Answer:
the charge of each small sphere, which is + Q / 2
Explanation:
For this exercise we must use the fact that a charged object induces charges on nearby bodies
Induced charge comes from the fact that charges of the same sign repel and charges of different signs attract.
In this case the large, fixed ball with a -Q charge induces a positive charge in the nearest part and the negative charges are repelled to the furthest point, but the net charge on the metallic sphere remains zero. It should be emphasized that since the charges are of different signs, there is an attractive force between the two spheres.
This first metallic sphere now has a negative charge on the back, this charge induces a positive charge on the second sphere, as the charges are of a different sign, they attract each other, which is why the force is attractive.
When the first sphere stops the second sphere hits it, at this moment the charge of the two spheres is equal, therefore the induced charge in the two spheres is + Q. When the two spheres are separated, the charge on each of them is half, that is, the sphere has a charge + Q / 2 and the second sphere has a charge + Q / 2.
Therefore the first sphere is subjected to two forces: an attractive force with the large sphere of charge -Q and a repulsive force with the second sphere of charge + Q / 2.
So the first sphere must approach the big ball and the second sphere must move away from the big sphere.
This is the process of the movement of this exercise, unfortunately the statements with which to compare this process do not appear, but one of the most common questions of what is the charge of each small sphere, which is + Q / 2
Help please Subject (general psychology)
Answer:
plz give the question... floor the proper answer
Explanation:
plz mark me add the brainliest
thank
me and follow ma
Determine the distance S to which the 90kg painter can climb without causing the 4m ladder to slip at it's lower end A. The top of the 15kg ladder has a small roller, and at the ground the coefficient of static friction is 0.25. The mass center of the painter is directly above her feet.
Explanation:
...................,..........
Newton's first law states that at rest the sum of the forces acting on a body is zero
The distance S to which the painter can climb is approximately 2.55 meters
The reason the above value for the distance is correct is as follows:
The given parameter are;
The mass of the painter, m₁ = 90 kg
Length of the ladder, l = 4 m
The mass of the ladder, m₂ = 15 kg
The coefficient pf static friction, μ = 0.25
Friction force, [tex]\mathbf{F_f}[/tex] = Normal reaction, N × μ
Normal reaction (flat surface), N = Weight of painter + Weight of ladder
N = m₁·g + m₂·g
Where;
g = The acceleration due to gravity, g = 9.81 m/s²
N = 90 kg × 9.81 m/s² + 15 kg × 9.81 m/s² = 1,030.05 N
∴ [tex]\mathbf{F_f}[/tex] = 1,030.05 N × 0.25 = 257.5125 N
Friction force, [tex]\mathbf{F_f}[/tex] = 257.5125 N (Horizontal force)
At equilibrium, the sum of the horizontal forces = 0
∴ Fₓ + [tex]\mathbf{F_f}[/tex] = 0
Fₓ = [tex]-F_f[/tex] = -257.5125 N (Acting opposite to the direction of the friction force)
The height of the top of the ladder from the ground, h = √(4²-1.5²) = √(13.75)
Taking moment about point A gives;
Clockwise moment = Anticlockwise moment
Fₓ × h = m₁·g·s·cos(θ) + m₂·g·(l/2)·cos(θ)
Therefore;
Fₓ × √(13.75) m = 90 kg × 9.81 m/s² × s × (1.5/4) + 15 kg × 9.81 m × 2 m × (1.5/4)
Fₓ × √(13.75) m = 331.0875 N × s + 110.3625 J
257.5125 N × √(13.75) m = 331.0875 N × s + 110.3625 J
s = (257.5125 N × √(13.75) m - 110.3625 J)/(331.0875 N) ≈ 2.55 m
The distance to which the 90 kg painter can climb without causing the 4 m ladder to slip at its lower end, A, distance S ≈ 2.55 meters
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3. A block of mass m is suspended by strings as shown in the figure. The tension in the horizontal string
is 36 N. The angle 0 is 60°. Find the mass of the block and the tension forces in string A and string
B
I am having trouble trying to answer this problem. Any help?
Answer:
mass of the block is 7.2Kg
Tension on string A = 72N
Tension on the string with an angle is = 108N
Explanation:
The Horizontal string will have the cos component of the force...
Hence, mg cos60 = 36
by keeping acceleration due to gravity = 10 m/s^2, we get,
m = 7.2Kg
The tension in String A = mg = 7.2*10 = 72N
since the whole string holding the block is pulled by another string with 36 N
the resultant force will be the tension of the string with an angle.
Hence, 72 +36 = 108 N
Please let me know whether I got this right or not...
Question
In order for work to be done, what three things are necessary
it is required that an object weighing 8N be moved up ward at an acceleration of 2m/s what is the upward force needed
Answer:
F = 1.6 N
Explanation:
Newton's Second Law of Motion states that whenever an unbalanced force is applied on a body. It produces an acceleration in the body, in its own direction. The magnitude of the force is given by the following formula:
F = ma
where,
F = Force Required to Move the Object Upward = ?
m = mass of object
a = upward acceleration = 2 m/s²
but, we know that weight of object is given as:
Weight = mg
8 N = m(9.8 m/s²)
m = 0.8 kg
using values in the equation:
F = (0.8 kg)(2 m/s²)
F = 1.6 N
What is the speed of an object that travels 20 meters in 4 seconds?
Given:-
Distance covered by the particle (s) = 20 mTime taken (t) = 4 secondsTo Find: Speed of it.
We know,
v = s/t
where,
v = Speed,s = Distance &t = Time taken.Putting the values,
v = (20 m)/(4 s)
→ v = 5 m/s (Ans.)
The speed of an object will be "5 m/s".
The given values are:
Distance,
d = 20 mTime,
t = 4 secondsAs we know, the formula:
→ [tex]Distance (d)= Speed (v)\times Time (t)[/tex]
or,
→ [tex]Speed (v) = \frac{Distance (d)}{Time (t)}[/tex]
By putting the values, we get
→ [tex]= \frac{20}{4}[/tex]
→ [tex]= 5 \ m/s[/tex]
Thus the above answer is correct.
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An object is released from rest from a top of a building 90 meters high. Neglect air friction.
What is the volume of its acceleration?
Calculate the time it takes to reach the floor?
With what velocity does it reach the floor?
How fast is it moving when it is 50 meter above the floor?
Answer:
1. a = 9.8 m/s²
2. t = 4.28 s
3. Vf = 42 m/s
4. Vf = 28 m/s
Explanation:
1.
Since, the body is under free fall motion. Therefore, the value of its acceleration shall be equal to the acceleration due to gravity.
a = 9.8 m/s²
2.
The time taken by the ball to reach the ground can be calculated by using second equation of motion:
h = Vi t + (1/2)gt²
where,
h = height = 90 m
Vi = initial velocity = 0 m/s
t = time taken = ?
Therefore,
90 m = (0 m/s)t + (1/2)(9.8 m/s²)t²
t = √(18.36 s²)
t = 4.28 s
3.
In order to find final velocity we use first equation of motion:
Vf = Vi + gt
Vf = 0 m/s + (9.8 m/s²)(4.28 s)
Vf = 42 m/s
4.
when the ball is at height of 50 m, it means it has covered:
h = 90 m - 50 m = 40 m
we use third equation of motion at this point:
2gh = Vf² - Vi²
(2)(9.8 m/s²)(40 m) = Vf² - (0 m/s)²
Vf = √(784 m²/s²)
Vf = 28 m/s
How many Newtons of force does it take to move a 53 kilogram object?
You stand at the top of a cliff while your friend stands on the ground below you. You drop a ball from rest and see that it takes 1.7s for the ball to hit the ground below. Your friend then picks up the ball and throws it up to you, such that it just comes to rest in your hand. What is the speed with which your friend threw the ball
Answer:
16.7m/s
Explanation:
Using v= u+ at
V = final velocity. U = initial velocity
So since the two movements are the same they will return with the same speed they went up
Then final velocity v= 0
So
0:= u +-9.8 x 1.7
U= 16.7m/s
A person drops two objects from the same height. One object weighs 15 N.
and the other weighs 10 N. How does the mass of the objects relate to the
force of gravity on them?
A. The 10 N object has the same mass as the 15 N object.
B. The 15 N object has twice the mass of the 10 N object.
C. The 15 N object has more mass than the 10 N object.
D. The 10 N object has more mass than the 15 N object.
Answer:
C
Explanation:
The 15N object has more mass than the 10N object.
A 15 N object has more mass than the 10 N object due to their weight differences.
What is Weight?
Weight is defined as the force exerted by the earth on an object on the surface of earth. The direction of this force is directed towards the center of earth. Mathematically -
W = mg
Where -
m is the mass of body
g is the acceleration due to gravity
Weight can also be written as -
W = ρVg [ρ is density and V is volume]
Given is a person who dropped two objects from the same height. One object weighs 15 N and the other weighs 10 N.
We can write -
Weight of Object 1 [W1] = 15 N
Weight of Object 2 [W2] = 10N
We know that -
W = m x g
We can write it as -
m = W/g
Therefore -
Mass of Object 1 [M1] = 15/9.8 = 1.53 Kg
Mass of Object 2 [M2] = 10/9.8 = 1.02 Kg
Clearly -
M1 > M2
So, the 15 N object has more mass than the 10 N object.
Therefore, a 15 N object has more mass than the 10 N object due to their weight differences.
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27.
In a graph showing how temperature of a material changes over time, temperature
change is the:
A. dependent variable
C. variable with the smallest range
B. independent variable
D. variable with the largest range
Answer:
A. Dependent variable
Explanation:
In a graph showing how temperature of a material changes over time, temperature is taken on y-axis and time is taken on x-axis. It shows how temerature altered as the time changes.
In temperature-time graph, temperature is directly dependent on time. With the increase in time, temperature rises, falls or remains constant. It implies that temperature is dependent variable that depend on time.
Hence, the correct option is (A).
Which kinds of objects emit visible light in the electromagnetic spectrum? A. all objects B. radioactive objects C. relatively cold objects D. relatively hot objects
Answer:
c
Explanation:
Answer:
the answer is c i just did it
Explanation:
How does deploying slats or slots on an airfoil affect CL max and Stall AOA?
a) Stall AOA increases, CL max increases.
b) Stall AOA decreases, CL max increases.
c) Stall AOA increases, CL max decreases.
d) Stall AOA decreases, CL max decreases.
Answer: a. Stall AOA increases, CL max increases.
Explanation:
Deploying slats or slots on an airfoil affect CL max and Stall AOA by increasing Stall AOA and also increasing CL max.
It should be noted that the coefficient of lift CL would rise through the use of slots due to increase in boundary later energy. The slots also leads to delay of stall by through increase in AOA.
Question 3
A box is being pulled by a rope that makes a 25 degree angle with the ground. The
pulling force is 100.0 N
along the rope. Find the horizontal and vertical components of the force vector.
A. 90.63 N, 42.26 N
B. 86.87 N, 32.17 N
C. 60.87 N, 75,63 N
D. 80.9 N, 45.5 N
What is the answer A,B,C or D?
Answer:
A
Explanation:
Suppose that a certain battery produces a voltage of 1.55V without a load connected (open circuit) and a current of 500mA when shorted. According to these specifications, if we were to model this as an ideal voltage source in series with a source resistance, what should the source voltage and internal resistance be? Justify your answer
Answer:
Explanation:
Let the internal resistance be r .
Since in open circuit the volt is 1.55 V , this will be the source voltage .
Source voltage = 1.55
If external resistance be R .
1.55 / (R + r ) = .500
R + r = 3.1 ohm
So sum of internal resistance and external resistance will be 3.1 ohm.
Question: A) A car is driving too fast on a flat non-banked curve. The car cannot stay on the road. What path will the car take as it leaves the road? See Image. (a) (b) B) The driver wants to go faster around the same curve mentioned above and not run off the road. He adds sandbags to his car to make it weigh more so that the friction force between the tires and the road will be increased. Will this work? Explain your answer.
Explanation:
(a) The car will leave tangentially to the road, so it will take path d.
(b) Sum of forces on the car in the vertical direction:
∑F = ma
N − mg = 0
N = mg
Sum of forces on the car in the centripetal direction:
∑F = ma
Nμ = m v²/r
mgμ = m v²/r
gμ = v²/r
v = √(grμ)
The maximum velocity is independent of mass, so adding sandbags will not work.
If you walk 5 km north and then 12 km east. Your resultant displacement is____.a. magnitude: 1.1km; direction: 53.1 degrees east of north.b. magnitude: 2.00; direction: 53.1 degrees east of north.c. magnitude: 2.00; direction: 36.9 degrees east of north.d. magnitude: 1.1km; direction: 36.9 degrees east of north.
Answer:
Resultant = 13km
Direction = 67.38° East of North
Explanation:
Given the following :
5km North ; 12km East
Resultant Displacement (r) :
r² = 5² + 12²
r² = 25 + 144
r² = 169
r = √169
r = 13
Direction:
Tangent = opposite / Adjacent
Tanθ = opposite / Adjacent
Opposite = 12 ; adjacent = 5
Tanθ = (12/5)
Tanθ = 2.4
θ = tan^-1(2.4)
θ = 67.38° east of north
In which direction does centripetal force act on an object? in the opposite direction of the tangential speed of the object in the same direction of the tangential speed of the object toward the outside of the circle in which the object is moving toward the center of the circle in which the object is moving
Answer:
towards the center of the circle
(which appears to be the last option in your list of possible answers)
Explanation:
The centripetal force points always towards the center of the circle described by the object moving.
The direction of centripetal force acting on an object is always towards the center of the circle in which the object is moving.
The given problem is based on the concept and fundamentals of centripetal force. When an object is known to move around a circular path, then there is a force comes to play acting towards the centre of circular path, known as centripetal force.
In other words, the centripetal force is also known termed as the force that tends to act on a body moving in a circular path and is directed towards the centre around which the body is moving.For example - Motion of merry go round and the motion of roller skates on the rink floor are the examples of centripetal force, where the force always acts in the direction, center of the circular track.Thus, we can conclude that the direction of centripetal force acting on an object is always toward the center of the circle in which the object is moving.
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The mere exposure effect is
What are the two main types of star clusters?
Answer:
Open and globular
Explanation:
The speed of x-rays is 300 000. m/s. In scientific notation, and with the correct number of significant figures, this speed is… *
Answer:
3.0 x 10 and the exponent on ten is 5.
Explanation:
You are moving the decimal point over to the left 5 times, making the exponent a positive 5 and then you would put a decimal after 3. Making it 3.0 x 10, and the exponent 5.
The speed of x-rays in scientific notation is 3.0 x 10⁶ m/s.
The given parameters;
speed of the x-ray, v = 300,000 m/s
The standard form of the given number represents the scientific notation of the number.
To represent a number in standard form, multiply the number in terms of power or exponent of 10. The value of the exponent represents the number of available zeros present.In scientific notation the speed of the x-ray can be expressed as follows;
300,000 m/s = 3.0 x 10⁶ m/s
Thus, the speed of x-rays in scientific notation is 3.0 x 10⁶ m/s.
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Suppose a fast-pitch softball player does a windmill pitch, moving her hand through a circular arc with her arm straight. She releases the ball at a speed of 21.3 m/s21.3 m/s (about 47.6 mph47.6 mph ). Just before the ball leaves her hand, the ball's radial acceleration is 871 m/s2871 m/s2 . What is the length of her arm from the pivot point at her shoulder?
Answer:
R = 52.08 cm
Explanation:
given data
tangential velocity v = 21.3 m/s
radial acceleration aR = 871 m/s²
solution
we will get here length of her arm from the pivot point at her shoulder so
here we know aR = [tex]\frac{v^2}{R}[/tex] ........................1
so put here value and we get
871 = [tex]\frac{21.3^2}{R}[/tex]
R = 0.5208840 m
R = 52.08 cm
Answer:
Explanation:
linear speed of ball v = 21.3 m /s
radial acceleration = 871 m /s²
radial acceleration = v² / R
where R is length of arm which acts as radius of circular path of the ball .
Putting the values
21.3² / R = 871
R = .52 m.
4. What is the momentum of a 78.0 N bowling ball with a velocity of 8.00 m/s?
Momentum = (mass) x (speed)
We don't know the mass of the bowling ball, but we know how much it weighs. So if we knew the acceleration of gravity in the place where the ball is, we could calculate its mass. The question doesn't tell us what planet the bowling ball is on. So if we only use the information given in the question, there's no way to find the answer, and we're stuck here.
But I urgently need the points, so I'm going to go out on a limb here and make a big assumption: I'll assume that the alley where this ball is rolling and bowling is located on planet Earth, where the acceleration of gravity is around 9.8 m/s² everywhere on the planet's surface. NOW I can go ahead and answer the question that I just invented.
weight = m g
Mass = (weight) / (g)
Mass = (78.0 N) / (9.8 m/s²)
Mass = 7.96 kilograms
Momentum = (mass) x (speed)
Momentum = (7.96 kg) x (8.00 m/s)
Momentum = 63.7 kg-m/s
Elmer Fudd is walking at 2.0 m/s. If he walks for 2.0 minutes, how far does he walk?
(Can someone help with the answer due in 5min)
Answer:
240 meters
Explanation:
He is walking 2 meters per second, and there are 60 seconds per minute. There are 120 seconds in two minutes. So, 120 x 2 = 240. That's your answer!
blutions math practice
Maddie Hicks: Attempt 1
07.5 ml
Question 4 (1 point)
You have a stock bottle of 3mM tetrodotoxin solution, but your protocol states that
you need 6ml of 1mM tetrodotoxin (MW=319.27 g/mol). How much of your 3mM
stock solution (liquid) should you use to make the 1mM solution?
Answer:
[tex]V_1=2mL[/tex]
Explanation:
Hello,
In this case, considering this as a dilution problem, the first step here is to consider that the moles of tetrodotoxin remains the same and just the volumes and concentrations are modified from the initial stock (1) and the required dilution (2):
[tex]V_1M_1=V_2M_2[/tex]
Whereas V is referred to volume and M to molar concentration, in this case in mM. In such a way, solving the V1 as the volume of the 3-mM solution we obtain:
[tex]V_1=\frac{V_2M_2}{M_1} \\\\V_1=\frac{6mL*1mM}{3mM} \\\\V_1=2mL[/tex]
It means you need 2 mL of the 3-mM solution.
Best regards.
A car moves from rest with an acceleration of 0.2ms
Find its velocity when it has moved a
distance of 50m
From the problem we notice that the question
does not involver. Therefore, the required equation
is)
Explanation:
Given:
Δx = 50 m
v₀ = 0 m/s
a = 0.2 m/s²
Find: v
The problem does not involve time. Therefore, the required equation is:
v² = v₀² + 2aΔx
Plug in values and solve:
v² = (0 m/s)² + 2 (0.2 m/s²) (50 m)
v = 4.47 m/s
The velocity of the car when it has moved a distance of 50 meters is approximately 4.47 m/s.
To find the velocity of the car when it has moved a distance of 50 meters, we can use the following equation of motion:
v² = u² + 2as
Where:
v = final velocity
u = initial velocity (which is 0 in this case as the car starts from rest)
a = acceleration
s = distance traveled
Plugging in the values into the equation:
v² = 0² + 2(0.2)(50)
v² = 0 + 20
v² = 20
Taking the square root of both sides:
v = √20
v = 4.47 m/s
Therefore, the velocity of the car when it has moved a distance of 50 meters is approximately 4.47 m/s.
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Suppose a radio signal (light) travels from Earth and through space at a speed of 3 × 108/ (this is the speed of light in vacuum). How far (in meters) into space did the signal travel during the first 10 minutes?
Answer:
18*10^10 meters
Explanation:
V= d/t 10 mins = 600 seconds
3*10^8 = d/600s
(3*10^8)*(6*10^2) = d
d = 18*10^10 m
Calculate the length of segment RS with midpoint, M , if RM = 5x and MS = x + 12.
A. 3
B. 25
C. 30
D. 36
E. 15
Answer:
the full segment is: 30 units long
which coincides with answer C in your list
Explanation:
If M is the midpoint, then it divides the segment in two equal parts. Then, we can say that:
RM = MS (equality among the two parts of the divided segment)
replacing RM with "5 x" and MS with "x + 12", we get:
5 x = x + 12
solving for x:
5 x - x = 12
4 x = 12
then x = 3
With this info, we can calculate the length of each half of the segment and consequently its full length:
RM = 5 x = 5 (3) = 15
then the full segment is: 30 units long
11. एक समान चुम्बकीय क्षेत्र में लम्बवत प्रवेश करने वाले किसी आवेशित कण द्वारा प्राप्त वृत्तीय पथ की
7 त्रिज्या का सूत्र ज्ञात कीजिए।
2
Find the formula of radius obtained by a charged particle entering perpendicularly
in a uniform magnetic field.
Answer:
r=mv/BeExplanation:
If a positive charge enters a magnetic field at 90 degrees the charge is deflected in a circular path by a force that acts perpendicular to it in line with Flemings right-hand rule
to derive the radius of the path of the charge we apply
F= mv^2/r=Bev
where
m= mass of the electronic charge
e=charge
B=magnetic field
v=average speed
r=radius
rearranging we have
r=mv^2/Bev
r=mv/Be
The mass of a string is 20 g and it has a length of 3.2 m. Assuming that the tension in the string is 2.5 N, what will be the wavelength of a travelling wave that is created by a sinusoidal excitation of this string with a frequency of 20 Hz. Provide the wavelength in units of m. Please note: You do not include the units in your answer. Just write in the number.
Answer:
The wavelength of the wave is 1 m
Explanation:
Given;
mass of the string, m = 20 g = 0.02 kg
length of the string, L = 3.2 m
tension on the string, T = 2.5 N
the frequency of the wave, f = 20 Hz
The velocity of the wave is given by;
[tex]v = \sqrt\frac{T}{\mu} {}[/tex]
where;
μ is mass per unit length = 0.02 kg / 3.2 m
μ = 6.25 x 10⁻³ kg/m
[tex]v = \sqrt{\frac{T}{\mu} } \\\\v = \sqrt{\frac{2.5}{6.25*10^{-3}} } \\\\v = 20 \ m/s[/tex]
The wavelength of the wave is given by;
λ = v / f
λ = (20 m/s )/ (20 Hz)
λ = 1 m
Therefore, the wavelength of the wave is 1 m