Answer:
312hz
Explanation:
thanks me later
Two particles are separated by 0.38 m and have charges of -6.25 x 10-°C and 2.91 x 10-°C. Use Coulomb's law to predict the force between the particles if the distance is cut in half. The equation for Coulomb's law is F = kqi 42, and the constant, k, equals 9.00 x 109 Nm2/C2 2
Answer:
-4.35 × 10^-6 N
Explanation:
i just answered it on ap3x :)
Which of the following is an instantaneous speed?
A: All of the above
B: 80 ft/s
C. 80 yds./min
D. 80 km/hr
Answer:
A: All of the above
Explanation:
The instantaneous speed of an object is simply the current seed of the object at any given time. The SI unit is m/S and it is a vector quantity.
Therefore, according to the given options, they all have SI units that are consistent with distance and time which makes them all an example of instantaneous speed.
According to the Law of Conservation of Energy, why does the first hill on a roller coaster always have to be the tallest of all the other hills?
a 150kg roller coaster SITTING ON THE TOP OF A 200M HILL HAS HOW MUCH POTETNTIAL ENERGY
Answer:
Epot = 294300 [J]
Explanation:
Potential energy is defined as the product of mass by height by gravitational acceleration. The height is measured with respect to the reference level. At this reference level the potential energy is equal to zero.
[tex]E_{pot}=m*g*h\\[/tex]
where:
m = mass = 150 [kg]
g = gravity acceleration [m/s²]
h = elevation = 200 [m]
[tex]E_{pot}=150*9.81*200\\E_{pot}=294300 [J][/tex]
Do scientific theories have math in them?
Answer:
Yes of course science has maths
A 20-kg object sitting at rest is struck elastically in a head-on collision with a 10-kg object initially moving at 3.0 m/s. Find the final velocity
Answer:
1 m/s
Explanation:
Using law of conservation of momentum
m1v1 + m2v2 = (m1 + m2)vf , where
m1 = mass of object at rest, 20 kg
v1 = initial velocity of object at rest, 0 m/s
vf = final velocity of the bodies
m2 = mass of object in motion, 10 kg
v2 = initial velocity of object in motion, 3 m/s
On substituting, we have
(20 * 0) + (10 * 3) = (20 + 10) vf
0 + 30 = 30 vf
vf = 30 / 30
vf = 1 m/s
Therefore, the velocity of the bodies after hitting each other is 1 m/s
what does the circulatory system consist CLASS 7
Answer:
the circulatory system consists of heart, blood, blood vessels.
hope this answer will help you
30 points? I have no clue
Answer:
The second graph, B
Explanation
A child blows a leaf from rest straight up in the air. the leaf has a constant upward acceleration of magnitude 1.0 m by s square. how much time does it take the leaf to displace 1.0m upwards?
Answer:
√2
Explanation:
From the question, we're given that the
Acceleration of the leaf is 1 m/s²
Change in displacement of the leaf is 1 m/s.
Again, from the question, we can tell that the initial velocity u = 0, since the object starts at rest
Now, to solve this, we don't the equation of motion to ur
S = ut + 1/2at², substituting the whole parameters, we then have
1 = 0 * t + 1/2 * 1 * t²
1 = 1/2 * t²
t²/2 = 1
t² = 2
t = √2 seconds
Therefore the time it takes the leaf to dislodge is 2 seconds
A student is performing an experiment that involves the charge on a metal sphere that is attached to a charged electroscope. A charged rod is brought near the sphere without touching it. As a result the leaves of the electroscope separate more. The rod is then removed, and the leaves return to their initial separated position. The student repeats the procedure, but this time the electroscope is grounded and the ground is removed before the rod is removed from near the sphere. The leaves again end up separated. What can be concluded about the charge on the separated leaves of the electroscope
Answer:
The leaves have a charge in each experiment, but the sign of the charge cannot be determined.
Explanation:
In the first experiment, A charged rod is brought near the sphere without touching it. As a result the leaves of the electroscope separate more.
Thus indicates that there are charges involved. Now, like charges would repel like what is happening here but we don't know if they are both positive or negative because in both cases, they will still repel.
Now for the second experiment, electroscope is grounded and the ground is removed before the rod is removed from near the sphere. The leaves end up being separated again.
Similar to the first time, it's clear there are charges but the charges repel. Thus, they are the same sign charges but we don't know if they are both positive or negative.
Thus, in both cases we can conclude that the leaves have charges but we don't know their signs.
What is the strength of the electric field in a region where the electric potential is constant?
Answer:
Where the electric potential is constant, the strength of the electric field is zero.
Explanation:
As a test charge moves in a given direction, the rate of change of the electric potential of the charge gives the potential gradient whose negative value is the same as the value of the electric field. In other words, the negative of the slope or gradient of electric potential (V) in a direction, say x, gives the electric field (Eₓ) in that direction. i.e
Eₓ = - dV / dx ----------(i)
From equation (i) above, if electric potential (V) is constant, then the differential (which is the electric field) gives zero.
Therefore, a constant electric potential means that electric field is zero.
List the four outer planets from smallest to largest.
WILL MARK BRAINLIEST! ASAP
Answer: Mercury, Mars, Venus, Earth, Neptune, Uranus, Saturn, and Jupiter.
Explanation:
That's all of the planets if you need them. Hope this helps!
A spacecraft on its way to Mars has small rocket engines mounted on its hull; one on its left surface and one on its back surface. At a certain time, both engines turn on. The one on the left gives the spacecraft an acceleration component in the x direction of ax = 5.10 m/s2, while the one on the back gives an acceleration component in the ydirection of ay = 7.30 m/s2. The engines turn off after firing for 675 s, at which point the spacecraft has velocity components of vx = 3630 m/s and vy = 4276 m/s.What was the magnitude and direction of the spacecraft's initial velocity, before the engines were turned on?Express the direction as an angle measured counterclockwise from the x -axis.
Answer:
v₀ = 677.94 m / s , θ = 286º
Explanation:
We can solve this exercise using the kinematic expressions, let's work on each axis separately.
X axis
has a relation of aₓ = 5.10 m / s², the motor is on for a time of t = 675 s, reaching the speed vₓ = 3630 m / s, let's use the relation
vₓ = v₀ₓ + aₓ t
v₀ₓ = vₓ - aₓ t
let's calculate
v₀ₓ = 3630 - 5.10 675
v₀ₓ = 187.5 m / s
Y Axis
[tex]v_{y}[/tex] = v_{oy} - a_{y} t
v_{oy} = v_{y} - a_{y} t
let's calculate
v_{oy} = 4276 - 7.30 675
v_{oy} = -651.5 m / s
we can give the speed starts in two ways
a) v₀ = (187.5 i ^ - 651.5 j ^) m / s
b) in the form of module and angle
Let's use the Pythagorean theorem
v₀ = [tex]\sqrt{v_{ox}^{2} + v_{oy}^{2} }[/tex]
v₀ = [tex]\sqrt{187.5^{2} +651.5^{2} }[/tex]
v₀ = 677.94 m / s
we use trigonometry
tan θ = [tex]\frac{v_{oy} }{v_{ox} }[/tex]
θ = tan⁻¹ \frac{v_{oy} }{v_{ox} }
θ = tan⁻¹ ([tex]\frac{-651.5}{187.5}[/tex])
θ = -73.94º
This angle measured from the positive side of the x-axis is
θ‘ = 360 - 73.94
θ = 286º
define alpha and beta
alpha is the excess return on an investment after adjusting for market related volatility and random fluctuations.
beta is a measure of volatility relative to a benchmark ,such as the S&P 500.
Explanation:
alpha and beta are two different parts of an equation used to explain the performance of stocks and investments funds. But in maths alpha and beta is the Greek alphabet
A truck covers 40.0 m in 9.50 s while uniformly slowing down to a final velocity of 2.75 m/s.
a. Find its original speed.
b. Find its acceleration.
Explanation:
Given that,
Distance covered, d = 40 m
Time, t = 9.5 s
Final velocity, v = 2.75 m/s
(a) Let u be the original speed of the truck. We can find it using first equation of motion.
[tex]v=u+at\\\\2.75=u+2.75\times 9.5\\\\2.75-26.125=u\\\\u=-23.375\ m/s[/tex]
(b) Acceleration = rate of change of velocity
[tex]a=\dfrac{v-u}{t}\\\\a=\dfrac{2.75-(-23.375)}{9.5}\\\\=2.75\ m/s^2[/tex]
So, the original speed is -23.375 and acceleration is 2.75 m/s².
You work at a garden store for the summer. You lift a bag of fertilizer with a force of 112 N, and it moves upward with an acceleration of 0.790 m/s^2.
a. What is the mass of the fertilizer bag?
b. How much does the fertilizer bag weigh?
Given :
Force provided, F = 112 N.
Acceleration of the bag, a = 0.79 m/s².
To Find :
a. What is the mass of the fertilizer bag?
b. How much does the fertilizer bag weigh?
Solution :
We know, force is given by :
F = ma
m = F/a
m = 112/0.79 kg
m = 141.77 kg
Now, weight is given by :
W = mg
W = 141.77 × 9.8 N
W = 1389.35 N
Therefore, the mass of fertilizer bag is 141.77 kg and weight us 1389.35 N.
A 15.0kg block is dragged over a rough, horizontal surface by a 70.0-N force acting at 20.0 degrees above the horizontal. The block is displaced 5.0 m, and the coefficient of kinetic friction is 0.3.
a. Draw a free body diagram. Draw your coordinate system and label the axes.
b. Calculate the work done on the block by the 70 N force.
c. Calculate the work done on the block by the normal force.
d. Calculate the work done on the block by the gravitational force.
e. Calculate the work done on the block by the force of friction.
Answer:
[tex]W=70 * 5cos20 = 328.89 J[/tex]
[tex]W_n = 0[/tex]
[tex]W_g=0[/tex]
[tex]W_f= -184.59J[/tex]
Work done is 0
Explanation:
From the question we are told that
Weight of block =15.0kg
Force acting on the block = 70.0N
At an angle of 20 degree
Displacement of block is 5m
Coefficient of kinetic friction 0.3
b) Generally work done by force is give by [tex]W=fdcos \theta[/tex]
therefore
[tex]W=70 * 5cos20 = 328.89 J[/tex]
c) there is no work done by the normal force in this scenario because
normal force in this case is perpendicular to the displacement of the motion
[tex]W_n = 0[/tex]
d) The displacement in the vertical direction is 0
Therefore the gravitational work done is 0 [tex]W_g=0[/tex]
e)Generally in finding work done by friction we first find frictional force
Mathematically the equation for frictional force is given [tex]f = \alpha N[/tex]
Given that
[tex]N=mg-Fsin20[/tex]
[tex]N= 15.0*9.8 - 70 sin20[/tex]
[tex]N=123 N[/tex]
[tex]f=0.3* 123.06 = 36.92N[/tex]
Mathematically solving to get work done by frictional force [tex]W_f[/tex]
[tex]W_f= -fd\\W_f = -36.92 * 5[/tex]
[tex]W_f= -184.59J[/tex]
the frictional force work done is [tex]W_f= -184.59J[/tex]
The energy of a photon is ________ proportional to its wavelength.
Answer:
The amount of energy is directly proportional to the photon's electromagnetic frequency and thus, equivalently, is inversely proportional to the wavelength. The higher the photon's frequency, the higher its energy. Equivalently, the longer the photon's wavelength, the lower its energy.
Explanation:
Plz mark brainliest thanks
A 2000 kg car moving at 100 km/h crosses the top of a hill with a radius of curvature of 100 m. What is the normal force exerted by the seat on the driver if the mass of the driver is 60 kg?
Answer:
The normal force the seat exerted on the driver is 125 N.
Explanation:
Given;
mass of the car, m = 2000 kg
speed of the car, u = 100 km/h = 27.78 m/s
radius of curvature of the hill, r = 100 m
mass of the driver, = 60 kg
The centripetal force of the driver at top of the hill is given as;
[tex]F_c = F_g - F_N[/tex]
where;
Fc is the centripetal force
[tex]F_g[/tex] is downward force due to weight of the driver
[tex]F_N[/tex] is upward or normal force on the drive
[tex]F_N = F_g-F_c\\\\F_N = mg - \frac{mv^2}{r} \\\\F_N = (60 \times 9.8) -\frac{60 \ \times \ 27.78^2 \ }{100} \\\\F_N = 588 \ N - 463 \ N\\\\F_N = 125 \ N[/tex]
Therefore, the normal force the seat exerted on the driver is 125 N.
The normal force exerted by the seat on the driver if the mass of the driver is 60 kg is of 125 N.
Given data:
The mass of car is, m' = 2000 kg.
The speed of car is, v = 100 km/h = 100 × 5/18 = 27.77 m/s.
The radius of curvature of path is, r = 100 m.
The mass of driver is, m = 60 kg.
In this case, the normal force on the driver is equal to the difference between weight of the driver and the centripetal force on the driver. Then the expression is given as,
[tex]N'= W - F\\\\N '= mg-\dfrac{mv^{2}}{r}[/tex]
Solving as,
[tex]N' = (60 \times 9.8)-\dfrac{60 \times 27.77^{2}}{100}\\\\N' = 125\;\rm N[/tex]
Thus, we can conclude that the normal force exerted by the seat on the driver if the mass of the driver is 60 kg is of 125 N.
Learn more about the centripetal force here:
https://brainly.com/question/14249440
In introductory physics laboratories, a typical Cavendish balance for measuring the gravitational constant G uses lead spheres with masses of 1.60 kg and 16.0 g whose centers are separated by about 3.30 cm. Calculate the gravitational force between these spheres, treating each as a particle located at the center of the sphere.
Answer:
The value is [tex]F = 1.568 *10^{-9} \ N[/tex]
Explanation:
From the question we are told that
The mass of the first lead sphere is [tex]m = 1.60 \ kg[/tex]
The mass of the second lead sphere is [tex]M = 16 \ g = 0.016 \ kg[/tex]
The separation between masses is [tex]r = 3.30 \ cm = 0.033 \ m[/tex]
Generally the gravitational force between each sphere is mathematically represented as
[tex]F = \frac{G * m * M }{r^2 }[/tex]
Here G is the gravitational constant with value [tex]G = 6.67 *10^{-11 } \ m^3 \cdot kg^{-1} \cdot s^{-2}[/tex]
[tex]F = \frac{6.67 *10^{-11 } * 1.60 * 0.016 }{0.033^2 }[/tex]
=> [tex]F = 1.568 *10^{-9} \ N[/tex]
Two identical blocks fall a distance H. One falls directly down, the other slides down a frictionless incline. Which has the larger speed at the bottom?
The question incomplete, the complete question is;
Two identical blocks fall a distance H. One falls directly down, the other slides down a frictionless incline.
(A)the one falling vertically
(B)the one on the incline
(C)Both have the same speed.
(D)cannot be determined
Answer:
(C)Both have the same speed.
Explanation:
When we consider the question closely, we will discover that an object falling down a frictionless incline is comparable to an object falling freely under gravity.
In both instances, the acceleration of objects is just the same irrespective of mass.
Hence, the object falling vertically and the object sliding down a frictionless plane will have the same speed at the bottom.
BERE
Which describes the positions on a horizontal number line?
0
O All points to the left of one are positive.
O All points to the right of one are positive.
O All points to the left of zero are negative.
O All points to the right of zero are negative.
Mark this and return
Save and Exit
Next
Submit
Answer:
All points to the left of zero are negative
Explanation:
Answer:
C
Explanation:
on edge
A hockey player strikes a puck that is initially at rest. The force exerted by the stick on the puck is 975 N, and the stick is in contact with the puck for 0.0049 s.
(a) What is the impulse imparted by the stick to the puck.
___________ kg m/s
(b) What is the speed of the puck (m= 1.67 kg)just after it leaves the hockey stick?
____________ m/s
Explanation:
Given that,
The force exerted by the stick on the puck is 975 N
The stick is in contact with the puck for 0.0049 s
Initial speed of the puck, u = 0 (at rest)
(a) We need to find the impulse imparted by the stick to the puck.
Impulse = Force × time
J = 4.7775 kg-m/s
(b) Mass of the puck, m = 1.76 kg
We need to find the speed of the puck just after it leaves the hockey stick.
Let the speed be v.
As impulse is equal to the change in momentum.
[tex]J=m(v-u)\\\\4.7775=1.67(v-0)\\\\v=\dfrac{4.7775}{1.67}\\\\v=2.86\ m/s[/tex]
So, when the puck leaves the hockey stick its speed is 2.86 m/s.
A tennis ball is hit with a vertical speed of 10 m/s and a horizontal speed of 30 m/s. How far will the ball travel horizontally before landing?
a. 10 m
b. 20 m
c. 40 m
d. 60 m
e. 80 m
Answer:
D) 60 m
Explanation:
We can use the constant acceleration equation that contains displacement, initial velocity, acceleration, and time. We want to solve for the time that the ball was in the air first.
Δx = v_i * t + 1/2at²Let's use this equation in terms of the y-direction.
Δx_y = (v_i)y * t + 1/2a_y * t²The vertical displacement will be 0 meters since the ball will be on the floor. The initial vertical velocity is 10 m/s, the vertical acceleration is g = 10 m/s², and we are going to solve for time t.
Let's set the upwards direction to be positive and the downwards direction to be negative. We must use -g to be consistent with our other values.
Plug the known values into the equation.
0 m = 10 m/s * t + 1/2(-10 m/s²) * t²Simplify the equation.
0 = -10t + 5t² 0 = 5t² - 10tFactor the equation.
0 = 5t(t - 2)Solve for t by setting both factors to 0.
5t = 0t - 2 = 0We get t = 0, t = 2. We must use t = 2 seconds because it is the only value for t that makes sense in the problem.
Now that we have the time that the ball was in the air, we can use the same constant acceleration equation to determine the horizontal displacement of the tennis ball. We will use this equation in terms of the x-direction:
Δx = v_i * t + 1/2at² Δx_x = (v_i)x * t + 1/2a_x * t²Plug the known values into the equation.
Δx_x = 30 m/s * 2 sec + 1/2(0 m/s²) * (2 sec)²We can eliminate the right side of the equation since anything multiplied by 0 outputs 0.
Δx_x = 30 * 2 Δx_x = 60The horizontal displacement of the ball is 60 meters. Therefore, the answer is D) 60 m.
Why do we perform stork stand test
Answer:
umm becuase it is a test and you need them
Explanation:
A student lifts a box of books 2 meters with a force of 45 N. He then carries the box 10 meters to the living room. What is the total amount of work done in this situation?
PLEASE ANSWER FAST
Answer:
90J
Explanation:
The only time work is being done is when he lifts the box off the ground. Therefore, using the work formula, 2 x 45, you get 90J. Hope this helps someone.
In schematic diagrams, currents are indicated using arrows. What do the arrows indicate? a) the direction of motion of the electrons b) the direction of the current vector c) the direction of motion of the charge carriers d) the direction that positive charge carriers would move e) nothing; they are just a convenient drawing tool
Answer:
D
Explanation:
The direction that positive charge would move
Which
type of energy transformation is taking place when natural gas is used to heat water?
O chemical energy into thermal energy
thermal energy into mechanical energy
mechanical energy into electromagnetic energy
electromagnetic energy into chemical energy
Answer:
O chemical energy into thermal energy
Hope this helped!!
Answer:
chemical energy into thermal energy
Explanation:
If a 0.750M solution exerts an osmotic pressure of 22.5atm, what must be the temperature (in Kelvin) of the solution
Answer:
T = 365.58 K
Explanation:
Given that,
The concentration of solution, C = 0.750M
Osmotic pressure, P = 22.5 atm
We need to find the temperature of the solution.
The formula for the osmotic pressure is given by :
[tex]P=CRT[/tex]
Where
R is gas constant, [tex]R=0.08206\ L\ atm/mol-K[/tex]
[tex]T=\dfrac{P}{CR}\\\\=\dfrac{22.5}{0.75\times 0.08206}\\\\=365.58\ K[/tex]
So, the temperature of the solution is 365.58 K.
a bird of mass 2350g is flying at a height of 20.0m above the ground with a speed of 10m/s2 calculate it's potential energy
Explanation:
p.e =mgh
given: m=2350g=2.35kg h=20 g=9.8m/s
p.e=mgh
=2.35kg×20.0m×9.8
=460.6j
I am not sure