Question: The coefficient of linear expansion of steel is 11 x 10⁻⁶ per °c . A steel ball has a volume of
exactly 100 cm³ at 0 C. When heated to 100 C, its volume becomes:
Answer:
100.11 cm³
Explanation:
From the question,
γ = (v₂-v₁)/(v₁Δt)...................... Equation 1
Where γ = coefficient of volume expansion, v₂ = final volume, v₁ = initial volume, Δt = change in temperature.
make v₂ the subject of the equation
v₂ = v₁+γv₁Δt..................... Equation 2
Given: v₁ = 100 cm³, γ = 11×10⁻⁶/°C, Δt = 100 °C.
Substitute into equation 2
v₂ = 100+100(11×10⁻⁶)(100)
v₂ = 100+0.11
v₂ = 100.11 cm³
That 85 kg paratrooper from the 50's was moving at constant speed of 56 m/s because the air was applying a frictional drag force to him that matched his weight. If he fell this way for 40 m, how much heat was generated by this frictional drag force in J
Answer:
46648 J
Explanation:
mass m= 85 Kg
velocity v = 56 m/s
distance covered s =40 m
According to Question,
frictional drag force to him that matched his weight
[tex]\Rightarrow F_d =mg\\=85\times9.81=833 N[/tex]
Therefore, work done by practometer against the drag force = heat was generated by this frictional drag force in J
W=Q= F_d×s
=833×56 = 46648 J
A swimmer of mass 64.38 kg is initially standing still at one end of a log of mass 237 kg which is floating at rest in water. He runs toward the other end of the log and dives off with a horizontal speed of 3.472 m/s relative to the water. What is the speed of the log relative to water after the swimmer jumps off
Answer:
0.9432 m/s
Explanation:
We are given;
Mass of swimmer;m_s = 64.38 kg
Mass of log; m_l = 237 kg
Velocity of swimmer; v_s = 3.472 m/s
Now, if we consider the first log and the swimmer as our system, then the force between the swimmer and the log and the log and the swimmer are internal forces. Thus, there are no external forces and therefore momentum must be conserved.
So;
Initial momentum = final momentum
m_l × v_l = m_s × v_s
Where v_l is speed of the log relative to water
Making v_l the subject, we have;
v_l = (m_s × v_s)/m_l
Plugging in the relevant values, we have;
v_l = (64.38 × 3.472)/237
v_l = 0.9432 m/s
A 4.5 kg ball swings from a string in a vertical circle such that it has constant sum of kinetic and gravitational potential energy. Ignore any friction forces from the air or in the string. What is the difference in the tension between the lowest and highest points on the circle
Answer:
88.29 N
Explanation:
mass of the ball = 4.5 kg
weight of the ball will be = mass x acceleration due to gravity(9.81 m/s^2)
weight W = 4.5 x 9.81 = 44.145 N
centrifugal forces Tc act on the ball as it swings.
At the top point of the vertical swing,
Tension on the rope = Tc - W.
At the bottom point of the vertical swing,
Tension on the rope = Tc + W
therefore,
difference in tension between these two points will be;
Net tension = tension at bottom minus tension at the top
= Tc + W - (Tc - W) = Tc + W -Tc + W
= 2W
imputing the value of the weight W, we have
2W = 2 x 44.145 = 88.29 N
How much force is needed to cause a 15 kilogram bicycle to accelerate at a rate of 10
meters per second per second?
O A. 15 newtons
OB. 1.5 newtons
C. 150 newtons
OD. 10 newtons
At a pressure of one atmosphere oxygen boils at −182.9°C and freezes at −218.3°C. Consider a temperature scale where the boiling point of oxygen is 100.0°O and the freezing point is 0°O. Determine the temperature on the Oxygen scale that corresponds to the absolute zero point on the Kelvin scale.
Answer: -254.51°O
Explanation:
Ok, in our scale, we have:
-182.9°C corresponds to 100° O
-218.3°C corresponds to 0°
Then we can find the slope of this relation as:
S = (100° - 0°)/(-182.9°C - (-218.3°C)) = 2.82°O/°C
So we can have the linear relationship between the scales is:
Y = (2.82°O/°C)*X + B
in this relation, X is the temperature in Celcius and Y is the temperature in the new scale.
And we know that when X = -182.9°C, we must have Y = 0°O
then:
0 = (2.82°O/°C)*(-182.9°C) + B
B = ( 2.82°O/°C*189.9°C) = 515.778°O.
now, we want to find the 0 K in this scale, and we know that:
0 K = -273.15°C
So we can use X = -273.15°C in our previous equation and get:
Y = (2.82°O/°C)*(-273.15°C) + 515.778°O = -254.51°O
A car is designed to get its energy from a rotating flywheel with a radius of 1.50 m and a mass of 430 kg. Before a trip, the flywheel is attached to an electric motor, which brings the flywheel's rotational speed up to 5,200 rev/min.
Required:
a. Find the kinetic energy stored in the flywheel.
b. If the flywheel is to supply energy to the car as would a 15.0-hp motor, find the length of time the car could run before the flywheel would have to be brought back up to speed.
Answer:
a
[tex]KE = 7.17 *10^{7} \ J[/tex]
b
[tex]t = 6411.09 \ s[/tex]
Explanation:
From the question we are told that
The radius of the flywheel is [tex]r = 1.50 \ m[/tex]
The mass of the flywheel is [tex]m = 430 \ kg[/tex]
The rotational speed of the flywheel is [tex]w = 5,200 \ rev/min = 5200 * \frac{2 \pi }{60} =544.61 \ rad/sec[/tex]
The power supplied by the motor is [tex]P = 15.0 hp = 15 * 746 = 11190 \ W[/tex]
Generally the moment of inertia of the flywheel is mathematically represented as
[tex]I = \frac{1}{2} mr^2[/tex]
substituting values
[tex]I = \frac{1}{2} ( 430)(1.50)^2[/tex]
[tex]I = 483.75 \ kgm^2[/tex]
The kinetic energy that is been stored is
[tex]KE = \frac{1}{2} * I * w^2[/tex]
substituting values
[tex]KE = \frac{1}{2} * 483.75 * (544.61)^2[/tex]
[tex]KE = 7.17 *10^{7} \ J[/tex]
Generally power is mathematically represented as
[tex]P = \frac{KE}{t}[/tex]
=> [tex]t = \frac{KE}{P}[/tex]
substituting the value
[tex]t = \frac{7.17 *10^{7}}{11190}[/tex]
[tex]t = 6411.09 \ s[/tex]
One kind of baseball pitching machine works by rotating a light and stiff rigid rod about a horizontal axis until the ball is moving toward the target. Suppose a 144 g baseball is held 82 cm from the axis of rotation and released at the major league pitching speed of 87 mph.
Required:
a. What is the ball's centripetal acceleration just before it is released?
b. What is the magnitude of the net force that is acting on the ball just before it is released?
Answer:
a. ac = 1844.66 m/s²
b. Fc = 265.63 N
Explanation:
a.
The centripetal acceleration of the ball is given as follows:
ac = v²/r
where,
ac = centripetal acceleration = ?
v = speed of ball = (87 mph)(1 h/ 3600 s)(1609.34 m / 1 mile) = 38.9 m/s
r = radius of path = 82 cm = 0.82 m
Therefore,
ac = (38.9 m/s)²/0.82 m
ac = 1844.66 m/s²
b.
The centripetal force is given as:
Fc = (m)(ac)
Fc = (0.144 kg)(1844.66 m/s²)
Fc = 265.63 N
In a contest, two tractors pull two identical blocks of stone thesame distance over identical surfaces. However, block A is moving twice as fast as block B when it crosses the finish line. Which statement is correct?a) Block A has twiceas much kinetic energy as block B.b) Block B has losttwice as much kinetic energy to friction as block A.c) Block B has losttwice as much kinetic energy as block A.d) Both blocks havehad equal losses of energy to friction.e) No energy is lostto friction because the ground has no displacement.
Answer:
d) Both blocks have had equal losses of energy to friction
Explanation:
As it is mentioned in the question that two tractors pull two same stone blocks having the identical distance over the same surfaces
Moreover, the block A is twice as fast than block B at the time of crossing the finish line
So based on the above information, it contains the losses of identical friction
And we also know that
Friction energy loss is
[tex]= \mu \times m \times g \times D[/tex]
It would be the same for both the blocks
hence, the option d is correct
The correct answer will be both blocks have had equal losses of energy to friction.
What is friction?Friction is defined as when any object is slides on a surface by means of any external force then the force in the opposite direction generated between the surface and the body restrict the motion of the body this force is called as the friction.
As it is mentioned in the question that two tractors pull two same stone blocks having the identical distance over the same surfaces.
Moreover, the block A is twice as fast as block B at the time of crossing the finish line.
So based on the above information, it contains the losses of identical friction.
And we also know that
Friction energy loss is
[tex]E_f=\mu m g D[/tex]
It would be the same for both the blocks
Hence both blocks have had equal losses of energy to friction.
To know more about friction, follow
https://brainly.com/question/24386803
A mass m slides down a frictionless ramp and approaches a frictionless loop with radius R. There is a section of the track with length 2R that has a kinetic friction coefficient of 0.5. From what height h must the mass be released to stay on the track
Answer:
h = 2 R (1 +μ)
Explanation:
This exercise must be solved in parts, first let us know how fast you must reach the curl to stay in the
let's use the mechanical energy conservation agreement
starting point. Lower, just at the curl
Em₀ = K = ½ m v₁²
final point. Highest point of the curl
[tex]Em_{f}[/tex] = U = m g y
Find the height y = 2R
Em₀ = Em_{f}
½ m v₁² = m g 2R
v₁ = √ 4 gR
Any speed greater than this the body remains in the loop.
In the second part we look for the speed that must have when arriving at the part with friction, we use Newton's second law
X axis
-fr = m a (1)
Y Axis
N - W = 0
N = mg
the friction force has the formula
fr = μ N
fr = μ m g
we substitute 1
- μ mg = m a
a = - μ g
having the acceleration, we can use the kinematic relations
v² = v₀² - 2 a x
v₀² = v² + 2 a x
the length of this zone is x = 2R
let's calculate
v₀ = √ (4 gR + 2 μ g 2R)
v₀ = √4gR( 1 + μ)
this is the speed so you must reach the area with fricticon
finally have the third part we use energy conservation
starting point. Highest on the ramp without rubbing
Em₀ = U = m g h
final point. Just before reaching the area with rubbing
[tex]Em_{f}[/tex] = K = ½ m v₀²
Em₀ = Em_{f}
mgh = ½ m 4gR(1 + μ)
h = ½ 4R (1+ μ)
h = 2 R (1 +μ)
A small, rigid object carries positive and negative 3.00 nC charges. It is oriented so that the positive charge has coordinates (−1.20 mm, 1.20 mm) and the negative charge is at the point (1.70 mm, −1.30 mm).
Required:
a. Find the electric dipole moment of the object.
b. The object is placed in an electric field E = (7.80 103 î − 4.90 103 ĵ). Find the torque acting on the object.
c. Find the potential energy of the object–field system when the object is in this orientation.
d. Assuming the orientation of the object can change, find the difference between the maximum and the minimum potential energies of the system,
Answer:
Umax = 105.8nJ
Umin =-105.8nJ
Umax-Umin = 211.6nJ
Explanation:
Two conductors made of the same material are connected across the same potential difference. Conductor A has seven times the diameter and seven times the length of conductor B. What is the ratio of the power delivere
Complete question:
Two conductors made of the same material are connected across the same potential difference. Conductor A has seven times the diameter and seven times the length of conductor B. What is the ratio of the power delivered to A to power delivered to B.
Answer:
The ratio of the power delivered to A to power delivered to B is 7 : 1
Explanation:
Cross sectional area of a wire is calculated as;
[tex]A = \frac{\pi d^2}{4}[/tex]
Resistance of a wire is calculated as;
[tex]R = \frac{\rho L}{A} \\\\R = \frac{4\rho L}{\pi d^2} \\\\[/tex]
Resistance in wire A;
[tex]R = \frac{4\rho _AL_A}{\pi d_A^2}[/tex]
Resistance in wire B;
[tex]R = \frac{4\rho _BL_B}{\pi d_B^2}[/tex]
Power delivered in wire;
[tex]P = \frac{V^2}{R}[/tex]
Power delivered in wire A;
[tex]P = \frac{V^2_A}{R_A}[/tex]
Power delivered in wire B;
[tex]P = \frac{V^2_B}{R_B}[/tex]
Substitute in the value of R in Power delivered in wire A;
[tex]P_A = \frac{V^2_A}{R_A} = \frac{V^2_A \pi d^2_A}{4 \rho_A L_A}[/tex]
Substitute in the value of R in Power delivered in wire B;
[tex]P_B = \frac{V^2_B}{R_B} = \frac{V^2_B \pi d^2_B}{4 \rho_B L_B}[/tex]
Take the ratio of power delivered to A to power delivered to B;
[tex]\frac{P_A}{P_B} = (\frac{V^2_A \pi d^2_A}{4\rho_AL_A} ) *(\frac{4\rho_BL_B}{V^2_B \pi d^2_B})\\\\ \frac{P_A}{P_B} = (\frac{V^2_A d^2_A}{\rho_AL_A} )*(\frac{\rho_BL_B}{V^2_B d^2_B})\\\\[/tex]
The wires are made of the same material, [tex]\rho _A = \rho_B[/tex]
[tex]\frac{P_A}{P_B} = (\frac{V^2_A d^2_A}{L_A} )*(\frac{L_B}{V^2_B d^2_B})\\\\[/tex]
The wires are connected across the same potential; [tex]V_A = V_B[/tex]
[tex]\frac{P_A}{P_B} = (\frac{ d^2_A}{L_A} )* (\frac{L_B}{d^2_B} )[/tex]
wire A has seven times the diameter and seven times the length of wire B;
[tex]\frac{P_A}{P_B} = (\frac{ (7d_B)^2}{7L_B} )* (\frac{L_B}{d^2_B} )\\\\\frac{P_A}{P_B} = \frac{49d_B^2}{7L_B} *\frac{L_B}{d^2_B} \\\\\frac{P_A}{P_B} =\frac{49}{7} \\\\\frac{P_A}{P_B} = 7\\\\P_A : P_B = 7:1[/tex]
Therefore, the ratio of the power delivered to A to power delivered to B is
7 : 1
Immediately outside a conducting sphere(i.e. on the surface) of unknown charge Q and radius R the electric potential is 190 V, and 10.0 cm further from the sphere, the potential is 140 V. What is the magnitude of the charge Q on the sphere
Answer:
Q = 5.9 nC (Approx)
Explanation:
Given:
Further distance = 10 cm
Electric potential(V) = 190 v
Potential difference(V1) = 140 v
Find:
Magnitude of the charge Q
Computation:
V = KQ / r
190 = KQ / r.............Eq1
V1 = KQ / (r+10)
140 = KQ / (r+10) ............Eq2
From Eq2 and Eq1
r = 28 cm = 0.28 m
So,
190 = KQ / r
190 = (9×10⁹)(Q) / 0.28
53.2 = (9×10⁹)(Q)
5.9111 = (10⁹)(Q)
Q = 5.9 nC (Approx)
1. Suppose a teenager puts her bicycle on its back and starts the rear wheel spinning from rest to a final angular velocity of 250 rpm in 5.00 s. Radius of tire is 50 cm. What angle did the tire move through in those 5 secs
Answer:
[tex]\theta=65.18rad[/tex]
Explanation:
The angle in rotational motion is given by:
[tex]\theta=\frac{w_o+w_f}{2}t[/tex]
Recall that the angular speed is larger than regular frequency (in rpm) by a factor of [tex]2\pi[/tex], so:
[tex]\omega_f=2\pi f\\\omega_f=2\pi*250rpm\\\omega_f=1570.80 \frac{rad}{min}[/tex]
The wheel spins from rest, that means that its initial angular speed is zero([tex]\omega_o[/tex]). Finally, we have to convert the given time to minutes and replace in the first equation:
[tex]t=5s*\frac{1min}{60s}=0.083min\\\theta=\frac{\omega_f}{2}t\\\theta=\frac{1570.800\frac{rad}{min}}{2}(0.083min)\\\theta=65.18rad[/tex]
The water level in identical bowls, A and B, is exactly the same. A contains only water; B contains floating ice as well as water. When we weigh the bowls, we find that Group of answer choices
Answer:
We know that the density of the ice is smaller than the density of the water (and this is why the ice floats in water).
Dw > Di
Da is the density of the water and Di is the density of the ice
Since in Bowl A we have a volume V, only of water, then the mass of the bowl A is:
Dw*V.
Now, in the bowl B we have a combination of water and ice, suppose that Vw is the volume of water and Vi is the volume of ice, and we know that:
Vw + Vi = V.
Then the mass in this second bowl is:
Dw*Vw + Di*Vi = Dw*(V - Vi) + Di*Vi = Dw*V + (Di - Dw)*Vi
and we know that Dw > Di, then the left term is a negative term, then the mass of bowl B is smaller than the mass of bowl A.
A 25 kg box is 220 N pulled at constant speed up a frictionless inclined plane by a force that is parallel to the incline. If the plane is inclined at an angle of 25o above the horizontal, the magnitude of the applied force is
Answer:
F = 103.54N
Explanation:
In order to calculate the magnitude of the applied force, you take into account that the forces on the box are the applied force F and the weight of the box W.
The box moves with a constant velocity. By the Newton second law you have that the sum of forces must be equal to zero.
Furthermore, you have that the sum of forces are given by:
[tex]F-Wsin\theta=0[/tex] (1)
F: applied force = ?
W: weight of the box = Mg = (25kg)(9.8m/s^2) = 245N
θ: degree of the incline = 25°
You solve the equation (1) for F:
[tex]F=Wsin\theta=(245N)sin(25\°)=103.54N[/tex] (2)
The applied force on the box is 103.54N
what is the orbital speed for a satellite 3.5 x 10^8m from the center of mars? Mars mass is 6.4 x 10^23 kg
Answer:
v = 349.23 m/s
Explanation:
It is required to find the orbital speed for a satellite [tex]3.5\times 10^8\ m[/tex] from the center of mass.
Mass of Mars, [tex]M=6.4\times 10^{23}\ kg[/tex]
The orbital speed for a satellite is given by the formula as follows :
[tex]v=\sqrt{\dfrac{GM}{r}} \\\\v=\sqrt{\dfrac{6.67\times 10^{-11}\times 6.4\times 10^{23}}{3.5\times 10^8}} \\\\v=349.23\ m/s[/tex]
So, the orbital speed for a satellite is 349.23 m/s.
Four fixed point charges are at the corners of a square with sides of length L. Q1 is positive and at (OL) Q2 is positive and at (LL) Q3 is positive and at (4,0) Q4 is negative and at (0,0) A) Draw and label a diagram of the described arrangement described above (include a coordinate system). B) Determine the force that charge Q1 exerts on charge Qz. C) Determine the force that charge Q3 exerts on charge Q2. D) Determine the force that charge Q4 exerts on charge Q2. E) Now assume that all the charges have the same magnitude (Q) and determine the net force on charge Q2 due to the other three charges. Reduce this to the simplest form (but don't put in the numerical value for the force constant).
Answer:
A) See Annex
B) Fq₁₂ = K * Q₁*Q₂ /16 [N] (repulsion force)
C) Fq₃₂ = K * Q₃*Q₂ /16 [N] (repulsion force)
D) Fq₄₂ = K * Q₄*Q₂ /32 [N] (attraction force)
E) Net force (its components)
Fnx = (2,59/64 )* K*Q² [N] in direction of original Fq₃₂
Fny =(2,59/64 )* K*Q² [N] in direction of original Fq₁₂
Explanation:
For calculation of d (diagonal of the square, we apply Pythagoras Theorem)
d² = L² + L² ⇒ d² = 2*L² ⇒ d = √2*L² ⇒ d= (√2 )*L
d = 4√2 units of length (we will assume meters, to work with MKS system of units)
B) Force of Q₁ exerts on charge Q₂
Fq₁₂ = K * Q₁*Q₂ /(L)² Fq₁₂ = K * Q₁*Q₂ /16 (repulsion force in the direction indicated in annex)
C) Force of Q₃ exerts on charge Q₂
Fq₃₂ = K * Q₃*Q₂ /(L)² Fq₃₂ = K * Q₃*Q₂ /16 (repulsion force in the direction indicated in annex)
D) Force of -Q₄ exerts on charge Q₂
Fq₄₂ = K * Q₄*Q₂ / (d)² Fq₄₂ = K * Q₄*Q₂ /32 (Attraction force in the direction indicated in annex)
E) Net force in the case all charges have the same magnitude Q (keeping the negative sign in Q₄)
Let´s take the force that Q₄ exerts on Q₂ and Q₂ = Q ( magnitude) and
Q₄ = -Q
Then the force is:
F₄₂ = K * Q*Q / 32 F₄₂ = K* Q²/32 [N]
We should get its components
F₄₂(x) = [K*Q²/32 ]* √2/2 and so is F₄₂(y) = [K*Q²/32 ]* √2/2
Note that this components have opposite direction than forces Fq₁₂ and
Fq₃₂ respectively, and that Fq₁₂ and Fq₃₂ are bigger than F₄₂(x) and F₄₂(y) respectively
In new conditions
Fq₁₂ = K * Q₁*Q₂ /16 becomes Fq₁₂ = K * Q²/ 16 [N] and
Fq₃₂ = K* Q₃*Q₂ /16 becomes Fq₃₂ = K* Q² /16 [N]
Note that Fq₁₂ and Fq₃₂ are bigger than F₄₂(x) and F₄₂(y) respectively
Then over x-axis we subtract Fq₃₂ - F₄₂(x) = Fnx
and over y-axis, we subtract Fq₁₂ - F₄₂(y) = Fny
And we get:
Fnx = K* Q² /16 - [K*Q²/32 ]* √2/2 ⇒ Fnx = K*Q² [1/16 - √2/64]
Fnx = (2,59/64 )* K*Q²
Fny has the same magnitude then
Fny =(2,59/64 )* K*Q²
The fact that Fq₁₂ and Fq₃₂ are bigger than F₄₂(x) and F₄₂(y) respectively, means that Fnx and Fny remains as repulsion forces
how much weight can a man lift in the jupiter if he can lift 100kg on the earth.calculate
Answer:
2479 NewtonSolution,
Mass=100 kg
Acceleration due to gravity(g)=24.79 m/s^2
Now,.
[tex]weight = m \times g \\ \: \: \: \: \: \: \: \: \: \: = 100 \times 24.79 \\ \: \: \: \: \: \: = 2479 \: newton[/tex]
hope this helps ..
Good luck on your assignment..
Question 4
3 pts
I am approaching a traffic light at a speed of 135 km/h when I suddenly notice that
the light is red. I slam on my brakes and come to a stop in 4.29 seconds. What is the
acceleration of the car as I screech to a complete stop? (Note that an object that slows down
simply has a negative acceleration.)
& show work please I want to also understand
Answer:
The deceleration of the car is [tex]\approx -0.065m/s^{2}[/tex]
Explanation:
to solve this, we will have to apply the knowledge that will be got from the equations of motion.
There are several equations of motion, and depending on the parameters given in the problem, we can choose the perfect equation that can best be used to solve the problem.
In this case, since we are given the velocity and time, and we are solving for the acceleration, we will use this formula
[tex]v = u +at[/tex]
where v= final velocity = 0
u = initial velocity = 135Km/h [tex]\approx 0.278 m/s[/tex]
t= time = 4.29 seconds.
[tex]a = \frac{v - u}{t}[/tex]
[tex]a =\frac{0-0.278}{4.29} \approx 0.065m/s^{2}[/tex]
Hence, the deceleration of the car is [tex]\approx -0.065m/s^{2}[/tex]
The figure shows an arrangement of four charged particles, with θ = 20.0° and d1 = 3.00 cm, which is the distance from the origin to a charge q1. Charge q1 is unknown, but q2= +7.00×10‒19 C and q3 = q4 = ‒2.00×10‒19 C. If there is no nett electrostatic force on q1 due to the other charges (the nett electrostatic force on q1 is zero), calculate the distance from the origin to q2, given by d2, in cm. Assume that all forces apart from the electrostatic forces in the system are negligible
Answer:
[tex]d_2=3.16cm[/tex]
Explanation:
So, in order to solve this problem, we must start by building a diagram of the problem itself. (See attached picture) And together with the diagram, we must build a free body diagram, which will include the forces that are being applied on the given charged particle together with their directions.
In this case we only care about the x-direction of the force, since the y-forces cancel each other. So if we do a sum of forces on the x-direction, we get the following:
[tex]\sum{F_{x}}=0[/tex]
so:
[Tex]-F_{12}+F_{13x}+F_{14x}=0[/tex]
Since [tex]F_{13x}=F_{14x}[/tex] we can simplify the equation as:
[tex]-F_{12}+2F_{13x}=0[/tex]
we can now solve this for [tex]F_{12}[/tex] so we get:
[tex]F_{12}=2F_{13x}[/tex]
Now we can substitute with the electrostatic force formula, so we get:
[tex]k_{e}\frac{q_{1}q_{2}}{r_{12}^{2}}=2k_{e}\frac{q_{1}q_{3}}{r_{13}^{2}}cos \theta[/tex]
We can cancel [tex]k_{e}[/tex] and [tex]q_{1}[/tex]
so the simplified equation is:
[tex]\frac{q_{2}}{r_{12}^{2}}=2\frac{q_{3}}{r_{13}^{2}}cos \theta[/tex]
From the given diagram we know that:
[tex]cos \theta = \frac{d_{1}}{r_{13}}[/tex]
so when solving for [tex]r_{13}[/tex] we get:
[tex]r_{13}=\frac{d_{1}}{cos\theta}[/tex]
and if we square both sides of the equation, we get:
[tex]r_{13}^{2}=\frac{d_{1}^{2}}{cos^{2}\theta}[/tex]
and we can substitute this into our equation:
[tex]\frac{q_{2}}{r_{12}^{2}}=2\frac{q_{3}}{d_{1}^{2}}cos^{3} \theta[/tex]
so we can now solve this for [tex]r_{12}[/tex] so we get:
[tex]r_{12}=\sqrt{\frac{d_{1}^{2}q_{2}}{2q_{3}cos^{3}\theta}}[/tex]
which can be rewritten as:
[tex]r_{12}=d_{1}\sqrt{\frac{q_{2}}{2q_{3}cos^{3}\theta}}[/tex]
and now we can substitute values.
[tex]r_{12}=(3cm)\sqrt{\frac{7x10^{-19}C}{2(2x10^{-19}C)cos^{3}(20^{o})}}[/tex]
which solves to:
[tex]r_{12}=6.16cm[/tex]
now, we must find [tex]d_{2}[/tex] by using the following equation:
[tex]r_{12}=d_{1}+d_{2}[/tex]
when solving for [tex]d_{2}[/tex] we get:
[tex]d_{2}=r_{12}-d_{1}[/tex]
when substituting we get:
[tex]d_{2}=6.16cm-3cm[/tex]
so:
[tex]d_{2}=3.16cm[/tex]
A bowling ball traveling with constant speed hits pins at the end of a bowling lane 16.5m long. The bowler hears the sound of the ball hitting the pins 2.65s after the ball is release from her hand. What is the speed of the ball down the lane, assuming that the speed of sound is 340.0m/s
Answer: The speed of the ball is 7.64 m/s.
Explanation:
The distance between the player and the pins is 16.5m
if the velocity of the ball is V, then the time in which the ball reaches the pins is:
T = 16.5/V
Now, after this point, the sound needs
T' = 16,5/340 = 0.049 seconds to reach the player, this means that the time in that the ball needs to reach te pins is:
2.65 s - 0.49s = 2.16s
Then we have:
T = 2.16s = 16.5/V
V = 16.5/2.16 m/s = 7.64 m/s
The robot HooRU is lost in space, floating around aimlessly, and radiates heat into the depths of the cosmos at the rate of 13.1 W. HooRU's surface area is 1.55 m2 and the emissivity of its surface is 0.287. Ignoring the radiation that HooRU absorbs from the cold universe, what is HooRU's temperature T?
Answer:
The temperature is [tex]T = 168.44 \ K[/tex]
Explanation:
From the question ewe are told that
The rate of heat transferred is [tex]P = 13.1 \ W[/tex]
The surface area is [tex]A = 1.55 \ m^2[/tex]
The emissivity of its surface is [tex]e = 0.287[/tex]
Generally, the rate of heat transfer is mathematically represented as
[tex]H = A e \sigma T^{4}[/tex]
=> [tex]T = \sqrt[4]{\frac{P}{e* \sigma } }[/tex]
where [tex]\sigma[/tex] is the Boltzmann constant with value [tex]\sigma = 5.67*10^{-8} \ W\cdot m^{-2} \cdot K^{-4}.[/tex]
substituting value
[tex]T = \sqrt[4]{\frac{13.1}{ 0.287* 5.67 *10^{-8} } }[/tex]
[tex]T = 168.44 \ K[/tex]
An object of mass 3.07 kg, moving with an initial velocity of 5.07 m/s, collides with and sticks to an object of mass 2.52 kg with an initial velocity of -3.11 m/s. Find the final velocity of the composite objec
Answer:
This is an inelastic collision. This means, unfortunately, that KE cannot save you, at least in the problem's current form.
Let's see what conservation of momentum in both directions does ya:
Conservation in the x direction:
Only 1 object here has a momentum in the x direction initally.
m1v1i + 0 = (m1 + m2)(vx)
3.09(5.10) = (3.09 + 2.52)Vx
Vx = 2.81 m/s
Explanation:
Conservation in the y direction:
Again, only 1 object here has initial velocity in the y:
0 + m2v2i = (m1 +m2)Vy
(2.52)(-3.36) = (2.52 + 3.09)Vy
Vy = -1.51 m/s
++++++++++++++++++++
Now that you have Vx and Vy of the composite object, you can find the final velocity by doing Vf = √Vx^2 + Vy^2)
Vf = √(2.81)^2 + (-1.51)^2
Vf = 3.19 m/s
A block slides down a ramp with friction. The friction experienced by the block is 21 N. The mass of the block is 8 kg. The ramp is 6 meters long (meaning, the block slides across 6 meters of ramp with friction). The block is originally 2 meters vertically above the ground (the bottom of the ramp). What is the change in energy of the block due to friction, expressed in Joules
Complete Question
The complete question is shown on the first uploaded image
Answer:
the change in energy of the block due to friction is [tex]E = -126 \ J[/tex]
Explanation:
From the question we are told that
The frictional force is [tex]F_f = 21 \ N[/tex]
The mass of the block is [tex]m_b = 8 \ kg[/tex]
The length of the ramp is [tex]l = 6 \ m[/tex]
The height of the block is [tex]h = 2 \ m[/tex]
The change in energy of the block due to friction is mathematically represented as
[tex]\Delta E = - F_s * l[/tex]
The negative sign is to show that the frictional force is acting against the direction of the block movement
Now substituting values
[tex]\Delta E = -(21)* 6[/tex]
[tex]\Delta E = -126 \ J[/tex]
Two astronauts, of masses 60 kg and 80 kg, are initially right next to each other and at rest in outer space. They suddenly push each other apart. What is their separation after the heavier astronaut has moved 12m
Answer:
The astronauts are separated by 28 m.
Explanation:
The separation of the astronauts can be found by conservation of linear momentum:
[tex] p_{i} = p_{f} [/tex]
[tex] m_{1}v_{1i} + m_{2}v_{2i} = m_{1}v_{1f} + m_{2}v_{2f} [/tex]
[tex] m_{1}*0 + m_{2}*0 = m_{1}v_{1f} + m_{2}v_{2f} [/tex]
[tex] m_{1}v_{1f} = -m_{2}v_{2f} [/tex]
[tex] v_{1f} = -\frac{m_{2}v_{2f}}{m_{1}} = -\frac{80v_{2f}}{60} [/tex]
Now, the distance (x) is:
[tex] x = \frac{v}{t} [/tex]
The distance traveled by the astronaut 1 is:
[tex] x_{1} = v_{1f}*t = -\frac{80v_{2f}}{60}*t [/tex] (1)
And, the distance traveled by the astronaut 2 is:
[tex] x_{2} = v_{2f}*t [/tex] (2)
From the above equation we have:
[tex] t = \frac{x_{2}}{v_{2f}} [/tex] (3)
By entering equation (3) into (1) we have:
[tex] x_{1} = -\frac{80v_{2f}}{60}*(\frac{x_{2}}{v_{2f}}) [/tex]
[tex] x_{1} = -\frac{4*12}{3} = -16 m [/tex]
The minus sign is because astronaut 1 is moving in the opposite direction of the astronaut 2.
Finally, the separation of the astronauts is:
[tex] x_{T} = |x_{1}| + x_{2} = (16 + 12)m = 28 m [/tex]
Therefore, the astronauts are separated by 28 m.
I hope it helps you!
The total separation between the two astronauts is 28m.
The given parameters:
masses of the astronauts, = 60 kg and 80 kgApply the principle of conservation of momentum to determine the final velocity of each astronauts as follows;
[tex]m_1u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2\\\\60(0) + 80(0) = 60(v_1) + 80(v_2)\\\\0 = 60v_1 + 80v_2\\\\-60v_1 = 80v_2\\\\v_1 = \frac{-80v_2}{60} \\\\v_1 = -1.333v_2[/tex]
Let the time when astronaut 2 moved 12 m = t
The distance traveled by astronaut 1 is calculated as;
[tex]x_1 = v_1 t\\\\x_1 = -1.333v_2t[/tex]
The distance traveled by astronaut 2 is calculated as;
[tex]x_2 = v_2 t\\\\12 = v_2t\\\\t = \frac{12}{v_2}[/tex]
Now solve for the distance of astronaut 1
[tex]x_1 = - 1.333v_2 \times t\\\\x_1 = -1.333 v_2 \times \frac{12}{v_2} \\\\x_1 = -16 \ m[/tex]
The total separation between the two astronauts is calculated as follows;
[tex]d = |x_1| + x_2\\\\d = 16 + 12\\\\d = 28 \ m[/tex]
Learn more about conservation of linear momentum here: https://brainly.com/question/24424291
Two red blood cells each have a mass of 9.0 x 10-14 kg and carry a negative charge spread uniformly over their surfaces. The repulsion from the excess charge prevents the cells from clumping together. One cell carries -2.5pC and the other -3.30 pC, and each cell can be modeled as a sphere 3.75 × 10-6 m in radius. If the red blood cells start very far apart and move directly toward each other with the same speed.
1. What initial speed would each need so that they get close enough to just barely touch?
2. What is the maximum acceleration of the cells as they move toward each other and just barely touch?
Answer:
Explanation:
Given that:
The mass of the cell is 9.0 x 10^-14 kg
The charges of the cell is -2.5pC and the other -3.30 pC
[tex]q_1=-2.5\times10^{-12}C \ \ and \ \ q_2=-3.75\times10^{-12}C[/tex]
Radius is 3.75 × 10-6 m
The final distance is twice the radius
i.e [tex]2*(3.75 \times 10^{-6}) = 7.5*10^{-6}m[/tex]
The formula for the velocity of the cell is
[tex]mv^2=\frac{q_1q_2}{4\pi \epsilon 2 r} \\[/tex]
[tex]v=\sqrt{\frac{q_1q_2}{4\pi \epsilon 2 r} }[/tex]
[tex]=\sqrt{\frac{(-2.5\times10^{-12})(-3.3\times10^{-12}}{4(3.14)(8.85\times10^{-112}(2\times3.75\times10^{-6})(9\times10^{-14})} } \\\\=\sqrt{\frac{(-8.25\times10^{-24})}{(7503.03\times10^{-32})} } \\\\=\sqrt{109955.5779} \\\\=331.60m/s[/tex]
The maximum acceleration of the cells as they move toward each other and just barely touch is
[tex]ma= \frac{q_1q_2}{4\pi \epsilon (2r)^2} \\\\a= \frac{q_1q_2}{4\pi \epsilon (2r)^2(m)}[/tex]
[tex]=\frac{(-2.5\times10^{-12})(-3.3\times10^{-12})}{4(3.14)(8.85\times10^{-12})(2\times3.75\times10^{-6})^2(9\times10^{-14})}[/tex]
[tex]=\frac{(-8.25\times10^{-24})}{(56272.725\times10^{-38})} \\\\=1.47\times10^{10}m/s^2[/tex]
The answers obtained are;
1. The initial speed of each of the red blood cells is [tex]v= 331.66\,m/s[/tex].
2. The maximum acceleration of the cells is [tex]a=1.47\times 10^{10}\,m/s^2[/tex].
The answer is explained as shown below.
We have, the mass of the red blood cell;
[tex]m=9\times 10^{-14}\,kg[/tex]Also, the charges of the cells are;
[tex]q_1=-2.5\times 10^{-12}\,C[/tex] and[tex]q_2=-3.30\times 10^{-12}\,C[/tex]The distance between the charges when they barely touch will be two times the radius of each charge.
[tex]r=2\times r\,'=2\times3.75\times10^{-6}\,m=7.5\times10^{-6}\,m[/tex]Kinetic Energy of moving charges1. As both the cells are negatively charged they will repel each other.
So, for the cells to come nearly close, their kinetic energies must be equal to the electric potential between them.[tex]\frac{1}{2}mv^2+ \frac{1}{2}mv^2=k\frac{q_1 q_2}{r^2}[/tex]Where, [tex]k=9\times10^9\,Nm^2/C^2[/tex] is the Coulomb's constant.Now, substituting all the known values in the equation, we get;
[tex](9\times 10^{-14}\,kg)\times v^2=9\times 10^9Nm^2/C^2\times\frac{(-2.5\times 10^{-12}\,C)\times(-3.30\times 10^{-12}\,C)}{7.5\times10^{-6}\,m}[/tex][tex]v^2=9\times 10^9Nm^2/C^2\times\frac{(-2.5\times 10^{-12}\,C)\times(-3.30\times 10^{-12}\,C)}{7.5\times10^{-6}\,m\times(9\times 10^{-14}\,kg)} =110000\,m^2/s^2[/tex]
[tex]\implies v=\sqrt{110000\,m^2/s^2}=331.66\,m/s[/tex]Electrostatic force between two charges2. Also as the force between them is repulsive, there must be an acceleration to make them barely touch each other.
[tex]ma=k\frac{q_1 q_2}{r^2}[/tex]Substituting the known values, we get;
[tex](9\times 10^{-14}\,kg)\times a=9\times 10^9Nm^2/C^2\times\frac{(-2.5\times 10^{-12}\,C)\times(-3.30\times 10^{-12}\,C)}{(7.5\times10^{-6}\,m)^2}[/tex]
[tex]\implies a=9\times 10^9Nm^2/C^2\times\frac{(-2.5\times 10^{-12}\,C)\times(-3.30\times 10^{-12}\,C)}{(7.5\times10^{-6}\,m)^2\times(9\times 10^{-14}\,kg) }[/tex]
[tex]a=1.47\times 10^{10}\,m/s^2[/tex]Find out more information about moving charges here:
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The larger the push, the larger the change in velocity. This is an example of Newton's Second Law of Motion which states that the acceleration an object experiences is
Answer:
According to Newtons 2nd law of motion ;
The acceleration an object experiences is as a result of the net force which is directly proportional to the magnitude of the net force, in the same direction as the net force, and inversely proportional to the mass of the object.
Explanation:
This law is simply saying ;
Force = Mass ×Acceleration
I Hope It Helps :)
A helium nucleus (charge = 2e, mass = 6.63 10-27 kg) traveling at 6.20 105 m/s enters an electric field, traveling from point circled A, at a potential of 1.50 103 V, to point circled B, at 4.00 103 V. What is its speed at point circled B?
Answer:
[tex]v_B=3.78\times 10^5\ m/s[/tex]
Explanation:
It is given that,
Charge on helium nucleus is 2e and its mass is [tex]6.63\times 10^{-27}\ kg[/tex]
Speed of nucleus at A is [tex]v_A=6.2\times 10^5\ m/s[/tex]
Potential at point A, [tex]V_A=1.5\times 10^3\ V[/tex]
Potential at point B, [tex]V_B=4\times 10^3\ V[/tex]
We need to find the speed at point B on the circle. It is based on the concept of conservation of energy such that :
increase in kinetic energy = increase in potential×charge
[tex]\dfrac{1}{2}m(v_A^2-v_B^2)=(V_B-V_A)q\\\\\dfrac{1}{2}m(v_A^2-v_B^2)={(4\times 10^3-1.5\times 10^3)}\times 2\times 1.6\times 10^{-19}=8\times 10^{-16}\\\\v_A^2-v_B^2=\dfrac{2\times 8\times 10^{-16}}{6.63\times 10^{-27}}\\\\v_A^2-v_B^2=2.41\times 10^{11}\\\\v_B^2=(6.2\times 10^5)^2-2.41\times 10^{11}\\\\v_B=3.78\times 10^5\ m/s[/tex]
So, the speed at point B is [tex]3.78\times 10^5\ m/s[/tex].
A nonuniform electric field is given by the expression = ay î + bz ĵ + cx , where a, b, and c are constants. Determine the electric flux (in the +z direction) through a rectangular surface in the xy plane, extending from x = 0 to x = w and from y = 0 to y = h. (Use any variable or symbol stated above as necessary.)
A 0.40-kg particle moves under the influence of a single conservative force. At point A, where the particle has a speed of 10 m/s, the potential energy associated with the conservative force is 40 J. As the particle moves from A to B, the force does 25 J of work on the particle. What is the value of the potential energy at point B
Answer:
The value of the potential energy of the particle at point B is 85 joules.
Explanation:
According to the Principle of Energy Conservation, the energy cannot be created nor destroyed, only transformed. The particle at point A has kinetic and potential energy and receives a work due to an external conservative force (Work-Energy Theorem), whose sum is equal to potential energy at point B. Mathematically speaking, the expression that describes the phenomenon is:
[tex]K_{A} + U_{A} + W_{A \rightarrow B} = U_{B}[/tex]
Where:
[tex]K_{A}[/tex] - Kinetic energy at point A, measured in joules.
[tex]U_{A}[/tex] - Potential energy at point A, measured in joules.
[tex]W_{A \rightarrow B}[/tex] - Work due to conservative force from A to B, measured in joules.
[tex]U_{B}[/tex] - Potential energy at point B, measured in joules.
The initial kinetic energy of the particle is:
[tex]K_{A} = \frac{1}{2}\cdot m \cdot v^{2}[/tex]
Where:
[tex]m[/tex] - Mass, measured in kilograms.
[tex]v[/tex] - Velocity, measured in meters per second.
If [tex]m = 0.4\,kg[/tex] and [tex]v = 10\,\frac{m}{s}[/tex], then:
[tex]K_{A} = \frac{1}{2}\cdot (0.4\,kg)\cdot \left(10\,\frac{m}{s} \right)^{2}[/tex]
[tex]K_{A} = 20\,J[/tex]
Finally, the value of the potential energy at point B is:
[tex]U_{B} = 20\,J + 40\,J + 25\,J[/tex]
[tex]U_{B} = 85\,J[/tex]
The value of the potential energy of the particle at point B is 85 joules.
The potential energy of the particle at point B is 85 J.
Given to us:
Mass of the particle, [tex]m=0.40\ kg[/tex]
velocity of the particle, [tex]v= 10\ m/s[/tex]
potential energy of the particle, [tex]PE= 40\ J[/tex]
Workdone from pt. A to B, [tex]WD_{(A\rightarrow B)} = 25\ J[/tex]
Calculating the kinetic energy of the particle,
[tex]\begin{aligned}KE&= \frac{1}{2}mv^2 \\\\&=\frac{1}{2}\times0.40\times (10)^2\\\\&=20 J\\\end{aligned}[/tex]
According to the Principle of Energy Conservation,
The energy cannot be created nor be destroyed, it can only be transformed from one form to another.Therefore,
Also,
Total Energy at point A ,
[tex]\begin{aligned}(TE)_A &= Kinetic\ Energy+ Potential\ Energy+ Internal Energy+ Workdone\\ &=KE_A+ PE_A+UE_A+ WD_{(0\rightarrow A)}\\&=20+40+0+0\\&=60\ J\end{aligned}[/tex]
Total Energy at point B,
[tex]\begin{aligned}(TE)_B &= Kinetic\ Energy+ Potential\ Energy+ Internal Energy+ Workdone\\ &=KE_B+ PE_B+UE_B+ WD_{(A\rightarrow B)}\end{aligned}[/tex]
As the total energy is conserved from point A to B and also an external work is done on the particle. we can write the above equation as,
[tex]\begin{aligned} TE_B&=KE_B+ PE_B+UE_B+ WD_{(A\rightarrow B)}\\&=(KE_B+ PE_B+UE_B)+ WD_{(A\rightarrow B)}\\&= TE_A+ WD_{(A\rightarrow B)}\\&=60+25\\&=85\ J\end{aligned}[/tex]
Therefore, the total energy for the particle at point B is 85 J but as the particle is not moving neither work is done at point B, the total energy of the particle is the potential energy of the particle.
Hence, the potential energy of the particle at point B is 85 J.
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