We are given:
the initial height of the object (h) = 100 m
initial velocity (u) = 0 m/s
we will let the value of g = 10 m/s/s
Speed of the object just before hitting the ground:
From the third equation of motion:
v² - u² = 2ah (where v is the final velocity)
replacing the variables, we get:
v² - (0)² = 2(10)(100)
v² = 2000
v = 10√20 = 44.7 m/s
Therefore, the speed of the object just before hitting the ground is 44.7 m/s
Which best explains a difference between Einstein’s general theory of relativity and his special theory of relativity?
His general theory includes uniform and accelerated motion, but his special theory applies only to uniform motion.
His general theory includes uniform and accelerated motion, but his special theory applies only to accelerated motion.
His general theory applies only to accelerated motion, but his special theory includes uniform and accelerated motion.
His general theory applies only to uniform motion, but his special theory includes uniform and accelerated motion.
Answer:
His general theory includes uniform and accelerated motion, but his special theory applies only to uniform motion.
Explanation:
According to Einstein's 1915 general theory of relativity, the force of gravity arises from the curvature of space and time.
According to theory of special relativity:
1. The laws of physics are the same for all non-accelerating observers
2. The speed of light in a vacuum was independent of the motion of all observers.
His general theory includes uniform and accelerated motion, but his special theory applies only to uniform motion.
Answer:
for those who dont like to read
the answer is A.
hope i helped
Explanation:
how are s waves and p waves simuliar?
A.they shake the ground
B.they travel through liquids
C. they arrive at the same time
D.they shake the ground from side to side
Answer:
A
Explanation:
hope this helps
Converting compound units
You would like to know whether silicon will float in mercury and you know that can determine this based on their densities. Unfortunately, you have the density of mercury in units of kilogram/meter3 and the density of silicon in other units: 2.33 gram/centimeter3. You decide to convert the density of silicon into units of kilogram/meter3 to perform the comparison. By which combination of conversion factors will you multiply 2.33 gram/centimeter3 to perform the unit conversion?
Answer:
Dividing the silicon density by 1000 and then multiply it by 1000000.
Explanation:
A kilogram equals 1000 grams and a cubic meter equals 1000000 cubic centimeters. Hence, we must divide the silicon density by 1000 and then multiply itby 1000000 to convert the value into kilograms per cubic centimeter. That is:
[tex]x = 2.33\,\frac{g}{cm^{3}}\times \frac{1\,kg}{1000\,g}\times \frac{1000000\,cm^{3}}{1\,m^{3}}[/tex]
[tex]x = 2330\,\frac{kg}{m^{3}}[/tex]
In a nutshell, we must multiply the density of silicon by 1000 to obtains its value in kilograms per cubic meter.
(iii) Why do right angle mirrors produce three images of the object?
Explanation:
The two mirrors inclined to each other formed the first two images with are of the same size as the object while the third mirror is produced from the intersection of rays that emanated during the production of the first two images to produce a third image which is smaller than the object and there making the total number of images to be 3.
Hence this mirrors produces 3 images due to the third image formed from the intersection of the rays that produces the first two images.
The formula that relates the image produced by inclined mirror and the angle of inclination is expressed as:
number of images n = 360/θ - 1
θ is the angle of inclination of the two mirrors
n is the number of images
If the mirrors are inclined at right angles, then θ = 90°
Substitute into the formula;
n = 360/90 -1
n = 36/9 -1
n = 4-1
n = 3
100 POINTS.
Please provide explanation.
Thank you
Answer:
(a) 0.829 m/s
(b) 3.27 m/s
(c) 0.000153 m²
55.8%
Explanation:
(a) Flow rate equals velocity times cross-sectional area. (1 L = 0.001 m³)
Q = vA
(0.001 m³ / 2.00 s) = v (48 × π (0.002 m)²)
v = 0.829 m/s
(b) Use Bernoulli equation. Choose point 1 to be the exit of the pump, and point 2 to be exit of the shower head. Choose 0 elevation to be at point 1.
P₁ + ½ ρ v₁² + ρgh₁ = P₂ + ½ ρ v₂² + ρgh₂
(1.50 atm × 1.0×10⁵ Pa/atm) + ½ (1000 kg/m³) v² + 0 = (1 atm × 1.0×10⁵ Pa/atm) + ½ (1000 kg/m³) (0.829 m/s)² + (1000 kg/m³) (10 m/s²) (5.50 m)
1.50×10⁵ Pa + (500 kg/m³) v² = 1×10⁵ Pa + 414.5 Pa + 55000 Pa
v = 3.27 m/s
(c) Flow rate is constant.
Q = vA
(0.001 m³ / 2.00 s) = (3.27 m/s) A
A = 0.000153 m²
Flow rate is proportional to the pressure difference and the radius raised to the fourth power.
Q ∝ ΔP r⁴
Q₂/Q₁ = (ΔP₂/ΔP₁) (r₂/r₁)⁴
Q₂/Q₁ = (1.120) (0.840)⁴
Q₂/Q₁ = 0.558
The flow decreases to 55.8% of the original value.
Answer:
Explanation:
Regarding the point of "Flow rate is proportional to the pressure difference and the radius raised to the fourth power", flow rate depends on pressure, cross-section area and speed. As speed also depends on cross-section area, flow rate becomes dependent on pressure and cross-section area squared.
In a round pipe like blood vessel, the cross-section area is equal to pi*radius squared. So flow rate is proportional to the pressure difference and (radius squared) squared; i.e. the radius raised to the fourth power.
The new flow rate = (1.12)*(0.84)^4
=0.5576 or 55.76% of the original flow rate
A charged isolated metal sphere of diameter 12 cm has a potential of 9200 V relative to V = 0 at infinity. Calculate the energy density in the electric field near the surface of the sphere.
Answer:
0.1 J/m³
Explanation:
We know that
V = k Q / R
We also know that
E = k Q / R²
Joining the two equations together, we have
E = V / R
To solve the question proper, we'd be using the formula
u = 1/2 E• E², substitute for E, we have
u = 1/2 E• (V/R)²
u = 1/2 * 8.85*10^-12 * (9000 / 0.06)²
u = 1/2 * 8.85*10^-12 * 150000²
u = 1/2 * 8.85*10^-12 * 2.25*10^10
u = 1/2 * 0.199125
u = 0.0996
u = 0.1 J/m³
The energy density is 0.1 J/m³
A 1400.0 kg car crests a 3200.0 m pass in the mountains and briefly comes to rest. The car descends 1000 m before climbing and cresting a 2800 m pass. (a) Neglecting friction, what should the speed of the car be at the top of the second pass? (b) Find the actual speed of the car if the work due to nonconservative forces is – 5 x106 J.
Answer:
a) v = 88.54 m/s
b) vf = 26.4 m/s
Explanation:
Given that;
m = 1400.0 kg
a)
by using the energy conservation
loss in potential energy is equal to gain in kinetic energy
mg × ( 3200-2800) = 1/2 ×m×v²
so
1400 × 9.8 × 400 = 0.5 × 1400 × v²
5488000 = 700v²
v² = 5488000 / 700
v² = 7840
v = √7840
v = 88.54 m/s
b)
Work done by all forces is equal to change in KE
W_gravity + W_non - conservative = 1/2×m×(vf² - vi²)
we substitute
1400 × 9.8 × ( 3200-2800) - (5 × 10⁶) = 1/2 × 1400 × (vf² -0 )
488000 = 700 vf²
vf² = 488000 / 700
vf² = 697.1428
vf = √697.1428
vf = 26.4 m/s
Bird A, with a mass of 2.2 kg, is stationary while Bird B, with a mass of 1.7 kg, is moving due north from Bird A at 3 m/s. What is the velocity of the center of mass for this system of two birds
Answer:
1.3 m/s
Explanation:
It is given that,
Mass of bird A, [tex]m_A=2.2\ kg[/tex]
Mass of bird B, [tex]m_B=1.7\ kg[/tex]
Initial speed of bird A is 0 as it was at rest
Initial speed of bird B is 3 m/s
We need to find the velocity of the center of mass for this system of two birds. Let it is V. so,
[tex]v_{cm}=\dfrac{m_Au_A+m_Bu_B}{m_A+m_B}\\\\v_{cm}=\dfrac{2.2\times 0+1.7\times 3}{2.2+1.7}\\\\v_{cm}=1.3\ m/s[/tex]
So, the center of mass for this system is 1.3 m/s.
I WILL MARK YOU AS BRAINLIEST IF RIGHT
What is the magnitude of the net force acting on this object?
Answer:
The net force on an object is the total force applied on the object after adding up all the forces
In the given diagram,
we can see that the 2 forces of 4N and 4N will cancel each other out since they are equal and in the opposite direction
Now, we are left with a force of 2N and 10N,
the net force will be the difference of these forces:
Net force = 10N - 2N
Net force = 8N downwards
Another way to do it:
The two 4N forces will be cancelled out,
and we are left with a 2N and a 10N force
(notice how we cancelled equal and opposite forces for the 4N)
We can divide the 10N force into (2N + 8N)
Since the 2N forces are equal and opposite, they will be cancelled out
and we will be left with a net force of 8N downwards
Please help
A student plans an investigation to determine the refractive index of glass. The student uses this equipments.
- a ray box
- a rectangular glass block
- a protractor
- a pencil
Describe how the student collect her data.
How much work is done lifting a 5 kg ball from a height of 2 m to a height of 6 m? (Use 10 m/s2 for the acceleration of gravity.)
A) 100 J B) 200 J C) 300 J D) 400 J
Answer:
B) 200 [J]
Explanation:
In order to solve this problem we must remember the definition of work which tells us that it is equal to the product of force by a distance, in this case, the force is the weight of the ball. The distance traveled is 4 [m] since 6-2 = 4[m]
F = m*g
where:
m = mass = 5 [kg]
g = gravity acceleration = 10 [m/s^2]
F = 5*10 = 50 [N]
w = F*d
where:
F = force = 50 [N]
d = 4 [m]
w = 50*4 = 200 [J]
If vector A = 6i - 2j + 3k, determine
(a) A vector in the same direction as A with magnitude 2A
(b) A unit vector in the direction of A
(c) a vector opposite to A with magnitude of 4 m
Answer:
(a) [tex]2\vec A=12\hat i-4\hat j+6\hat k[/tex]
(b) [tex]\displaystyle \vec{U_A}=12/7\hat i-4/7\hat j+6/7\hat k[/tex]
(c) [tex]-4\vec{U_A}=-48/7\hat i+16/7\hat j-24/7\hat k[/tex]
Explanation:
Vectors
Given a vector
[tex]\vec A=6\hat i-2\hat j+3\hat k[/tex]
We must determine the following:
a) A vector in the same direction as A with double magnitude 2A.
If the vector goes in the same direction but has a different magnitude, we only need to multiply each component by a common factor, in this case, by 2. Thus, the required vector is:
[tex]2\vec A=12\hat i-4\hat j+6\hat k[/tex]
b) A unit vector in the same direction of A.
The unit vector needs to compute the magnitude of the vector:
[tex]\mid A\mid=\sqrt{6^2+2^2+3^2}[/tex]
[tex]\mid A\mid=\sqrt{36+4+9}=\sqrt{49}=7[/tex]
[tex]\mid A\mid=7[/tex]
The unit vector is:
[tex]\displaystyle \vec{U_A}=\frac{\vec A}{\mid \vec A\mid}[/tex]
[tex]\displaystyle \vec{U_A}=\frac{12\hat i-4\hat j+6\hat k}{7}[/tex]
[tex]\displaystyle \vec{U_A}=12/7\hat i-4/7\hat j+6/7\hat k[/tex]
c) A vector opposite to A with magnitude 4 m. We assume the original vector is also expressed in m.
The opposite vector to A is obtained simply by multiplying the unit vector by -1. To make its magnitude equal to 4, also multiply by 4. In all, we multiply the unit vector by -4:
[tex]-4\vec{U_A}=-4(12/7\hat i-4/7\hat j+6/7\hat k)[/tex]
[tex]-4\vec{U_A}=-48/7\hat i+16/7\hat j-24/7\hat k[/tex]
how much power is used if it takes frank (a 450 N boy ) 3 seconds to run 2 meters ?
Answer:
300
Explanation:
450Newton × 2Meter ÷ 3sec
A pile of bricks of mass M is being raised to the tenth floor of a building of height H = 4y above the ground by a crane that is on top of the building. During the first part of the lift, the crane lifts the bricks a vertical distance h1=3y in a time t1=4T. During the second part of the lift, the crane lifts the bricks a vertical distance h2=y in t2=T. Which of the following correctly relates the power P1 generated by the crane during the first part of the lift to the power P2 generated by the crane during the second part of the lift?
A. P2=4P1
B. P2=43P1
C. P2=P1
D. P2=34P1
E. P2=13P1
Answer:
The correct option is B
Explanation:
From the question we are told that
The mass of the pile is M
The height is H = 4 y
The vertical distance achieve during the first lift is [tex]h_1 = 3 y[/tex]
The time taken is [tex]t_1 = 4T [/tex]
The vertical distance achieve during the second lift is [tex]h_2 = y[/tex]
The time taken is [tex] t_2 = T [/tex]
Generally the velocity of the crane during the first lift is
[tex]v _1 = \frac{h_1}{t_1 }[/tex]
=> [tex]v _1 = \frac{3 y}{4T }[/tex]
Generally the velocity of the crane during the second lift is
[tex]v _2 = \frac{h_2}{t_2 }[/tex]
=> [tex]v _2 = \frac{ y}{T}[/tex]
Generally the power generated by the crane during the first lift is
[tex]P_1 = F_1 * v_1[/tex]
Here [tex]F_1[/tex] is the weight of the brick which is mathematically represented as
[tex]F_1 = M * g [/tex] , g is the acceleration due to gravity
So
[tex]P_1 = Mg * \frac{3y}{4T}[/tex]
Generally the power generated by the crane during the first lift is
[tex]P_1 = F_2 * v_2[/tex]
Here [tex]F_2[/tex] is the weight of the brick which is mathematically represented as
[tex]F_2 = M * g [/tex] , g is the acceleration due to gravity
So
[tex]P_1 = Mg * \frac{y}{T}[/tex]
The ratio of the first power generated to the second power is
[tex]\frac{P_1}{P_2} = \frac{Mg * \frac{3y}{4T} }{ Mg * \frac{y}{T} }[/tex]
=> [tex]\frac{P_1}{P_2} = \frac{3}{4}[/tex]
=> [tex]P_2 = \frac{4}{3} P_1[/tex]
What does the principle of superposition help scientists determine?
A) The super powers of a rock layer
B) The exact and absolute age of a rock layer
C) The relative age of a rock layer
D) The position of a fossil
Answer:
B
Explanation:
the exact and absolute age of a rock layer
Answer:
The relative age of a rock layer.
Explanation:
The answer is C.
An object, initially at rest, is subject to an acceleration of 45 m/s^2. How long will it take that object to travel 1000m? Round to one decimal place.
Answer:
6.7 seconds
Explanation:
d=(1/2)at^2
equation
1000=(1/2)45t^2.
substitute
2000=45t^2.
multiply by 2 for both sides
44.44=t^2.
divide both sides by 45
6.7=t
take the square root of both sides
Please provide an explanation.
Thank you!!
Answer:
(a) 22 kN
(b) 36 kN, 29 kN
(c) left will decrease, right will increase
(d) 43 kN
Explanation:
(a) When the truck is off the bridge, there are 3 forces on the bridge.
Reaction force F₁ pushing up at the first support,
reaction force F₂ pushing up at the second support,
and weight force Mg pulling down at the middle of the bridge.
Sum the torques about the second support. (Remember that the magnitude of torque is force times the perpendicular distance. Take counterclockwise to be positive.)
∑τ = Iα
(Mg) (0.3 L) − F₁ (0.6 L) = 0
F₁ (0.6 L) = (Mg) (0.3 L)
F₁ = ½ Mg
F₁ = ½ (44.0 kN)
F₁ = 22.0 kN
(b) This time, we have the added force of the truck's weight.
Using the same logic as part (a), we sum the torques about the second support:
∑τ = Iα
(Mg) (0.3 L) + (mg) (0.4 L) − F₁ (0.6 L) = 0
F₁ (0.6 L) = (Mg) (0.3 L) + (mg) (0.4 L)
F₁ = ½ Mg + ⅔ mg
F₁ = ½ (44.0 kN) + ⅔ (21.0 kN)
F₁ = 36.0 kN
Now sum the torques about the first support:
∑τ = Iα
-(Mg) (0.3 L) − (mg) (0.2 L) + F₂ (0.6 L) = 0
F₂ (0.6 L) = (Mg) (0.3 L) + (mg) (0.2 L)
F₂ = ½ Mg + ⅓ mg
F₂ = ½ (44.0 kN) + ⅓ (21.0 kN)
F₂ = 29.0 kN
Alternatively, sum the forces in the y direction.
∑F = ma
F₁ + F₂ − Mg − mg = 0
F₂ = Mg + mg − F₁
F₂ = 44.0 kN + 21.0 kN − 36.0 kN
F₂ = 29.0 kN
(c) If we say x is the distance between the truck and the first support, then using our equations from part (b):
F₁ (0.6 L) = (Mg) (0.3 L) + (mg) (0.6 L − x)
F₂ (0.6 L) = (Mg) (0.3 L) + (mg) (x)
As x increases, F₁ decreases and F₂ increases.
(d) Using our equation from part (c), when x = 0.6 L, F₂ is:
F₂ (0.6 L) = (Mg) (0.3 L) + (mg) (0.6 L)
F₂ = ½ Mg + mg
F₂ = ½ (44.0 kN) + 21.0 kN
F₂ = 43.0 kN
Answer:
a. Left support = Right support = 22 kNb. Left support = 36 kN Right support = 29 kNc. Left support force will decrease Right support force will increase.d. Right support = 43 kNExplanation:
given:
weight of bridge = 44 kN
weight of truck = 21 kN
a) truck is off the bridge
since the bridge is symmetrical, left support is equal to right support.
Left support = Right support = 44/2
Left support = Right support = 22 kN
b) truck is positioned as shown.
to get the reaction at left support, take moment from right support = 0
∑M at Right support = 0
Left support (0.6) - weight of bridge (0.3) - weight of truck (0.4) = 0
Left support = 44 (0.3) + 21 (0.4)
0.6
Left support = 36 kN
Right support = weight of bridge + weight of truck - Left support
Right support = 44 + 21 - 36
Right support = 29 kN
c)
as the truck continues to drive to the right, Left support will decrease
as the truck get closer to the right support, Right support will increase.
d) truck is directly under the right support, find reaction at Right support?
∑M at Left support = 0
Right support (0.6) - weight of bridge (0.3) - weight of truck (0.6) = 0
Right support = 44 (0.3) + 21 (0.6)
0.6
Right support = 43 kN
There are 5.5 L of a gas present at -38.0 C. What is the temperature if the volume of the gas has changed to 1.30 L?
Answer:
We are given:
V1 = 5.5L T1 = -38 C or 235 k
V2 = 1.3L T2 = T
From the gas equation:
PV = nRT
Since the pressure (P) , number of moles (n) and the universal gas constant (R) are constants, we can write the same equation as:
V / T = k (where k is a constant)
so a bit more insight, since the values noted above are constant, when multiplied by each other, they will provide us with a constant number irrespective of the value of the variables
Changing the variables for the first case:
V1 / T1 = k (where k is the same constant) ----------------(1)
Similarly,
V2 / T2 = k (again, k has the same value)----------------(2)
From (1) and (2):
k is the common value
V1 / T1 = V2 / T2
Replacing the variables
5.5 / 235 = 1.3 / T
T = 1.3 * 235 / 5.5
T = 55.54 k
Therefore, at 55.54 K the gas will have a volume of 1.3L
An object is dropped from rest and falls freely 20 m to Earth. When is the speed of the object 9.8 m/s?
At the end of the first second of its fall
At the end of the first second of its fall
During the entire time of its fall
During the entire time of its fall
At no time is the speed 9.8 m/s
At no time is the speed 9.8 m/s
During the entire first second of its fall
During the entire first second of its fall
After it has fallen 9.8 meters
A student creates an electromagnetic wave and then reverses the direction of the current. Which of the following will happen to the magnetic field?
Answer:
I believe the electromagnetic field should be reversed.
Explanation:
When a student creates an electromagnetic wave and then reverses the direction of the current, the direction of the magnetic field will be reversed.
What is an Electromagnetic wave?An electromagnetic wave may be defined as a type of wave that is significantly created as a result of vibrations between an electric field and a magnetic field. These waves are composed of oscillating magnetic and electric fields.
According to the context of this question, when an individual is constructing an electromagnetic wave and then reverses the direction of the current, it will eventually affect the direction of the magnetic field in the same direction with respect to the current. So, if the direction of the current is reversed, the direction of the magnetic field would also be reversed.
Therefore, when a student creates an electromagnetic wave and then reverses the direction of the current, the direction of the magnetic field will be reversed.
To learn more about Magnetic fields, refer to the link:
https://brainly.com/question/14411049
#SPJ6
Your question seems incomplete. The most probable complete question is as follows:
A student creates an electromagnetic wave and then reverses the direction of the current. Which of the following will happen to the magnetic field?
The direction of the magnetic field will be reversed. The magnetic field will expand.The magnetic field would be canceled out and disappear.The magnetic field will cause the voltage of the battery to be reduced.A car is stopped for a traffic signal. When the light turns green, the car accelerates, increasing its speed from zero to 7.63 m/s in 3.94 s. What is the magnitude of the linear impulse experienced by a 73.7 kg passenger in the car during this time? Submit Answer Tries 0/20 What is the average force experienced by the passenger?
Answer:
1. p = 562.3 kg*m/s
2. F = 142.7 N
Explanation:
1. The linear impulse (p) is given by:
[tex] p = mv [/tex]
Where:
m: is the passenger's mass = 73.7 kg
v: is the speed = 7.63 m/s
[tex] p = mv = 73.7 kg*7.63 m/s = 562.3 kg*m/s [/tex]
Hence, the magnitude of the linear impulse experienced by a passenger is 562.3 kg*m/s.
2. The average force can be calculated using the following equation:
[tex] F = \frac{m(v_{f} - v_{0})}{t} = \frac{73.7 kg(7.63 m/s - 0)}{3.94 s} = 142.7 N [/tex]
Therefore, the average force experienced by the passenger is 142.7 N.
I hope it helps you!
Two identical items, object 1 and object 2, are dropped from the top of a 50.0m50.0m building. Object 1 is dropped with an initial velocity of 0m/s0m/s, while object 2 is thrown straight downward with an initial velocity of 13.0m/s13.0m/s. What is the difference in time, in seconds rounded to the nearest tenth, between when the two objects hit the ground
Answer:
Δt = 1.1 s
Explanation:
Given information:
H= 50.0 m
g= 9.8 m/s²
Object 1v₀ = 0[tex]H = \frac{1}{2} * g* t^{2}[/tex]
Solving for t, we get:
t₁= 3.2 s
Object 2v₀ = 13 m/sWe can find the final velocity for the object when it hits the ground, using the following expression:
[tex]v_{f}^{2} - v_{o}^{2} = 2*g*H[/tex]
Solving for vf, we get:
vf = 33.9 m/s
Applying the definition of acceleration, being this acceleration the one due to gravity (g), we can write the following equation:
[tex]v_{f} = v_{o} + g*t[/tex]
(Assuming the downward direction to be positive).
Solving for t, we get:
t₂ = 2.1 s
So the difference in time when both objects hit the ground, it's simply
Δt = t₂ - t₁ = 3.2 s - 2.1 s = 1.1 s
Before the development of quantum theory, Ernest Rutherford's experiments with gold atoms led him to propose the so-called Rutherford Model of atomic structure. The basic idea is that the nucleus of the atom is a very dense concentration of positive charge, and that negatively charged electrons orbit the nucleus in much the same manner as planets orbit a star. His experiments appeared to show that the average radius of an electron orbit around the gold nucleus must be about 10−1010−10 m. Stable gold has 79 protons and 118 neutrons in its nucleus.
What is the strength of the nucleus' electric field at the orbital radius of the electrons?
What is the kinetic energy of an electron in a circular orbit around the gold nucleus?
Answer:
1. [tex] E = 1.14 \cdot 10^{13} N/C [/tex]
2. [tex]E_{k} = 9.1 \cdot 10^{-17} J[/tex]
Explanation:
1. The strength of the nucleus' electric field (E):
[tex]E = \frac{kq}{r^{2}}[/tex]
Where:
k: is the Coulomb constant = 9x10⁹ Nm²/C²
q: is the proton charge = 1.6x10⁻¹⁹ C
r: is the radius = 10⁻¹⁰ m
[tex]E = \frac{kq}{r^{2}} = \frac{9\cdot 10^{9} Nm^{2}/C^{2}*79*1.6 \cdot 10^{-19} C}{(10^{-10} m)^{2}} = 1.14 \cdot 10^{13} N/C[/tex]
2. The kinetic energy (Ek) of an electron is the following:
[tex] E_{k} = \frac{1}{2}mv^{2} [/tex]
Where:
m is the electron's mass = 9.1x10⁻³¹ kg
v: is the speed of the electron
We can find the speed of the electron by equaling the centripetal force (Fc) and the electrostatic force (Fe):
[tex] F_{c} = F_{e} [/tex]
[tex] \frac{mv^{2}}{r} = \frac{kq^{2}}{r^{2}} = qE [/tex]
[tex] v^{2} = \frac{qEr}{m} = \frac{1.6 \cdot 10^{-19} C*1.14 \cdot 10^{13} N/C*10^{-10} m}{9.1 \cdot 10^{-31} kg} = 2.00 \cdot 10^{14} m^{2}/s^{2} [/tex]
Now, we can find the kinetic energy:
[tex] E_{k} = \frac{1}{2}mv^{2} = \frac{1}{2}9.1 \cdot 10^{-31} kg*2.00 \cdot 10^{14} m^{2}/s^{2} = 9.1 \cdot 10^{-17} J [/tex]
I hope it helps you!
A force of 15 newtons is used to push a box along the floor a distance of 3 meters. How much work was done?
Answer:
The answer is 45 JExplanation:
The work done by an object can be found by using the formula
workdone = force × distanceFrom the question
distance = 3 meters
force = 15 newtons
We have
workdone = 15 × 3
We have the final answer as
45 JHope this helps you
A solid nonconducting sphere of radius R carries a charge Q distributed uniformly throughout its volume. At a certain distance rl (r
(A) E/8
(B) E 78.
(C) E/2
(D) 2E
(E) 8E
Answer:
A ) E/8
Explanation:
If the sphere of radius R carries charge Q, then the volumetric charge density is:
ρ₁ = [Q/ (4/3)*π*R³]
Therefore the net charge inside r ( r < R ) is:
q₁ = ρ * (4/3)*π*r³
And E = K * q₁/r K = 9,98 *10⁹ [N*m²/C²]
E = K * ρ * (4/3)*π*r³/r
E = K * ρ * (4/3)*π*r²
If now the charge is distributed over a sphere of radius 2R
ρ₂ = [Q/ (4/3)*π*(2R)³]
ρ₂ = [Q/ (4/3)*π*8*R³]
Then ρ₂ < ρ₁ in fact ρ₂ = (1/8)*ρ₁
The electric field depends on the net charge enclosed by a gaussian surface, and the distance between the net charge and the considered point, ( considering the net charge as being at the center of the gaussian surface) In this case, there was no distance change then
E₂ = E₁/8
The right answer is lyrics A ) E/8
What measurements would you make (assuming you have the money, time, & equipment) to determine a star’s surface temperature? Explain your answer.
Answer:
use special filters on the telescope
Explanation:
Assuming you have access to a very high-grade telescope you would need to use special filters on the telescope that allows you to view the star's color spectrum. The color spectrum represents different levels of heat that a star is generating. This spectrum ranges from red to blue. Therefore in order to calculate the surface temperature, you would need to apply both a blue and red filter onto the telescope. Once you have these measurements you would need to compare them in order to pinpoint the correct variation of color which would give a close enough estimate of the surface temperature of the star.
A certain superconducting magnet in the form of a solenoid of length 0.300 m can generate a magnetic field of 8.90 T in its core when its coils carry a current of 95 A. Find the number of turns in the solenoid.
Answer:
The number of turns in the solenoid is 22366.
Explanation:
The number of turns in the solenoid can be found using the following equation:
[tex] B = \mu_{0} I\frac{N}{L} [/tex]
Where:
B: is the magnetic field = 8.90 T
L: is the solenoid's length = 0.300 m
N: is the number of turns =?
I: is the current = 95 A
μ₀: is the magnetic constant = 4π×10⁻⁷ H/m
By solving equation (1) for N we have:
[tex] N = \frac{BL}{\mu_{0} I} = \frac{8.90 T*0.300 m}{4\pi \cdot 10^{-7} H/m*95 A} = 22366 turns [/tex]
Therefore, the number of turns in the solenoid is 22366.
I hope it helps you!
You are hired to lift 30 kg crates a vertically 0.90 m from the ground onto a truck. How many crates would you have to load onto the truck in 1 minute for your average power output in lifting the crates to be 100 W
Answer:
22 crates
Explanation:
Power = Force ×Distance/time taken
Power = m×a×d/t
Power = 30×9.81×0.9/60 (1min was converted to second)
Power = 264.87/60
Power = 4.4145Watts
If my average power output us 100aw, then;
Number of crate to load = average power/4.4145
Number of crates to load = 100/4.4146
Number of crates to load = 22.6
Hence I will have to load about 22 crates onto the truck in 1 minute for my average power output in lifting the crates to be 100 W.
What cause objects to move? In three to five sentences .
Answer:
Gravity can affect the motion of objects as the force pulls objects closer to earth. Kinetic energy also causes movement in objects as that is energy in motion coming from stored energy known as (potential energy). With almost most importantly is needed is a force as without a force acting upon an objects is moving will continue to move and an object at rest will remain at rest as a gravitational pull or kinetic and potential energy for example are forces.
Explanation:
Have a great day :)
The equation that governs the period of a pendulum’s swinging. T=2π√L/g
Where T is the period, L is the length of the pendulum and g is a constant, equal to 9.8 m/s2. The symbol g is a measure of the strength of Earth’s gravity, and has a different value on other planets and moons.
On our Moon, the strength of earth’s gravity is only 1/6th of the normal value. If a pendulum on Earth has a period of 4.9 seconds, what is the period of that same pendulum on the moon?
Answer:
The period of that same pendulum on the moon is 12.0 seconds.
Explanation:
To determine the period of that same pendulum on the moon,
First, we will determine the value of g (which is a measure of the strength of Earth's gravity) on the Moon. Let the value of g on the Moon be [tex]g_{M}[/tex].
From the question, the strength of earth’s gravity is only 1/6th of the normal value. The normal value of g is 9.8 m/s²
∴ [tex]g_{M}[/tex] = [tex]\frac{1}{6} \times 9.8 m/s^{2}[/tex]
[tex]g_{M}[/tex] = 1.63 m/s²
From the question, T=2π√L/g
[tex]T = 2\pi \sqrt{\frac{L}{g} }[/tex]
We can write that,
[tex]T_{E} = 2\pi \sqrt{\frac{L}{g_{E} } }[/tex] .......... (1)
Where [tex]T_{E}[/tex] is the period of the pendulum on Earth and [tex]g_{E}[/tex] is the measure of the strength of Earth's gravity
and
[tex]T_{M} = 2\pi \sqrt{\frac{L}{g_{M} } }[/tex] .......... (2)
Where [tex]T_{M}[/tex] is the period of the pendulum on Moon and [tex]g_{M}[/tex] is the measure of the strength of Earth's gravity on the Moon.
Since we are to determine the period of the same pendulum on the moon, then, [tex]2\pi[/tex] and [tex]L[/tex] are constants.
Dividing equation (1) by (2), we get
[tex]\frac{T_{E} }{T_{M} } = \sqrt{\frac{g_{M} }{g_{E} } }[/tex]
From the question,
[tex]T_{E} = 4.9secs[/tex]
[tex]g_{E}[/tex] = 9.8 m/s²
[tex]g_{M}[/tex] = 1.63 m/s²
[tex]T_{M}[/tex] = ??
From,
[tex]\frac{T_{E} }{T_{M} } = \sqrt{\frac{g_{M} }{g_{E} } }[/tex]
[tex]\frac{4.9}{T_{M} } = \sqrt{\frac{1.63}{9.8} }[/tex]
[tex]\frac{4.9}{T_{M} } = 0.40783[/tex]
[tex]T_{M} =\frac{4.9}{0.40783 }[/tex]
[tex]T_{M} = 12.01 secs[/tex]
∴ [tex]T_{M} = 12.0secs[/tex]
Hence, the period of that same pendulum on the moon is 12.0 seconds.
Answer:
The period of that same pendulum on the moon is 12.0 s
Explanation:
Given;
period of a pendulum’s swinging, T=2π√L/g
the strength of earth’s gravity on moon, g₂ = ¹/₆(g₁)
period of pendulum on Earth, T₁ = 4.9 s
period of pendulum on moon, T₂ = ?
The length of the pendulum is constant, make it the subject of the formula;
[tex]T = 2\pi \sqrt{\frac{L}{g} }\\\\\frac{T}{2\pi} = \sqrt{\frac{L}{g}}\\\\(\frac{T}{2\pi} )^2 =\frac{L}{g}\\\\\frac{T^2}{4\pi^2} = \frac{L}{g}\\\\ L = \frac{gT^2}{4\pi^2}\\\\L_1 = L_2\\\\\frac{g_1T_1^2}{4\pi^2}= \frac{g_2T_2^2}{4\pi^2}\\\\g_1T_1^2 = g_2T_2^2\\\\T_2^2 = \frac{g_1T_1^2}{g_2} \\\\T_2 = \sqrt{\frac{g_1T_1^2}{g_2}}\\\\ T_2 = \sqrt{\frac{g_1T_1^2}{g_1/6}}\\\\ T_2 = \sqrt{\frac{6*g_1T_1^2}{g_1}}\\\\T_2 = \sqrt{6T_1^2}\\\\ T_2 = T_1\sqrt{6} \\\\T_2 = (4.9)\sqrt{6}\\\\ T_2 = 12.0 \ s[/tex]
Therefore, the period of that same pendulum on the moon is 12.0 s