Answer:
[tex]V=\frac{4}{3}\pi R^{3}[/tex]
Explanation:
The volume of a body is defined as the capacity of the object. The amount of matter that object contains is called its volume.
All the three dimensional objects have volume.
The SI unit of volume is m^3. The volume of liquids is measured by teh unit litre or milli litre.
The volume of the sphere is given by
[tex]V=\frac{4}{3}\pi R^{3}[/tex]
where, R is the radius of the sphere.
To provide the pulse of energy needed for an intense bass, some car stereo systems add capacitors. One system uses a 2.4F capacitor charged to 24 V, double the normal 12 V provided by the car's battery. How much energy does the capacitor store at 12 V? How much energy does the capacitor store at 24 V?
Answer:
Explanation:
Energy stored in a capacitor
= 1/2 CV²
C is capacitance and V is potential of the capacitor .
When capacitor is charged to 24 V ,
E₁ = 1/2 x 2.4 x 24 x24 = 691.2 J
When it is charged to 12 volt
E₂ = 1/2 CV²
.5 X 2.4 X 12 X12
= 172.8 J
Show that a sinusoidal wave propagating to the left along x-axis is a solution of the differential wave equation.
Answer:
The wave equation is [tex]\frac{d^{2}u }{dt^{2} }[/tex] = [tex]c^{2}[/tex] [tex]\frac{d^{2}u }{dx^{2} }[/tex]
a sinusoidal wave can be u = Acos( ax + bt) + B*sin(ax + bt)
where A, a, B and b are real constants. (here you also can add a phase to the arguments of the sin and cosine)
then [tex]\frac{d^{2}u }{dt^{2} }[/tex] = [tex]b^{2}[/tex]*( -Acos(ax + bt) - B*sin(ax + bt))
and [tex]c^{2}[/tex] [tex]\frac{d^{2}u }{dx^{2} }[/tex]= [tex]ac^{2}[/tex]*( -Acos(ax + bt) - B*sin(ax + bt))
then if a*c = b, this is a solution of the wave equation.
A snowboarder glides down a 48-m-long, 15° hill. She then glides horizontally for 10 m before reaching a 30° upward slope. Assume the snow is frictionless. What is her velocity at the bottom of the hill?
How far can she travel up the 30° slope?
Answer:
Her velocity at the bottom of the hill is 15.61m/s and she travel up the 30° slope 24.85m
Explanation:
For simplicity purpose, we can analyze the section of the snowboarder's travel in the hill, in the horizontal surface and in the slope separately.
In the hill, we will say that the x-axis is parallel to the hill, and y-axis is perpendicular. Using geometry, we can see that the angle of the snowboarder's weight force from the y-axis is 15°. The velocity of the snowboarder will increase in the direction parallel to the hill, in a constant acceleration motion:
[tex]F_x: W_x = ma\\W*sin(15) = ma\\mg*sin(15) = ma\\a = g*sin(15) = 9.81m/s^2 * sin(15) = 2.54m/s^2[/tex]
With the acceleration, we can use the equations for constant acceleration motion:
[tex]v_f^2 - v_o^2 = 2a*d\\v_f^2 - (0m/s)^2=2*2.54m/s^2*48m\\vf = \sqrt{2*2.54m/s^2*48m}=15.61 m/s[/tex]
This would be her velocity at the bottom of the hill.
As there is no friction, she would reach the bottom of the slope with this velocity.
In the slope, the line of reasoning is similar as in the hill, with the difference that the acceleration will oppose velocity.
[tex]F_x: -W_x = ma\\-W*sin(30) = ma\\-mg*sin(30) = ma\\a = -g*sin(30) = -9.81m/s^2 * sin(30) = -4.905m/s^2[/tex]
[tex]v_f^2-v_o^2=2a*d\\d=\frac{(v_f^2-v_o^2)}{2a}=\frac{((0m/s)^2-(15.61m/s)^2)}{2(-4.905m/s^2)} = 24.85m[/tex]
The speed of the snowboarder at the bottom of the hill is 15.62 m/s.
The distance the snowboarder travel up the 30° slope is 24.9 m.
Acceleration of the snowboarder on 15⁰ hillThe acceleration of the snowboarder is calculated as follows;
[tex]W sin\theta - F_f = ma\\\\mgsin\theta - 0 = ma\\\\mg sin(\theta) = ma\\\\a = g(sin\theta)\\\\a = 9.8 \times sin(15)\\\\a = 2.54 \ m/s^2[/tex]
The speed of the snowboarder at the bottom of the hill is calculated as follows;
[tex]v^2 = u^2 + 2ah\\\\v^2 = 0 + 2ah\\\\v = \sqrt{2ah} \\\\v = \sqrt{2(2.54)(48)} \\\\v = 15.62 \ m/s[/tex]
The acceleration of the snowboarder up 30° slope is calculated;
[tex]-Wsin(\theta)- F_f = ma\\\\-Wsin(\theta) -0 = ma\\\\-mgsin(\theta)= ma\\\\-gsin(\theta) = a\\\\a = -9.8 \times sin(30)\\\\a = -4.9 \ m/s^2[/tex]
The distance the snowboarder travel up the 30° slope is calculated as follows;
[tex]v^2 = u^2 - 2ah\\\\-2ah = v^2- u^2\\\\-2ah = v^2 -0\\\\-2ah = v^2\\\\h = \frac{v^2}{-2a} \\\\h = \frac{(15.62)^2}{-2(-4.9)} \\\\h = 24.9 \ m[/tex]
Learn more about net force on inclined here: https://brainly.com/question/14408327
Assume Young’s modulus for bone is 1.50 3 1010 N/m2. The bone breaks if stress greater than 1.50 3 108 N/m2 is imposed on it. (a) What is the maximum force that can be exerted on the femur bone in the leg if it has a minimum effective diameter of 2.50 cm? (b) If this much force is applied compressively, by how much does the 25.0-cm-long bone shorten?
Answer:
[tex]F_{max}=2.95*10^{5}N[/tex]
[tex]\Delta l=0.25cm[/tex]
Explanation:
E=1.50x10^10 N/m2 Young's modulus of bone
σmax=1.50x10^8 N/m2 Max stress tolerated by the bone
Relation between stress and Force:
[tex]\sigma=\frac{F}{A}=\frac{F}{\pi*d^{2}/4}[/tex]
[tex]F_{max}=\sigma_{max}*\pi*d^{2}/4}=1.50*10^{8}*\pi*(2.5*10^{-2})^{2}=2.95*10^{5}N[/tex]
Relation between stress and strain:
Young's modulus is defined by the ratio of longitudinal stress σ , to the longitudinal strain ε:
[tex]E=\frac{\sigma}{\epsilon}[/tex]
[tex]\epsilon=\frac{\Delta l}{l}[/tex]
We solve these equations to find the bone compression:
[tex]\Delta l=l*\frac{\sigma}{E}=25*\frac{1.50*10^{8}}{1.50*10^{10}}=0.25cm[/tex]
Final answer:
To find the maximum force that can be exerted on the femur bone in the leg, we need to calculate the stress using Young's modulus. The maximum force is approximately 7.35 * 10^7 N. To find the amount by which the bone shortens, we need to calculate the strain using Young's modulus and the stress.
Explanation:
To find the maximum force that can be exerted on the femur bone in the leg, we need to calculate the stress. The stress is equal to the force divided by the cross-sectional area of the bone. The cross-sectional area can be determined using the diameter of the bone.
(a) The diameter of the bone is 2.50 cm, which is equal to 0.025 m. The cross-sectional area can be calculated using the formula for the area of a circle: π * (radius)^2. In this case, the radius is half of the diameter, so the area is approximately 0.4909 m2. To find the maximum force, we can use the formula for stress: stress = force / area. Rearranging the formula, we have: force = stress * area. Plugging in the values, we get: force = (1.50 * 108 N/m2) * (0.4909 m2). The maximum force that can be exerted on the femur bone is approximately 7.35 * 107 N.
(b) To find the amount by which the bone shortens, we need to calculate the strain. The strain is equal to the change in length divided by the original length of the bone. The change in length can be determined using the formula for strain: strain = change in length / original length. Rearranging the formula, we have: change in length = strain * original length. We can calculate the strain using the stress and Young's modulus: strain = stress / Young's modulus. Plugging in the values, we get: strain = (1.50 * 108 N/m2) / (1.50 * 1010 N/m2). The change in length can be calculated using the formula: change in length = strain * original length. Plugging in the values, we get: change in length = (1.50 * 108 N/m2) / (1.50 * 1010 N/m2) * 25.0 cm.
If an arrow's mass is doubled and the speed is halved, the momentum is changed by a factor of: a) 1 b) 2 c) 0.25 d) 0.5
Explanation:
The momentum of an object is given by :
[tex]p=m\times v[/tex]............(1)
m is the mas of the object
v is the speed of the object
According to question, arrow's mass is doubled and the speed is halved. So,
m' = 2m
v' = v/2
The new momentum becomes :
[tex]p'=2m\times \dfrac{v}{2}[/tex]
p' = mv
p' = p
So, the momentum remains the same. The momentum is changed by a factor of 1. Hence, this is the required solution.
An electron moving to the right at 7.5 x 10^5 m/s enters a uniform electric field parallel to its direction of motion. If the electron is to be brought to rest in the space of 7.0 cm . What is the strength of the field?
Answer:
The strength of the field is 22.84 N/C.
Explanation:
Given that,
Speed [tex]v= 7.5\times10^{5}\ m/s[/tex]
Distance = 7.0 cm
We need to calculate the acceleration
Using equation of motion
[tex]v^2-u^2=2as[/tex]
Put the value in the equation
[tex]0-(7.5\times10^{5})^2=2\times a\times7.0\times10^{-2}[/tex]
[tex]a =-\dfrac{(7.5\times10^{5})^2}{2\times7.0\times10^{-2}}[/tex]
[tex]a =-4.017\times10^{12}\ m/s^2[/tex]
We need to calculate the strength of the field
Using newton's second law and electric force
[tex]F = ma = qE[/tex]
[tex]-qE=-ma[/tex]
[tex]E=\dfrac{ma}{q}[/tex]
Put the value into the formula
[tex]E=\dfrac{9.1\times10^{-31}\times4.017\times10^{12}}{1.6\times10^{-19}}[/tex]
[tex]E=22.84\ N/C[/tex]
Hence, The strength of the field is 22.84 N/C.
Final answer:
The strength of the electric field that can stop an electron moving at 7.5 x 10^5 m/s within a distance of 7 cm can be calculated using the work-energy principle, where the work done by the electric field is equal to the change in kinetic energy of the electron.
Explanation:
The strength of the electric field required to bring an electron moving to the right at 7.5 x 105 m/s to rest in a distance of 7.0 cm can be found using the work-energy principle and knowing the force exerted by an electric field on a charge. The work done by the electric field is equal to the change in kinetic energy of the electron, which is initially ½ mv2 and becomes zero when the electron is at rest.
To calculate the strength of the field, we can use:
Work (W) = Electric field (E) x Charge (e) x Distance (d)
½ mv2 = E * e * d
where m is the mass of the electron, v is its initial velocity, e is the elementary charge (approximately 1.6 x 10−12 C), and d is the distance (7.0 cm or 0.07 m). We solve for E to find the strength of the electric field
If a mile is 5280 ft long and a yard contains 3 ft, how many yards are there in a mile?
To determine how many yards are in a mile, knowing that a mile equals 5280 feet and a yard contains 3 feet, divide the total feet in a mile by the feet in a yard, resulting in 1760 yards in a mile.
Explanation:To find how many yards are there in a mile, given that a mile is 5280 feet long and a yard contains 3 feet, we can divide the total number of feet in a mile by the number of feet in a yard. Using the formula for conversion, we calculate:
Yards in a mile = Total feet in a mile ÷ Feet in a yard
By substituting the given values:
Yards in a mile = 5280 ft ÷ 3 ft
Yards in a mile = 1760
This calculation clearly shows that there are 1760 yards in a mile. This example emphasizes the importance of understanding unit conversions in mathematics, allowing us to easily switch between units of measurement.
A solid sphere of uniform density has a mass of 8.4 × 104 kg and a radius of 4.0 m. What is the magnitude of the gravitational force due to the sphere on a particle of mass 9.8 kg located at a distance of (a) 19 m and (b) 0.52 m from the center of the sphere
Answer:
a) [tex]F_a=0.152 \mu N[/tex]
b) [tex]F_b=203.182 \mu N[/tex]
Explanation:
The center of mass of an homogeneous sphere is its center, therefore you can use Newton's universal law of gravitation to find both questions.
[tex]F_g=G\frac{m_1m_2}{d}[/tex]
[tex]G=6.674*10^{-11} NmKg^{-2}[/tex]
a) d = 19m
[tex]F_a = G\frac{8.4*10^{4}*9.8}{19^2}[/tex]
[tex]F_a=0.152 \mu N[/tex]
b) d = 0.52
[tex]F_b = G\frac{8.4*10^{4}*9.8}{0.52^2}[/tex]
[tex]F_b=203.182 \mu N[/tex]
Answer:
(a) GF = 1.522 x (10 ^ -7) N
(b) GF = 2.032 x (10 ^ -4) N
Explanation:
The magnitude of the gravitational force follows this equation :
GF = (G x m1 x m2) / (d ^ 2)
Where G is the gravitational constant universal.
G = 6.674 x (10 ^ -11).{[N.(m^ 2)] / (Kg ^ 2)}
m1 is the mass from the first body
m2 is the mass from the second body
And d is the distance between each center of mass
m2 is a particle so m2 it is a center of mass itself
The center of mass from the sphere is in it center because the sphere has uniform density
For (a) d = 19 m
GF = {6.674 x (10 ^ -11).{[N.(m ^ 2)] / (Kg ^ 2)} x 8.4 x (10 ^ 4) Kg x 9.8 Kg} / [(19 m)^ 2]
GF = 1.522 x (10 ^ -7) N
For (b) d = 0.52 m
GF = 2.032 x (10 ^ -4) N
Notice that we have got all the data in congruent units
Also notice that the force in (b) is bigger than the force in (a) because the distance is shorter
The speed you compute by dividing your car's mileage by the time of travel is the ______ speed of the car, whereas the speed your speedometer reads is the car's ________speed.
Answer:
average and instant
Explanation:
The average speed is the ratio of the total path traveled and the time it took to travel that path, that is why the first space must be average speed, this because it takes into account the total amount of distance, and the total amount of time.
Instant speed, is the speed an objet (in this case a car) has in a particular moment in time, for this speed it doesn't matter the distance or the time that the car has traveled, it only matters the speed in that moment, that is what the speedometer measures, thus the second blank space must be instant speed.
A subway train starts from rest at a station and accelerates at a rate of 1.68 m/s2 for 14.2 s. It runs at constant speed for 68.0 s and slows down at a rate of 3.70 m/s2 until it stops at the next station. What is the total distance covered in kilometers?
Answer:
total distance = 1868.478 m
Explanation:
given data
accelerate = 1.68 m/s²
time = 14.2 s
constant time = 68 s
speed = 3.70 m/s²
to find out
total distance
solution
we know train start at rest so final velocity will be after 14 .2 s is
velocity final = acceleration × time ..............1
final velocity = 1.68 × 14.2
final velocity = 23.856 m/s²
and for stop train we need time that is
final velocity = u + at
23.856 = 0 + 3.70(t)
t = 6.44 s
and
distance = ut + 1/2 × at² ...........2
here u is initial velocity and t is time for 14.2 sec
distance 1 = 0 + 1/2 × 1.68 (14.2)²
distance 1 = 169.37 m
and
distance for 68 sec
distance 2= final velocity × time
distance 2= 23.856 × 68
distance 2 = 1622.208 m
and
distance for 6.44 sec
distance 3 = ut + 1/2 × at²
distance 3 = 23.856(6.44) - 0.5 (3.70) (6.44)²
distance 3 = 76.90 m
so
total distance = distance 1 + distance 2 + distance 3
total distance = 169.37 + 1622.208 + 76.90
total distance = 1868.478 m
Answer:
1868.5 m
Explanation:
For AB :
u = 0 m/s
a = + 1.68 m/s^2
t = 14.2 s
Let the distance is s1 and the velocity at B is v.
Use first equation of motion
v = u + at
v = 0 + 1.68 x 14.2 = 23.856 m/s
Use third equation of motion
[tex]v^{2}=u^{2}+2as_{1}[/tex]
[tex]23.856^{2}=0^{2}+2\times1.68\times s_{1}[/tex]
s1 = 169.38 m
For BC:
Let the distance is s2.
s2 = v x t
s2 = 23.856 x 68 = 1622.21 m
For CD:
u = 23.856 m/s
a = - 3.7 m/s^2
v = 0
Let the distance is s3.
Use third equation of motion
[tex]v^{2}=u^{2}+2as_{3}[/tex]
[tex]0^{2}=23.856^{2}-2\times 3.7 \times s_{3}[/tex]
s3 = 76.91 m
The total distance traveled is
s = s1 + s2 + s3
s = 169.38 + 1622.21 + 76.91 = 1868.5 m
Thus, the total distance traveled is 1868.5 m.
A 0.350kg bead slides on a curved fritionless wire,
startingfrom rest at point A. At point B the bead
collideselastically with a 0.530kg ball at rest. find distance call
risesas it moves up the wire. Point A is 2.20 m from ground andfree
fall accel is 9.80 m/s. round answer to 3 significantfigures.
Answer:
h2 = 0.092m
Explanation:
From a balance of energy from point A to point B, we get speed before the collision:
[tex]m1*g*h-\frac{m1*V_B^2}{2}=0[/tex] Solving for Vb:
[tex]V_B=\sqrt{2gh}=6.56658m/s[/tex]
Since the collision is elastic, we now that velocity of bead 1 after the collision is given by:
[tex]V_{B'}=V_B*\frac{m1-m2}{m1+m2} = \sqrt{2gh}* \frac{m1-m2}{m1+m2}=-1.34316m/s[/tex]
Now, by doing another balance of energy from the instant after the collision, to the point where bead 1 stops, we get the distance it rises:
[tex]m1*g*h2-\frac{m1*V_{B'}^2}{2}=0[/tex] Solving for h2:
h2 = 0.092m
A fullback preparing to carry the football starts from rest and accelerates straight ahead. He is handed the ball just before he reaches the line of scrimmage. Assume that the fullback accelerates uniformly (even during the handoff), reaching the line with a velocity of 7.60 m/s. If he takes 1.07 s to reach the line, how far behind it did he start?
Answer:
x=4.06m
Explanation:
A body that moves with constant acceleration means that it moves in "a uniformly accelerated movement", which means that if the velocity is plotted with respect to time we will find a line and its slope will be the value of the acceleration, it determines how much it changes the speed with respect to time.
When performing a mathematical demonstration, it is found that the equations that define this movement are as follows.
Vf=Vo+a.t (1)\\\\
{Vf^{2}-Vo^2}/{2.a} =X(2)\\\\
X=Xo+ VoT+0.5at^{2} (3)\\
Where
Vf = final speed
Vo = Initial speed
T = time
A = acceleration
X = displacement
In conclusion to solve any problem related to a body that moves with constant acceleration we use the 3 above equations and use algebra to solve
for this problem
Vf=7.6m/s
t=1.07
Vo=0
we can use the ecuation number one to find the acceleration
a=(Vf-Vo)/t
a=(7.6-0)/1.07=7.1m/s^2
then we can use the ecuation number 2 to find the distance
{Vf^{2}-Vo^2}/{2.a} =X
(7.6^2-0^2)/(2x7.1)=4.06m
A proton is released in a uniform electric field, and it experiences an electric force of 2.07 x 10^-14 N toward the south. Part A) What is the magnitude of the electric field? Part B) What is the direction of the electric field? O west O east O south O north
Answer:
The magnitude of the electric field is 129375 N/C toward south.
Explanation:
Given that,
Electric force [tex]F=2.07\times10^{-14}\ N[/tex]
(A). We need to calculate the magnitude of the electric field
Using formula of electric field
[tex]F = qE[/tex]
[tex]E=\dfrac{F}{q}[/tex]
Where, q = charge of proton
E = electric field
[tex]E=\dfrac{2.07\times10^{-14}}{1.6\times10^{-19}}[/tex]
[tex]E=129375\ N/C[/tex]
(B). The direction of the electric field is toward the direction of the force.
So, The direction of the electric field is toward south
Hence, The magnitude of the electric field is 129375 N/C toward south.
For an object of mass 1 Kg, estimate the number of protons and neutrons in the material, noting that a proton or neutron weighs 2000 times more than an electron. Assume that the number of neutrons is equal to the number of protons. Also assume that the absolute value of the net charge is less than 1Coulomb. Estimate the number of electrons in this object.
Answer:
The answer is very close to [tex]N_e=2.989\times10^{26}[/tex], where [tex]N_e[/tex] is the number of electrons.
Explanation:
First we take into account that the block weighs 1Kg, and that the number of protons and electrons is the same. As the electron mass is tiny even compared to that of the proton and neutrons we can neglect it in our considerations.
Let's start by equating the mass of all protons and neutrons to the mass of the of the object:
[tex]N_p m_p+N_n m_n=1[/tex]
Where [tex]N_p[/tex] and [tex]N_n[/tex] is the number of protons and neutrons respectively. [tex]m_p[/tex] and [tex]m_n[/tex] is the mass of an proton and a neutron respectively. Because the number of protons and neutrons is equal we can say the following [tex]N_p=N_n=N[/tex], thus we have:
[tex]N_p m_p+N_n m_n=N(m_e+m_p)=1Kg \implies N=\frac{1}{m_p+m_n} [/tex]
On the other hand we have that the sum of all charges is less than the absolute value of 1C, we can express this by the following:
[tex]-1<N_p\cdot e-N_e\cdot e<1[/tex]
[tex]\implies -1<N\cdot e-N_e\cdot e<1[/tex]
Where [tex]e[/tex] is the proton charge (same as for the electron). We continue with the inequality:
[tex]-N\cdot e-1<N_p\cdot e-N_e\cdot e<-N\cdot e+1[/tex]
[tex]\implies \frac{N\cdot e+1}{e}>N_e>\frac{N\cdot e-1}{e}[/tex]
[tex]\implies \frac{(m_e+m_p)^{-1}\cdot e+1}{e}>N_e>\frac{(m_e+m_p)^{-1}\cdot e-1}{e}[/tex]
We have the estimated number of electrons bound. Because
[tex](m_e+m_p)^{-1}\cdot e>>1[/tex] We can neglect the ones on the rightmost and leftmost parts of the inequality. We then have
[tex]N_e\approx\frac{(m_p+m_n)^{-1}\cdot e}{e}[/tex]
Using the table values of the mass of the proton, mass of the neutron and the electron charge e we get
[tex]N_e\approx\frac{(1.672\times 10^{-27}+1.674\times 10^{-27})^{-1}\cdot e}{e}=(1.672\times 10^{-27}+1.674\times 10^{-27})^{-1}[/tex]
[tex]\, =2.989\times 10^{26}[/tex] electrons
Explain why Earth is not spherical in shape, but bulges at
theEquator?
Explanation:
Every rotating body experiences centrifugal force. Due to this force the body tends to bulge out around it mid point and gets flattened at the poles. Same is applicable to Earth as well. Since the Earth is rotating at a very high speed, its equator gets bulged out due to centrifugal force. Because of this bulged equator, Earth's pole to pole diameter and equatorial diameter has difference of around 42.76 km. It is flatter on the poles. This also proves that Earth is not a perfect sphere.
Answer and Explanation:
The reason for the not being perfectly spherical ad bulging out at the equator is that The centripetal force acting toward's the earth gravitational center tries to keep the Earth in perfect spherical shape.
Also the angular momentum of the orbiting planet influences the bulge,
The greater angular momentum results in more bulge while the lower value of it results in lesser bulge and more perfect spherical shape.
Also, a greater amount of force directed towards the center and acting on the object at the equator results in the bulges at the equator whereas at poles this force is not required and hence radius is lower in that region.
A pendulum on Earth has a period of 1.2 seconds. The same pendulum on Mercury (dark side) has a period of 1.95 seconds. What is the free-fall acceleration of Mercury (in m/s^2)
Answer:
[tex]g'=3.71\ m/s^2[/tex]
Explanation:
Given that,
Time period of a pendulum on the earth's surface, T₁ = 1.2 s
Time period of the same pendulum on Mercury, T₂ = 1.95 s
The time period of the pendulum is given by :
[tex]T=2\pi \sqrt{\dfrac{l}{g}}[/tex]
On earth :
[tex]T_1=2\pi \sqrt{\dfrac{l}{g}}[/tex]
[tex]1.2=2\pi \sqrt{\dfrac{l}{9.8}}[/tex].............(1)
Let g' is the acceleration due to gravity on Mercury. So,
[tex]1.95=2\pi \sqrt{\dfrac{l}{g'}}[/tex]............(2)
From equation (1) and (2) :
[tex]\dfrac{1.2}{1.95}=\sqrt{\dfrac{g'}{9.8}}[/tex]
[tex]g'=(\dfrac{1.2}{1.95})^2\times 9.8[/tex]
[tex]g'=3.71\ m/s^2[/tex]
So, the acceleration due to gravity on the mercury is [tex]3.71\ m/s^2[/tex]. Hence, this is the required solution.
A horizontal sheet of negative charge has a uniform electric field E = 3000N/C. Calculate the electric potential at a point 0.7m above the surface.
Answer:
Electric potential, E = 2100 volts
Explanation:
Given that,
Electric field, E = 3000 N/C
We need to find the electric potential at a point 0.7 m above the surface, d = 0.7 m
The electric potential is given by :
[tex]V=E\times d[/tex]
[tex]V=3000\ N/C\times 0.7\ m[/tex]
V = 2100 volts
So, the electric potential at a point 0.7 m above the surface is 2100 volts. Hence, this is the required solution.
A 7450 kg rocket blasts off vertically from the launch pad with a constant upward acceleration of 2.35 m/s2 and feels no appreciable air resistance. When it has reached a height of 520 m , its engines suddenly fail so that the only force acting on it is now gravity.
(a) What is the maximum height this rocket will reach above the launch pad?
(b) How much time after engine failure will elapse before the rocket comes crashing down to the launch pad?
(c) How fast will it be moving just before it crashes?
Answer:
a) 520m
b) 10.30 s
c) 100,95 m/s
Explanation:
a) According the given information, the rocket suddenly stops when it reach the height of 520m, because the engines fail, and then it begins the free fall.
This means the maximum height this rocket reached before falling was 520 m.
b) As we are dealing with constant acceleration (due gravity) [tex]g=9.8 \frac{m}{s^{2}}[/tex] we can use the following formula:
[tex]y=y_{o}+V_{o} t-\frac{gt^{2}}{2}[/tex] (1)
Where:
[tex]y_{o}=520 m[/tex] is the initial height of the rocket (at the exact moment in which it stops due engines fail)
[tex]y=0[/tex] is the final height of the rocket (when it finally hits the launch pad)
[tex]V_{o}=0[/tex] is the initial velocity of the rocket (at the exact moment in which it stops the velocity is zero and then it begins to fall)
[tex]g=9.8m/s^{2}[/tex] is the acceleration due gravity
[tex]t[/tex] is the time it takes to the rocket to hit the launch pad
Clearing [tex]t[/tex]:
[tex]0=520 m+0-\frac{9.8m/s^{2} t^{2}}{2}[/tex] (2)
[tex]t^{2}=\frac{-520 m}{-4.9 m/s^{2}}[/tex] (3)
[tex]t=\sqrt{106.12 s^{2}[/tex] (4)
[tex]t=10.30 s[/tex] (5) This is the time
c) Now we need to find the final velocity [tex]V_{f}[/tex] for this rocket, and the following equation will be perfect to find it:
[tex]V_{f}=V_{o}-gt[/tex] (6)
[tex]V_{f}=0-(9.8 m/s^{2})(10.30 s)[/tex] (7)
[tex]V_{f}=-100.95 m/s[/tex] (8) This is the final velocity of the rocket. Note the negative sign indicates its direction is downwards (to the launch pad)
A car cruises with constant velocity on a (low traffic) freeway at 70 mph (about 31 m/s). Wind resistance opposes the car's motion with a force of 5000 N. Intuitively is the forward force on the car less than 5000N, equal to 5000 N or more than 5000 N. Explain your reasoning.
Answer:
Explanation:
The car is moving with uniform velocity . Hence there is no acceleration in the car .It indicates that net force on the car is zero . Since force in backward direction is exerted by the wind, to make net force zero , the forward push by the car must be equal to backward force by the wind. In other words
Forward force by car = backward force by wind = 5000 N.
The terminal velocity of a person falling in air depends upon the weight and the area of the person facing the fluid. Find the terminal velocity (in meters per second and kilometers per hour) of an 84.0 kg skydiver falling in a pike (headfirst) position with a surface area of 0.160 m2. (Assume that the density of air is 1.21 kg/m3 and the drag coefficient of a skydiver in a pike position is 0.7.)
Answer:
110.27 m/s or 396.972 km/h
Explanation:
F = Force
m = Mass = 84 kg
g = Acceleration due to gravity = 9.81 m/s²
C = Drag coefficient = 0.7
ρ = Density of air = 1.21 kg/m³
A = Surface area = 0.16 m²
v = Terminal velocity
F = ma
[tex]F=\frac{1}{2}\rho CAv^2\\\Rightarrow ma=\frac{1}{2}\rho CAv^2\\\Rightarrow v=\sqrt{2\frac{ma}{\rho CA}}\\\Rightarrow v=\sqrt{2\frac{84\times 9.81}{1.21\times 0.7\times 0.16}}\\\Rightarrow v=110.27\ m/s[/tex]
Converting to km/h
[tex]\frac{110.27}{1000}\times 3600=396.972\ km/h[/tex]
Terminal velocity of the skydiver is 110.27 m/s or 396.972 km/h
The sun is 150,000,000 km from earth; its diameter is 1,400,000 km. A student uses a 5.2-cm-diameter lens with f = 10 cm to cast an image of the sun on a piece of paper. Where should the paper be placed relative to the lens to get a sharp image?
To get a sharp image of the Sun, the paper should be placed at the focal length of the lens, which is 10 centimeters away from the lens. The image is in focus at this point because the light rays from the Sun are effectively parallel when they reach the lens, which then focuses these rays at its focal point.
Explanation:Given the sun is so far away, the light it emits is nearly parallel by the time it reaches Earth. When using a lens to cast an image of the Sun, the point where the image is in focus, that is, the focal point, is also the focal length of the lens.
In this case, the student uses a lens with f = 10 cm, meaning the focal length of the lens is 10 centimeters. To get a sharp image, the paper on which the image is being projected should be placed 10 cm away from the lens, or at the focal length of the lens. This is because the light is in sharp focus at this distance, creating a clear image on the paper.
An important concept here is that the Sun is an astronomical unit away, so the light rays from the Sun are essentially parallel when they reach the lens. The lens then focuses these parallel rays to its focal point, forming a sharp image at a distance equal to its focal length.
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When we throw an object vertically upwards its initial velocity: a. It will be less than the final
b. It will be greater than the final
c. It will be equivalent to the final
d. It remains constant until reaching its maximum height
Answer:
Its initial velocity will be greater than final velocity so option (b) will be correct option
Explanation:
As we throw the any object vertically the motion of the object will be opposes by the gravity.
And as the velocity of object is opposes by gravity, the final velocity goes on decreasing and finally it becomes zero.
So the initial velocity is always greater than final velocity when the object is thrown vertically upward.
So option (b) will be the correct option
A place-kicker must kick a football from a point 36.0 m (about 40 yards) from the goal. Half the crowd hopes the ball will clear the crossbar, which is 3.05 m high. When kicked, the ball leaves the ground with a speed of 23.6 m/s at an angle of 45.0° to the horizontal. (a) By how much does the ball clear or fall short (vertically) of clearing the crossbar? (Enter a negative answer if it falls short.)
_______m
(b) Does the ball approach the crossbar (and cross above or beneath it) while still rising or while falling?
rising or falling? _________
Answer:
Part (a) 10.15 m
Part (b) Rising
Explanation:
Given,
Initial speed of the ball = u = 23.6 m/sHeight of the crossbar = h = 3.05 mDistance between the ball and the cross bar = r = 36.0 mAngle of projection = [tex]\theta\ =\ 45.0^o[/tex]Initial velocity of the ball in the horizontal direction = [tex]u_x\ =\ ucos\theta[/tex]Initial velocity of the ball in the vertical direction = [tex]u_y\ =\ usin\theta[/tex]part (a)
Let 't' be the time taken to reach the ball to the cross bar,
In x-direction,
[tex]\therefore r\ =\ u_xt\\\Rightarrow t\ =\ \dfrac{r}{u_x}\ =\ \dfrac{r}{ucos\theta}\\\Rightarrow t\ =\ \dfrac{36.0}{23.6cos45^o}\\\Rightarrow t\ =\ 2.15\ sec[/tex]
Let y be the height attained by the ball at time t = 2.15 sec,
[tex]y\ =\ u_yt\ \ -\ \dfrac{1}{2}gt^2\\\Rightarrow y\ =\ usin\theta t\ -\ \dfrac{1}{2}gt^2\\\Rightarrow y\ =\ 23.6\times sin45^o\times 2.15\ -\ 0.5\times 9.81\ 2.15^2\\\Rightarrow y\ =\ 13.205\ m[/tex]
Now Let H be the height by which the ball is clear the crossbar.
[tex]\therefore H\ =\ y\ -\ h\ =\ 13.205\ -\ 3.05\ =\ 10.15\ m[/tex]
part (b)
At the maximum height the vertical velocity of the ball becomes zero.
i,e, [tex]v_y\ =\ 0[/tex]
Let h be the maximum height attained by the ball.
[tex]\therefore v_y^2\ =\ u_y^2\ -\ 2gh\\\Rightarrow 0\ =\ (usin\theta)^2\ -\ 2gh\\\Rightarrow h\ =\ \dfrac{(usin\theta)^2}{2g}\\\Rightarrow h\ =\ \dfrac{23.6\times sin45.0^o)^2}{2\times 9.81}\\\Rightarrow h\ =\ 14.19\ m[/tex]
Hence at the cross bar the ball attains the height 13.205 m but the maximum height is 14.19 m. Therefore the ball is rising when it reaches at the crossbar.
A flat disk of radius 0.50 m is oriented so that the plane of the disk makes an angle of 30 degrees with a uniform electric field. If the field strength is 713.0 N/C find the electric Tiux through the surface A) 560 Nm2/C B) 620 N·m2/C C) 160 n N.m2/C D) 280 N.m2/C
Answer:
electric flux is 280 Nm²/C
so correct option is D 280 Nm²/C
Explanation:
radius r = 0.50 m
angle = 30 degree
field strength = 713 N/C
to find out
the electric flux through the surface
solution
we find here electric flux by given formula that is
electric flux = field strength × area× cos∅ .......1
here area = πr² = π(0.50)²
put here all value in equation 1
electric flux = field strength × area× cos∅
electric flux = 713 × π(0.50)² × cos60
we consider the cosine of the angle between the direction of the field and the normal to the surface of the disk
so we use cos60
electric flux = 280 Nm²/C
so correct option is D 280 Nm²/C
If the potential due to a point charge is 500 V at a distance of 15.0 m, what are the sign and magnitude of the charge?
Answer:
[tex]q=+8.34*10^{-7}C}[/tex]
Explanation:
The potential V due to a charge q, at a distance r, is:
[tex]V=k\frac{q}{r}[/tex]
k=8.99×109 N·m^2/C^2 :Coulomb constant
We replace the values in order to find q:
[tex]q=\frac{V*r}{k}=\frac{500*15}{8.99*10^{9}}=8.34*10^{-7}C[/tex]
Answer:
i apolagize im late but yeah bois 700 points
Explanation:
Ships A and B leave port together. For the next two hours, ship A travels at 28 mph in a direction 32° west of north while ship B travels 24° east of north at 35 mph . -- What is the distance between the two ships two hours after they depart? -- What is the speed of ship A as seen by ship B?
Answer:
Explanation:
We shall write velocities in vector form
Ship A travels in the direction of 32 °west of north with velocity 28 mph
V₁ = - 28 Sin 32 i + 28 Cos 32 j
Ship B travels in the direction of 24 ° east of north with velocity 35 mph
V₂ = 35 Sin 24 i + 35 Cos 24 j
Their relative velocity
= V₁ -V₂ = - 28 Sin 32 i + 28 Cos 32 j - (35 Sin 24 i + 35 Cos 24 j )
-14.83 i - 14.23 i + 23.74 j - 31.97 j
= - 29.06 i - 8.23 j
Distance between them = relative velocity x time
- 29.06 i - 8.23 j x 2
= - 58.12 i - 16.46 j
magnitude² =( 58.12 ) ² + ( 16.46)² = 60.40²
magnitude = 60.40 km
Speed of ship A as seen by ship B
= Relative velocity of A wrt B
= - 28 Sin 32 i + 28 Cos 32 j - (35 Sin 24 i + 35 Cos 24 j )
= - 29.06 i - 8.23 j
A hollow sphere of inner radius 8.82 cm and outer radius 9.91 cm floats half-submerged in a liquid of density 948.00 kg/m^3. (a) What is the mass of the sphere? (b) Calculate the density of the material of which the sphere is made.
Answer:
a) 0.568 kg
b) 474 kg/m³
Explanation:
Given:
Inner radius = 8.82 cm = 0.0882 m
Outer radius = 9.91 cm = 0.0991 m
Density of the liquid = 948.00 Kg/m³
a) The volume of the sphere = [tex]\frac{4\pi}{3}\times(0.0991^2-0.0882^2)[/tex]
or
volume of sphere = 0.0012 m³
also, volume of half sphere = [tex]\frac{\textup{Total volume}}{\textup{2}}[/tex]
or
volume of half sphere = [tex]\frac{\textup{0.0012}}{\textup{2}}[/tex]
or
Volume of half sphere =0.0006 m³
Now, from the Archimedes principle
Mass of the sphere = Weight of the volume of object submerged
or
Mass of the sphere = 0.0006× 948.00 = 0.568 kg
b) Now, density = [tex]\frac{\textup{Mass}}{\textup{Volume}}[/tex]
or
Density = [tex]\frac{\textup{0.568}}{\textup{0.0012}}[/tex]
or
Density = 474 kg/m³
Emergency Landing. A plane leaves the airport in Galisteo and flies 170 km at 68° east of north and then changes direction to fly 230 km at 48° south of east, after which it makes an immediate emergency landing in a pasture. When the airport sends out a rescue crew, in which direction and how far should this crew fly to go directly to this plane?
To locate the emergency landed plane, we add the vectors representing the plane's two separate legs, breaking them into components using trigonometry. Summing these vectors gives us the direct path for the rescue crew in terms of both distance and bearing from the airport to the plane.
Explanation:To assist in this emergency landing scenario, we need to compute the resulting position vector by analyzing the two separate motions of the plane. The first motion has the plane fly 170 km at 68° east of north, and the second has it flying 230 km at 48° south of east. By representing these movements as vectors and adding them, we find the direct path the rescue crew should take.
This vector addition can be done graphically or by using trigonometry to break each leg of the plane's journey into its horizontal (east-west) and vertical (north-south) components. After determining the components, we can find the direct distance and bearing from the airport to the plane's location. The past examples and explanations equip us with strategies to calculate the required velocity of the plane relative to the ground and the direction the pilot must head by accounting for the known wind velocities, when necessary.
In summary, to find the direction and distance for the rescue crew, we add the vectors representing the plane's path, utilizing trigonometry to solve the components and then applying vector sum principles to find the result. This procedure allows us to efficiently direct the rescue efforts.
As a science project, you drop a watermelon off the top of the Empire State Building. 320 m above the sidewalk. It so happens that Superman flies by at the instant you release the watermelon. Superman is headed straight down with a constant speed of 30 m/s. A) How much time passes before the watermelon has the same velocity? B) How fast is the watermelon going when it passes Superman?C) How fast is the watermelon traveling when it hits the ground?
Answer:
3.06 seconds time passes before the watermelon has the same velocity
watermelon going at speed 59.9 m/s
watermelon traveling when it hits the ground at speed is 79.19 m/s
Explanation:
given data
height = 320 m
speed = 30 m/s
to find out
How much time passes before the watermelon has the same velocity and How fast is the watermelon going and How fast is the watermelon traveling
solution
we will use here equation of motion that is
v = u + at ....................1
here v is velocity 30 m/s and u is initial speed i.e zero and a is acceleration i.e 9.8 m/s²
put the value and find time t
30 = 0 + 9.8 (t)
t = 3.06 s
so 3.06 seconds time passes before the watermelon has the same velocity
and
we know superman cover distance is = velocity × time
so distance = 30 × t
and distance formula for watermelon is
distance = ut + 0.5×a×t² .............2
here u is initial speed i.e 0 and a is acceleration i.e 9.8 m/s² and h is 30 × t
30 × t = 0 + 0.5×9.8×t²
t = 6.12 s
so by equation 1
v = u + at
v = 0 + 9.8 ( 6.12)
v = 59.9 m/s
so watermelon going at speed 59.9 m/s
and
watermelon traveling speed formula is by equation of motion
v² - u² = 2as ......................3
here v is speed and u is initial speed i.e 0 and a is acceleration i.e 9.8 m/s² and s is distance i.e 320 m
v² - 0 = 2(9.8) 320
v = 79.19 m/s
so watermelon traveling when it hits the ground at speed is 79.19 m/s
An object is 30 cm in front of a converging lens with a focal length of 10 cm. Use ray tracing to determine the location of the image. Is the image upright or inverted? Is it real or virtual?
Answer:
Inverted
Real
Explanation:
u = Object distance = 30 cm
v = Image distance
f = Focal length = 10 cm
Lens Equation
[tex]\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\\\Rightarrow \frac{1}{f}-\frac{1}{u}=\frac{1}{v}\\\Rightarrow \frac{1}{v}=\frac{1}{10}-\frac{1}{30}\\\Rightarrow \frac{1}{v}=\frac{1}{15}\\\Rightarrow v=15\ cm[/tex]
As, the image distance is positive the image is real and forms on the other side of the lens
[tex]m=-\frac{v}{u}\\\Rightarrow m=-\frac{-15}{30}\\\Rightarrow m=-0.5[/tex]
As, the magnification is negative the image is inverted
By applying the lens equation, we calculate that the image is formed 15 cm behind the lens. This is a real image as it forms on the opposite side of the lens, and in the case of a converging lens, it will be inverted.
Explanation:First, we use the lens equation, which is 1/f = 1/do + 1/di, where f is the focal length, do is the object distance, and di is the image distance. In this case, the object distance 'do' is 30cm and the focal length 'f' is 10cm. Solving for 'di', we find that the image is located 15 cm behind the lens (i.e., on the opposite side from the object).
Since the image forms on the opposite side of the lens from where the object is, this indicates it's a real image. A positive image distance indicates a real image and a negative image distance indicates a virtual image.
For a converging lens, a real image is always inverted, and a virtual image is always upright. Therefore, in this case, the image would be inverted.
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