Answer:
Theory
Explanation:
Theory is a term that is used often in academic work or scientific research to explain certain things or conditions established on universal principles or laws.
It is used to describe the "why and how" or the reason behind the occurrence of a situation.
Hence, it is correct to conclude that THEORY is "an explanation of the relationships among particular phenomena."
Answer:
E) Theory
Explanation:
Edge 2020
Brainliest?
A man walks south at a speed of 2.00 m/s for 60.0 minutes. He then turns around and walks north a distance 3000 m in 25.0 minutes. What is the average velocity of the man during his entire motion?
Answer:
v = 0.823 m/s
Explanation:
A man walks south at a speed of 2.00 m/s for 60.0 minutes.
The distance covered in South = 60 × 60 × 2 = 7200 m
He then turns around and walks north a distance 3000 m in 25.0 minutes.
As they moved in opposite direction, net displacement will be : 7200 - 3000 = 4200 m
Average velocity of the man = net displacement/time
[tex]v=\dfrac{4200\ m}{(60+25)\times 60}\\\\=0.823\ m/s[/tex]
So, the average velocity of the man is 0.823 m/s.
A spinning ice skater will slow down if she extends her arms away from her body. Which of the following statements explain this phenomenon
A) circular motion is always uniform
B) A centripetal force always points outward
C) Angular momentum is always conserved
D) Centripetal acceleration cannot change
Marking brainliest
Answer:
B, which is why ice skaters often keep their arms close to their body when doing spins and jumps to minimize resistance.
premium
A student releases a marble from the top of a ramp. The marble increases speed
while on the ramp then continues across the floor. The marble travels a total of
150cm in 4.80s.
What was the marble's final speed?
Explanation:
the formula of speed is distance traveled by time it work
31.25 cm/sec
__________________________________________________________
Explanation:We are given:
Distance travelled = 150 cm
Time taken = 4.8 seconds
Final Speed of the Marble:
Speed of the marble = Distance travelled / Time taken
Speed of the marble = 150 / 4.8
Speed of the marble = 31.25 cm/sec
A projector lens projects an image from a 6.35 cm wide LCD screen onto a
screen 3.25 m wide. If the focal length of the projector lens is 13.8 cm, the screen
must be how far from the projector
Answer:
For any given projector, the width of the image (W) relative to the throw distance (D) is know as the throw ratio D/W or distance over width. So for example, the most common projector throw ratio is 2.0. This means that for each foot of image width, the projector needs to be 2 feet away or D/W = 2/1 = 2.0.
Is a parked car potential or kinetic ?
Answer:
Potential energy is the energy that is stored in an object. ... When you park your car at the top of a hill, your car has potential energy because the gravity is pulling your car to move downward; if your car's parking brake fails, your vehicle may roll down the hill because of the force of gravity.
During the stretching routine who used the medicine ball for support.
The guy
The guy
The Girl
The Girl
Both people used a ball for support
Both people used a ball for support
No one used it
No one used it
An object in FREE-FALL on the MOON would experience which of the following
FORCES?
O a. Weight
O b. Normal
O c. Air Resistance
d. a and c
O e. None of these
Answer:
e. none of these
Explanation:
An object in FREE-FALL on the MOON would experience only acceleration
Why evaporation takes place from the Surface?
Answer:
in the water cycle evaporation occurs when the sunlight warms the surface of the water the heat from the sun makes the water molecules move faster and faster until they move so fast they can escape as a gas once evaporated a molecule from water that vapor spends about 10 days in the air
A 4.8-g particle is moving toward a stationary 7.4-g particle at 3.0 m/s. What percentage of the original kinetic energy is convertible to internal energy?
Answer:
60.185 percent of the original kinetic energy is convertible to internal energy.
Explanation:
Let suppose that collision between both particles is entirely inellastic. If there is no external forces exerted on any of the particles, then we can apply the Principle of Linear Momentum Conservation. That is:
[tex]m_{A}\cdot v_{A,o} + m_{B}\cdot v_{B,o} = (m_{A}+m_{B})\cdot v[/tex]
[tex]v = \frac{m_{A}\cdot v_{A,o}+v_{B}\cdot v_{B,o}}{m_{A}+m_{B}}[/tex] (1)
Where:
[tex]m_{A}[/tex] - Mass of the 4.8-g particle, measured in kilograms.
[tex]m_{B}[/tex] - Mass of the 7.4-g particle, measured in kilograms.
[tex]v_{A,o}[/tex] - Initial speed of the 4.8-g particle, measured in meters per second.
[tex]v_{B,o}[/tex] - Initial speed of the 7.4-g particle, measured in meters per second.
[tex]v[/tex] - Final speed of the collided particles, measured in meters per second.
If we know that [tex]m_{A} = 4.8\times 10^{-3}\,kg[/tex], [tex]m_{B} = 7.4\times 10^{-3}\,kg[/tex], [tex]v_{A,o} = 3\,\frac{m}{s}[/tex] and [tex]v_{B,o} = 0\,\frac{m}{s}[/tex], then the final speed of the system is:
[tex]v = \frac{(4.8\times 10^{-3}\,kg)\cdot \left(3\,\frac{m}{s} \right)+(7.4\times 10^{-3}\,kg)\cdot \left(0\,\frac{m}{s} \right)}{4.8\times 10^{-3}\,kg+7.4\times 10^{-3}\,kg}[/tex]
[tex]v = 1.180\,\frac{m}{s}[/tex]
During the collision part of the initial energy is dissipated in the form of heat, which is related to the internal energy ([tex]\Delta U[/tex]), measured in joules. According to the Principle of Energy Conservation, we have the following model:
[tex]\Delta U = K_{A}+K_{B}-K[/tex] (2)
Where:
[tex]K_{A}[/tex], [tex]K_{B}[/tex] - Initial translational kinetic energies of each particle, measured in joules.
[tex]K[/tex] - Final translational kinetic energy of the collided particles, measured in joules.
By applying the definition of translational kinetic energy, we expand and simplify the equation above:
[tex]\Delta U = \frac{1}{2}\cdot m_{A}\cdot v_{A,o}^{2}+\frac{1}{2}\cdot m_{B}\cdot v_{B,o}^{2} -\frac{1}{2}\cdot (m_{A}+m_{B})\cdot v^{2}[/tex] (3)
If we get that [tex]m_{A} = 4.8\times 10^{-3}\,kg[/tex], [tex]m_{B} = 7.4\times 10^{-3}\,kg[/tex], [tex]v_{A,o} = 3\,\frac{m}{s}[/tex], [tex]v_{B,o} = 0\,\frac{m}{s}[/tex] and [tex]v = 1.180\,\frac{m}{s}[/tex], the internal energy associated with the system is:
[tex]\Delta U = \frac{1}{2}\cdot (4.8\times 10^{-3}\,kg)\cdot \left(3\,\frac{m}{s} \right)^{2}+ \frac{1}{2}\cdot (7.4\times 10^{-3}\,kg)\cdot \left(0\,\frac{m}{s} \right)^{2}-\frac{1}{2}\cdot (4.8\times 10^{-3}\,kg+7.4\times 10^{-3}\,kg)\cdot \left(1.180\,\frac{m}{s} \right)^{2}[/tex]
[tex]\Delta U = 0.013\,J[/tex]
And the initial energy of both particles is:
[tex]E_{o} = \frac{1}{2}\cdot (4.8\times 10^{-3}\,kg)\cdot \left(3\,\frac{m}{s}\right)^{2}+\frac{1}{2}\cdot (7.4\times 10^{-3}\,kg)\cdot \left(0\,\frac{m}{s} \right)^{2}[/tex]
[tex]E_{o} = 0.0216\,J[/tex]
Lastly, the percentage of the original kinetic energy that is convertible to internal energy is: ([tex]\Delta U = 0.013\,J[/tex], [tex]E_{o} = 0.0216\,J[/tex])
[tex]\%e = \frac{\Delta U}{E_{o}}\times 100\,\%[/tex] (4)
[tex]\%e = \frac{0.013\,J}{0.0216\,J}\times 100\,\%[/tex]
[tex]\%e = 60.185\,\%[/tex]
60.185 percent of the original kinetic energy is convertible to internal energy.
Aluminum wire with a diameter of 0.8650 mm is wound onto a spool. The wire is insulated, but you have access to both ends. The resistivity of aluminum at 20.0 °C is 2.65 x 10^-8 Ω-m. You measure the resistance of the wire at that temperature, and it is 2.48 Ω. What is the length of the wire?
a. 8.10 x 10^4 m
b. 22.0 m
c. 5.68 m
d. 0.111 m
e. 55.0 m
Answer:
e. 55.0 m
Explanation:
Given;
diameter of the aluminum wire, d = 0.865 mm
radius of the wire, r = d/2 = 0.4325 mm = 0.4325 x 10⁻³ m
resistivity of the wire, ρ = 2.65 x 10⁻⁸ Ω-m
resistance of the wire, R = 2.48 Ω
The resistance of a wire is given by;
[tex]R = \frac{\rho \ L}{A} \\\\[/tex]
where;
L is length of the wire
A is area of the wire = πr² = π(0.4325 x 10⁻³ )² = 5.877 x 10⁻⁷ m²
Substitute the givens and solve for L,
[tex]L = \frac{RA}{\rho} \\\\L = \frac{(2.48)(5.877*10^{-7})}{2.65*10^{-8}}\\\\L = 55.0 \ m[/tex]
Therefore, the length of the wire is 55.0 m
A ball is kicked off the ground reaching a maximum height of 60m and lands 80m away. Calculate the initial speed and the angle above the horizontal of the ball when it was kicked
Answer:
36.87°
Explanation:
Given
Maximum height = 60m
Horizontal distance (range) = 80,m
Required
Initial speed U
Angle of launch
To get the speed, we will use the range formula;
R = U √2H/g
80 = U√2(60)/9.8
80 = U√12.25
80 = 3.5U
U = 80/3.5
U = 22.86m/s
Get the angle of launch
Using the formula
Theta = tan^-1(y/x)
y is the vertical distance
x is the horizontal distance
Theta = tan^-1(60/80)
Theta = tan^-1(0.75)
Theta = 36.87°
Hence the angle of launch is 36.87°
A freshly caught catfish is placed on a spring scale, and it oscillates up and down with a period of 0.19 s. If the spring constant of the scale is 2330 N/m, what is the mass of the catfish?
Answer:
The mass of the catfish is 2.13 kg
Explanation:
Period of oscillation, T = 0.19 s
spring constant, k = 2330 N/m
The period of oscillation of the spring is given by;
[tex]T = 2\pi \sqrt{\frac{m}{k} }\\\\\frac{T}{2\pi} = \sqrt{\frac{m}{k} }\\\\\frac{T^2}{4\pi^2} = \frac{m}{k}\\\\m = \frac{kT^2}{4\pi^2}[/tex]
where;
m is mass of the catfish
substitute the given values and solve for m;
[tex]m = \frac{kT^2}{4\pi^2} \\\\m = \frac{(2330)(0.19)^2}{4\pi^2} \\\\m = 2.13 \ kg[/tex]
Therefore, the mass of the catfish is 2.13 kg
Jared walks 120 m east, 150 m south, and then 40 m west. Find the total
distance traveled by Jared
Answer:
310 m
Explanation:
120+150+40=310
Please answer my question
Answer:
Answer is (b) Mercury, venus and Mars.
Explanation:
i think b is correct!!
;-) :-) :-) :-)
2. Which bicyclist was traveling the fastest at the end of the race?
Answer:
This question is incomplete
Explanation:
This question is incomplete. However, to determine the bicyclist that traveled the fastest at the end of the race, the speed of the bicyclists at the end of the race will determine this (not the bicyclist that came first nor there overall speed). The speed of the bicyclist at the end of the race can be determined by using the formula below
s = d ÷ t
Where s is the speed of each bicyclist at the end of the race
d is the specific distance covered by the bicyclist at the end of the race
t is the time taken for the bicyclist to complete that distance
It should be noted that to get an accurate result, the distance covered at the end of the race must be the same for all the bicyclists.
"2.40 A pressure of 4 × 106N/m2 is applied to a body of water that initially filled a 4300 cm3 volume. Estimate its volume after the pressure is applied."
Answer:Final volume after pressure is applied=4,292cm3
Explanation:
Using the bulk modulus formulae
We have that The bulk modulus of waTer is given as
K =-V dP/dV
Where K, the bulk modulus of water = 2.15 x 10^9N/m^2
2.15 x 10^9N/m^2= - 4,300 x 4 × 106N/m2 / dV
dV = - 4,300 x 4 × 10^6N/m^2/ 2.15 x 10^9N/m^2
dV (change in volume)= -8.000cm^3
Final volume after pressure is applied,
V= V+ dV
V= 4300cm3 + (-8.000cm3)
=4300cm3 - 8.000cm3
Final Volume, V =4,292cm3
A large pizza is cut into 8 even slices. A person orders 4 large pizzas from a restaurant. How many total slices of pizza did the person order?
Answer:
32 slicesExplanation:
Step one:
given data
we are told that 1 large pizza can be cut into 8 even slices
Required
we want to find how many slices are there in 4 large pizzas
Step two:
so if 1 pizza has 8 slices
4 pizza will have x
cross multiply we have
x= 8*4
x=32 slices
Two objects attract each other with a gravitational force of magnitude 1.01 10-8 N when separated by 19.9 cm. If the total mass of the two objects is 5.11 kg, what is the mass of each?
Answer:
m₂ = 1.17 kg
Explanation:
Given that,
Force between two objects, [tex]F=1.01\times 10^{-8}\ N[/tex]
Mass of object 1, [tex]m_1=5.11\ kg[/tex]
Distance between masses, r = 19.9 cm = 0.199 m
The gravitational force between two masses is given by :
[tex]F=\dfrac{Gm_1m_2}{r^2}[/tex]
m₂ is the mass of object 2
[tex]m_2=\dfrac{Fr^2}{Gm_1}\\\\m_2=\dfrac{1.01\times 10^{-8}\times (0.199)^2}{6.67\times 10^{-11}\times 5.11}\\\\=1.17\ kg[/tex]
So, the mass of second object is 1.17 kg.
While riding a multispeed bicycle, the rider can select the radius of the rear sprocket that is fixed to the rear axle. The front sprocket of a bicycle has radius 12.0 cm. If the angular speed of the front sprocket is 0.600 rev/s, what is the radius of the rear sprocket for which the tangential speed of a point on the rim of the rear wheel will be 5.00 m/s?
Answer:
2.9 cm
Explanation:
Assuming that the rear wheel has a radius of 0.330 m
Given that
r(a) = 12 cm -> 0.12 m
w(a) = 0.6 rev/s -> 3.77 rad/s
v = 5 m/s
r(w) = 0.330 m
The speed on any point on the rim at the sprocket in the front is
v(a) = w(a).r(a) = 3.77 * 0.12 = 0.4524 m/s
Also,
v(a) = speed at any point on the chain
v(b) = speed at any point on the rim of the rear sprocket
v(a) = v(b)
where v(b) = w(b).r(b)
Recall that the speed at any point on the rear wheel is v, where
v = w(b).r(w)
5 = w(b) * 0.330
w(b) = 5/0.330
w(b) = 15.15 rad/s
On substituting this in the equation, we have
v(b) = w(b).r(b).
Remember also, that v(a) = v(b), so
0.4524 = 15.15 * r(b)
r(b) = 0.4524 / 15.15
r(b) = 0.029 m -> 2.9 cm
Therefore, the radius of the rear sprocket needed is 2.9 cm
A baseball is thrown across the field. The ____________is measured from where the ball is thrown to where landed was 75 feet.
motion
direction
distance
reference point
Answer:
distance i think
Explanation:
The earliest mineral observed to showmagnetic properties is called
A leadstone
Blodestone
Cloadstone
Dnone of the above
E all of the above
Answer:
B: lodestone
Explanation:
Each magnet has its magnetic poles, north (N) and south (S). Diversified ones are attracted and reptiles of the same name are repelled, similarly to charges, so it was considered possible to separate the magnet at the north and south poles.
Magnetic properties can be lost if the magnet is exposed to high temperatures if it falls or due to some mechanical shocks.
A cheetah can maintain a maximum constant velocity of 34.2 m/s for 8.70 s. What is
the displacement the cheetah covered at that velocity?
Answer:
297.54mExplanation:
step one:
given data
velocity v=34.2m/s
time t= 8.7s
Step two
Required is the distance the cheetah has covered on the condition
we know that speed= distance/time
make distance subject of formula we have
distance= velocity *time
distance= 34.2*8.7
distance = 297.54m
Therefore the displacement the cheetah covered at that velocity
is 297.54m
Two blocks with different masses are dropped, hitting the ground with the same velocity. Which of the following is true?
They have same change in velocity but different changes in kinetic energy
The lighter object started at a smaller height.
The heavier object started at a smaller height
They started at the same height
They have same change in kinetic energy but different changes in velocity
Answer: • They have same change in velocity but different changes in kinetic energy
•They started at the same height.
Explanation:
First and foremost, we need to note that both balls have thesame acceleration due to gravity and due to this, even though they've different masses, they'll fall at same speed.
Also, since kinetic energy that's, the energy relating to motion of a mass, us dependent on mass and speed, their kinetic energy will be different.
Therefore, based in the explanation, the correct options are:
• They have same change in velocity but different changes in kinetic energy
•They started at the same height.
which of the following elements is the most reactive? Chlorine Bromine Fluorine Helium
Answer:
Fluorine is the most reactive
Explanation:
Among the halogens, fluorine, chlorine, bromine, and iodine, fluorine is the most reactive one. It forms compounds with all other elements except the noble gases helium (He), neon (Ne) and argon (Ar), whereas stable compounds with krypton (Kr) and xenon (Xe) are formed.
A repeated back and forth or up and down motion is called a
Answer:
A vibration is a repeated back-and-forth or up-and-down motion.
Explanation:
Waves carry energy through empty space or through a medium without transporting matter.
A ball is thrown vertically upward with an initial velocity of 23 m/s. What are its position and velocity after 2 s?
Answer:
The position of the ball after 2 s is 26.4 mThe velocity of the ball after 2 s is 3.4 m/sExplanation:
Given;
initial velocity of the ball, u = 23 m/s
time of motion, t = 2 s
The position of the ball after 2 s is given by;
h = ut - ¹/₂gt²
h = (23 x 2) - ¹/₂ x 9.8 x 2²
h = 46 - 19.6
h = 26.4 m
The velocity of the ball after 2 s is given by;
v² = u² + 2(-g)h
v² = u² - 2gh
v² = 23² - (2 x 9.8 x 26.4)
v² = 529 - 517.44
v² = 11.56
v = √11.56
v = 3.4 m/s
How much work would be done on a particle with 5.0 C of charge on it if it moved from an equipotential line at 5.5 volts to another equipotential line at 3.5 volts?
Answer:
10J
Explanation:
In this question we have the following information
The charge of the particle is q = 5 C
The equipotenetial level is V1 = 5.5 v
and also the
equipotenetial level is V2 = 3.5 v
So we calculate the
work done W=q x (v1-v2)
workdone = 5 x (5.5-3.5)
= 5x2
=10 J
Workdone = 10 J
So we conclude that the workdone on a particle with these information is 10j
The x and y coordinates of a particle at any time t are x = 5t - 3t2 and y = 5t respectively, where x and y are in meter and t in second. The speed of the particle at t = 1 second is
Answer:
[tex]v=\sqrt{26}~m/s[/tex]
Explanation:
Parametric Equation of the Velocity
Given the position of the particle at any time t as
[tex]r(t) = (x(t),y(t))[/tex]
The instantaneous velocity is the first derivative of the position:
[tex]v(t)=(v_x(t),v_y(t))=(x'(t),y'(t))[/tex]
The speed can be calculated as the magnitude of the velocity:
[tex]v=\sqrt{v_x^2+v_y^2}[/tex]
We are given the coordinates of the position of a particle as:
[tex]x=5t-3t^2[/tex]
[tex]y=5t[/tex]
The coordinates of the velocity are:
[tex]v_x(t)=(5t-3t^2)'=5-6t[/tex]
[tex]v_y(t)=(5t)'=5[/tex]
Evaluating at t=1 s:
[tex]v_x(1)=5-6(1)=-1[/tex]
[tex]v_y(1)=5[/tex]
The velocity is:
[tex]v=\sqrt{(-1)^2+5^2}[/tex]
[tex]v=\sqrt{1+25}[/tex]
[tex]\mathbf{v=\sqrt{26}~m/s}[/tex]
A projectile is shot straight up from the earth's surface at a speed of 11,000 km/hr. How high does it go? ________km?
Taken from "Physics for Scientists and Engineers by Randall D. Knight 2nd Edition. Chapter 13 #34. There is an answer in the database already, but I do not understand it.
Answer:
476.35 km
Explanation:
The following data were obtained from the question:
Initial velocity (u) = 11000 km/hr
Final velocity (v) = 0 km/hr (at maximum height)
Acceleration due to gravity (g) = 9.8 m/s²
Maximum height (h) = ?
Next, we shall convert 9.8 m/s² to km/hr². This is illustrated below:
1 m/s² = 12960 km/hr²
Therefore,
9.8 m/s² = 9.8 m/s² × 12960 km/hr² / 1 m/s²
9.8 m/s² = 127008 km/hr²
Thus, 9.8 m/s² is equivalent to 127008 km/h²
Finally, we shall determine the maximum height reached by the projectile.
This is illustrated below:
Initial velocity (u) = 11000 km/hr
Final velocity (v) = 0 km/hr (at maximum height)
Acceleration due to gravity (g) = 127008 km/hr²
Maximum height (h) = ?
v² = u² – 2gh (since the projectile is going against gravity)
0² = 11000² – (2 × 127008 × h)
0 = 121×10⁶ – 254016h
Collect like terms
0 – 121×10⁶ = – 254016h
– 121×10⁶ = – 254016h
Divide both side by – 254016
h = – 121×10⁶ / – 254016
h = 476.35 km
Thus, the maximum height reached by the projectile is 476.35 km
What is the current in the wire now?
Answer:
220v
Explanation:
Sorry, the question is incomplete
Answer:
on the potential difference applied and on the resistance of the wire.
Explanation:
Ohms law state that the current through a conductor between two points is directly proportional to the potential difference across the two points. Imtroducing the comstant of proportionality, the resistance, one arrives at the usual athematical equation that describes this relationship: I = V/R.