Answer:
100 N
Explanation:
The following data were obtained from the question:
Mass (m) = 2 Kg
Velocity (v) = 5 m/s
Radius (r) = 50 cm
Force (F) =.?
Next, we shall convert 50 cm to metre (m). This can be obtained as follow:
100 cm = 1 m
Therefore,
50 cm = 50 cm × 1 m / 100 cm
50 cm = 0.5 m
Therefore, 50 cm is equivalent to 0.5 m.
Finally, we shall determine the magnitude of the net force on the ball by using the following formula:
F = mv²/r
Mass (m) = 2 Kg
Velocity (v) = 5 m/s
Radius (r) = 0.5 m
Force (F) =.?
F = mv²/r
F = 2 × 5²/ 0.5
F = 2 × 25/ 0.5
F = 50 / 0.5
F = 100 N
Therefore, the magnitude of the net force on the ball is 100 N.
The magnitude of the net force on the ball will be "100 N".
Force and speedAccording to the question,
Mass, m = 2 kg
Velocity, v = 5 m/s
Radius, r = 50 cm or,
= 50 × [tex]\frac{1}{100}[/tex]
= 0.5 m
We know the relation,
Force, F = [tex]\frac{mv^2}{r}[/tex]
By substituting the values, we get
= [tex]\frac{2\times (25)^2}{0.5}[/tex]
= [tex]\frac{2\times 25}{0.5}[/tex]
= [tex]\frac{50}{0.5}[/tex]
= 100 N
Thus the above response is appropriate.
Find out more information about force here:
https://brainly.com/question/6504879
a car is moving eastward and speeding up. the momentum of the car is
A toy rocket is launched vertically from ground level (y = 0 m), at time t = 0.0 s. The rocket engine provides constant upward acceleration during the burn phase. At the instant of engine burnout, the rocket has risen to 98 m and acquired a velocity of The rocket continues to rise in unpowered flight, reaches maximum height, and falls back to the ground. The upward acceleration of the rocket during the burn phase is closest to:
29 m/s2
31 m/s2
33 m/s2
30 m/s2
32 m/s2
Explanation:
The question is incomplete. Here is the complete question.
A toy rocket is launched vertically from ground level (y = 0 m), at time t = 0.0 s. The rocket engine provides constant upward acceleration during the burn phase. At the instant of engine burnout, the rocket has risen to 98 m and acquired a velocity of 30m/s. The rocket continues to rise in unpowered flight, reaches maximum height, and falls back to the ground. The upward acceleration of the rocket during the burn phase is closest to...
Given
initial velocity of rocket u = 0m/s
final velocity of rocket = 30m/s
Height reached by the rocket = 98m
Required
upward acceleration of the rocket
Using the equation of motion below to get the acceleration a:
[tex]v^2 = u^2+2as\\30^2 = 0^2 + 2(a)(98)\\900 = 196a\\a = \frac{900}{196}\\a = 4.59m/s^2[/tex]
Hence upward acceleration of the rocket during the burn phase is closest to 5m/s²
Note that the velocity used in calculation was assumed.
A plane flying horizontally at a speed of 40.0 m/s and at an elevation of 160 m drops a package. Two seconds later it drops a second package. How far apart will the two packages land on the ground?
Answer:
Package 1 will land at 228.0 m, package 2 will land at 308.0 m, and the distance between them is 80.0 m.
Explanation:
To find the distance at which the first package will land we need to calculate the time:
[tex] Y_{f} = Y_{0} + V_{0y}t - \frac{1}{2}gt^{2} [/tex]
Where:
Y(f) is the final position = 0
Y(0) is the initial position = 160 m
V(0y) is initial speed in "y" direction = 0
g is the gravity = 9.81 m/s²
t is the time=?
[tex] 0 = 160 m + 0t - \frac{1}{2}9.81 m/s^{2}t^{2} [/tex]
[tex] t = \sqrt{\frac{2*160 m}{9.81 m/s^{2}}} = 5.7 s [/tex]
Now we can find the distance of the first package:
[tex] X_{1} = V_{0x}*t = 40.0 m/s*5.7 s = 228.0 m [/tex]
Then, after 2 seconds the distance traveled by plane is (from the initial position):
[tex] X_{p} = V_{0x}*t = 40.0 m/s*2 s = 80.0 m [/tex]
Now, the distance of the second package is:
[tex] X _{2} = X_{1} + X_{p} = 228.0 m + 80.0 m = 308.0 m [/tex]
The distance between the packages is:
[tex] X = X_{2} - X_{1} = 308.0 - 228.0 m = 80.0 m [/tex]
Therefore, package 1 will land at 228.0 m, package 2 will land at 308.0 m and the distance between them is 80.0 m.
I hope it helps you!
Suppose that your favorite AM radio station broadcasts at a frequency of 1600 kHz. What is the wavelength in meters of the radiation from the station
Answer:
187.5
Explanation:
Colored lights are called additive colors. Why do you think this is so?
What is the speed of a wave that has a frequency of 2,400 Hz and a wavelength of 0.75
Answer:
1800 m/s
Explanation:
The equation is v = fλ
λ= 0.75
f = 2400 Hz
V = 2400 × 0.75
V = 1800 m/s
[ you did not give units for wavelength, I assumed it would be m/s]
Which property of a substance can be determined using a pH indicator.
Answer:
Acidity
Explanation:
A pH indicator measures how acidic or basic a substance is.
Hope this helped :)
WHAT IS ACCURACY, PRECISION, AND REPRODUCIBILITY? AND WHY ARE THEY SO NECESSARY IN CONDUCTING/DESIGNING EXPERIMENTS? 30 POINTS AND WILL MARK BRAINLIEST
Answer:
Explanation:
Accuracy can be said to mean the degree to which the particular result of a measurement, or calculation, and even possibly specification agrees or is the same with respect to the correct value or an established standard. Succinctly put, it's is how close a value is to the actual value it ought to be.
Precision on the other hand, is a change in a measurement, or calculation, and even as far as specification, much especially as represented by the number of digits that has been established. In other words, it is the proximity of two or more measurements with respect to one another.
Reproducibility occurs when a measurement(for example) is made by another person, or a different instrument is used. Yet, the same values are obtained.
They are very important in design because they account for very important part of an experiment. Neglecting these quantities means exposing an instrument to unknown danger to the factory and even the personnels.
Also, neglect in taking note of accuracy, precision and reproducibility can lead to poor data processing and even human errors.
Now, vote brainliest, will you? :)
All biomes don’t have the same level of biodiversity. What seems to be the optimal conditions for high biodiversity?
Answer:
See the answer below
Explanation:
The optimal conditions for high biodiversity seem to be a warm temperature and wet climates.
The tropical areas of the world have the highest biodiversity and are characterized by an average annual temperature of above 18 [tex]^oC[/tex] and annual precipitation of 262 cm. The areas are referred to as the world's biodiversity hotspots.
Consequently, it follows logically that the optimal conditions for high biodiversity would be a warm temperature of above 18 [tex]^oC[/tex] and wet environment with annual precipitation of not less than 262 cm.
The variation in temperature and precipitation across biomes can thus be said to be responsible for the variation in the level of biodiversity in them.
A particle moves along a path described by y=Ax^3 and x = Bt, where tt is time. What are the units of A and B?
Answer:
In a nutshell, units of A and B are [tex]\frac{1}{[l]^{2}}[/tex] and [tex]\frac{[l]}{[t]}[/tex], respectively.
Explanation:
From Dimensional Analysis we understand that [tex]x[/tex] and [tex]y[/tex] have length units ([tex][l][/tex]) and [tex]t[/tex] have time units ([tex][t][/tex]). Then, we get that:
[tex][l] = A\cdot [l]^{3}[/tex] (Eq. 1)
[tex][l] = B\cdot [t][/tex] (Eq. 2)
Now we finally clear each constant:
[tex]A = \frac{[l]}{[l]^{3}}[/tex]
[tex]A = \frac{1}{[l]^{2}}[/tex]
[tex]B = \frac{[l]}{[t]}[/tex]
In a nutshell, units of A and B are [tex]\frac{1}{[l]^{2}}[/tex] and [tex]\frac{[l]}{[t]}[/tex], respectively.
A motorboat is a lot heavier than a pebble. Why does the boat float?
Answer:
The boat has more buoyancy
Explanation:
Weight of a body becomes greater at the pole than that at the equator . why ?
It takes a minimum distance of 48.96 m to stop a car moving at 12.0 m/s by applying the brakes (without locking the wheels). Assume that the same frictional forces apply and find the minimum stopping distance when the car is moving at 25.0 m/s.
Answer:
102 m
Explanation:
Given that It takes a minimum distance of 48.96 m to stop a car moving at 12.0 m/s by applying the brakes (without locking the wheels). Assume that the same frictional forces apply and find the minimum stopping distance when the car is moving at 25.0 m/s.
Let the stopping distance be equal to S.
According to the definition of speed,
Speed = distance / time.
make time the subject of the formula
Time = distance / speed
then, the equivalent time is:
48.96 / 12 = S / 25
Cross multiply
12S = 48.96 x 25
12S = 1224
S = 1224 / 12
S = 102 m
Therefore, the stopping distance is 102 m
ionic bonds form when electrons?
Answer:
when the electron transferred permanently to another atom
dimensional formula of stress is same as
Dimensional formula of stress is same as pressure.
A wire loop with 3030 turns is formed into a square with sides of length ss . The loop is in the presence of a 1.20 T1.20 T uniform magnetic field B⃗ B→ that points in the negative yy direction. The plane of the loop is tilted off the x-axisx-axis by θ=15∘θ=15∘ . If i=1.10 Ai=1.10 A of current flows through the loop and the loop experiences a torque of magnitude 0.0256 N⋅m0.0256 N⋅m , what are the lengths of the sides ss of the square loop, in centimeters?
Answer:
2.59 cm
Explanation:
The torque τ on a current carrying loop of wire is given by τ = NiABsinθ where N = number of turns of loop, i = current in loop, A = area of loop and B = magnetic field.
Now, given that τ = 0.0256 Nm, i = 1.10 A, B = 1.20 T,N = 30 and since the loop is tilted 15° off the x-axis and the magnetic field points in the negative y- direction, the angle between the normal to the loop and the magnetic field is thus 90° - 15° = 75°. So, θ = 75°.
We now find the area of the loop A from
τ = NiABsinθ
A = τ/NiBsinθ
substituting the values of the variables, we have
A = 0.0256 Nm/30 × 1.10 A × 1.20 T × sin75°
A = 0.0256 Nm/38.25
A = 6.69 × 10⁻⁴ m²
Since the loop is a square, with length of side L, its area A = L² and
L = √A
= √(6.69 × 10⁻⁴ m²)
= 2.59 × 10⁻² m
converting to cm, we have
L = 2.59 × 10⁻² m × 100 cm/m
L = 2.59 cm
So, the lengths of sides of the loop is 2.59 cm
m_Cu * sh_CuA system consists of a copper tank whose mass is 13 kilogram , 4 kilogram of liquid water, and an electrical resistor of negligible mass. The system is insulated on its outer surface. Initially, the temperature of the copper is 27 degC and the temperature of the water is 50 degC . The electrical resistor transfers 100 kilojoule to the system. Eventually the system comes to equilibrium. Determine the final equilibrium temperature, in ∘C.
Answer:
T₂ = 49.3°C
Explanation:
Applying law of conservation of energy to the system we get the following equation:
Energy Supplied by Resistor = Energy Absorbed by Tank + Energy Absorbed by Water
E = mC(T₂ - T₁) + m'C'(T'₂ - T'₁)
where,
E = Energy Supplied by Resistor = 100 KJ = 100000 J
m = mass of copper tank = 13 kg
C = Specific Heat of Copper = 385 J/kg.°C
T₂ = Final Temperature of Copper Tank
T₁ = Initial Temperature of Copper Tank = 27°C
T'₂ = Final Temperature of Water
T'₁ = Initial Temperature of Water = 50°C
m' = Mass of Water = 4 kg
C' = Specific Heat of Water = 4179.6 K/kg.°C
Since, the system will come to equilibrium finally. Therefor: T'₂ = T₂
Therefore,
(100000 J) = (13 kg)(385 J/kg.°C)(T₂ - 27°C) + (4 kg)(4179.6 J/kg.°C)(T₂ - 50°C)
100000 J = (5005 J/°C)T₂ - 135135 J + (16718.4 J/°C)T₂ - 835920 J
100000 J + 135135 J + 835920 J = (21723.4 J/°C)T₂
(1071055 J)/(21723.4 J/°C) = T₂
T₂ = 49.3°C
Sometimes we will want to write vectors in terms of a coordinate grid. To show a vector points
horizontally (along the x-axis), place an x after the magnitude of the vector. To show a vector point
vertically (along the y-axis), place a y after the magnitude.
4) Using the notation above,
i. How would you write d1?
ii. How would you write d2?
iii. How would you write dtotal?
d1=(0,5)
d2=(5,5)
Answer:
III) [tex]d_{1}+ d_{2}=d_{t}[/tex]
Explanation:
I) coordinate (0,5) is the head for [tex]d_{1}[/tex] I will put the tail coordinate as (0,0) but it could be any other number in the x just not in the 5 with the the y being any other value.
II) coordinate (5,5) is the head for [tex]d_{2}[/tex] the tail needs to be in the head of [tex]d_{1}[/tex] being (0,5)
III) coordinates for [tex]d_{t}[/tex] is connecting the tail from [tex]d_{1}[/tex] and the head of [tex]d_{2}[/tex] making it (0,0)[tex](tail)[/tex] and (0,5)[tex](head)[/tex] and is written as [tex]d_{1}+ d_{2}=d_{t}[/tex]
(i) using coordinate grid notation to represent d₁, d₁ = 5y
(ii) using coordinate grid notation to represent d₂, d₂ = 5x + 5y
(ii) The sum of d₁ and d₂ is written as 5x + 10y
In order to show the horizontal direction of a vector, we will place x after the magnitude of the vector.
Also, to show the vertical direction of a vector, we will place a y after the magnitude of the vector.
(i) Using coordinate grid to represent d₁ = (0, 5)
[tex]d_1 = 0(x) + 5(y)\\\\d_1 = 5y[/tex]
(ii) Using coordinate grid to represent d₂ = (5, 5)
[tex]d_2 = 5x + 5y[/tex]
(iii) The total vector is written as;
[tex]d_1 + d_2 = 5y + (5x + 5y)\\\\d_1 + d_2 = 5y + 5x + 5y\\\\d_1 + d_2 = 5x + 10y[/tex]
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In a pickup game of dorm shuffleboard, students crazed by final exams use a broom to propel a calculus book along the dorm hallway. If the 4.7 kg book is pushed from rest through a distance of 0.85 m by the horizontal 42 N force from the broom and then has a speed of 1.0 m/s, what is the coefficient of kinetic friction between the book and floor
Answer:
μ_k = 0.851
Explanation:
We are given;
Mass of book; m_book = 4.7 kg
Horizontal force; F_horiz = 42 N
Distance; d = 0.85 m
Speed; v = 1 m/s
First of all let's find the acceleration using Newton's equation of motion;
v² = u² + 2ad
u is initial velocity and it's 0 m/s in this case.
Thus;
1² = (2 × 0.85)a
1 = 1.7a
a = 1/1.7
a = 0.5882 m/s²
Now, resolving forces along the vertical direction, we have;
W - N = 0
Thus,W = N
Where W is weight = mg and N is normal force
Thus; N = mg = 4.7 × 9.81 = 46.107 N
Now, resolving forces along the horizontal direction, we have;
F_horiz - ((μ_k)N) = ma
Where μ_k is coefficient of kinetic friction.
Thus;
42 - 46.107(μ_k) = 4.7 × 0.5882
42 - 46.107(μ_k) = 2.76454
μ_k = (42 - 2.76454)/46.107
μ_k = 0.851
An object from a certain height falls freely. which of the following happens PE and KE when the object is half on its way down
Answer:
A. Loses PE and gains KE
Explanation:
Statement is incomplete. Complete statement of problem is:
1. An object from a certain height falls freely. Which of the following happens to PE and KE when the object is half on its way down?
A. Loses PE and gains KE
B. Gains PE and loses KE
C. Loses both PE and KE
D. Gains both PE and KE
If we neglect the effects of any conservative force, the application of the Principle of Energy Conservation is reduced to a sum of gravitational potential ([tex]U_{g}[/tex]) and translational kinetic energies, measured in joules. That is:
[tex]U_{g,1}+K_{1} =U_{g,2}+K_{2}[/tex] (Eq. 1)
Let assume that an object falls from a height [tex]h[/tex] with a speed of zero. By definitions of gravitational potential and translational kinetic energies the previous is expanded. If final height is the half of initial value, then:
[tex]m\cdot g\cdot h = 0.5\cdot m\cdot g\cdot h +K_{2}[/tex] (Eq. 1b)
[tex]K_{2} = 0.5\cdot m\cdot g\cdot h[/tex]
[tex]K_{2} = 0.5\cdot U_{g,1}[/tex]
In a nutshell, translational kinetic energy is increased at the expense of diminishing gravitational potential energy. The correct answer is A.
Consider the force field and circle defined below. F(x, y) = x2 i + xy j x2 + y2 = 121 (a) Find the work done by the force field on a particle that moves once around the circle oriented in the clockwise direction.
Answer: the work done by the force is 0
Explanation:
F (x², xy)
121 = 11²
so R = x² + y² = 11²
p = x². Q = xy
Δp/Δy = 0, ΔQ/Δx
using Green's theorem
woek = c_∫F.Δr = R_∫∫ ΔQ/Δx - Δp/Δy) ΔA
= (x² + y² = 121)_∫∫ yΔA
now let x = rcosФ, y = rsinФ
ΔA = rΔrΔФ
so r from 0 to 11
and Ф from 0 to 2π
= 0_∫^2π 0_∫^11 rsinФ × rΔrΔФ
= 0_∫^2π SinФΔФ 0_∫^11 r²Δr
= [ -cosФ]^2π_0 [r³/3]₀¹¹ = ( -cos2π + cos0) (11³/3) = 0
therefore the work done by the force is 0
A 0.75 m3 rigid tank initially contains air whose density is 1.18 kg/m3. The tank is connected to a high-pressure supply link through a valve. The valve is opened, and air is allowed to enter the tank until the density in the tank rises to 4.95 kg/m3. Determine, in kg, the mass of air that has entered the tank..
Answer:
2.83kg
Explanation:
Answer:
2.83kg
Explanation:
Given
initial density = 1.18 kg/m3
Final density in the tank = 4.95 kg/m3.
Let us write the mass balance first.
Change in the mass of the system=mass of the air entering the system - Mass of air out the system
Mass that entered= M2 - M1
But DENSITY= MASS/ VOLUME
Mass= volume × Density
We can expressed the mass in terms of density since density is given in the question.
Mass that entered= (volume × density)2 - ( volume × density)1
= (V ρ)2 - (V ρ)1
But V1= V2 the volume remains the same
= ( ρ2 - ρ1)v
= (4.95 kg/m3 - 1.18 kg/m3) 0.75 m3
= 3.77× 0.75
= 2.8275kg
Mass that entered= 2.83kg
therefore, mass of air that has entered the tank= 2.83kg
Suppose astronomers discover a radio message from a civilization whose planet orbits a star 35 light-years away. Their message encourages us to send a radio answer, which we decide to do. Suppose our governing bodies take 2 years to decide whether and how to answer. When our answer arrives there, their governing bodies also take two of our years to frame an answer to us. How long after we get their first message can we hope to get their reply to ours? Enter your answer in years.
Answer:
The duration is [tex]T =72 \ years /tex]
Explanation:
From the question we are told that
The distance is [tex]D = 35 \ light-years = 35 * 9.46 *10^{15} = 3.311 *10^{17} \ m [/tex]
Generally the time it would take for the message to get the the other civilization is mathematically represented as
[tex]t = \frac{D}{c}[/tex]
Here c is the speed of light with the value [tex]c = 3.08 *10^{8} \ m/s[/tex]
=> [tex]t = \frac{3.311 *10^{17} }{3.08 *10^{8}}[/tex]
=> [tex]t = 1.075 *10^9 \ s[/tex]
converting to years
[tex]t = 1.075 *10^9 * 3.17 *10^{-8} [/tex]
[tex]t = 1.075 *10^9 * 3.17 *10^{-8} [/tex]
[tex]t = 34 \ years [/tex]
Now the total time taken is mathematically represented as
[tex]T = 2* t + 2 + 2[/tex]
=> [tex]T = 2* 34 + 2 + 2[/tex]
=> [tex]T =72 \ years /tex]
cameron drives his car 15 km north. He stops for lunch and then drives 12 km south. What is his displacement?
Answer:
Displacement is 3 km North
Explanation:
I NEED HELP PLEASEE ITS AN ECONOMICS QUESTION ABOVE
Answer:
I believe the answer is Property taxes
Explanation:
Answer: I'm pretty sure property taxes
Explanation:
In the absence of a gravitational field, you could determine the mass of an object (of unknown composition) by:
A) applying a known force and measuring it's acceleration.
B) measuring the volume.
C) weighing it.
Answer:
A) By applying a known force, and measuring it's acceleration.
Explanation:
This is actually something that astronauts do in space as a mathmatical exercise when calculating the mass of an object since F = m × a.
Once the force, and acceleration are applied, the only unknown is the mass which can be solved by dividing force over acceleration. This is because inertial mass is equal to gravitational mass.
Part D
Next, we'll examine magnetic force. Bring the ends of your two magnets together. Explore the three
possible combinations. In two of the combinations, the two ends are the same. In one combination, the
two ends are different. Describe the force you feel in each combination
Answer:
i. The magnetic force of repulsion.
ii. The magnetic force of attraction.
Explanation:
A magnet is a material that has the attraction and repulsion capability. Magnets has two poles, north and south, thus would attract or repel another magnet in its neighborhood. It can either be a permanent or temporal magnet, and attracts ferrous metals.
i. In the case of two combinations where two ends are the same, it could be observed that the two ends (poles) repels each other. Thus since like poles repels, magnetic force of repulsion is felt.
ii. In the case of one combination in which the two ends are different, the two ends (poles) attract. Since unlike poles attracts, magnetic force of attraction is observed.
Which statement best describes an atom? (2 points)
оа
Protons and neutrons grouped in a specific pattern
Ob
Protons and electrons spread around randomly
ос
A group of protons and neutrons that are surrounded by electrons
Od
A ball of electrons and neutrons surrounded by protons
Answer:
A group of protons and neutrons that are surrounded by electrons I think that's the answer...
Explanation:
A negative charge -Q is placed inside the cavity of a hollow metal solid. The outside of the solid is grounded by connecting a conducting wire between it and the earth. Is any excess charge induced on the inner surface of the metal? Is there any excess charge on the outside surface of the metal? Why or why not? Would someone outside the solid measure an electric field due to the charge -Q? Is it reasonable to say that the grounded conductor has shielded the region outside the conductor from the effects of the charge -Q? In principle, could the same thing be done for gravity? Why or why not?
Answer:
a) + Q charge is inducce that compensates for the internal charge
b) There is no excess charge on the external face q_net = 0
c) E=0
Explanation:
Let's analyze the situation when a negative charge is placed inside the cavity, it repels the other negative charges, leaving the necessary positive charges to compensate for the -Q charge. The electrons that migrated to the outer part of the sphere, as it is connected to the ground, can pass to the earth and remain on the planet; therefore on the outside of the sphere the net charge remains zero.
With this analysis we can answer the specific questions
a) + Q charge is inducce that compensates for the internal charge
b) There is no excess charge on the external face q_net = 0
c) If we create a Gaussian surface on the outside of the sphere the net charge on the inside of this sphere is zero, therefore there is no electric field, on the outside
d) If it is very reasonable and this system configuration is called a Faraday Cage
e) We cannot apply this principle to gravity since there are no particles that repel, in all cases the attractive forces.
what chemical diverged from trees
a resin and turpentine
b sodium
c lead
d marcotting
Answer:
a. resin and turpentine
Explanation:
The chemicals that diverged from trees are resin and turpentine.
Resin are produced by special cells in trees, most times we see them when a tree is damaged or cut. They are usually derived from pines and firs.
Turpentine is obtained by distilling resin.
Turpentine has an antiseptic property that has different uses. They are used as cleansing agents and for producing sanitary materials.