Please delete my answer. I made a mistake
A student on a new planet wants to determine the value of gravity on that planet. Luckily for them they brought equipment that they can use to set up an oscillating spring or an oscillating pendulum. Which procedure would allow the student to determine the value of gravity on the new planet
Answer:
By measure the effective length and the time period of the pendulum.
Explanation:
Let the student take the oscillating pendulum at the planet.
He measure the time period of the pendulum by using the stop watch or the ordinary watch.
Then measure the effective length of the pendulum which is the distance between the center of gravity of the bob and the point of suspension of the pendulum.
Now, use the formula of the time period of the pendulum,
[tex]T =2\pi\sqrt\frac{L}{g}[/tex]
Here, L is the effective length of the pendulum, g is the acceleration due to gravity at the planet and T is time period of the pendulum.
By rearranging the terms, we get
[tex]T =2\pi\sqrt\frac{L}{g}\\\\T^{2}=4\pi^2\times\frac{L}{g}\\\\g =\frac{4\pi^2L}{T^2}[/tex]
Here, by substituting the values of L and T, the student get the value of acceleration due to gravity at that planet.
When the Glen Canyon hydroelectric power plant in Arizona is running at capacity, 690 m3m3 of water flows through the dam each second. The water is released 220 mm below the top of the reservoir. If the generators that the dam employs are 90% efficient, what is the maximum possible electric power output?
Answer:
The output electric power is 1338876 W.
Explanation:
Volume, V = 690 cubic meter
height, h = 220 mm = 0.22 m
efficiency = 90 %
time , t = 1 s
Let the mass is m.
m = volume x density
m = 690 x 1000 = 690000 kg
The input power is
P = m g h = 690000 x 9.8 x 0.22 = 1497640 W
The electric power out put is
[tex]P' = 90 % of 1487640\\\\\\P' = 1338876 W[/tex]
Which one of the following pairs of units may not be added together, even after the appropriate unit conversions have been made?
a. grams and milligrams
b. miles and kilometers
c. kilometers and kilograms
d. centimeters and yards
Answer:
c. kilometers and kilograms
Explanation:
If two units represent different things, like for example grams and joules, where the first one is a unit for mass, and the second one a unit for energy we can not add them together. (
Let's analyze each one of the given options:
a: grams and milligrams
both grams and milligrams are units of mass, so after the appropriate unit conversions, we can add them.
b: miles and kilometers
Both are units for distance, after the appropriate unit conversions, we can add them.
c: kilometers and kilograms
"Kilometer" is a unit for distance, and "kilogram" a unit for mass, we can not add distance and mass, so these two can not be added together.
d: centimeters and yards
Both are units of distance, after the appropriate unit conversions, we can add them.
Then the only one that we can not add together is option c: kilometers and kilograms
Answer: The correct option is C (kilometers and kilograms).
Explanation:
UNITS is the quantity of a constant magnitude which is used to measure the magnitudes of other quantities of the same nature. The magnitude of a physical quantity is expressed as, Physical quantity = (numerical value) × (unit). There are six basic units of measurement which includes:
--> metre (m) - unit of length.
--> kilograms (kg) - unit of mass.
--> second (s) - unit of time.
--> ampere (A) - unit of electrical current.
--> kelvin (K) - unit of temperature and
--> mole (mol) - unit of the amount of substance.
Mass is the quantity of matter present in a body which is constant and is measured in KILOGRAMMES or GRAMS. While position is the distance and direction of an object from a particular reference point which is known to vary and is measured in METERS OR KILOMETRES.
From the above definition, KILOGRAMMES is used to measure the magnitude of a different quantity from KILOMETERS and therefore should not be added up after conversation.
For an object with a given mass on Earth, calculate the weight of the object with the mass equal in magnitude to the number representing the day given in part 3 in kilograms using the formula F=W=mg. On the surface of the Earth g=9.8m/s^2
Answer: The weight of the object is 29.4 N
Explanation:
To calculate the weight of the object, we use the equation:
[tex]W=m\times g[/tex]
where,
m = mass of the object = 3 kg
g = acceleration due to gravity = [tex]9.8m/s^2[/tex]
Putting values in above equation, we get:
[tex]W=3kg\times 9.8m/s^2\\\\W=29.4N[/tex]
Hence, the weight of the object is 29.4 N
two factor of a number are 5 and 6 .what is the number show working
Answer:
30
Explanation:
since [tex]\frac{30}{5}[/tex]=6
[tex]\frac{30}{6}[/tex]=5
then both 5 and 6 are factors of 30
Have a nice day
Can someone help me
Btw the last one say current
If the temperature stays constant, which change would decrease the amount
of thermal energy in an object?
A. Decreasing its density
B. Increasing its velocity
c. Decreasing its mass
D. Increasing its mass
The motion that is not repeated in regular interval of time is called....?
please help me!
Answer:
Motion which repeats itself after regular intervals of time is known as periodic motion.How does the magnitude of the normal force exerted by the ramp in the figure compare to the weight of the static block? The normal force is:______ a. greater than the weight of the block. b. possibly greater than or less than the weight of the block, depending on whether or not the ramp surface is smooth. c. equal to the weight of the block. d. possibly greater than or equal to the weight of the block, depending on whether or not the ramp surface is smooth. less than the weight of the block.
Answer:
less than the weight of the block.
Explanation:
From the free body diagram, we get.
The normal force is N = Mg cosθ
The tension in the string is T = Mg sinθ
Wight of the block when the block is static, W = Mg
Now since the magnitude of cosθ is in the range of : 0 < cosθ < 1,
therefore, the normal force is less than the weight of the static block.
3.1Chất điểm chuyển động thẳng với phương trình: x = – 1 + 3t2
– 2t
3
(hệ SI, với t ≥ 0). Chất điểm dừng lại để
đổi chiều chuyển động tại vị trí có tọa độ:
Answer:
eqcubuohwehuuc
Explanation:
the the the the the the the the the the the the the the the the dhueirrjhrhjrirjrheh3jeiiruj fnrhfjjjrjrj fjfiirrjejrjejej jrkrjrjrjrjrjdjdjfjrhruriruru rjridjhwjjsjd
You have been hired to design a spring-launched roller coaster that will carry two passengers per car. The car goes up a 13-m-high hill, then descends 20 m to the track's lowest point. You've determined that the spring can be compressed a maximum of 2.4 m and that a loaded car will have a maximum mass of 430 kg. For safety reasons, the spring constant should be 13 % larger than the minimum needed for the car to just make it over the top. Part A What spring constant should you specify
Answer:
22.15 N/m
Explanation:
As we know potential energy = m*g*h
Potential energy of spring = (1/2)kx^2
m*g*h = (1/2)kx^2
Substituting the given values, we get -
(400)*(9.8)*(10) = (0.5)*(k)*(2.0^2)
k = 39200/2.645
k = 19600 N/m
For safety reasons, this spring constant is increased by 13 % So the new spring constant is
k = 19600 * 1.13 = 22148 N/m = 22.15 N/m
The object has a mass of 100kg. The Tension is 200N[U]. What is the acceleration of this elevator? *
A) 8m/s/s
B) 8m/s/s[D]
C) 9.8m/s/s[D]
D) 0.5m/s/s[D]
Answer:
So the answer is B. A is wrong because negative answer = deceleration
Use a variation model to solve for the unknown value. Use as the constant of variation. The stopping distance of a car is directly proportional to the square of the speed of the car. (a) If a car travelling has a stopping distance of , find the stopping distance of a car that is travelling . (b) If it takes for a car to stop, how fast was it travelling before the brakes were applied
Complete question is;
Use a variation model to solve for the unknown value.
The stopping distance of a car is directly proportional to the square of the speed of the car.
a. If a car traveling 50 mph has a stopping distance of 170 ft, find the stopping distance of a car that is traveling 70 mph.
b. If it takes 244.8 ft for a car to stop, how fast was it traveling before the brakes were applied?
Answer:
A) d = 333.2 ft
B) 60 mph
Explanation:
Let the stopping distance be d
Let the speed of the car be v
We are told that the stopping distance is directly proportional to the square of the speed of the car. Thus;
d ∝ v²
Therefore, d = kv²
Where k is constant of variation.
A) Speed is 50 mph and stopping distance of 170 ft.
v = 50 mph
d = 170 ft = 0.032197 miles
Thus,from d = kv², we have;
0.032197 = k(50²)
0.032197 = 2500k
k = 0.032197/2500
k = 0.0000128788
If the car is now travelling at 70 mph, then;
d = 0.0000128788 × 70²
d = 0.06310612 miles
Converting to ft gives;
d = 333.2 ft
B) stopping distance is now 244.8 ft
Converting to miles = 0.046363636 miles
Thus from d = kv², we have;
0.046363636 = 0.0000128788(v²)
v² = 0.046363636/0.0000128788
v² = 3599.99658
v = √3599.99658
v ≈ 60 mph
A team of people who traveled to the North Pole by dogsled lived on butter because they needed to consume 6 000 dietitian's Calories each day. Because the ice there is lumpy and irregular, they had to help the dogs by pushing and lifting the load. Assume they had a 16-hour working day and that each person could lift a 500-N load. How many times would a person have to lift this weight 1.00 m upwards in a constant gravitational field, where (g = 9.80m/s2) where to do the work equivalent to 6 000 Calories?
Answer:
The right solution is "50200 days".
Explanation:
Given:
Calories intake,
= 6000 kcal,
or,
= [tex]2.52\times 10^7 \ J[/tex]
Force,
= 500 N
As we know,
⇒ [tex]Work \ done = Force\times distance[/tex]
Or,
⇒ [tex]distance = \frac{Work \ done}{Force}[/tex]
By putting the values, we get
[tex]=\frac{2.52\times 10^7}{500}[/tex]
[tex]=0.502\times 10^5[/tex]
[tex]=50200 \ m[/tex]
hence,
The number of days will be:
= [tex]\frac{50200}{1}[/tex]
= [tex]50200 \ days[/tex]
Question 9 of 10
What causes the different seasons on Earth?
A. The angles at which the suns rays strike the Earth
Ο Ο Ο
B. The distance between Earth and the sun
C. The speed at which the Earth rotates on its axis
O
D. Increasing levels of carbon dioxide in the atmosphere.
SUBMIT
Answer:
B
Explanation:
The seasons are measured in how far or close the earth is to the sun.
A system consists of a disk of mass 2.0 kg and radius 50 cm upon which is mounted an annular cylinder of mass 1.0 kg with inner radius 20 cm and outer radius 30 cm (see below). The system rotates about an axis through the center of the disk and annular cylinder at 10 rev/s. (a) What is the moment of inertia of the system
Complete Question
A system consists of a disk of mass 2.0 kg and radius 50 cm upon which is mounted an annular cylinder of mass 1.0 kg with inner radius 20 cm and outer radius 30 cm (see below). The system rotates about an axis through the center of the disk and annular cylinder at 10 rev/s.
(a) What is the moment of inertia of the system
(b) What is its rotational kinetic energy? axis 50 cm 30 cm 20 cm
Answer:
a) [tex]M.I=0.32kg.m^2[/tex]
b) [tex]K.E=621.8J[/tex]
Explanation:
From the question we are told that:
Mass [tex]m=2.0kg[/tex]
Disk Radius [tex]r_d=50cm=0.5m[/tex]
Mass of annular cylinder [tex]M_c=1.0kg[/tex]
Inner Radius of cylinder [tex]R_i=20cm=0.2m[/tex]
Outer Radius of cylinder [tex]R_o=0.3m[/tex]
Angular Velocity [tex]\omega=10rev/s=2 \pi*10=62.83[/tex]
Generally the equation for moment of inertia is mathematically given by
[tex]M.I=0.5M r_d^2+0.5M_c(R_i^2+R_o^2)[/tex]
[tex]M.I=0.5(2*0.50)^2)+0.5*1*(0.2^2+0.30^)[/tex]
[tex]M.I=0.32kg.m^2[/tex]
Generally the equation for Rotational Kinetic Energy is mathematically given by
[tex]K.E=0.5M.I \omega^2[/tex]
[tex]K.E=0.5 *0.32*62.83[/tex]
[tex]K.E=621.8J[/tex]
An object is positively charged if it has more what
Answer:
An Object is positively charged if it has more Positive Electrons in that object
A 97.6-kg baseball player slides into second base. The coefficient of kinetic friction between the player and the ground is μk = 0.555.
(a) What is the magnitude of the frictional force?
(b) If the player comes to rest after 1.22 s, what is his initial speed?
Answer:
[tex]v=6.65m/sec[/tex]
Explanation:
From the Question we are told that:
Mass [tex]m=97.6[/tex]
Coefficient of kinetic friction [tex]\mu k=0.555[/tex]
Generally the equation for Frictional force is mathematically given by
[tex]F=\mu mg[/tex]
[tex]F=0.555*97.6*9.8[/tex]
[tex]F=531.388N[/tex]
Generally the Newton's equation for Acceleration due to Friction force is mathematically given by
[tex]a_f=-\mu g[/tex]
[tex]a_f=-0.555 *9.81[/tex]
[tex]a_f=-54455m/sec^2[/tex]
Therefore
[tex]v=u-at[/tex]
[tex]v=0+5.45*1.22[/tex]
[tex]v=6.65m/sec[/tex]
Charlotte throws a paper airplane into the air, and it lands on the ground. Which best explains why this is an example of projectile motion? The paper airplane’s motion is due to horizontal inertia and the vertical pull of gravity. A force other than gravity is acting on the paper airplane. The paper airplane’s motion can be described using only one dimension. A push and a pull are the primary forces acting on the paper airplane.
highschool physics, not college physics
Answer:
Answer:
A). The paper airplane’s motion is due to horizontal inertia and the vertical pull of gravity.
Explanation:
Edge.
Answer:
The motion of the paper airplane is best explained by horizontal inertia and vertical pull of gravity.
Explanation:
What is horizontal inertia and vertical pull of gravity?Inertia is the property by which the body wants to remain in its position unless any external for is applied. Here horizontal inertia is inertia of motion which is acting horizontally .
While vertical pull is due to the earth .
In a paper airplane , four forces act .these forces provide it flight.These forces are horizontal inertia , vertical pull downwards , lift by air and drag.Hence horizontal inertia and vertical pull best explain the projectile motion of paper airplane.
Also read it;
https://brainly.com/question/11049671
#SPJ2
A projectile of mass m is fired horizontally with an initial speed of v0 from a height of h above a flat, desert surface. Neglecting air friction, at the instant before the projectile hits the ground, find the following in terms of m, v0, h, and g : Are any of the answers changed if the initial angle is changed?
Complete question is;
A projectile of mass m is fired horizontally with an initial speed of v0 from a height of h above a flat, desert surface. Neglecting air friction, at the instant before the projectile hits the ground, find the following in terms of m, v0, h and g:
(a) the work done by the force of gravity on the projectile,
(b) the change in kinetic energy of the projectile since it was fired, and
(c) the final kinetic energy of the projectile.
(d) Are any of the answers changed if the initial angle is changed?
Answer:
A) W = mgh
B) ΔKE = mgh
C) K2 = mgh + ½mv_o²
D) No they wouldn't change
Explanation:
We are expressing in terms of m, v0, h, and g. They are;
m is mass
v0 is initial velocity
h is height of projectile fired
g is acceleration due to gravity
A) Now, the formula for workdone by force of gravity on projectile is;
W = F × h
Now, Force(F) can be expressed as mg since it is force of gravity.
Thus; W = mgh
Now, there is no mention of any angles of being fired because we are just told it was fired horizontally.
Therefore, even if the angle is changed, workdone will not change because the equation doesn't depend on the angle.
B) Change in kinetic energy is simply;
ΔKE = K2 - K1
Where K2 is final kinetic energy and K1 is initial kinetic energy.
However, from conservation of energy, we now that change in kinetic energy = change in potential energy.
Thus;
ΔKE = ΔPE
ΔPE = U2 - U1
U2 is final potential energy = mgh
U1 is initial potential energy = mg(0) = 0. 0 was used as h because at initial point no height had been covered.
Thus;
ΔKE = ΔPE = mgh
Again like a above, the change in kinetic energy will not change because the equation doesn't depend on the angle.
C) As seen in B above,
ΔKE = ΔPE
Thus;
½mv² - ½mv_o² = mgh
Where final kinetic energy, K2 = ½mv²
And initial kinetic energy = ½mv_o²
Thus;
K2 = mgh + ½mv_o²
Similar to a and B above, this will not change even if initial angle is changed
D) All of the answers wouldn't change because their equations don't depend on the angle.
Steve pushes a crate 20 m across a level floor at a constant speed with a force of 200 N, this time on a frictionless floor. The velocity of the crate is in the direction of the force Steve is applying to the crate. What is the net work done on the crate
Answer:
The correct answer is "4000 J".
Explanation:
Given that,
Force,
= 200 N
Displacement,
= 20 m
Now,
The work done will be:
⇒ [tex]Work=Force\times displacement[/tex]
By putting the values, we get
[tex]=200\times 20[/tex]
[tex]=4000 \ J[/tex]
Baseball runner with a mass of 70kg, moving at 2.7m/s and collides head-on into a shortstop with a mass of 85kg and a velocity of 1.6m/s. What will be the resultant velocity of the system when they make contact with each other
Answer:
The speed of the combined mass after the collision is 2.1 m/s.
Explanation:
mass of runner, m = 70 kg
speed of runner, u = 2.7 m/s
mass of shortstop, m' = 85 kg
speed of shortstop, u' = 1.6 m/s
Let the velocity of combined system is v.
Use conservation of momentum
Momentum before collision = momentum after collision
m u + m' u' = (m + m') v
70 x 2.7 + 85 x 1.6 = (70 + 85) v
189 + 136 = 155 v
v = 2.1 m/s
A car is travelling at a speed of 30m/s on a straight road. what would be the speed of the car in km
Answer:
[tex] = \frac{30 \times {10}^{ - 3} }{1} \\ = 0.03 \: km \: per \: second[/tex]
Answer:
108 km/hr or 0.03 km/s
Explanation:
conversion factor for m/s to km/hr is 5/18
conversion factor for m/s to km/s is 1/1000
The number 0.00325 × 10-8 cm can be expressed in millimeters as A) 3.25 × 10-11 mm. B) 3.25 × 10-10 mm. C) 3.25 × 10-12 mm. D) 3.25 × 10-9 mm.
Answer:
Option B. 3.25×10¯¹⁰ mm.
Explanation:
Measurement (cm) = 0.00325×10⁻⁸ cm
Measurement (mm) =?
The measurement in mm can be obtained as follow:
1 cm = 10 mm
Therefore,
0.00325×10⁻⁸ cm = 0.00325×10⁻⁸ cm × 10 mm / 1 cm
0.00325×10⁻⁸ cm = 3.25×10¯¹⁰ mm
Thus, 0.00325×10⁻⁸ cm is equivalent to 3.25×10¯¹⁰ mm.
The conversion from centimeter to millimeter of the number 0.00325*10^-8cm is 3.25*10^-10mm
The number given is in standard form and can be written as 3.25*10^-11 cm.
To convert this from centimeter to millimeter, we have to multiply this value by 10.
Conversion Units1 cm - 10mm100cm = 1m1000m = 1kmSo, let's 3.25*10^-11 by 10 and get our value in mm
[tex]3.25*10^-^1^1 * 10 = 3.25*10^-^1^0[/tex]
From the calculation above, we can see that option B is the right answer since it carries [tex]3.25*10^-^1^0mm[/tex]
Learn more about conversion of units here;
https://brainly.com/question/8426032
The first and second coils have the same length, and the third and fourth coils have the same length. They differ only in the cross-sectional area. According to theory, what should be the ratio of the resistance of the second coil to the first coil and the fourth coil to the third
Answer:
The ratio of the resistances of second coil to the first coil is the ratio of square of radius of the first coil to the square of radius of second coil.
And
The ratio of the resistances of fourth coil to the third coil is the ratio of square of radius of the third coil to the square of radius of fourth coil.
Explanation:
The resistance of the coil is directly proportional to the length of the coil and inversely proportional to the area of coil and hence inversely proportional to the square of radius of the coil.
So, the ratio of the resistances of second coil to the first coil is the ratio of square of radius of the first coil to the square of radius of second coil.
And
The ratio of the resistances of fourth coil to the third coil is the ratio of square of radius of the third coil to the square of radius of fourth coil.
Phát biểu nào sau đây là SAI?
A. Cường độ điện trường là đại lượng
đặc trưng cho điện trường về phương
diện tác dụng lực.
B. Điện trường tĩnh là điện trường có
cường độ E không đổi tại mọi điểm.
C. Đơn vị đo cường độ điện trường là
vôn trên mét (V/m).
D. Trong môi trường đẳng hướng,
cường độ điện trường giảm lần so với
trong chân không
Answer:
B.
Explanation:
sana makatulong sayo
What is the efficiency of a ramp that is 5.5 m long when used to move a 66 kg object to a height of 110 cm when the object is pushed by a 150 N force .
Answer and I will give you brainiliest
Explanation:
Energy input = F×d = (150 N)(5.5 m) = 825 J
Energy output = mgh = (66 kg)(9.8 m/s^2)(1.10 m) = 711 J
efficiency = [tex]\dfrac{\text{output}}{\text{input}}[/tex]×100% = 86.2%
To accurately describe the wind, the measurement should include
A) a direction, but not a speed
B)a speed, but not a direction
C) both a speed and a direction
D) neither a speed nor a direction
Answer:
C. both a speed and a direction
Speed of light in water
Answer:
225,000 kilometers per second
Explanation:
Have a nice day
Kirchhoff's junction rule is a statement of: Group of answer choices the law of conservation of momentum the law of conservation of charge the law of conservation of energy. the law of conservation of angular momentum. Newton's second law
Answer:
the law of conservation of energy.
Explanation:
An electric circuit can be defined as an interconnection of electrical components which creates a path for the flow of electric charge (electrons) due to a driving voltage.
Generally, an electric circuit consists of electrical components such as resistors, capacitors, battery, transistors, switches, inductors, etc.
In Physics, a point where at least three circuit paths (wires) meet is referred to as a junction.
The Kirchhoff’s circuit laws are two(2) equations first published by Gustav Kirchhoff in 1845.
Fundamentally, they address the conservation of energy and charge in the context of electrical circuits.
One of the laws known as Kirchoff's Current Law (KCL) deals with the principle of application of conserved energy in electrical circuits.
Kirchoff's Current Law (KCL) states that the sum of all currents entering a junction must equal the sum of all currents leaving the junction.
This simply means that the algebraic sum of currents in a network of conductors(wires) meeting at a point is equal to zero.
The Law of Conservation of Energy states that energy cannot be destroyed but can only be transformed or converted from one form to another.
This ultimately implies that, Kirchhoff's junction rule is a statement of the law of conservation of energy.
