Answer: [tex]68.75\ kg, 0.06[/tex]
Explanation:
Mary applies a force of 73 N to create an acceleration of [tex]0.48\ m/s^2[/tex]
When She increases force to 84 N, it creates an acceleration of [tex]0.64\ m/s^2[/tex]
Friction opposes the motion of box
[tex]\Rightarrow 73-f=m\times 0.48\quad \ldots(i)\\\Rightarrow 84-f=m\times 0.64\quad \ldots(ii)[/tex]
Subtract (i) from (ii)
[tex]\Rightarrow 11=m(0.64-0.48)\\\Rightarrow m=68.75\ kg[/tex]
Therefore friction is
[tex]\Rightarrow f=73-68.75\times 0.48\\\Rightarrow f=73-33\\\Rightarrow f=40\ N[/tex]
Here, friction is kinetic friction which is given by
[tex]\Rightarrow f=\mu_kmg\\\Rightarrow 40=\mu_k 68.75\times 9.8\\\Rightarrow \mu_k=0.061[/tex]
Falls often cause injuries, so one of the significant aspects of falls is the displacement and motion of a body. It is defined as how far a body moves vertically during the fall. Accordingly, there are three important factors that should be known to compute the kinetic energy (KE) of a falling body. These factors are:
Answer:
We need, mass, gravity and height.
Explanation:
When a body falls freely from a height, its initial velocity is zero, but due to the height it has some potential energy at the top and the kinetic energy is zero.
As it falls, the potential energy is gradually converted in to the kinetic energy so that the total energy of the falling body is conserved.
At the time as the body strikes the ground, the entire potential energy is converted into the kinetic energy.
Potential energy is given by
U = m g h
where, m is the mass, g is the gravity and h is the height,
So, to get the kinetic energy we require mass, gravity and height of the body.
Estimate the force a person must exert on a massless string attached to a 0.15 kg ball to make the ball revolve in horizontal circle of radius 0.6 m. The ball makes 2 revolutions per second.
Answer:
[tex]F = 14.2 N[/tex]
Explanation:
From the question we are told that:
Mass [tex]m=0.15kg[/tex]
Radius [tex]r=0.6[/tex]
Angular Velocity [tex]\omega=2rev/s[/tex]
[tex]\omega= =2x2 \pi rad/s=>4 \pi rad/s[/tex]
Generally the equation for Force applied is mathematically given by
[tex]F =mrw2[/tex]
[tex]F=0.15*0.6* (4*x3.14^)2[/tex]
[tex]F = 14.2 N[/tex]
A merry-go-round at a playground is a circular platform that is mounted parallel to the ground and can rotate about an axis that is perpendicular to the platform at its center. The angular speed of the merry-go-round is constant, and a child at a distance of 1.4 m from the axis has a tangential speed of 2.2 m/s. What is the tangential speed of another child, who is located at a distance of 2.1 m from the axis?
(a) 1.5 m/s
(b) 3.3 m/s
(c) 2.2 m/s
(d) 5.0 m/s
(e) 0.98 m/s
Answer:
[tex]V_2=3.3m/s[/tex]
Explanation:
From the question we are told that:
Distance [tex]d_1=1.4m[/tex]
Tangential speed [tex]V=2.2m/s[/tex]
Distance 2 [tex]d_2=2.1m[/tex]
Generally the equation for Angular velocity is mathematically given by
[tex]w=\frac{v}{r}[/tex]
Therefore
[tex]\frac{v_1}{r_1}=\frac{v_2}{r_2}[/tex]
[tex]V_2=\frac{2.2*2.1}{1.4}[/tex]
[tex]V_2=3.3m/s[/tex]
A string of length 3m and total mass of 12g is under a tension of 160N. A transverse harmonic wave with wavelength 25cm and amplitude 2cm travels to the right along the string. It is observed that the displacement at x=0 at t=0 is 0.87cm. a) What is the wave? b) Wrote the wave function, y(x,t) c) Find the particle velocity at position x=0 at time t=10s. What is the maximum particle velocity?
Answer:
What is the answer bro idont now
How fast much an 816kg Volkswagen travel to have the same momentum as (a) a 2650kg Cadillac going 16.0 km/h? (b) a 9080-kg truck also going 16.0 km/hr?
Answer:
(a) v₁ = 51.96 km/h
(b) v₁ = 178 km/h
Explanation:
(a)
For having the same momentum:
m₁v₁ = m₂v₂
where,
m₁ = mass of Volkswagen = 816 kg
v₁ = speed of Volkswagen = ?
m₂ = mass of Cadillac = 2650 kg
v₂ = speed of Cadillac = 16 km/h
Therefore, using these values in the equation, we get:
[tex](816\ kg)v_1 = (2650\ kg)(16\ km/h)\\\\v_1 = (16\ km/h)(\frac{2650\ kg}{816\ kg})[/tex]
v₁ = 51.96 km/h
(b)
For having the same momentum:
m₁v₁ = m₂v₂
where,
m₁ = mass of Volkswagen = 816 kg
v₁ = speed of Volkswagen = ?
m₂ = mass of Truck = 9080 kg
v₂ = speed of Truck = 16 km/h
Therefore, using these values in the equation, we get:
[tex](816\ kg)v_1 = (9080\ kg)(16\ km/h)\\\\v_1 = (16\ km/h)(\frac{9080\ kg}{816\ kg})[/tex]
v₁ = 178 km/h
How does exhailing remove waste from the body? Explain the systems that make this happen‚ using complete sentences
When we exhale, 90% waste material is Carbon Dioxide ( CO2 ) , so, it gets exhaled out in the form of CO2 rich air and it gets removed from the body, therefore our internal body becomes more pure and helps in making our internal temperature constant at a suitable level.
. If force (F), work (W) and velocity (v) are taken as fundamental quantities.
What is the dimensional formula of time (T)?
Answer:
∴ [T]=[WF−1V−1]
Hope this answer is right!!
Answer :
[T] = [W(F)^-1(V)^-1]
How does the density of water change when: (a) it is heated from 0o
C to
4o
C; (b) it is heated from 4o
C to 10o
C ?
Answer:
[b] it id heated from 4o
Explanation:
explain how a lever can act as a force multiplier
Answer:
Example:Opening of a bottle cap by tool
when we hold a tool and open the bottle cap this is because , force x tool force .
The load arm is usually shorter than the effort arm in second order levers. Moving a large weight hence requires less effort. A force multiplier lever or effort multiplier lever is the name for this kind of lever. A boat's oars, for instance, can increase the force.
What is second order levers?Second-order levers are devices with the input force farthest from the fulcrum and the output force on the same side of the fulcrum. A wheelbarrow is an excellent illustration of a second-order lever.
A second-order lever will have an output force greater than an input force, similar to first-order levers. The output journey, however, will be shorter than the input length. Both the input and output forces in this situation will move in the same direction.
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The sound from a trumpet radiates uniformly in all directions in 20C air. At a distance of 5.00 m from the trumpet the sound intensity level is 52.0 dB. The frequency is 587 Hz. (a) What is the pressure amplitude at this distance
Answer:
The answer is below
Explanation:
The intensity level (B) of a sound wave is given by:
B = 10log(I/I₀);
where I₀ is the threshold intensity = 1 * 10⁻¹² W/m², I is the intensity at distance 5 m, B is the intensity level = 52 dB
Substituting gives:
[tex]52=10log(\frac{I}{10^{-12}} )\\\\log(\frac{I}{10^{-12}} )=5.2\\\\I=1.58*10^{-7}\ W/m^2[/tex]
The pressure is given by:
[tex]I=\frac{p_{max}^2}{2\rho v} \\\\\rho=air\ density=1.2\ kg/m^3,v=speed\ of\ sound\ in\ air=344\ m/s,p_{max}=pressure:\\\\p_{max}=\sqrt{2\rho vI}=\sqrt{2*1.58*10^{-7}*1.2*344} =1.14*10^{-2}Pa[/tex]
A rock is thrown from the edge of the top of a 51 m tall building at some unknown angle above the horizontal. The rock strikes the ground a horizontal distance of 74 m from the base of the building 8 s after being thrown. Assume that the ground is level and that the side of the building is vertical. Determine the speed with which the rock was thrown.
Answer:
The speed of projection is 34 m/s.
Explanation:
Height of building, h = 51 m
horizontal distance, d = 74 m
time, t = 8 s
Let the angle is A and the speed is u.
d = u cos A x t
74 = u cos A x 8
u cos A = 9.25 .... (1)
Use second equation of motion
[tex]h = u sin A t - 0.5 gt^2\\\\-51 = u sinA \times 8 - 0.5\times 9.8\times8\times 8\\\\u sin A = 32.8 .... (2)[/tex]
Squaring and adding both the equations
[tex]u^2 = 9.25^2 + 32.8^2 \\\\u = 34 m/s[/tex]
A factory worker pushes a 32.0 kg crate a distance of 7.0 m along a level floor at constant velocity by pushing horizontally on it. The coefficient of kinetic friction between the crate and floor is 0.26.
Required:
a. What magnitude of force must the worker apply?
b. How much work is done on the crate by this force?
c. How much work is done on the crate by friction?
d. How much work is done on the crate by the normal force? By gravity?
e. What is the total work done on the crate?
Answer:
(a) 81.54 N
(b) 570.75 J
(c) - 570.75 J
(d) 0 J, 0 J
(e) 0 J
Explanation:
mass of crate, m = 32 kg
distance, s = 7 m
coefficient of friction = 0.26
(a) As it is moving with constant velocity so the force applied is equal to the friction force.
F = 0.26 x m x g = 0.26 x 32 x 9.8 = 81.54 N
(b) The work done on the crate
W = F x s = 81.54 x 7 = 570.75 J
(c) Work done by the friction
W' = - W = - 570.75 J
(d) Work done by the normal force
W'' = m g cos 90 = 0 J
Work done by the gravity
Wg = m g cos 90 = 0 J
(e) The total work done is
Wnet = W + W' + W'' + Wg = 570.75 - 570.75 + 0 = 0 J
Kinematics equations tells us the position of an object under constant acceleration increases linearly with time.
A. True
B. False
Answer:
False.
Explanation:
Suppose that we have an object that moves with constant acceleration A.
Then the acceleration of the object is defined by the equation:
a(t) = A
The acceleration is the rate of change of the velocity, then the velocity equation is given by the integration of the acceleration equation, we will get:
v(t) = A*t + V₀
Where V₀ is the velocity of the object at the time t = 0s.
Now, if we integrate it again, we will get the position equation:
p(t) = (1/2)*A*t^2 + V₀*t + P₀
Where P₀ is the initial position equation.
Here, we can see that the position equation is a quadratic equation (not a linear equation), then the statement is false.
(4.56 x 10^-13)-(1.17 x 10^-13)
What type of Earth scientist would be interested in understanding volcanic eruptions on
the seafloor?
Answer:
Marine geologists learn about the rocks and geologic processes of the ocean basins.
An airplane flies between two points on the ground that are 500 km apart. The destination is directly north of the point of origin of the flight. The plane flies with an airspeed of 120 m/s. If a constant wind blows at 10 m/s toward the west during the flight, what direction must the plane fly relative to the air to arrive at the destination
Answer:
The right solution is "4.8° east of north".
Explanation:
Given:
Distance,
= 500 km
Speed,
[tex]\vec{v}=120 \ m/s[/tex]
Wind (towards west),
[tex]v_0=10 \ m/s[/tex]
According to the question, we get
The angle will be:
⇒ [tex]\Theta=Cos^{-1}(\frac{v_0}{v_1} )[/tex]
[tex]=Cos^{-1}(\frac{10}{120} )[/tex]
[tex]=85.21[/tex] (north of east)
hence,
The direction must be:
⇒ [tex]\Theta'=90-85.21[/tex]
[tex]=4.79^{\circ}[/tex]
or,
[tex]=4.8^{\circ}[/tex] (east of north)
What is the internal resistance of a current source with an EMF of 12 V if, when a resistor with an unknown resistance is connected to it, a current of 2 A flows through the circuit? A voltmeter connected to the source terminals shows 8 V.
What happens to the water when you throw rock into a pond
Answer:
The water usually rushes back too enthusiastically, causing a splash – and the bigger the rock, the bigger the splash. The splash then creates even more ripples that tend to move away from where the rock went into the water.
A cat with a mass of 5.00 kg pushes on a 25.0 kg desk with a force of 50.0N to jump off. What is the force on the desk?
Answer:
First of all the formula is F= uR,( force= static friction× reaction)
mass= 5+25=30
F= 50
R= mg(30×10)=300
u= ?
F=UR
u= F/R
u= 50/300=0.17N
At a playground, Maryam a 3-year old girl and Zahirah a 6-year old girl are playing with the swings. Maryam is sitting while Zahirah is standing on the swing. Both of them were given the same push by their mother. Choose the CORRECT statements:
A. Maryam is swinging faster than Zahirah.
B. Zahirah is swinging faster than Maryam.
C. Both swings at the same pace.
D. Maryam is swinging faster since she is younger.
E. Zahirah is swing faster since she is older.
Answer:
both swings at the same place
Explanation:
because there mother is giving same amount of force to both.
what was the significance of jumping a.keep the snake b.keep feet cleans c.avoid the hot water d.avoid the Bumbo stick
Answer:
D I think I’m not for sure
Explanation:
You are a venture capitalist that is asked to invest in a startup company that claims it will be able to launch tiny "micro space probes" into space at close to the speed of light using a massive electromagnetic rail gun system2. You are cynical about their cost estimates and decide to analyze the problem in more detail before you invest in their company. Neglect air resistance for this worksheet.
1. A typical payload they claim to launch will weigh 1 kilogram and be accelerated to 90% the speed of light. How much electrical energy will the rail gun require to launch the probe, assuming it is 20% efficient at converting electrical energy to projectile kinetic energy?
2. Given that typical electrical costs are about 5 cents/MJ, how much would this launch cost? Would you invest in this company?
43. You are also concerned about safety. What happens if this projectile were to hit an airplane that is flying overhead and dissipate all of its kinetic energy in the collision? To give you a sense of scale, a large nuclear explosion generates about 1015 J of energy.
4. The system must be able to launch probes to all parts in the sky and must be transportable on a ship. Assume that the railgun is mounted on a frigate-class navy ship (weight = 4,000 metric tons).
a. Will the recoil momentum of the ship be relativistic? Justify your argument.
b. At what speed will the ship recoil after it launches a probe? Do you think that this is a problem for the ship?
Answer:
1. 5.825 × 10¹⁷ J
2. i. $ 29.125 billion ii. I would not invest in the company
3. A nuclear reaction would occur if the payload hits an airplane flying overhead and dissipates all its kinetic energy in the collision.
4. a i. The momentum will not be relativistic
ii. This is because objects with large masses do not move at relativistic speeds
b i. 155 m/s
ii. This speed wouldn't be a problem for the ship.
Explanation:
1. A typical payload they claim to launch will weigh 1 kilogram and be accelerated to 90% the speed of light. How much electrical energy will the rail gun require to launch the probe, assuming it is 20% efficient at converting electrical energy to projectile kinetic energy?
The kinetic energy of the payload is K = (γ - 1)mc² where m = mass of payload = 1 kg, c = speed of light = 3 × 10⁸ m/s and γ = 1/√(1 - β²) where β = 0.9 (since the payload moves at 90 % speed of light)
So, K = (γ - 1)mc²
= (1/√(1 - β²) - 1)mc²
= (1/√(1 - (0.9)²) - 1) × 1 kg × (3 × 10⁸ m/s)²
= (1/√(1 - 0.81) - 1) × 1 kg × (3 × 10⁸ m/s)²
= (1/√0.19 - 1) × 1 kg × (3 × 10⁸ m/s)²
= (1/0.436 - 1) × 1 kg × (3 × 10⁸ m/s)²
= (2.294 - 1) × 1 kg × (3 × 10⁸ m/s)²
= 1.294 × 1 kg × 9 × 10¹⁶ m²/s²
= 11.65 × 10¹⁶ kgm²/s²
= 1.165 × 10¹⁷ J
Let E be the total electrical energy of the rail gun. Since 20 % of this energy is converted to kinetic energy of the payload, we have
20 % of E = K
0.2E = K
E = K/0.2
= 1.165 × 10¹⁷ J/0.2
= 5.825 × 10¹⁷ J
2 Given that typical electrical costs are about 5 cents/MJ, how much would this launch cost? Would you invest in this company?
i. how much would this launch cost?
Since the total energy required is E = 5.825 × 10¹⁷ J = 5.825 × 10¹¹ MJ and it costs 5 cent/MJ. So the total cost of energy will be total energy rate = 5.825 × 10¹¹ MJ × 5 cent/MJ = 29.125 × 10¹¹ = 2.9125 × 10¹² cents. Converting this to dollars, we have 2.9125 × 10¹² cents/100 cents/dollar = 2.9125 × 10¹⁰ dollars = 29.125 × 10⁹ dollars = 29.125 billion dollars = $ 29.125 billion
ii. Would you invest in this company?
I would not invest in the company
3. You are also concerned about safety. What happens if this projectile were to hit an airplane that is flying overhead and dissipate all of its kinetic energy in the collision? To give you a sense of scale, a large nuclear explosion generates about 1015 J of energy.
Since the kinetic energy of the payload is 1.165 × 10¹⁷ J and a nuclear explosion generates about 10¹⁵ J of energy, then a nuclear reaction would occur if the payload hits an airplane flying overhead and dissipates all its kinetic energy in the collision.
4. The system must be able to launch probes to all parts in the sky and must be transportable on a ship. Assume that the rail gun is mounted on a frigate-class navy ship (weight = 4,000 metric tons).
a. Will the recoil momentum of the ship be relativistic? Justify your argument.
i. Will the recoil momentum of the ship be relativistic?
The momentum will not be relativistic.
ii. Justify your argument.
This is because objects with large masses do not move at relativistic speeds. Since the speed cannot be relativistic, its momentum which is the product of mass and speed is non-relativistic
b. At what speed will the ship recoil after it launches a probe? Do you think that this is a problem for the ship?
i. At what speed will the ship recoil after it launches a probe?
Since the total energy of the payload E' = K + mc² = 1.165 × 10¹⁷ J + 1 kg × (3 × 10⁸ m/s)² = 1.165 × 10¹⁷ J + 1 kg × 9 × 10¹⁶ m²/s² = 11.65 × 10¹⁶ J + 9 × 10¹⁶ J = 20.65 × 10¹⁶ J
Also, E'² = (pc)² + (mc²)² where p = momentum of payload
So, making p subject of the formula, we have
(pc)² = E'² - (mc²)²
pc = √[E'² - (mc²)²]
p = √[E'² - (mc²)²]/c
substituting the values of the variables into the equation, we have
p = √[E'² - (mc²)²]/c
p = √[(20.65 × 10¹⁶ J)² - 1kg × (3 × 10⁸ m/s²)²]/3 × 10⁸ m/s
p = √[(20.65 × 10¹⁶ J)² - (1kg × 9 × 10⁸ m²/s²)²]/3 × 10⁸ m/s
p = √[426.4225 × 10³² J² - 81 × 10³² J²]/3 × 10⁸ m/s
p = √[345.4225 × 10³² J²]/3 × 10⁸ m/s
p = 18.59 × 10¹⁶/3 × 10⁸ m/s
p = 6.20 × 10⁸ kgm/s
From the law of conservation, this momentum of the payload equals the momentum of recoil of the ship.
So, p = m'v where m' = mass of navy ship = 4,000 metric tons = 4,000 × 1000 kg = 4 × 10⁶ kg and v = speed of navy ship
So, v = p/m'
= 6.20 × 10⁸ kgm/s ÷ 4 × 10⁶ kg
= 1.55 × 10² m/s
= 155 m/s
ii. Do you think that this is a problem for the ship?
Since the ship's speed is 155 m/s, which is small for an object with such a large mass, this speed wouldn't be a problem for the ship.
It is said that a gas fills all the space available to it. Why then doesn't the atmosphere go off into space?
How much time will it take for a person to walk the length of a football field (100 yards)
at a constant speed of 5 ft/s ?
The speed is in feet per seconds so change the length of the field from yards to feet.
1 yard = 3 feet
100 yards x 3 = 300 feet
The field is 300 feet long
Time = distance / speed
Time = 300 feet / 5 feet per second = 60 seconds = 1 minute
It will take 1 minute
Answer:
A person will take 60 Seconds to walk the distance of 100 yards.
Explanation:
Data Given ;
Distance ( d ) = 100 yards = 300 Ft
Speed ( v ) = 5 Ft/s
Time ( t ) = ?
What is speed ?The distance travelled in unit time is called speed.
formula ; [tex]v = \frac{d}{t} \\[/tex]
On putting values,
[tex]5 = \frac{300}{t}[/tex]
[tex]t =\frac{300}{5}[/tex]
[tex]t = 60 sec[/tex]
Hence the time taken by the person is 60 sec.
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Assignment: 06.05 Infections and Health
A electron gains electric potential energy as it moves from point 1 to point 2. Which of the following is true regarding the electric potential at points 1 and 2?
a. V1 = V2
b. V1 > V2
c. V1 < V2
Answer:
We know that the change in electric potential energy is defined as:
q*ΔV = ΔP
So, the change in the electric potential energy is the charge times the change in the electric potential.
For the case of an electron gas, we have:
q = -e
where -e is the charge of an electron (remember that is negative).
So, if the electron gains electric potential then:
ΔP > 0
this means that the final potential energy is larger than the initial one, then we have:
-e*ΔV > 0
This means that ΔV must be negative.
V₂ = electric potential at point 2, so it is the final electric potential
V₁ = electric potential at point 1, so it is the final electric potential
Then we should get:
ΔV = V₂ - V₁ < 0.
This means that:
V₂ < V₁
The correct option is b.
A man standing on a frictionless ice throws a 1.00kg mass at 20m/s at an angle elevation of 40.0 degrees. What was the magnitude of the mans momentum immediately after the the throl
Answer:
Explanation:
1.00kg×20m/s×cos40=15.3
A horizontal force is applied to a 4.0 kg box. The box starts from rest, moves a horizontal distance of 10.0 meters, and obtains a velocity of 7.0 m/s. The change in the kinetic energy is:_____.
Answer:
98 J
Explanation:
Applying,
Change in kinetic energy = Final kinetic energy- initial kinetic energy
ΔK.E = mv²/2-mu²/2..............Equation 1
Where ΔK.E = Change in kinetic energy, m = mass of the box, u = initial velocity of the box, v = final velocity of the box.
From the question,
Given: m = 4.0 kg, u = 0 m/s, v = 7 ,0 m/s
Substitute these values into equation 1
ΔK.E = (4(7²)/2)-(4(0²)/2)
ΔK.E = (2×49)-0
ΔK.E = 98 J
Hence the change in kinetic energy 98 J
Which describes a characteristic of metallic bonds?
Answer:
arge number of electrons free to move between the charged ions in the lattice.
Explanation:
The metallic bond occurs when an atom with few electrons is united in its last level, therefore the best way to decrease the total energy of the system is to lose all its electrons to remain with the configuration of a noble gas. The electrons that it loses cannot be acquired by other atoms since they all have few electrons, thus leaving a large number of electrons free to move between the charged ions in the lattice.
Some important characteristics emerge from this description of the metallic bond:
* It has many free electrons therefore its electrical conductivity is high
* As the charged ions are fixed, the material can be malleable, bent without breaking since the free electrons create the bond that keeps the system stable.
* As the electrons are free when heating a part of the material, these electrons acquire energy and rapidly propagate it to the other side, giving a high thermal conductivity
* As the temperature increases, the electrons acquire more kinetic energy, which is why there are more collisions between them and consequently the resistivity of the material increases.
The moon Phobos orbits Mars
(mass = 6.42 x 1023 kg) at a distance
of 9.38 x 106 m. What is its period of
orbit?
Answer:
The moon Phobos orbits Mars (m = 6.42 x 1023 kg) at a distance of 9.38 x 106 m.
