwave?
O heat
O longitudinal
sound
transverse
Answer:
Microwave is a types of a electromagnetic radiation
Answer:
Transvers
Explanation:
Because microwave is electromagnetic waves and all electromagnetic waves are transvers.
what is the distance time how can we find the speed of an object from its distance time graph
Answer:
speed is the gradient of the graph
Answer:
Speed is the slope of a distance time graph.
Explanation:
Speed= d/t
Slope is equal to rise/run
If the rise of the graph is the distance and the run is the time, calculating slope is the equivalent of calculating average speed.
In a game of pool, the cue ball moves at a speed of 2 m/s toward the eight ball. When the cue ball hits the eight ball, the cue ball bounces off with a speed of 0.8 m/s at an angle of 20', as shown in the diagram below. Both balls have a mass of 0.6 kg.
a) what is the momentum of the system before the collision
b) what is the momentum after the collision
c) what angle dose the right ball travel after the collision
d) what is the magnitude of the eight balls velocity after the collision
Answer:
a) p₀ = 1.2 kg m / s, b) p_f = 1.2 kg m / s, c) θ = 12.36, d) v_{2f} = 1.278 m/s
Explanation:
For this exercise we define a system formed by the two balls, which are isolated and the forces during the collision are internal, therefore the moment is conserved
a) the initial impulse is
p₀ = m v₁₀ + 0
p₀ = 0.6 2
p₀ = 1.2 kg m / s
b) as the system is isolated, the moment is conserved so
p_f = 1.2 kg m / s
we define a reference system where the x-axis coincides with the initial movement of the cue ball
we write the final moment for each axis
X axis
p₀ₓ = 1.2 kg m / s
p_{fx} = m v1f cos 20 + m v2f cos θ
p₀ = p_f
1.2 = 0.6 (-0.8) cos 20+ 0.6 v_{2f} cos θ
1.2482 = v_{2f} cos θ
Y axis
p_{oy} = 0
p_{fy} = m v_{1f} sin 20 + m v_{2f} cos θ
0 = 0.6 (-0.8) sin 20 + 0.6 v_{2f} sin θ
0.2736 = v_{2f} sin θ
we write our system of equations
0.2736 = v_{2f} sin θ
1.2482 = v_{2f} cos θ
divide to solve
0.219 = tan θ
θ = tan⁻¹ 0.21919
θ = 12.36
let's look for speed
0.2736 = v_{2f} sin θ
v_{2f} = 0.2736 / sin 12.36
v_{2f} = 1.278 m / s
I’m a photoelectric effect, which property of the incident light determines how much kinetic energy the ejected electrons have ?
A) brightness
B) frequency
C) size of the beam
D) none of the above
Answer:
b = frequency
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
A capacitor consists of a set of two parallel plates of area A separated by a distance d. This capacitor is connected to a battery and charged until its plates carry charges If the separation between the plates is doubled, the electrical energy stored in the capacitor will
Answer:
The electrical energy stored in the capacitor will be cut in half.
Explanation:
The energy in a capacitator is given by E=C[tex]V^{2}[/tex]/2 and the formula for the Capacitance in a capacitator is C= [tex]\frac{Q}{V}[/tex] = ε[tex]\frac{A}{d}[/tex] .
So if we replace C = ε[tex]\frac{A}{d}[/tex] in the first equation we have:
E = ε[tex]\frac{AV^{2} }{2d}[/tex]
An astronaut on the moon drops a rock from rest. The rock falls 0.8m in one second of falling time. If the dropped rock fell for a total of two seconds of time instead of 1 second, then the distance traveled would be:____.
A) The same.
B) Doubled.
C) Tripled.
D) Quadruple.
E) None of the above.
Answer:
D) Quadruple.
Explanation:
We will use the second equation of motion to solve this problem:
[tex]s = v_it + \frac{1}{2}gt^2[/tex]
where,
s = distance travelled by the rock
vi = initial speed of rock = 0 m/s
t = time taken
g = acceleration due to gravity on the surface of the moon
Therefore,
[tex]s = (0\ m/s)t+\frac{1}{2}gt^2\\\\s =\frac{1}{2}gt^2[/tex]----------- equation (1)
Now, we double the time:
[tex]s' = \frac{1}{2}g(2t)^2\\\\s' = 4(\frac{1}{2}gt^2)[/tex]
using equation (1):
s' = 4s
Hence, the correct option is:
D) Quadruple.
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
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.
The armature of an AC generator has 200 turns, which are rectangular loops measuring 5 cm by 10 cm. The generator has a sinusoidal voltage output with an amplitude of 18 V. If the magnetic field of the generator is 300 mT, with what frequency does the armature turn
Answer:
[tex]f=9.55Hz[/tex]
Explanation:
From the question we are told that:
Number of Turns [tex]N=200[/tex]
Length [tex]l=5cm to 10cm[/tex]
Voltage [tex]V=18V[/tex]
Magnetic field [tex]B=300mT[/tex]
Generally, the equation for Frequncy of an amarture is mathematically given by
[tex]f =\frac{ V}{(N B A * 2 pi )}[/tex]
[tex]f =\frac{ 18}{(200 300*10^{-3} (10*10^-2)(5*10^{-2}) * 2 *3.142 )}[/tex]
[tex]f=9.55Hz[/tex]
