Answer:
C
Explanation:
The value of the impulse of his skateboard will be 190 kg-m/s. Then the correct option is C.
What is an impulse?The change in momentum is equal to the product of impact force applied while colliding and the time for that impact.
The impulse is given as the product of the mass and the difference between the final velocity and initial velocity.
Impulse = Mass x (Final velocity - initial velocity)
John coasts on his skateboard at a speed of 1 m/s.
He uses his foot to increase his speed to 3 m/s. If he has a mass of 95 kg.
Then the value of the impulse of his skateboard will be given as,
Impulse = 95 x (3 - 1)
Impulse = 95 x 2
Impulse = 190 kg-m/s
Thus, the value of the impulse of his skateboard will be 190 kg-m/s.
Then the correct option is C.
Learn more about impulse.
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A train station bell gives off a fundamental tone of 500 Hz as the train approaches the station at a speed of 20 m/s. If the speed of sound in air is 334 m/s, what will be the apparent frequency of the bell to an observer riding the train
Answer: 529.9 Hz
Explanation:
Here we need to use the Doppler equation, so we have:
f' = f*(v + v0)/(v - vs)
Here, f is the frequency = 500Hz
v is the velocity of the wave, = 334m/s
v0 is the velocity of the observer = 20m/s
vs is the velocity of the source = 0m/s
Then we have:
f' = 500Hz*(334m/s + 20m/s)/(334m/s) = 529.9 Hz
Which of the following statements is accurate? A) Compressions and rarefactions occur throughout a transverse wave. B) The wavelength of both transverse and longitudinal waves is measured parallel to the direction of the travel of the wave. C) Sound waves passing through the air will do so as transverse waves, which vibrate vertically and still retain their horizontal positions. D) Amplitude of longitudinal waves is measured at right angles to the direction of the travel of the wave and represents the maximum distance the molecule has moved from its normal position.
Answer:
B) The wavelength of both transverse and longitudinal waves is measured parallel to the direction of the travel of the wave
Answer:
B) The wavelength of both transverse and longitudinal waves is measured parallel to the direction of the travel of the wave.
Explanation: hope this helps ;)
A speed skater moving across frictionless ice at 9.2 m/s hits a 5.0 m wide patch of rough ice. She slows steadily, then continues on at 5.8 m/s. What is her acceleration on the rough ice?
Answer:
a = -5.10 m/s^2
her acceleration on the rough ice is -5.10 m/s^2
Explanation:
The distance travelled on the rough ice is equal to the width of the rough ice.
distance d = 5.0 m
Initial speed u = 9.2 m/s
Final speed v = 5.8 m/s
The time taken to move through the rough ice can be calculated using the equation of motion;
d = 0.5(u+v)t
time t = 2d/(u+v)
Substituting the given values;
t = 2(5)/(9.2+5.8)
t = 2/3 = 0.66667 second
The acceleration is the change in velocity per unit time;
acceleration a = ∆v/t
a = (v-u)/t
Substituting the values;
a = (5.8-9.2)/0.66667
a = -5.099974500127
a = -5.10 m/s^2
her acceleration on the rough ice is -5.10 m/s^2
The potential difference between two parallel conducting plates in vacuum is 165 V. An alpha particle with mass of 6.50×10-27 kg and charge of 3.20×10-19 C is released from rest near the positive plate. What is the kinetic energy of the alpha particle when it reaches the other plate? The distance between the plates is 40.0 cm.
Answer:
kinetic energy (K.E) = 5.28 ×10⁻¹⁷
Explanation:
Given:
Mass of α particle (m) = 6.50 × 10⁻²⁷ kg
Charge of α particle (q) = 3.20 × 10⁻¹⁹ C
Potential difference ΔV = 165 V
Find:
kinetic energy (K.E)
Computation:
kinetic energy (K.E) = (ΔV)(q)
kinetic energy (K.E) = (165)(3.20×10⁻¹⁹)
kinetic energy (K.E) = 528 (10⁻¹⁹)
kinetic energy (K.E) = 5.28 ×10⁻¹⁷
A child has a toy car on a horizontal platform. The car starts from rest and reaches a maximum speed in 4 s. If the mass of the car is
0.1 kg and engine has an effective pull of 0.4 N Find the acceleration of the car.
Answer:
a=4m/s²
Explanation:
F=ma
0.4=0.1a
Answer:
a=4m/s
Explanation:
F=ma
0.4=0.1a
[tex] \frac{0.4}{0.1} = \frac{0.1}{0.1} [/tex]
a =4m/ s
If a sample of 346 swimmers is taken from a population of 460 swimmers,
the population mean, w, is the mean of how many swimmers' times?
Answer:
It is the mean of 460 swimmers
Explanation:
In this question, we are concerned with knowing the mean of the population w
Now, according to the question at hand, we have a total population of 460 swimmers and we have taken out 346 swimmers for a study
The population mean in this case is simply the mean of the swimming times of the 460 swimmers
There is another related thing here called the sample mean. For the sample mean, we only make a reference to the mean of the 346 swimmers who were taken out from the population to conduct a separate study
So conclusively, the population mean w is simply the mean of the total 460 swimmers
A particle with charge 7.76×10^(−8)C is moving in a region where there is a uniform 0.700 T magnetic field in the +x-direction. At a particular instant, the velocity of the particle has components Vx=−1.68×10^4m/s, Vy=−2.61×10^4m/s, and Vz=5.85×10^4m/s. What is the z-component of the force on the particle at this time?
Answer:
The z-component of the force is [tex]\= F_z = 0.00141 \ N[/tex]
Explanation:
From the question we are told that
The charge on the particle is [tex]q = 7.76 *0^{-8} \ C[/tex]
The magnitude of the magnetic field is [tex]B = 0.700\r i \ T[/tex]
The velocity of the particle toward the x-direction is [tex]v_x = -1.68*10^{4}\r i \ m/s[/tex]
The velocity of the particle toward the y-direction is
[tex]v_y = -2.61*10^{4}\ \r j \ m/s[/tex]
The velocity of the particle toward the z-direction is
[tex]v_y = -5.85*10^{4}\ \r k \ m/s[/tex]
Generally the force on this particle is mathematically represented as
[tex]\= F = q (\= v X \= B )[/tex]
So we have
[tex]\= F = q ( v_x \r i + v_y \r j + v_z \r k ) \ \ X \ ( \= B i)[/tex]
[tex]\= F = q (v_y B(-\r k) + v_z B\r j)[/tex]
substituting values
[tex]\= F = (7.7 *10^{-8})([ (-2.61*10^{4}) (0.700)](-\r z) + [(5.58*10^{4}) (0.700)]\r y)[/tex]
[tex]\= F= 0.00303\ \r j +0.00141\ \r k[/tex]
So the z-component of the force is [tex]\= F_z = 0.00141 \ N[/tex]
Note : The cross-multiplication template of unit vectors is shown on the first uploaded image ( From Wikibooks ).
If 62.9 cm of copper wire (diameter = 1.15 mm, resistivity = 1.69 × 10-8Ω·m) is formed into a circular loop and placed perpendicular to a uniform magnetic field that is increasing at the constant rate of 8.43 mT/s, at what rate is thermal energy generated in the loop?
Answer:
The answer is "[tex]\bold{7.30 \times 10^{-6}}[/tex]"
Explanation:
length of the copper wire:
L= 62.9 cm
r is the radius of the loop then:
[tex]r=\frac{L}{2 \pi}\\[/tex]
[tex]=\frac{62.9}{2\times 3.14}\\\\=\frac{62.9}{6.28}\\\\=10.01\\[/tex]
area of the loop Is:
[tex]A_L= \pi r^2[/tex]
[tex]=100.2001\times 3.14\\\\=314.628[/tex]
change in magnetic field is:
[tex]=\frac{dB}{dt} \\\\ = 0.01\ \frac{T}{s}[/tex]
then the induced emf is: [tex]e = A_L \times \frac{dB}{dt}[/tex]
[tex]=314.628 \times 0.01\\\\=3.14\times 10^{-5}V[/tex]
resistivity of the copper wire is: [tex]\rho =[/tex] 1.69 × 10-8Ω·m
diameter d = 1.15mm
radius (r) = 0.5mm
[tex]= 0.5 \times 10^{-3} \ m[/tex]
hence the resistance of the wire is:
[tex]R=\frac{\rho L}{\pi r^2}\\[/tex]
[tex]=\frac{1.69 \times 10^{-8}(62.9)}{3.14 \times (0.5 \times 10^{-3})^2}\\\\=\frac{1.69 \times 10^{-8}(62.9)}{3.14 \times 0.5 \times 0.5 \times 10^{-6}}\\\\=\frac{1.69 \times 10^{-8}(62.9)}{3.14 \times 0.25 \times 10^{-6}}\\\\=135.41 \times 10^{-2}\\=1.35\times 10^{-4}\\[/tex]
Power:
[tex]P=\frac{e^2}{R}[/tex]
[tex]=\frac{3.14\times 10^{-5}\times 3.14\times 10^{-5}}{1.35 \times 10^{-4}}\\\\=7.30 \times 10^{-6}[/tex]
The final answer is: [tex]\boxed{7.30 \times 10^{-6} \ W}[/tex]
What is the approximate value of the gravitational force between a 73 kg astronaut and a 7.1×104 kg spacecraft when they're 89 m apart?
Answer:
[tex]F = 4.3671 * 10^{-8}\ Newtons[/tex]
Explanation:
The gravitational force between two corpses is given by the following equation:
[tex]F = GMm/d^2[/tex]
Where F is the force, G is the gravitational constant
([tex]G = 6.67408*10^{-11}\ m^3kg^{-1}s^{-2}[/tex]), M and m are the masses of the corpses and d is the distance between them.
So we have that:
[tex]F = 6.67408*10^{-11} * 7.1*10^4 * 73/89^2[/tex]
[tex]F = 4.3671 * 10^{-8}\ Newtons[/tex]
The aorta pumps blood away from the heart at about 40 cm/s and has a radius of about 1.0 cm. It then branches into many capillaries, each with a radius of about 5 x 10−4 cm carrying blood at a speed of 0.10 cm/s.
How many capillaries are there?
Answer:
n = 1.6*10^9 capillaries
Explanation:
In order to calculate the number of capillaries, you take into account that the following relation must be accomplished:
[tex]A_1v_1=nA_2v_2[/tex] (1)
A1: area of the aorta
v1: speed of the blood in the aorta = 40cm/s
n: number of capillaries = ?
A2: area of each capillary
v2: speed of the blood in each capillary
For the calculation of A1 and A2 you use the formula for the cross sectional area of a cylinder, that is, the area of a circle:
[tex]A=\pi r^2\\\\A_1=\pi r_1^2=\pi(1.0cm)^2=3.1415 cm^2\\\\A_2=\pi r_2^2=\pi (5*10^{-4}cm)^2=7.85*10^{-7}cm^2[/tex]
Where you have used the values of the radius for the aorta and the capillaries.
Next, you solve the equation (1) for n, and replace the values of all parameters:
[tex]n=\frac{A_1v_1}{A_2v_2}=\frac{(3.1415cm^2)(40cm/s)}{(7.85*10^{-7}cm^2)(0.10cm/s)}=1.6*10^9[/tex]
Then, the number of capillaries is 1.6*10^9
Julie is playing with a toy car and is pushing it around on the floor. The little car has a mass of 6.3 g. The car has a velocity of 2.5 m/s. What is the car's momentum?
Answer:
Momentum of the car = [tex]1.575\times 10^{-2}[/tex] kg meter per second
Explanation:
Julie is playing with a car which has mass = 6.3 g = [tex]6.3\times 10^{-3}[/tex] kg
Velocity of the car is 2.5 meter per second
Since formula to calculate the momentum of an object is,
p = mv
Where, p = momentum of the object
m = mass of the object
v = velocity of the object
By substituting these values in the formula,
p = [tex](6.3\times 10^{-3})\times 2.5[/tex]
= [tex]1.575\times 10^{-2}[/tex] Kg meter per second
Therefore, momentum of the car will be [tex]1.575\times 10^{-2}[/tex] Kg meter per second.
If you could see stars during the day, the drawing above shows what the sky would look like at noon on a given day. The Sun is near the stars of the constellation Taurus. Which constellation will be highest in the sky at sunset?
A. Pisces
B. Cancer
C. Gemini
D. Taurus
E. Aries
Answer:
Taurus constellation
Explanation:
Because Taurus constellation is mostly seen In the Northern Hemisphere, the bull passes through the sky from November to March, but the constellation's at its most visible in January. Taurus covers 797 square degrees.
Right ascension: 4 hours
Declination: 15 degrees
Best visible between latitudes 90 degrees and minus 65 degrees
A uniform thin rod of mass ????=3.41 kg pivots about an axis through its center and perpendicular to its length. Two small bodies, each of mass m=0.249 kg , are attached to the ends of the rod. What must the length L of the rod be so that the moment of inertia of the three-body system with respect to the described axis is ????=0.929 kg·m2 ?
Answer:
The length of the rod for the condition on the question to be met is [tex]L = 1.5077 \ m[/tex]
Explanation:
The Diagram for this question is gotten from the first uploaded image
From the question we are told that
The mass of the rod is [tex]M = 3.41 \ kg[/tex]
The mass of each small bodies is [tex]m = 0.249 \ kg[/tex]
The moment of inertia of the three-body system with respect to the described axis is [tex]I = 0.929 \ kg \cdot m^2[/tex]
The length of the rod is L
Generally the moment of inertia of this three-body system with respect to the described axis can be mathematically represented as
[tex]I = I_r + 2 I_m[/tex]
Where [tex]I_r[/tex] is the moment of inertia of the rod about the describe axis which is mathematically represented as
[tex]I_r = \frac{ML^2 }{12}[/tex]
And [tex]I_m[/tex] the moment of inertia of the two small bodies which (from the diagram can be assumed as two small spheres) can be mathematically represented as
[tex]I_m = m * [\frac{L} {2} ]^2 = m* \frac{L^2}{4}[/tex]
Thus [tex]2 * I_m = 2 * m \frac{L^2}{4} = m * \frac{L^2}{2}[/tex]
Hence
[tex]I = M * \frac{L^2}{12} + m * \frac{L^2}{2}[/tex]
=> [tex]I = [\frac{M}{12} + \frac{m}{2}] L^2[/tex]
substituting vales we have
[tex]0.929 = [\frac{3.41}{12} + \frac{0.249}{2}] L^2[/tex]
[tex]L = \sqrt{\frac{0.929}{0.40867} }[/tex]
[tex]L = 1.5077 \ m[/tex]
A metal such as copper is a(n) _______________ because it provides a pathway for electric charges to move easily. A material such as rubber is a(n) _______________ because it _______________ the flow of electric charges. A material that partially conducts electric current is a(n) _______________. These materials include _______________ elements.
Explanation:
A metal such as copper is a conductor because it provides a pathway for electric charges to move easily. A material such as rubber is an insulator because it resists the flow of electric charges. A material that partially conducts electric current is a semiconductor. These materials include group 3 and group 5 elements.
Answer:
conductor
insulator
resists
semiconductor
group 3 and group 5
Explanation:
A WOMAN HAS A MASS OF 75.0 kg What is her weight on earth?
Answer:
735 N
Explanation:
If a woman has mass 75kg and u know that mass is constant everywhere then just apply the formula W=mg...as gravity on earth is 9.8 m/s2 , so her weight will be 735 N...hope it helps...
If a woman has a mass of 75 kilograms then her weight on the earth would be 735.75 Newtons, because the weight of the woman is the multiplication of the mass of the woman and the acceleration due to the gravity of the earth.
What is gravity?It can be defined as the force by which a body attracts another body toward its center as the result of the gravitational pull of one body and another,
As given in the problem if a woman has a mass of 75 kilograms we have to find the weight of the woman,
The weight of the woman = mass of the woman × acceleration due to the gravity of the earth
= 75 kilograms × 9.81
=735.75 Newtons
Thus, the weight of a woman who has a mass of 75 kilograms would be 735.75 Newtons
Learn more about gravity here, refer to the link given below;
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A rigid tank A of volume 0.6 m3 contains 5 kg air at 320K and the rigid tank B is 0.4 m3 with air at 600 kPa, 360 K. They are connected to a piston cylinder initially empty with closed valves. The pressure in the cylinder should be 800 kPa to float the piston. Now the valves are slowly opened and the entire process is adiabatic. The internal energy of the mixture at final state is:_____.
a. 229 k/kg.
b. 238 kJ/kg
c. 257 kg
d. cannot be determined.
Answer:
the internal energy of the mixture at final state = 238kJ/kg
Explanation:
Given
V= 0.6m³
m=5kg
R=0.287kJ/kg.K
T=320 K
from ideal gas equation
PV = nRT
where P is pressure, V is volume, n is number of mole, R is ideal gas constant , T is the temperature.
Recall, mole = mass/molar mass
attached is calculation of the question.
Two Earth satellites, A and B, each of mass m = 980 kg , are launched into circular orbits around the Earth's center. Satellite A orbits at an altitude of 4100 km , and satellite B orbits at an altitude of 12100 km The radius of Earth RE is 6370 km.
(a) What is the ratio of the potential energy of satellite B to that of satellite A, in orbit?
(b) What is the ratio of the kinetic energy of satellite B to that of satellite A, in orbit?
(c) Which satellite has the greater total energy if each has a mass of 14.6 kg?
(d) By how much?
Answer:
Do u have a picture of the graph?
Explanation:
I can solve it with refraction
Austin is interested in working for an intelligence branch of the government. A polygraph test is part of the interview
process. Though Austin intends to be honest, he is worried that the polygraph will say he is lying when he is not.
Austin's friend, Gabe, assures Austin that polygraph tests are infallible. Is Gabe right?
Answer:
Gabe is wrong in concluding that the polygraph tests are infallible. Even though the polygraph machine is a man-made machine, there is a small margin of error factored into it.
This error could be caused as a result of the anxiety of the test taker, the heart beat rate, the blood pressure etc which would lead to the answers provided by the person to be flagged as false despite being true.
In most cases, the test result will be declared inconclusive if there was an error in the readings.
Explanation:
Please help! Which statements correctly describe the effect of distance in determining the gravitational force and the electrical force? Check all that apply.
There are six statements on the list.
The first 2 are true, and the last 2 are true.
The 2 in the middle aren't true. They are false.
Calculate the total current in parallel circuit using Kirchhoff’s Current Law? anyone give example and solution...
Answer:
please check attachment and follow the steps.
A negative charge of -0.550 μC exerts an upward 0.900-N force on an unknown charge that is located 0.300 m directly below the first charge.
Required:
a. What is the value of the unknown charge (magnitude and sign)?
b. What is the magnitude of the force that the unknown charge exerts on the -0.590 μC charge?
c. What is the direction of this force?
Answer:
a. q2 = 16.4μC, positive charge
b. F = 0.900N
c. downward
Explanation:
a. In order to calculate the charge of the unknown charge you use the following formula, for the electric force between two charges:
[tex]F_e=k\frac{q_1q_2}{r^2}[/tex] (1)
k: Coulomb's constant = 8.98*10^9Nm^2/C^2
r: distance between the charges = 0.300m
q1: charge 1 = -0.550 μC = 0.550*10^-6C
q2: charge 2 = ?
Fe: electric force = 0.900N
The force exerted in the second charge points upward, then, the sign of the second charge is positive because this charge is getting closer to the first one.
You solve the equation (1) for the second charge ans replace the values of the other parameters:
[tex]q_2=\frac{r^2F_e}{kq_1}=\frac{(0.300m)^2(0.900N)}{(8.98*10^9Nm^2/C^2)(0.550*10^{-6}C)}\\\\q_2=1.64*10^{-5}C\\\\q_2=16.4*10^{-6}C=16.4*10\mu C[/tex]
The values of the second charge is 1.64 μC
b. By the third Newton Law, you have that the force exerted in the second charge is equal to the force exerted by the first charge on the second one.
The force exerted on the first charge is 0.900N
c. The charges are attracting between them, then, the force exerted on the first charge is pointing downward.
toy rocket engine is securely fastened to a large puck that can glide with negligible friction over a horizontal surface, taken as the x-y plane. The 4.00 kg puck has a velocity of 3.00 i m/s at one instant. Eight seconds later, its velocity is (8.00 i 10.00 j) m/s. Assuming the rocket engine exerts a constant force, find (a) the components of the force and (b) its magnitude.
Answer:
Fx = 2.5 N
Fy = 5 N
|F| = 5.59 N
Explanation:
Given:-
- The mass of puck, m = 4.0 kg
- The initial velocity of puck, u = 3.00 i m/s
- The final velocity of puck, v = ( 8.00 i + 10.00 j ) m/s
- The time interval for the duration of force, Δt = 8 seconds
Find:-
the components of the force and (b) its magnitude.
Solution:-
- We will set up a coordinate system ( x - y ) plane. With unit vectors i and j along x and y axes respectively.
- To model the situation we will seek help from Newton's second law of motion. Defined by the rate of change of linear momentum of the system.
[tex]F_net = \frac{m*( v - u ) }{dt}[/tex]
Where,
Fnet: The net force that acts on the puck-rocket system
- Here we will assume that the mass of rocket is negligible compared to the mass of the puck. The only force ( F ) acting on the puck is due to the thrust produced of the rocket. The dry and air frictions are both neglected for the analysis.
- We will apply the newton's second law of motion in component forms. And determine the components of force F, as ( Fx ) and ( Fy ) as follows:
[tex]F_x = \frac{m* ( v_x - u_x)}{dt} \\\\F_x = \frac{4* ( 8 - 3)}{8} \\\\F_x = 2.5 N\\\\F_y = \frac{m* ( v_y - u_y)}{dt} \\\\F_y = \frac{4* ( 10 - 0)}{8} \\\\F_y = 5 N\\\\[/tex]
- We will apply the Pythagorean theorem and determine the magnitude of the thrust force produced by the rocket with which the puck accelerated:
[tex]| F | = \sqrt{( F_x)^2 + ( F_y)^2} \\\\| F | = \sqrt{( 2.5)^2 + ( 5)^2} \\\\| F | = \sqrt{31.25} \\\\| F | = 5.590[/tex]
Answer: the magnitude of the thrust force is F = 5.59 N
3. Two spherical objects at the same altitude move with identical velocities and experience the same drag force at a time t. If Object 1 has twice (2x) the diameter of Object 2, which object has the larger drag coefficient? Explain your answer using the drag equation.
Answer:
Object 2 has the larger drag coefficient
Explanation:
The drag force, D, is given by the equation:
[tex]D = 0.5 c \rho A v^2[/tex]
Object 1 has twice the diameter of object 2.
If [tex]d_2 = d[/tex]
[tex]d_1 = 2d[/tex]
Area of object 2, [tex]A_2 = \frac{\pi d^2 }{4}[/tex]
Area of object 1:
[tex]A_1 = \frac{\pi (2d)^2 }{4}\\A_1 = \pi d^2[/tex]
Since all other parameters are still the same except the drag coefficient:
For object 1:
[tex]D = 0.5 c_1 \rho A_1 v^2\\D = 0.5 c_1 \rho (\pi d^2) v^2[/tex]
For object 2:
[tex]D = 0.5 c_2 \rho A_2 v^2\\D = 0.5 c_2 \rho (\pi d^2/4) v^2[/tex]
Since the drag force for the two objects are the same:
[tex]0.5 c_1 \rho (\pi d^2) v^2 = 0.5 c_2 \rho (\pi d^2/4) v^2\\4c_1 = c_2[/tex]
Obviously from the equation above, c₂ is larger than c₁, this means that object 2 has the larger drag coefficient
Find acceleration. Will give brainliest!
Answer:
16200 km/s
270 km/min
4.5 km/h
Explanation:
Acceleration Formula: Average Acceleration = Δv/Δt (change in velocity over change in time)
Simply plug in our known variables and solve:
a = (45.0 - 0)/10
a = 45.0/10
a = 4.5 km/h
Answer:
[tex]\boxed{\mathrm{4.5 \: kmph/s \: or \: 1.25 \: m/s^2 }}[/tex]
Explanation:
[tex]\displaystyle \mathrm{acceleration = \frac{change \: in \: velocity}{time \: taken}}[/tex]
[tex]\displaystyle \mathrm{a = \frac{v - u}{t}}[/tex]
[tex]\displaystyle \mathrm{v=final \: velocity}\\\displaystyle \mathrm{u=initial \: velocity}[/tex]
[tex]\displaystyle \mathrm{a = \frac{45- 0}{10}}[/tex]
[tex]\displaystyle \mathrm{a = \frac{45}{10}}[/tex]
[tex]\displaystyle \mathrm{a = 4.5}[/tex]
[tex]\mathrm{4.5 \: kmph/s = 1.25 \: m/s^2 }[/tex]
A 100 kg lead block is submerged in 2 meters of salt water, the density of which is 1096 kg / m3. Estimate the value of the hydrostatic pressure.
Answer:
21,920 Pascals
Explanation:
P = ρgh
P = (1096 kg/m³) (10 m/s²) (2 m)
P = 21,920 Pa
a crate b of mass 40kg is raised by the rope of crane from the hold of a ship. mark and name forces on the crate . find acceleration if tension is 480N
Find the acceleration, a .
Formula used:-Force = Mass × Acceleration
Solution:-We know that ,
Force = Mass × Acceleration
★ Substituting the values in the above formula,we get:
⇒ 480 = 40 × Acceleration
⇒ Acceleration, a = 480/40
⇒ Acceleration,a = 12 m/s
Thus,the acceleration of a body is 12 metres per seconds.
What does the vertical polarization axis of polarized sunglasses indicate about the direction of polarization of light bouncing off a horizontal surface, such as a wet road or lake surface
Answer:
it is desired that the lenses stop this ray, its polarization must be vertical
Explanation:
To answer this exercise, let's analyze the rays of light reflected on a horizontal surface, when the incident light that we consider non-polarized is reflected on a surface, the electric field of light moves the electrons on the surface horizontally and this re-emits the radiation same shape, that is horizontal.
The other vertical direction the atoms have a lot of movement restricted by the attraction on the surface, so for the reflected ray this polarization is attenuated, this does not stop the transmitted ray where the two polarizations are transmitted.
Total polarizations only for one angle, but in general as we approach dominant polarization it horizontal. Specifically the angle for full polarization is
n = tan teaP
Now we can analyze what polarization the lenses have, if the ray that comes is polarized horizontally and it is desired that the lenses stop this ray, its polarization must be vertical
The lower the value of the coefficient of friction, the____the resistance to sliding
Answer: lower
There are a number of factors that can affect the coefficient of friction, including surface conditions.
Values of the coefficient of sliding friction can be a good reference for specific combinations of materials. The frictional force and normal reaction are directly proportion but an increase or decrease in coefficient of friction will cause an increase or decrease in the resistance of sliding respectively
If an ant is starting from 0 radians, and travels all the way to 4.5 radians in 18 seconds, what is angular velocity? If the same ant then slows down back to 0 rad/s in 3 seconds what is its angular acceleration?
Answer:
Angular velocity is 0.25 rad/sAngular acceleration is 0.017 rad/s²Explanation:
Given;
initial angular displacement, θ₁ = 0 radians
final angular displacement, θ₂ = 4.5 radians
Angular velocity is calculated as;
[tex]\omega = \frac{\delta \theta}{\delta t}= \frac{4.5 -0}{18} \\\\\omega = 0.25 \ rad/s[/tex]
Angular acceleration is calculated as;
[tex]\alpha = \frac{\delta \omega}{\delta t} = \frac{\omega _f - \omega_i}{t_2 -t_1}[/tex]
where;
[tex]\omega_f[/tex] is the final angular velocity = 0 rad/s
[tex]\omega _i[/tex] is the initial angular velocity = 0.25 rad/s
t₂ is the final time of the motion = 3 seconds
t₁ is the initial time of the motion = 18 seconds
[tex]\alpha =\frac{\omega _f - \omega_i}{t_2 -t_1} \\\\\alpha = \frac{0 - 0.25}{3-18} \\\\\alpha = 0.017 \ rad/s^2[/tex]
An RLC circuit has a resistance of 200 Ω and an inductance of 15 mH. Its oscillation frequency is 7000 Hz. At time t = 0, the current is 25 mA, and there is no charge on the capacitor. After five complete cycles, the current is
Answer: 2.13 × 10^-4 A
Explanation:
Given that the RLC circuit has a resistance R = 200 Ω and an inductance L = 15 mH.
Its oscillation frequency F = 7000 Hz
The initial current I = 25 mA = 25/1000 or 25 × 10^-3 A
Since there is no charge on the capacitor, the current after complete 5 circle will be achieved by using the formula in the attached file.
Please find the attached file for the remaining explanation for the solution.
The current in the RLC circuit after five (5) complete cycles is equal to [tex]2.15 \times 10^{-4} \; Amperes[/tex]
Given the following data:
Resistance = 200 ΩInductance = 15 mH = 0.015 HOscillation frequency = 7000 HzCurrent = 25 mA = 0.025 ATo determine the current in the RLC circuit after five (5) complete cycles, we would use the following formula:
[tex]I_t = I_o e^{\frac{Rt}{2L} coswt}[/tex]Note: There is no electrical charge on the capacitor.
After five (5) complete cycles, the formula becomes:
[tex]I_5 = I_o e^{\frac{-R}{2L} \frac{5}{F} }[/tex]Substituting the given parameters into the formula, we have;
[tex]I_5 = 0.025 e^{\frac{-200}{2\; \times \;0.015} \times \frac{5}{7000} }\\\\I_5 = 0.025 e^{\frac{-200}{0.03} \times \frac{5}{7000} }\\\\I_5 = 0.025 e^{\frac{-1000}{210} }\\\\I_5 = 0.025 e^{-4.7619}\\\\I_5 = 0.025 \times 0.0086\\\\I_5 = 0.00215 \\\\I_5 = 2.15 \times 10^{-3} \; Amps[/tex]
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