Answer:
Evaluate the following numerical expressions.
6 + 3 • 4 =
18
(6 + 3) ÷ (4 – 5) =
Explanation:
Evaluate the following numerical expressions.
6 + 3 • 4 =
18
(6 + 3) ÷ (4 – 5) =
A spring is attached to an inclined plane as shown in the figure. A block of mass m = 2.27 kg is placed on the incline at a distance d = 0.327 m along the incline from the end of the spring. The block is given a quick shove and moves down the incline with an initial speed v = 0.750 m/s. The incline angle is = 20.0°, the spring constant is k = 455 N/m, and we can assume the surface is frictionless. By what distance (in m) is the spring compressed when the block momentarily comes to rest?
Answer:
0.1351 m
Explanation:
*my number calculations may be wrong, but the process is correct*
W = Wg - Ws = mgh - [tex]\frac{1}{2}[/tex]kx^2 = mg(d + x)sinθ - [tex]\frac{1}{2}[/tex]kx^2
Δk = 0 - [tex]\frac{1}{2\\}[/tex]mv[tex]_{i}[/tex]^2
W = Δk
-[tex]\frac{1}{2}[/tex]mv[tex]_{i}[/tex]^2 = mg(d + x)sinθ - [tex]\frac{1}{2}[/tex]kx^2
Now plug in your given values:
m=2.27, g=9.8, d=0.327, θ=20, k=455, v=0.750
Now rearrange the equation so it is in the form ax^2 + bx + c = 0 where x is the unknown distance you are looking for.
~if my math is correct it should be:~
0 = -227.5x^2 + 7.608580108x + 3.126443195
Now plug your numbers into the quadratic formula and the positive x value you get is the answer.
~if my math is correct it should be:~
x = 0.1351376847 m = 0.1351 m
A hockey puck is given an initial speed of 5.30 m/s. If the coefficient of friction between the puck and ice is 0.052, how far does the puck slide before coming to rest
Let m be the mass of the puck. Then the upward normal force exerted by the ice has magnitude n = mg (since Newton's law says the net force on the puck in the vertical direction is ∑ F = n - mg = 0).
Then the frictional force applies a force of magnitude f = 0.052mg, and Newton's second law says the net horizontal force acting on the puck is
∑ F = -f = ma
Solve for the acceleration a :
-0.052mg = ma
a = -0.052 (9.8 m/s²)
a ≈ -0.510 m/s²
Assuming constant friction, the puck slides to a rest over a distance x such that
0² - (5.30 m/s)² = 2ax
Solve for x :
x = -(5.30 m/s)²/(2a)
x = -(5.30 m/s)² / (2 (-0.510 m/s²))
x ≈ 27.6 m
Answer:
27.47 meters
Explanation:
The change in kinetic energy of the puck is ΔK = [tex]K_{f} -K_{i} = -mv{_1}^{2} /2[/tex]
The work done on the puck by the force of friction is
[tex]W_{f} = F_{f}dcos(180) = (Mu)Nd = - (Mu)mgd[/tex]
Knowing that
[tex]W_{f}[/tex] = ΔK
obtain
[tex]-(Mu)mgd = -mv_{i}^{2} /2[/tex]
which is:
[tex]d = v_{i}^{2} /2(Mu)g = (5.30)^{2} / 2(0.052)(9.8) =[/tex] 27.47 meters
In the circuit, the capacitor is fully charged when switch is closed. Calculate the time needed for the potential energy stored by the circuit to be equally distributed between the capacitor and inductor. The capacitance is =20.0 mF and inductance is =45.0 H .
We have that for the Question "In the circuit, the capacitor is fully charged when switch is closed. Calculate the time needed for the potential energy stored by the circuit to be equally distributed between the capacitor and inductor. The capacitance is =20.0 mF and inductance is =45.0 H ." it can be said that the time required is
[tex]T=0.745s[/tex]From the question we are told
In the circuit, the capacitor is fully charged when switch is closed. Calculate the time needed for the potential energy stored by the circuit to be equally distributed between the capacitor and inductor. The capacitance is =20.0 mF and inductance is =45.0 H .
Generally the equation for the Time is mathematically given as
[tex]T=\sqrt{LC}*\frac{\pi}{4}[/tex]
[tex]T=\sqrt{20*10^{-3}*45}*\frac{\pi}{4}[/tex]
[tex]T=0.745s[/tex]
For more information on this visit
https://brainly.com/question/23379286
Why do heavier objects roll down slopes faster than lighter objects? (Assuming they are the same shape)
Answer:
because of the friction against the object
Explanation:
A person walks 32∘ north of west for 94m and 35∘ east of south for 122m. What is the magnitude of his displacement?
Answer:
n
Explanation:
What are last two commandments? Please write them down.
The Second Commandment refers to: The second of the Ten Commandments, which is: "Thou shalt have no other gods before me. Thou shalt not make unto thee any graven image" under the Talmudic division of the third-century Jewish Talmud.
Answer:
Thou shalt love the Lord thy God with all thy heart, and with all thy soul, and with all thy mind. This is the first and great commandment. And the second is like unto it, Thou shalt love thy neighbor as myself.
Explanation:
Two forces of 68 pounds and 66 pounds act simultaneously on an object. The angle
between the two forces is 77º. Find the magnitude of the resultant, to the nearest
pound. Find the measure of the angle between the resultant and the smaller force, to
the nearest degree.
Resultant:
Angle:
Submit Answer
Answer:
Explanation:
Let the x axis lie along the 66 lb force with the force vector tail at the origin.
The 68 lb force would lie along a radial 77° CCW from the x axis
The resultant would have an x and y values of
Fx = 66cos0 + 68cos77 = 81.3 lb
Fy = 66sin0 + 68sin77 = 66.3 lb
F = √(81.3² + 66.3²) = 104.87 = 105 lb
θ = arctan (66.3/81.3) = 39°
Explain in your own words what is the fourth commandment.
the fourth commandment is to remember my Sabbath day which is to not do any labor on the 7th day which is Sunday.
Explanation:
hope this helps I don't know if you meant from the Bible or not.
Answer:
From Wikipedia, the free encyclopedia. The Fourth Commandment of the Ten Commandments may refer to: "Remember the sabbath day, to keep it holy", under the Philonic division used by Hellenistic Jews, Greek Orthodox and Protestants except Lutherans, or the Talmudic division of the third-century Jewish Talmud
Explanation:
I hope this helps
PLEASE GIVE ME BRAINLEST i really need this
Một xe có khối lượng 20000kg chuyển động dần đều dưới tác dụng của lực 600N,vận tốc ban đầu của xe bằng 15m/s.hỏi :
a,gia tốc của xe
b,sau bao lâu xe dừng lại
c,đoạn đường xe đã chạy được kể từ lúc hàm cho đến khi dừng hẳn.
Answer:
please vote and heart please and follow me because I have no followers yet
If you're under stress what is the best way to recharge and calm the brain
Answer:
Take slow, deep breaths. Or try other breathing exercises for relaxation. ...
Soak in a warm bath.
Listen to soothing music.
Practice mindful meditation. The goal of mindful meditation is to focus your attention on things that are happening right now in the present moment. ...
Write. ...
Use guided imagery.
A 0.25 kg ball is suspended from a light 0.65 m string as shown. The string makes an angle of 31° with the vertical. Let U = 0 when the ball is at its lowest point (θ = 0).
a) What is the gravitational potential energy, in joules, of the ball before it is released?
b) What will be the speed of the ball, in meters per second, when it reaches the bottom?
Explanation:
a) The height of the ball h with respect to the reference line is
[tex]h = L - L\cos{31°} = L(1 - \cos{31°})[/tex]
so its initial gravitational potential energy [tex]U_0[/tex] is
[tex]U = mgh = mgL(1 - \cos{31°})[/tex]
[tex]\:\:\:\:\:=(0.25\:\text{kg})(9.8\:\text{m/s}^2)(0.65\:\text{m})(1 - \cos{31})[/tex]
[tex]\:\:\:\:\:=0.23\:\text{J}[/tex]
b) To find the speed of the ball at the reference point, let's use the conservation law of energy:
[tex]\Delta{K} + \Delta{U} = 0 \Rightarrow K_0 + U_0 = K + U[/tex]
We know that the initial kinetic energy [tex]K_0,[/tex] as well as its final gravitational potential energy [tex]U[/tex] are zero so we can write the conservation law as
[tex]mgL(1 - \cos{31°}) = \frac{1}{2}mv^2[/tex]
Note that the mass gets cancelled out and then we solve for the velocity v as
[tex]v = \sqrt{2gL(1 - \cos{31°})}[/tex]
[tex]\:\:\:\:\:= \sqrt{2(9.8\:\text{m/s}^2)(0.65\:\text{m})(1 - \cos{31°})}[/tex]
[tex]\:\:\:\:\:= 1.3\:\text{m/s}[/tex]
Answer:
a. 0.23J
b. 1.35 m/s
Explanation:
a. U = mgh where where m = mass of the object, g = acceleration due to gravity, and h = height
h = L - Lcos(θ) where L = length of the rope, and θ = angle with respect to vertical.
Therefore, U = mg(L - Lcos(θ))
U = 0.25 * 9.8 (0.65m - 0.65cos(31° ))
U = 0.2275 ≈ 0.23J
The gravitational potential energy of the ball before it is released = 0.23J
b. To determine the velocity of the object at the bottom of its motion, all of the energy has gone from gravitational potential into kinetic since at the bottom, the problem says that U = 0. The kinetic energy of an object is given by the following equation:
[tex]K.E=\frac{I}{2}*mv^{2}[/tex]
where m = mass of the object and v = velocity of the object. Since we know that all of the energy was transferred into kinetic energy at the bottom, we can conclude that:
[tex]0.2275=\frac{1}{2} *0.25*v^{2}[/tex]
[tex]v^{2}=\frac{2*0.2275}{0.25}[/tex]
[tex]v^{2}=1.82[/tex]
[tex]v=\sqrt{1.82}=1.3491\\[/tex] ≈ [tex]1.35m/s[/tex]
Therefore, the speed of the ball when it reaches the bottom = 1.35m/s
Select all of the following that are involved in translation:
amino acids
tRNA
mRNA
ribonucleotides
deoxyribonucleotides
RNA Polymerase
Answer:
Amino Acids
tRNA
mRNA
ribonucleotides
Explanation:
Amino Acids
- During translation, an mRNA sequence is read using the genetic code, which is a set of rules that defines how an mRNA sequence is to be translated into the 20-letter code of amino acids, which are the building blocks of proteins.
tRNA
- At the beginning of translation, the ribosome and a tRNA attach to the mRNA. The tRNA is located in the ribosome's first docking site. This tRNA's anticodon is complementary to the mRNA's initiation codon, where translation starts. The tRNA carries the amino acid that corresponds to that codon. ransfer ribonucleic acid (tRNA) is a type of RNA molecule that helps decode a messenger RNA (mRNA) sequence into a protein. tRNAs function at specific sites in the ribosome during translation, which is a process that synthesizes a protein from an mRNA molecule.
mRNA
- Translation is the process by which a protein is synthesized from the information contained in a molecule of messenger RNA (mRNA). ... Then a transfer RNA (tRNA) molecule carrying the amino acid methionine binds to what is called the start codon of the mRNA sequence.
ribonucleotides
- During transcription, a ribonucleotide complementary to the DNA template strand is added to the growing RNA strand and a covalent phosphodiester bond is formed by dehydration synthesis between the new nucleotide and the last one added.
When converted to a household measurement, 9 kilograms is approximately equal to a) 9000 grams. b) 9000 ounces. 19.8 ounces. d) 19.8 pounds.
Answer:
9000 grams which is also very close to 19.8 pounds
so answers a) and d) are correct
Explanation:
9 kg(1000 gm/kg) = 9000 gms
9 kg(2.205 lb/kg) = 19.845 lb
I guess it rather depends on whose household you are in. One in the US would probably use 19.8 lb while one in Canada might use 9000 gm.
100 POINTS!!!!!!!!!!!!!!! (EASY SCIENCE 10 QUESTION!!!!!!)
In a GUT (Grand Unified Theory), what is "unified"?
Question 9 options:
subatomic particles
galaxies
fundamental forces
matter
Answer:
AccordingTo GUT(Grand Unified Theory)subatomic particles is Unified
A trapeze artist performs an aerial maneuver. While in a tucked position, as shown in figure A, she rotates about her center of mass at a rate of 6.27 rad/s. Her moment of inertia about this axis is 17.9 kg⋅m2. A short time later, the aerialist is in the straight position, as shown in figure B. If the moment of inertia about her center of mass in this position is now 33.1 kg⋅m2, what is her rotational speed in revolutions per minute (rpm)?
The rotational speed of the trapeze artist is 32.372 revolutions per second.
In this question we should apply the principle of angular momentum conservation to determine the final angular speed of the trapeze artist, since there are no external forces acting on the trapeze artist:
[tex]I_{A}\cdot \omega_{A} = I_{B}\cdot \omega_{B}[/tex] (1)
Where:
[tex]I_{A}[/tex], [tex]I_{B}[/tex] - Initial and final moments of the trapeze artist, in kilograms-square meter.
[tex]\omega_{A}[/tex], [tex]\omega_{B}[/tex] - Initial and final angular speeds, in radians per second.
If we know that [tex]I_{A} = 17.9\,kg\cdot m^{2}[/tex], [tex]I_{B} = 33.1\,kg\cdot m^{2}[/tex] and [tex]\omega_{A} = 6.27\,\frac{rad}{s}[/tex], then the final angular speed is:
[tex]\omega_{B} = \frac{I_{A}}{I_{B}}\times \omega_{A}[/tex]
[tex]\omega_{B} = \frac{17.9\,kg\cdot m^{3}}{33.1\,kg\cdot m^{2}}\times 6.27\,\frac{rad}{s}[/tex]
[tex]\omega_{B} = 3.390\,\frac{rad}{s}[/tex] ([tex]32.372\,rpm[/tex])
The rotational speed of the trapeze artist is 32.372 revolutions per second.
We kindly invite to check this question on conservation of rotational momentum: https://brainly.com/question/13183618
A 20 kg block (4) rests on a frictionless table;
a cord attached to the block extends
horizontally to a pulley at the edge of the
table. A 10 kg mass (B) hangs at the end of the
cord.
a. Draw and label the force vectors acting on
each object.
b. Calculate the acceleration of the block and
mass.
c. Calculate the tension in the cord.
Answer:
Try again?
Explanation:
Sin is the ____________________________ of the law. 1 John 3:4.
Answer:
transgression
Explanation:
find resultant vector and direction
Answer:
Explanation:
V = √(15.4² + 15.4²) = 21.77888886054... 21.8 m/s
θ = arctan(15.4/15.4) = 45.0°
Thermal is to heat as ___ is to water.
drain
waste
dilute
absorb
Answer:Thermal is to heat as absorb is to water.
Explanation:
Water is able to absorb heat - without increasing much in temperature - better than many substances. This is because for water to increase in temperature, water molecules must be made to move faster within the water; this requires breaking hydrogen bonds, and the breaking of hydrogen bonds absorbs heat.
please mark me as brainliest.Thank you so much
The LANDSAT C Earth resources satellite has a nearly circular orbit with an eccentricity of 0.00132. At perigee the satellite is at an altitude (measured from the earth's surface) of 417km. a) Calculate its altitude at apogee. b) Calculate its period. c) Calculate the velocity at perigee.
We have that for the Question it can be said that
altitude at apogee = [tex]434.96km[/tex]period = [tex]5585.41s[/tex]velocity at perigee = [tex]7.67km/s[/tex]From the question we are told
The LANDSAT C Earth resources satellite has a nearly circular orbit with an eccentricity of 0.00132. At perigee the satellite is at an altitude (measured from the earth's surface) of 417km.
Generally the equation for perigee radius is mathematically given as
[tex]r_p = R_E+Z_P\\\\=6378+417\\\\=6795km[/tex]
the equation for orbit is
[tex]r_p = \frac{h^2}{u}*\frac{1}{1+e}\\\\6795 = \frac{h^2}{398600}*\frac{1}{1+0.00132}\\\\h = 52077.5km^2/s[/tex]
the equation for velocity at perigee is mathematically given as
[tex]V_p = \frac{h}{r_p}\\\\= \frac{52077.5}{67.95}\\\\=7.67km/s[/tex]
the equation for Apagee radius is mathematically given as
[tex]r_a = \frac{h^2}{u} * \frac{1}{1-e}\\\\= \frac{52077.5^2}{398600} * \frac{1}{1-0.00132}\\\\= 6812.97km[/tex]
the equation for altitude at Apagee is mathematically given as
[tex]Z_a = r_a - R_E\\\\=6812.97-6378\\\\=434.96km[/tex]
the equation for period
[tex]Period T = \frac{2\pi}{u^2} * (\frac{h}{\sqrt1-e^2})^3\\\\= \frac{2\pi}{398600^2} * (\frac{52077.5}{\sqrt1-0.00132^2})^3\\\\= 5585.41s[/tex]
For more information on this visit
https://brainly.com/question/23379286
What is the gravitational potential energy of a circus performer with a mass of 50 kg walking on a tightrope 10 m above the ground
Answer:
5000 J
Explanation:
GPE = mgh
50*10*10
= 5000 J
The gravitational potential energy of the circus performer walking on the tightrope above the ground is 4900J.
Given the data in the question;
Mass of circus performer; [tex]m = 50kg[/tex]Distance from ground level; [tex]h = 10m[/tex]Gravitational potential energy; [tex]U = ?[/tex]Gravitational Potential EnergyGravitational potential energy is simply the potential energy an object possesses relative to another object due to gravity.
It is expressed as;
[tex]U = mgh[/tex]
Where m is mass, h is height or distance from the other massive object and g is the gravitational field ( [tex]g = 9.8m/s^2[/tex] )
To determine the gravitational potential energy of circus performer, we substitute our values into the expression above.
[tex]U = mgh\\\\U = 50kg\ *\ 9.8m/s^2\ *\ 10m\\\\U = 4900kgm^2/s^2\\\\U = 4900J[/tex]
Therefore, the gravitational potential energy of the circus performer walking on the tightrope above the ground is 4900J.
Learn more about gravitational potential energy here: https://brainly.com/question/3910603
A constant force of 8.0 N is exerted for 4.0 s on a 16-kg object initially at rest. The change in speed of this object will be:
The change in speed of this object is 3m/s
According to Newton's second law;
F = ma
F = mv/t
Given the following parameters
Force F = 8.0N
mass m = 16kg
time t = 4.0s
Required
speed v
Substitute the given parameters into the formula
v = Ft/m
v = 8 * 6/16
v = 48/16
v = 3m/s
Hence the change in speed of this object is 3m/s
Learn more here: https://brainly.com/question/19072061
car is traveling with a velocity of 40 m/s and has a mass of 1120 kg. The car has
_______________ energy.
Answer:896000 J
Explanation: K.E = 1/2*m*v^2
Answer:
The car has Kinetic energy. 896000 Joules
Explanation:
Kinetic energy formula: 1/2 (mass)(speed)^2
First you multiply 1120 by 1/2 and get 560
Then you take the square root of 40 ([tex]40x^{2}[/tex]) and get 1600
Finally to get the answer multiply 560 by 1600 and your final answer should be 896000 Joules
9. Two spheres, one of mass 4.0 x 107 kg and the other of mass 8.0 x 107 kg are separated by a
distance of 400.0 m. The force of gravity between them is
a) 1.3 N b) 530 N c) 1.8 x 104 N
d) 5.1 x 1021 N
e) 5.3 x 1022 N
Hi there!
We can use the gravitational force equation:
[tex]F_g = \frac{Gm_1m_2}{r^2}[/tex]
Where:
G = Gravitational Constant
m₁ = mass of one sphere (kg)
m₂ = mass of the other sphere (kg)
r = distance between the spheres (m)
Plug in the given values into the equation:
[tex]F_g = \frac{(6.67*10^{-11})(4.0*10^7)(8.0*10^7)}{400^2} = \boxed{1.334 N}[/tex]
Thus, the correct answer is A.
How many different values can digital signals take on?
Any value .
Two values : and 1.
Only specific values .
Any value between 1 and 10.
Answer:
Option d: value between 1 and 10
The data are continuous if they can take on any value within a range.
not complete sure if it is correct but yea
.
Explanation:
A power plant running at 39 % efficiency generates 330 MW of electric power. Part A At what rate (in MW) is heat energy exhausted to the river that cools the plant
516.154 megawatts of heat are exhausted to the river that cools the plant.
By definition of energy efficiency, we derive an expression for the energy rate exhausted to the river ([tex]Q_{out}[/tex]), in megawatts:
[tex]Q_{out} = Q_{in} - W[/tex]
[tex]Q_{out} = \left(\frac{1}{\eta}-1 \right)\cdot W[/tex](1)
Where:
[tex]\eta[/tex] - Efficiency.[tex]W[/tex] - Electric power, in megawatts.If we know that [tex]\eta = 0.39[/tex] and [tex]W = 330\,MW[/tex], then the energy rate exhausted to the river is:
[tex]Q_{out} = \left(\frac{1}{0.39}-1 \right)\cdot (330\,MW)[/tex]
[tex]Q_{out} = 516.154\,MW[/tex]
516.154 megawatts of heat are exhausted to the river that cools the plant.
We kindly to check this question on first law of thermodynamics: https://brainly.com/question/3808473
The Israelites grumbled 11 times against God. True or False?
Answer:
This is not a question about physics.
Anyway. Eleven times seems underrated.
Answer:
True
Explanation:
Because they didn't always appreciate the work of God
during the spin cycle of a washing machine the cloth stick to the outer wall of the barrel as it spins at a speed of 65 m per second. the radius of the barrel is 0.35 m. what is the magnitude and direction of centripetal acceleration of the clothes which are located on the wall of the barrel
Hi there!
Centripetal acceleration can be written as:
[tex]a_c = \frac{v^2}{r}[/tex]
v = tangential velocity (m/s)
r = radius (m)
Plug in the given values:
[tex]a_c = \frac{65^2}{0.35} = \boxed{12071.43 \text{ m}/s^2}}}[/tex]
The centripetal acceleration vector ALWAYS points towards the CENTER POINT of the barrel.
He drives 150 meters in 18 seconds. Assuming constant speed, what is his speed in meters per second?
Answer:
Explanation:
Givens
d = 150 meters
t = 18 seconds
r (rate) = ?
Formula
r = d/t
Solution
r = 150/18
r = 8.33 m/s
A slender rod of length 80.0 cm and 0.600 kg has its center of gravity at its geometrical center. But its density is not uniform; it increases by the same amount from the center of the rod out to either end. You want to determine the moment of inertia Icm of the rod for an axis perpendicular to the rod at its center, but you don't know its density as a function of distance along the rod, so you can't use an integration method to calculate Icm. Therefore, you make the following measurements: You suspend the rod about an axis that is a distance d (measured in meters) above the center of the rod and measure the period T (measured in seconds) for small-amplitude oscillations about the axis. You repeat this for several values of d. When you plot your data as T2−4π2d/g versus 1/d, the data lie close to a straight line that has slope 0.470 m⋅s2. What is the value of Icm for the rod?
Answer:
Icm = 0.0701 kgm^2
Explanation:
