Answer:
b. 48.0 g.
Explanation:
Given;
mass of the exoplanet, [tex]M_p = 3M_e[/tex]
radius of the exoplanet, [tex]r_p = \frac{1}{4} r_e[/tex]
The acceleration due to gravity of the planet is calculated as;
[tex]g_p = \frac{GM_p}{r_p^2} \\\\for \ Earth's \ surface\\\\g = \frac{GM_e}{r_e^2} \\\\G = \frac{gr_e^2}{M_e} = \frac{g_pr_p^2}{M_p} \\\\\frac{gr_e^2}{M_e} = \frac{g_p(\frac{r_e}{4}) ^2}{3M_e} \\\\\frac{gr_e^2}{M_e} = \frac{g_pr_e ^2}{16\times 3M_e} \\\\g = \frac{g_p}{48} \\\\g_p = 48 \ g[/tex]
Therefore, the correct option is b. 48.0 g
Consider the heaviest box of 150 lb that you can push at constant speed across a level floor, where the coefficient of kinetic friction is 0.45, and estimate the maximum horizontal force that you can apply to the box. A box sits on a ramp that is inclined at an angle of 60.0° above the horizontal. The coefficient of kinetic friction between the box and the ramp is 0.45.
If you apply the same magnitude force, now parallel to the ramp, that you applied to the box on the floor, what is the heaviest box (in pounds) that you can push up the ramp at constant speed? (In both cases assume you can give enough extra push to get the box started moving.)
Maximum horizontal force that can be applied on the box is 300.32 N.
Mass of the heaviest box that can be pushed on the ramp at constant speed is 105.16 pound.
What is meant by kinetic friction ?Kinetic friction is defined as the opposing force exerted by the surface on an object in contact with it, when there is relative motion between the two surfaces.
Here,
Mass of the box, m = 150 lb = 68.1 kg
Coefficient of kinetic friction, μ = 0.45
Maximum horizontal force that can be applied on the box is the kinetic frictional force. Frictional force,
F(k) = μmg
F(k) = 0.45 x 68.1 x 9.8
F(k) = 300.32 N
Now, the box sits on a ramp inclined at 60°
Coefficient of kinetic friction, μ = 0.45
The net force here acting on the box placed in the ramp is due to the kinetic frictional force and the weight of the box.
So,
Frictional force, F(k)' = μmgcosθ
F(k)' = 0.45 x M x 9.8 x cos 60
F(k)' = 2.2M
Weight of the box acting horizontally,
W = Mgsinθ
W = M x 9.8 x sin60
W = 8.5M
Therefore, net force,
Fn = W - F(k)'
Fn = 8.5M - 2.2M
Fn = 6.3M
The total force acting on the box is
F = F(k) - Fn
ma = 300.32 - 6.3M
Since, the box is moving with constant speed, the acceleration, a = 0
Therefore,
300.32 - 6.3M = 0
6.3M = 300.32
M = 300.32/6.3
M = 47.7 kg = 105.16 pound
Hence,
Maximum horizontal force that can be applied on the box is 300.32 N.
Mass of the heaviest box that can be pushed on the ramp at constant speed is 105.16 pound.
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While riding his bike through a neighborhood, Joe notices a red sign out of
the corner of his eye. He quickly comes to a stop. Which active reading
strategy is he using?
A. Asking questions
B. Making mental images
C. Summarizing
D. Using visual cues
SUBMIT
Answer:
B
Explanation:
because it take to the thought of a situation
Answer:
d i think
Explanation:
and are u using IXL???
(d) A glass prism made from a material of refractive index 1.86 has a refracting angle of 60. The prism is immersed in a water of refractive index 1.33. Determine the minimum deviation for a parallel beam of light transversing the prism
Answer:
The angle of minimum deviation is 28.72°
Explanation:
u welcome
A car traveling at 14 m/s accelerates at 3.5 m/s² for 5 seconds. How much distance does it travel during that time?
Answer: 113.75
Explanation:
You know
acceleration = a = 3.5 m/s²
time = t = 5 seconds
initial velocity = u = 14 m/s
Unknown is distance = s = ?
Use equation: s = ut + [tex]\frac{1}{2}[/tex] at²
Substitute all the known values inside the equation:
s = (14*5) + 0.5 * 3.5 * 5²
s = 70 + 43.75 = 113.75 m
The car travels 113.75 metres.
A wheelbarrow can be used to help lift a load, such as a pile of dirt, and then push the load across a distance.
A man pushes a wheelbarrow.
Which simple machines make up a wheelbarrow?
a pulley and an inclined plane
a wheel and axle and a lever
a pulley and a wheel and axle
a lever and a wedge
Answer:
a wheel and axle and a lever
The wheelbarrow has a wheel and the then the barrow. The wheel is an wheel and axle machine while the other part is a lever.
What is a simple machine?A simple machine is a device that is used to make work easier. It ensures that less effort is applied to overcome a large load.
The wheelbarrow has a wheel and the then the barrow. The wheel is an wheel and axle machine while the other part is a lever.
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Which one is the dependent variable in distance, force, or work
Answer:
Distance
Explanation:
Work can be defined as the energy transferred to a physical object by exertion of a force on the object to cause a displacement of the object. Thus, work is typically done when a person or simple machine move an object over a distance through the application of a force.
Mathematically, work done is given by the formula;
[tex] W = F * d[/tex]
Where,
W is the work done
F represents the force acting on a body.
d represents the distance covered by the body.
A dependent variable is the event expected to change when an independent variable is manipulated.
Hence, distance is the dependent variable because its value changes with respect to the amount of force exerted on an object.
The angle between the reflected ray and the normal line is called the________________
Answer:
Angle of reflection
Explanation:
The angle between the reflected ray and the normal line is called the angle of reflection
Answer:
Activity 2 Directions: Solve the following problems using Charles' Law,
1. A 3.8 mL of a gas in a container has a temperature of 2.0K. What would the volume be if the temperature becomes 8.0K?
2. If at 27°C the volume of a dry gas is 3.0 ml. What is the new temperature if the volume increased to 7.5 ml?
3. The temperature of the gas is 460,0 mL at 3.5°C. Calculate the new volume at 1.5°C
Speed of light in water
Answer:
225,000 kilometers per second
Explanation:
Have a nice day
You have been hired to design a spring-launched roller coaster that will carry two passengers per car. The car goes up a 13-m-high hill, then descends 20 m to the track's lowest point. You've determined that the spring can be compressed a maximum of 2.4 m and that a loaded car will have a maximum mass of 430 kg. For safety reasons, the spring constant should be 13 % larger than the minimum needed for the car to just make it over the top. Part A What spring constant should you specify
Answer:
22.15 N/m
Explanation:
As we know potential energy = m*g*h
Potential energy of spring = (1/2)kx^2
m*g*h = (1/2)kx^2
Substituting the given values, we get -
(400)*(9.8)*(10) = (0.5)*(k)*(2.0^2)
k = 39200/2.645
k = 19600 N/m
For safety reasons, this spring constant is increased by 13 % So the new spring constant is
k = 19600 * 1.13 = 22148 N/m = 22.15 N/m
Can someone help me
Btw the last one say current
The formula to convert oF to oC =5/9( F -32)
A 3.0 kg pendulum swings from point A of height ya = 0.04 m to point B of height yb = 0.12 m, as seen in the diagram below.
Answer:
3.0−0.12=2.88 or 2.88÷0.04=72
0.04×3.0=0.12 and 0.04+3.0=3.04
Answer: 2.4 J
Explanation: Khan Academy
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 caterpillar climbs up a one-meter a wall. For every 2 cm it climbs up, it slides down 1 cm. It takes 10 minutes for the
caterpillar to climb to the top. What is the distance traveled? (Round the number to the nearest hundred.)
cm
Answer:
1.5 m is ur answer
Explanation:
Monica pulls her daughter Jessie in a bike trailer. The trailer and Jessie together have a mass of 25 kg. Monica starts up a 100-m-long slope that's 4.0 m high. On the slope, Monica's bike pulls on the trailer with a constant force of 9.0 N. They start out at the bottom of the slope with a speed of 5.0 m/s.
What is their speed at the top of the slope?
Answer:
Explanation:
From the given information:
total mass = 25 kg
distance d = 100 m
height = 4.0 m
Force F = 9.0 N
The speed at (bottom) u = 5.0 m/s
Using the concept of energy conservation;
[tex]\dfrac{1}{2}mu^2 + W = \dfrac{1}{2}mv^2 + mgh[/tex]
divide both sides by m
[tex]\dfrac{1}{2}u^2 + \dfrac{W }{m}= \dfrac{1}{2}v^2 + gh[/tex]
multiply both sides by 2
[tex]\dfrac{1}{2}u^2\times 2 + \dfrac{W }{m}\times 2= \dfrac{1}{2}v^2\times 2 + gh\times 2[/tex]
[tex]u^2 +2 \dfrac{W }{m}=v^2 + 2gh[/tex]
[tex]v^2 =u^2 - 2gh+ 2 \dfrac{W }{m}[/tex]
Recall that:
W = Fd
∴
[tex]v^2 =u^2 - 2gh+ 2 \dfrac{Fd }{m}[/tex]
[tex]v^2 =(5.0 \ m/s)^2 - 2(9.81 \ m/s)(4.0 \ m)+ 2 \dfrac{( 9.0 \ N \times 100 \ m) }{25 \ kg}[/tex]
[tex]v^2 =(25.0 ) - 78.48 +72[/tex]
[tex]v^2 = 18.52 \ m^2/s^2[/tex]
[tex]v = \sqrt{18.52 \ m^2/s^2}[/tex]
v = 4.30 m/s
what are parasites? Give some example
Answer:
parasites are creatures the gain benefit off of other animals usually harming them eg:ticks on dogs
Alice's friends Bob and Charlie are having a race to a distant star 10 light years away. Alice is the race official who stays on Earth, and her friend Darien is stationed on the star where the race ends. Bob is in a rocket that can travel at 0.7c; whereas Charlie's rocket can reach a speed of 0.866c. Bob and Charlie start at the same time. Draw space- diagrams from each perspective.
Required:
Estimate how long it takes Bob and Charlie to finish the race from each perspective.
Solution :
The distance between the starting point and the end point, [tex]L_0[/tex] = 10 light years
But due to the relativistic motion of Bob and Charlie, the distance will be reduced following the Lorentz contraction. The contracted length will be different since they are moving with different speeds.
For Bob,
Speed of Bob's rocket with respect to Alice, [tex]L_b = 0.7 \ c[/tex]
So the distance appeared to Bob due to the length contraction,
[tex]$L_b=L_0\sqrt{1-\frac{V_b^2}{c^2}$[/tex]
[tex]$L_b=10\times \sqrt{1-0.49} \ Ly$[/tex]
[tex]$=7.1 \ Ly$[/tex]
Therefore, the time required to finish the race by Bob is
[tex]$t_b = \frac{L_b}{V_b}$[/tex]
[tex]$=\frac{7.1 \ c}{0.7 \ c}$[/tex]
= 10.143 year
For Charlie,
Speed of Charlie's rocket with respect to Alice, [tex]L_c = 0.866 \ c[/tex]
So the distance appeared to Charlie due to the length contraction,
[tex]$L_b=L_0\sqrt{1-\frac{V_c^2}{c^2}$[/tex]
[tex]$L_b=10\times \sqrt{1-0.75} \ Ly$[/tex]
[tex]$=5 \ Ly$[/tex]
The time required to finish the race by Charlie is
[tex]$t_b = \frac{L_c}{V_c}$[/tex]
[tex]$=\frac{5 \ c}{0.866 \ c}$[/tex]
= 5.77 year
How many feet does a No. 2 pencil last?
Answer:45,000
Explanation:
:TIL the average pencil holds enough graphite to draw a line about 35 miles long or to write roughly 45,000 words
Verify that your equation has the masses and the velocities before and after the collision. Solve equation for the initial velocity of the projectile, Vo. As the bob swings upward from h1 to a max of h2, what is happening to the kinetic energy of the system?
Answer:
Decrease occur in kinetic energy.
Explanation:
The kinetic energy decreases when the the Bob swings reaches to the maximum height because the motion of the bob slower down. At maximum height, the kinetic energy decreases whereas the value of potential energy is the highest. The main reason of higher potential energy is that it depends on height of an object while on the other hand, kinetic energy depends on the motion of an object so that's why the value of kinetic energy decreases and potential energy increases at maximum height of the bob.
a motor car reaches a velocity of 15m/s in 6s from rest on a perfect test track . what is the average acceleration
Answer:
[tex]{ \tt{initial \: velocity, \: u = 0}} \: (at \: rest) \\ { \tt{final \: velocity, \: v = 15 { {ms}^{ - 1} }}} \\ { \tt{time, \: t = 6s}} \\ { \bf{from \: first \: newtons \: equation \: of \: motion : }} \\ { \bf{v = u + at}} \\ { \tt{15 = 0 + (a \times 6)}} \\ { \tt{6a = 15}} \\ { \tt{acceleration, \: a = 2.5 \: {ms}^{ - 2} }}[/tex]
can someone please help me
F = 153 N
[tex]\theta = 11.3°[/tex]
Explanation:
Let us define first our directional convention. Anything pointing up or to the right is considered positive and anything pointing down or to the left is considered negative. Now let's look at the components [tex]F_{x}[/tex] and [tex]F_{y}[/tex]:
[tex]F_{x}[/tex] = 350 N - 200 N = 150 N
[tex]F_{y}[/tex] = 180 N - 150 N = 30 N
The magnitude of the resultant force F is given by
[tex]F = \sqrt{F_{x}^{2}+F_{y}^{2}}[/tex]
[tex]\:\:\:\:\:\:= \sqrt{(150\:N)^{2}+(30\:N)^{2}}[/tex]
[tex]\:\:\:\:\:\:=153\:N[/tex]
To find the direction [tex]\theta[/tex], we use
[tex]\tan \theta = \dfrac{F_{y}}{F_{x}}=\dfrac{30\:N}{150\:N}=0.2[/tex]
or
[tex]\theta = \tan^{-1}(0.2) = 11.3°[/tex]
Certain rifles can fire a bullet with a speed of 950 m/s just as it leaves the muzzle (this speed is called the muzzle velocity). The muzzle is 75.0 cm long and the bullet is accelerated uniformly from rest within it. What is the acceleration (in {g}'s) of the bullet in the muzzle? If, when this rifle is fired vertically, the bullet reaches a maximum height {H}, what would be the maximum height (in terms of H) for a new rifle that produced half the muzzle velocity of this one?
Answer:
a) By [tex]v^2 = u^2 + 2as[/tex] => a= 70291.70.
(b)By [tex]v = u + at[/tex] => t= 1.58 ms.
(c)By [tex]v^2 = u^2 - 2gh[/tex] => H = 46045.92 m.
Explanation:
a) By [tex]v^2 = u^2 + 2as[/tex]
[tex](950)^2 = 0 + 2 \times a \times 0.75\\a = 601666.67 m/s^2\\a/g = 688858.70/9.8 = 70291.70[/tex]
(b)By [tex]v = u + at[/tex]
[tex]950 = 0 + 601666.67 \times t\\t = 1.58 x 10^-3 sec = 1.58 ms[/tex]
(c)By [tex]v^2 = u^2 - 2gh[/tex]
[tex]0 = (950)^2 - 2 \times9.8 \times H\\H = 46045.92 m[/tex]
What will be resultant force ?
Answer:
It will go to the right way 500N.
Explanation:
Because, up and down is 300 and 300 so it minus each other and get zero, that made the resultant force will never go up or down. And now we are focusing on the left and right. 900 is more than 400, so it will go to the right. But we also pull to the left too!! That make the resultant force is 500 to the right.
A car of mass 500 kg is moving at a speed of 1.2 m/s. A man pushes the car,
increasing the speed to 2 m/s. How much work did the man do?
A. 640 J
B. 360 J
C. 1360 J
D. 1000 J
Work done by man will be A. 640 J
What is work energy theorem?
The work-energy theorem states that the net work done by the forces on an object equals the change in its kinetic energy.
according to work energy theorem
Work done = final Kinetic energy - initial kinetic energy
= KE (final) - KE (initial )
= 1/2 m ([tex]v^{2}[/tex]) - 1/2 m ([tex]u^{2}[/tex])
= 1/2 m ([tex]v^{2}[/tex] - [tex]u^{2}[/tex])
= 1/2 * 500 * ( [tex]2^{2}[/tex] - [tex]1.2^{2}[/tex])
= 250 * 2.56 = 640 J
correct answer is A. 640 J
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A system consists of a disk of mass 2.0 kg and radius 50 cm upon which is mounted an annular cylinder of mass 1.0 kg with inner radius 20 cm and outer radius 30 cm (see below). The system rotates about an axis through the center of the disk and annular cylinder at 10 rev/s. (a) What is the moment of inertia of the system
Complete Question
A system consists of a disk of mass 2.0 kg and radius 50 cm upon which is mounted an annular cylinder of mass 1.0 kg with inner radius 20 cm and outer radius 30 cm (see below). The system rotates about an axis through the center of the disk and annular cylinder at 10 rev/s.
(a) What is the moment of inertia of the system
(b) What is its rotational kinetic energy? axis 50 cm 30 cm 20 cm
Answer:
a) [tex]M.I=0.32kg.m^2[/tex]
b) [tex]K.E=621.8J[/tex]
Explanation:
From the question we are told that:
Mass [tex]m=2.0kg[/tex]
Disk Radius [tex]r_d=50cm=0.5m[/tex]
Mass of annular cylinder [tex]M_c=1.0kg[/tex]
Inner Radius of cylinder [tex]R_i=20cm=0.2m[/tex]
Outer Radius of cylinder [tex]R_o=0.3m[/tex]
Angular Velocity [tex]\omega=10rev/s=2 \pi*10=62.83[/tex]
Generally the equation for moment of inertia is mathematically given by
[tex]M.I=0.5M r_d^2+0.5M_c(R_i^2+R_o^2)[/tex]
[tex]M.I=0.5(2*0.50)^2)+0.5*1*(0.2^2+0.30^)[/tex]
[tex]M.I=0.32kg.m^2[/tex]
Generally the equation for Rotational Kinetic Energy is mathematically given by
[tex]K.E=0.5M.I \omega^2[/tex]
[tex]K.E=0.5 *0.32*62.83[/tex]
[tex]K.E=621.8J[/tex]
If we use 1 millimeter to represent 1 light-year, how large in diameter is the Milky Way Galaxy?
Answer:
if 1 light year was one millimeter then 105,700 light years = 105,700 mm, (or 105.7 meters in case you needed to simplify or something)
Ocean waves of wavelength 26 m are moving directly toward a concrete barrier wall at 4.0 m/s . The waves reflect from the wall, and the incoming and reflected waves overlap to make a lovely standing wave with an antinode at the wall. (Such waves are a common occurrence in certain places.) A kayaker is bobbing up and down with the water at the first antinode out from the wall.
Required:
How far from the wall is she? What is the period of her up-and-down motion?
Answer:
a) the distance between her and the wall is 13 m
b) the period of her up-and-down motion is 6.5 s
Explanation:
Given the data in the question;
wavelength λ = 26 m
velocity v = 4.0 m/s
a) How far from the wall is she?
Now, The first antinode is formed at a distance λ/2 from the wall, since the separation distance between the person and wall is;
x = λ/2
we substitute
x = 26 m / 2
x = 13 m
Therefore, the distance between her and the wall is 13 m
b) What is the period of her up-and-down motion?
we know that the relationship between frequency, wavelength and wave speed is;
v = fλ
hence, f = v/λ
we also know that frequency is expressed as the reciprocal of the time period;
f = 1/T
Hence
1/T = v/λ
solve for T
Tv = λ
T = λ/v
we substitute
T = 26 m / 4 m/s
T = 6.5 s
Therefore, the period of her up-and-down motion is 6.5 s
An object is positively charged if it has more what
Answer:
An Object is positively charged if it has more Positive Electrons in that object
Một điện tích q = 6.10-6 C đặt tại tâm của một hình lập phương. Tính thông lượng điện trường gửi qua mỗi mặt hình lập phương.
b. Một điện tích q = 4.10-8 C đặt tại tâm của một hình vuông. Tính thông lượng điện trường gửi qua mỗi mặt của hình vuông.
Answer:
mmmlbdhdjdkekkdnxnfjkckckcklclglglvkglglvkgkvkvkvkvkvkvkvkvkvkvkbkkbkbkkbbkkbkbkbkkbkbkblbkkbkb
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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