Answer:
200 N
Explanation:
Applying,
The force a golie must exert on the ball is,
F = ma...................... Equation 1
Where m = mass of the ball, a = acceleration of the ball.
But,
a = Δv/t............... Equation 2
Where Δv = change in velocity, t = time.
Substitute equation 2 into equation 1
F = m(Δv/t)............... Equation 3
From the question,
Given: m = 0.8 g, t = 0.1 s, Δv = 25 m/s
Substitute these values into equation 3
F = 0.8×25/0.1
F = 200 N
Describe two 'beneficial' aspects from the last 10,000 years of humanity, as well as two 'dis-advantageous' aspects. Agriculture should be your guiding point here.
Answer:
Two beneficial aspects of the last 10,000 years of mankind have been the development of writing and the development of civilized communities.
Thus, writing has allowed the recording of the different historical events that have taken place during the development of humanity, setting precedents for subsequent generations, allowing documenting the progress that humanity has been achieving. On the other hand, the development of civilized communities around urban systems has allowed for greater economic and social development on the part of human groups, which through cooperation have been acquiring an increasingly higher quality of life.
On the other hand, two negative aspects of these years have been the world overpopulation and the pollution of the planet.
Thus, the amount of the world's population has been getting closer and closer to the limit of our planet's carrying capacity, which could lead humanity to a point where resources are definitely scarce, causing famines and various catastrophes. On the other hand, the pollution arising from the development of industrial societies has notably undermined the health of individuals and has damaged their standard of living, while causing irreparable damage to the natural environment.
effieiency of simple machine is always less than 100% why
Answer:
efficiency of a machine is less than 100% because some part is energy is utilized to overcome some opposing forces like friction which is wasted as heat ,sound energy etc
Explanation:
After You Read
Mini Glossary
atomic number: the number of protons in an atom of
an element
mass number: the sum of the number of protons and
neutrons in an atom
average atomic mass: the average mass of the element's
isotopes, weighted according to the abundance of each isotope
nuclear decay: a process that occurs when an unstable atomic
nucleus changes into another more stable nucleus by
emitting radiation
ion: an atom that is no longer neutral because it has gained or
lost electrons
radioactive: spontaneously emits radiation
isotope: an atom of the same element that has a different
number of neutrons
1. Review the terms and their definitions in the Mini Glossary. Write a sentence that
explains how to determine the number of neutrons in an isotope that has 6 protons and
a mass number of 13.
Answer:
protons+neutrons= mass number, so if the mass number is 13 and protons are 6 its 13-6=7 neutrons
Explanation:
mass number: the sum of the number of protons and
neutrons in an atom
this is key as it explains that protons+neutrons= mass number, so if the mass number is 13 and protons are 6 its 13-6=7 neutrons
Jesse drives 120km to a farm. His trip takes 2 1/2 hoursWhat is his speed?
Speed = distance / time
Speed = 120 km / 2 1/2 hours
Speed = 48 km per hour
People who do very detailed work close up, such as jewellers, often can see objects clearly at much closer distance than the normal 25 cm. a. What is the power in D of the eyes of a woman who can see an object clearly at a distance of only 8.5 cm
This question is incomplete, the complete question is;
People who do very detailed work close up, such as jewelers, often can see objects clearly at much closer distance than the normal 25 cm.
a) What is the power in D of the eyes of a woman who can see an object clearly at a distance of only 8.50 cm? (Assume the lens-to-retina distance is 2.00 cm.)
b) What is the size in mm of an image of a 8.00 mm object, such as lettering inside a ring, held at this distance? (Include the sign of the value in your answer.) __ mm
Answer:
1) the power in D of the eyes of a woman is 61.7647 D
2) the size in mm of an image of a 8.00 mm object is -1.882 mm
Explanation:
Given the data in the question;
a) power in D of the eyes of woman who can see an object clearly at a distance of only 8.5 cm and the lens-to-retina distance is 2.00 cm,
so
u = 8.5 cm = ( 8.5 / 100 )m = 0.085 m
v = 2.00 cm = ( 2 / 100 )m = 0.02 m
Now, we know that power of lens p = 1 / u + 1 / v
so we substitute
p = ( 1 / 0.085 ) + ( 1 / 0.02 )
p = 11.7647 + 50
p = 61.7647 D
Therefore, the power in D of the eyes of a woman is 61.7647 D
b) What is the size in mm of an image of a 8.00 mm object, such as lettering inside a ring, held at this distance? (Include the sign of the value in your answer.)
we know that;
m = -v / u
we substitute
m = -0.02 / 0.085
m = -0.2353
since H₀ = 8.0 mm
H[tex]_i[/tex] = m × H₀
H[tex]_i[/tex] = -0.2353 × 8.0
H[tex]_i[/tex] = -1.882 mm
the size in mm of an image of a 8.00 mm object is -1.882 mm
A 9.2 resistor and a 5.2 resistor are connected in series with a battery. The potential difference across the 5.2 resistor is measured as 12 V. Find the potential difference across the battery.
Answer:
V = 33.12 V
Explanation:
Given that,
A 9.2 resistor and a 5.2 resistor are connected in series with a battery.
The potential difference across the 5.2 resistors is measured as 12 V.
In series combination of resistors, the current is same throughout the circuit. Let I is the current. Using Ohm's law,
[tex]I=\dfrac{V}{R}[/tex]
Put all the values,
[tex]I=\dfrac{12}{5.2}\\\\I=2.30\ A[/tex]
The equivalent resistance is :
R = 9.2 + 5.2
= 14.4 ohms
Let V os the potential difference across the battery. So,
[tex]V=2.3\times 14.4\\\\V=33.12\ V[/tex]
So, the potential difference across the battery is equal to 33.12 V.
Last stage of a medium mass star
White Dwarf Stage
This ring is called a planetary nebula. when the last of the helium atoms in the core are fused into carbon atoms, the medium size star begins to die. Gravity causes the last of the star's matter to collapse inward and compact. This is the white dwarf stage.A mass m, which is connected to a spring of spring constant k, is released from x = A to perform
a simple harmonic motion. Another mass 2m, which is connected to another spring of the same
spring constant k, is also released from x = A to perform a simple harmonic motion. Compare the
values of total mechanical energy stored in these two spring-mass systems.
the boiling point of F2 much lower than the boiling point of NH3
Answer:yeah it A
Explanation:
A 50 kg child sits on the left side of the bathtub. A small toy boat of 0.5 kg is on the right side of the bathtub. Which part of the bathtub has the greatest pressure
Answer:
Option 2
Explanation:
The complete question is
A 50 kg child sits on the left side of the bathtub. A small toy boat of 0.5 kg is on the right side of the bathtub. Which part of the bathtub has the greatest pressure
TopBottomLeftRightSolution
It is the bottom of the bucket that will high pressure because of the additional weight of 50 Kg boy along with the weight of the water and the tub itself.
Pressure acts in the down ward direction and is equal to the force/weight divided by the area.
Hence, option 2 is correct
Parallel rays from a distant object are traveling in air and then are incident on the concave end of a glass rod with a radius of curvature of 15.0 cm. The refractive index of the glass is 1.50. What is the distance between the vertex of the glass surface and the image formed by the refraction at the concave surface of the rod? Is the image in the air or in the glass?
Answer:
the distance of image from the vertex is 45 cm and the image formed is in the glass.
Explanation:
distance of object, u = - infinity
radius of curvature, R = - 15 cm
refractive index, n = 1.5
Let the distance of image is v.
Use the formula
[tex]-\frac{n1}{u}+\frac{n2}{v}=\frac{n2- n1}{R}\\\\-\frac{1}{\infty }+\frac{1.5}{v}=\frac{1.5-1}{-15}\\\\v=45 cm[/tex]
The image is in the glass.
Iron is a solid phase of iron still unknown to science. The only difference between it and ordinary iron is that Iron forms a crystal with an fcc unit cell and a lattice constant . Calculate the density of Iron.
The question is incomplete. The complete question is :
Iron β is a solid phase of iron still unknown to science. The only difference between it and ordinary iron is that Iron β forms a crystal with an fcc unit cell and a lattice constant, a = 0.352 nm. Calculate the density of Iron β.
Solution :
The density is given by :
[tex]$\rho = \frac{ZM}{a^3N_0} \ \ g/cm^3$[/tex] ..................(i)
Here, Z = number of atoms in a unit cell
M = atomic mass
[tex]$N_0$[/tex] = Avogadro's number = [tex]$6.022 \times 10^{23}$[/tex]
a = edge length or the lattice constant
Now for FCC lattice, the number of atoms in a unit cell is 4.
So, Z = 4
Atomic mass of iron, M = 55.84 g/ mole
Given a = 0.352 nm = [tex]$3.52 \times 10^{-8}$[/tex] cm
From (i),
[tex]$\rho = \frac{ZM}{a^3N_0} $[/tex]
[tex]$\rho = \frac{4 \times 55.84}{(3.52 \times 10^{-8})^3 \times 6.022 \times 10^{23}} $[/tex]
[tex]$= 8.51 \ \ g \ cm ^{-3}$[/tex]
Therefore, the density of Iron β is [tex]$ 8.51 \ \ g \ cm ^{-3}$[/tex].
Fill in the blanks. Power station produce electricity at __________.
Answer:
heat
Explanation:
heat is the answer hdhhdhdhdhdhdhdhdhdhd
What is the period of revolution of a satellite with mass m that orbits the earth in a circular path of radius 7880 km (about 1500 km above the surface of the earth)
Answer:
Explanation:
For time period of revolution , the expression is as follows .
T² = 4π² R³ /GM , M is mass of the earth.
Putting the values
T² = 4π² (7880 x 10³)³ /(6.67 x 10⁻¹¹ )( 5.97 x 10²⁴ )
T² = 4.846 x 10⁷ s
T = 6.961 x 10³ s
= 6961 s
= 116 minutes .
Please helppppppp!!!!!!!!!!!!!!
Answer:
circuit breaker
Explanation:
A circuit breaker is a device used for electrical safety. It consists of a switch designed to protect an electrical circuit from damage that may result from heating due to overload in the circuit.
Its basic function is to interrupt current flow through its switch that consists of metal stripe which bends when it gets hot.
Fuse has similar action with circuit breaker, the only difference is that fuse can only be used once because it melts when it gets hot.
Therefore, the correct answer is "circuit breaker"
Two people that have identical weight are holding onto a massless pole while standing on horizontal frictionless ice. 1)If the guy on the left starts to pull on the pole, where do they meet
Answer:
Explanation:
From the missing image attached below, it is obvious that there no external force. This implies that they cannot change their position by merely just pulling the ropes. As a result, there will be no movement and no net force will exist.
So, if there is no external force;
The center of mass of the two people is:
[tex]X_{cm}= \dfrac{m_1x_1+m_2x_2}{m_1+m_2} \\ \\ X_{cm}= \dfrac{m(-3m)+m(+3m)} {m+m}\\ \\ X_{cm}= \dfrac{0}{2m} \\ \\ X_{cm} =0[/tex]
Thus, In the system, no movement occurs and all forces remain the same.
In some situations, matter demonstrates wave behavior rather than particle behavior. This is best illustrated by which phenomenon? A. Emission spectra of atoms B. Blackbody radiation C. Interference patterns of electrons D. Photoelectric effect
Answer:
In some situations, matter demonstrates wave behavior rather than particle behavior. This is best illustrated by which phenomenon is:
C. Interference patterns of electrons.
Answer:
C. Interference patterns of electrons
a body thrown vertically upwards from grounf with inital vel 40m/s then time taken by it to reach max hieght is?
Answer:
t = 4.08 s
Explanation:
if the body is thrown upward, it has negative gravity. Knowing through the International System that the earth's gravity is 9.8 m/s²
Data:
Vo = 40 m/sg = -9.8 m/s²t = ?Use formula:
[tex]\boxed{\bold{t=\frac{-(V_{0})}{g}}}[/tex]Replace and solve:
[tex]\boxed{\bold{t=\frac{-(40\frac{m}{s})}{-9.8\frac{m}{s^{2}}}}}[/tex][tex]\boxed{\boxed{\bold{t=4.08\ s}}}[/tex]Time taken by it to reach max height is 4.08 seconds.
Greetings.
Which statements are true of noble gases?
Check all that apply.
A. They are metalloids.
B. Their valence shells are full of electrons.
C. They are not very reactive.
D. All of the noble gases have at least two electron shielding layers.
Consider an electron confined in a region of nuclear dimensions (about 5 fm). Find its minimumpossible kinetic energy in MeV. Treat this problem as one-dimensional, and use the relativistic relationbetweenEandp. Give your answer to 2 significant figures. (The large value you will find is a strongargument against the presence of electrons inside nuclei, since no known mechanism could contain anelectron with this much energy.)
Answer:
39.40 MeV
Explanation:
Determine the minimum possible Kinetic energy
width of region = 5 fm
From Heisenberg's uncertainty relation below
ΔxΔp ≥ h/2 , where : 2Δx = 5fm , Δpc = hc/2Δx = 39.4 MeV
when we apply this values using the relativistic energy-momentum relation
E^2 = ( mc^2)^2 + ( pc )^2 = 39.4 MeV ( right answer ) because the energy grows quadratically in nonrelativistic approximation,
Also in a nuclear confinement ( E, P >> mc )
while The large value will portray a Non-relativistic limit as calculated below
K = h^2 / 2ma^2 = 1.52 GeV
After your school's team wins the regional championship, students go to the dorm roof and start setting off fireworks rockets. The rockets explode high in the air and the sound travels out uniformly in all directions. If the sound intensity is 1.62 10-6 W/m2 at a distance of 165 m from the explosion, at what distance from the explosion is the sound intensity half this value
Answer:
required distance is 233.35 m
Explanation:
Given the data in the question;
Sound intensity [tex]I[/tex] = 1.62 × 10⁻⁶ W/m²
distance r = 165 m
at what distance from the explosion is the sound intensity half this value?
we know that;
Sound intensity [tex]I[/tex] is proportional to 1/(distance)²
i.e
[tex]I[/tex] ∝ 1/r²
Now, let r² be the distance where sound intensity is half, i.e [tex]I[/tex]₂ = [tex]I[/tex]₁/2
Hence,
[tex]I[/tex]₂/[tex]I[/tex]₁ = r₁²/r₂²
1/2 = (165)²/ r₂²
r₂² = 2 × (165)²
r₂² = 2 × 27225
r₂² = 54450
r₂ = √54450
r₂ = 233.35 m
Therefore, required distance is 233.35 m
Air in a thundercloud expands as it rises. If its initial temperature is 292 K and no energy is lost by thermal conduction on expansion, what is its temperature when the initial volume has tripled
Answer:
Explanation:
It is a case of adiabatic expansion .
[tex]T_1V_1^{\gamma-1}=T_2V_2^{\gamma-1}[/tex]
T₁ , T₂ are initial and final temperature , V₁ and V₂ are initial and final volume.
Given ,
V₂ = 3 V₁ and T₁ = 292 . γ for air is 1.4 .
[tex]( 3 )^{\gamma-1}= \frac{292}{ T_2}[/tex]
[tex]( 3 )^{1.4-1}= \frac{292}{ T_2}[/tex]
1.552 = 292 / T₂
T₂ = 188 K .
Consider a solid sphere and a solid disk with the same radius and the same mass. Explain why the solid disk has a greater moment of inertia than the solid sphere, even though it has the same overall mass and radius.
Answer:
Explanation:
In a Solid sphere; the moment of inertia around its geometrical axis can be expressed by using the formula:
[tex]\mathtt{I_s = \dfrac{2}{5} M_s R^2_s}[/tex]
For the solid disk; the moment of inertia around the central axis is:
[tex]\mathtt{I_D= \dfrac{1}{2}M_DR_D^2}[/tex]
Suppose [tex]M_D = M_S[/tex]; then we can say both to be equal to M
As well as [tex]R_D = R_S[/tex]; then that too can be equal to R
Now;
[tex]\mathtt{I_s = \dfrac{2}{5} M R^2} --- (1)[/tex]
[tex]\mathtt{I_D= \dfrac{1}{2}MR^2}---(2)[/tex]
Multiplying equation (1) by 2, followed by dividing it by 2; we have:
[tex]\mathtt{I_s= \dfrac{2}{5}MR^2} \times \dfrac{2}{2}[/tex]
[tex]I_s = \dfrac{4}{5} \times \dfrac{1}{2}MR^2 \\ \\ I_s = \dfrac{4}{5}\times I_D \\ \\ I_s > I_D[/tex]
Keesha is looking at a beetle with a magnifying glass. She wants to see an upright, enlarged image at a distance of 25 cm. The focal length of the magnifying glass is +5.0 cm. Assume that Keesha's eye is close to the magnifying glass.
(a) What should be the distance between the magnifying glass and the beetle?
(b) What is the angular magnification?
Answer:
a) p = 4.167 cm, b) m = + 6
Explanation:
a) For this exercise we must use the equation of the constructor
[tex]\frac{1}{f} = \frac{1}{p} + \frac{1}{q}[/tex]
where f is the focal length, p and q are the distance to the object and the image, respectively
In this case the distance to the image q = 25 cm and the focal length is f = 5.0 cm
Since the object and its image are on the same side of the lens, the distance to the image by the sign convention must be negative.
[tex]\frac{1}{p } = \frac{1}{f} - \frac{1}{q}[/tex]
[tex]\frac{1}{p} = \frac{1}{5} - \frac{1}{-25}[/tex]
[tex]\frac{1}{ p}[/tex] = 024
p = 4.167 cm
b) angular magnification
m = h ’/ h = - q / p
m = - (-25) /4.167
m = + 6
the positive sign indicates that the image is straight and enlarged
Once a disk forms around a star, the process of planetary formation can begin. Rank the evolutionary stages for the formation of planets from earliest to latest.
a. Small clumps of matter stick together via the process of accrection to form plantesimals a few hundred kilometers in diameter
b. Dust keeps matter inside the disk cool long enough for planet formation to start
c. Planetisimals begin to accrete, forming protoplanets
d. Dust grains form condensation nuclei on which surrounding atoms condense to form small clumps of matter
e. A collection of a few planet-sized protoplanets remain in a fairly cleared out disk around the star
Answer: See explanation
Explanation:
The evolutionary stages for the formation of planets from earliest to latest will be:
1. Dust keeps matter inside the disk cool enough for planet formation to start
2. Dust grains form condensation nuclei on which surrounding atoms condense to form small clumps of matter.
3. Small clumps of matter stick together via the process of accretion to form planetesimals a few hundred kilometers in diameter.
4. Planetesimals begin to accrete, forming protoplanets.
5. A collection of a few planet-sized protoplanets remain in a fairly cleared out disk around the star
What is the momentum of a lab cart with a mass of 0.60 [kg] and a speed of 2.2 [m/s]?
-3.67 [kg m/s)
-0.27 [kg m/s]
-1.32 [kg m/s)
-0.82 [kg m/s)
Answer:
1.32kgm/s
Explanation:
use the formula: p=mv
The momentum of a lab cart = -1.32 kg m/s
What is momentum?It is measure of the inertia of a body/ object .It can be calculated by multiplying mass with velocity .
General formula for momentum = M = m * v
given
mass = 0.60 kg
speed = 2.2 m/s
velocity = - 2.2 m/s ( answer is in negative , since mass is a scaler quantity but velocity is a vector quantity hence , velocity can be negative )
momentum = mass * velocity
= 0.60 * (-2.2 ) = -1.32 kg m/s
The momentum of a lab cart =c) -1.32 kg m/s
learn more about momentum
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a) Find the current in the 1 Ω resistor.
b ) Find the current in the 8 Ω resistor.
c ) Find the current in the 5 Ω resistor.
PLEASE HELP I NEED THIS TODAY
Answer:
a)I=V/R
39.5 amp
Explanation:
because the voltage in serious with 1ohm resistor
A metallic circular plate with radius r is fixed to a tabletop. An identical circular plate supported from above by a cable is fixed in place a distance d above the first plate. Assume that d is much smaller than r. The two plates are attached by wires to a battery that supplies voltage V.
Required:
a. What is the tension in the cable?
b. Compute the energy stored in the electric field after the top plate was raised.
Answer:
A) F = V²E_o•πr²/2d²
B) U = E_o•Aπr²V²/2d
Explanation:
A) Since we have two circular plates, the formula for the electric field is expressed as;
E = V/d
Where;
V is voltage
d is distance
However, the net electric field produced is given by;
E' = V/2d
The tension in the cable can then be expressed as;
F = qE'
Where q is charge
Thus;
F = qV/2d - - - (eq 1)
We also know that;
C = q/V = E_o•A/d
A is area = πr²
Thus;
q/V = E_o•πr²/d
q = VE_o•πr²/d
Let's put VE_o•πr²/d for q in eq 1 to get;
F = V²E_o•πr²/2d²
B) formula for the energy stored in the electric field is;
U = ½CV²
From earlier, we saw that; C = E_o•A/d
Thus;
U = ½E_o•AV²/d
A = πr²
Thus;
U = E_o•Aπr²V²/2d
A 0.50-kg mass is attached to a spring of spring constant 20 N/m along a horizontal, frictionless surface. The object oscillates in simple harmonic motion and has a speed of 1.5 m/s at the equilibrium position. At what location are the kinetic energy and the potential energy the same
Given :
A 0.50-kg mass is attached to a spring of spring constant 20 N/m along a horizontal, frictionless surface.
The object oscillates in simple harmonic motion and has a speed of 1.5 m/s at the equilibrium position.
To Find :
At what location are the kinetic energy and the potential energy the same.
Solution :
Let, at location x from the equilibrium position the kinetic energy and the potential energy the same.
So,
[tex]P.E = K.E\\\\\dfrac{kx^2}{2} = \dfrac{mv^2}{2}\\\\x = v\sqrt{\dfrac{m}{k}}\\\\x = 1.5 \times \sqrt{\dfrac{0.5}{20}}\ m\\\\x = 0.238 \ m[/tex]
Hence, this is the required solution.
Engineers are working on a design for a cylindrical space habitation with a diameter of 7.50 km and length of 29.0 km. The habitation will simulate gravity by rotating along its axis. With what speed (in rad/s) should the habitation rotate so that the acceleration on its inner curved walls equals 8 times Earth's gravity
Answer:
The speed will be "0.144 rad/s".
Explanation:
Given that,
Diameter,
d = 7.50 km
Radius,
R = [tex]\frac{7.5}{2} \ Km[/tex]
Acceleration on inner curve,
= 8 times
Now,
As we know,
⇒ [tex]\omega^2R=8g[/tex]
or,
⇒ [tex]\omega=\sqrt{\frac{8g}{R} }[/tex]
On substituting the values, we get
⇒ [tex]=\sqrt{\frac{8\times 9.8}{\frac{7.5}{2}\times 10^3 } }[/tex]
⇒ [tex]=\sqrt{\frac{78.4}{3750} }[/tex]
⇒ [tex]=\sqrt{0.0209}[/tex]
⇒ [tex]=0.144 \ rad/s[/tex]