Answer:
Explanation:
Archimedes principle states that the upward buoyant foce exrted on a body is equal to th wight o the liquid displaced.
Now, the buoyant force on the boat is given by:
[tex](m+m_b)g=V\rho g[/tex]
[tex]V[/tex] is the volum [tex]\rho[/tex] is the density [tex]m_b[/tex] is the mass of the boat and [tex]m[/tex] is the mass added to the boat.
[tex](m+m_b)g=(Sd)\rho[/tex]
[tex]S[/tex] is the surface area and [tex]d[/tex] is the depth.
[tex]m=Sd\rho - m_b...(1)[/tex]
The equation for thebest fit linis,
[tex]d=(0.374m/kg)m+0.11m[/tex]
Re-arrangethis equati for [tex]m[/tex]
[tex]m=\frac{d}{(0.374m/kg)}-\frac{0.11m}{0.374m/kg}...(2)[/tex]
From equations(1) and (2),
[tex]Sd\rho=\frac{d}{0.374m/kg}[/tex]
the density is,
[tex]\rho=\frac{1}{S(0.0374m/kg)}=\frac{1}{(25cm^2)(\frac{1m^2}{10^4cm^2})(0.374m/kg)}=1.069\times 10^3 kg/m^3[/tex]
Therefore, the density of the liquid is
[tex]\rho=1.07\times 10^3 kg/m^3[/tex]
To determine the density of the liquid, one needs to get the slope of the graph from the data given. Using this slope in the formula ρ = k/g, where 'g' is the gravity, gives the density of the liquid.
Explanation:
Let's first understand this concept with the help of Archimedes' Principle, which states that, the upward buoyant force exerted on a body immersed in a fluid, whether fully or partially submerged, is equal to the weight of the fluid that the body displaces. To determine the density of the liquid, we need the slope of the line from the graph.
Let's assume that the slope of the line is 'k' (which you will obtain from your graph). The slope of the line of best fit in the graph of m versus d will be m/d = ρVg/(Ag), where 'm' is the mass added, 'd' is the depth, 'ρ' is the density of the fluid, 'V' is the volume of the fluid displaced, 'A' is the cross-sectional area of the boat, and 'g' is the acceleration due to gravity.
We can write Volume 'V' = Ad, so the equation simplifies to k = ρg and hence the density ρ = k/g, where 'k' is the obtained slope and 'g' (assuming you are on the earth) is 9.81 m/s². Therefore, once you obtain the value of 'k', you can easily calculate the density of the liquid.
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A Porsche challenges a Honda to a 400-m race. Because the Porsche's acceleration of 3.5 m/s² is larger than the Honda's 3.0 m/s², the Honda gets a 1.0 s head start. Who wins, and by how much time?
The Porsche, with a higher acceleration, wins the 400-meter race, beating the Honda by approximately 0.2 seconds, despite the Honda's 1-second head start.
Explanation:To solve this problem, we need to calculate the times it would take for both the Porsche and the Honda to cross the 400m line. We can do that by using one of the equations of motion: d = ut + 1/2at², where d is the distance, u is the initial velocity, a is the acceleration, and t is the time.
For the Porsche, u = 0 (as it started from rest), a = 3.5m/s², and d = 400m. Substituting these values into the equation, we get: 400 = 0*t + 1/2*3.5*t², which simplifies to t² = 400/(0.5*3.5), giving t = sqrt(228.57), so t ≈ 15.1 seconds.
Now for the Honda, u = 0, a = 3.0m/s², and d = 400m. The same calculation gives t ≈ 16.3 seconds. However, the Honda had a head start of 1 second, so the actual time for the Honda would be 16.3 - 1 = 15.3 seconds.
Therefore, the Porsche wins by approximately 0.2 seconds.
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If we monitor a point on a wire where there is a current for a certain time interval, which gives the charge that moves through the point in that interval?
a. the product of the current and the time interval
b. the ratio of the current to the time interval
c. the ratio of the time interval to the current
Answer:
a. the product of the current and the time interval
Explanation:
the basic formula is: [tex]Q = It[/tex]
current means the time rate of flow of charge: [tex]I =\frac{Q}{t}[/tex]
where Q - charge
I - current
t - time
You throw a small rock straight up from the edge of a highway bridge that crosses a river. The rock passes you on its way down, 8.00 s after it was thrown. What is the speed of the rock just before it reaches the water 25.0 m below the point where the rock left your hand?
Explanation:
First let us find the initial velocity,
We have after 8 seconds the displacement is zero,
We have equation of motion s = ut + 0.5 at²
Initial velocity, u = ?
Acceleration, a = -9.81 m/s²
Time, t = 8 s
Displacement,s = 0 m
Substituting
s = ut + 0.5 at²
0 = u x 8 + 0.5 x -9.81 x 8²
u = 39.24 m/s
Initial velocity is 39.24 m/s.
Now this case is similar to case where a rock is thrown at 39.24 m/s downward.
We have equation of motion v² = u² + 2as
Initial velocity, u = 39.24 m/s
Acceleration, a = 9.81 m/s²
Final velocity, v = ?
Displacement, s = 25 m
Substituting
v² = u² + 2as
v² = 39.24² + 2 x 9.81 x 25
v = 45.06 m/s
The speed of the rock just before it reaches the water 25.0 m below the point where the rock left your hand is 45.06 m/s
The _______ includes all elements existing outside the boundary of the organization that have the potential to affect it. a. general environment b. task environment c. internal environment d. organizational environment
Answer:
The correct option was not given that is External organizational environment.
Explanation:
It's the external organizational environment which contains the entities that exist outside of it but still have a significant impact upon it's development and growth etc.
A rectangular wooden block of weight W floats with exactly one-half of its volume below the waterline.Masses are stacked on top of the block until the top of the block is level with the waterline. This requires 20 g of mass. What is the mass of the wooden block?A.) 40 gB.) 20 gC.) 10 g
Answer:
b) M=20g
Explanation:
For this exercise we must use the Archimedes principle that states that the thrust that a body receives is equal to the weight of the dislodged liquid.
B = ρ g V
Let's use balance healing for this case
Initial.
B - W = 0
The weight of the body can be related to its density
W = ρ V_body g
ρ_liq g (½ V_body) = m g
Final
Some masses were added
M = 20 g = 0.020 kg
B - W - W₂ = 0
ρ_liq g V_Body = m g + M g
Let's replace and write the system of equations
½ ρ_liq V_body = m
ρ V_body = m + M
We solve the equations
2 m = m + M
m = M
m = 20 g
The answer is b
A computer software update involved updating the flash memory of a hardware component. The update failed. A phone technician said the component required replacement, but based on knowledge of how flash memory worked, the user suggested manually downloading the software, which worked.
A. True
B. False
Answer:
Yes, it's true. Computers do work that way. It's experienced by one of the authors of the book how computers work.
Explanation:
A. True, Flash memory is a type of non-volatile computer memory that can be electrically erased and reprogrammed. If a software update fails due to issues with the flash memory, it's possible that the memory itself is corrupted or malfunctioning. However, it's also possible that the software update process itself encountered errors.
Given the scenario, if the flash memory is the only issue and the hardware component is otherwise functional, manually downloading the software directly onto the flash memory could potentially bypass any issues encountered during the automated update process.
Therefore, the user's suggestion of manually downloading the software could indeed solve the problem, making the statement true.
On a scale where the Sun is about the size of a grapefruit and the Earth is about 15 meters away, how far away are the nearest stars besides the Sun?
The distance between the Sun and the nearest star, besides the Sun, is about 40,000 kilometers (25,000 miles), or 4 times the circumference of the Earth.
Explanation:The distance between the Sun and the nearest star, besides the Sun, is incredibly vast. To put it in perspective, if the Sun is the size of a grapefruit and the Earth is 15 meters away, the nearest star would be about 40,000 kilometers away. This is equivalent to approximately 25,000 miles or about 4 times the circumference of the Earth.
It is important to note that the actual distance to the nearest star can vary as there are multiple stars that could be considered the nearest depending on how we define 'nearest.' The star Proxima Centauri, located in the Alpha Centauri system, is one of the closest stars to our Solar System, and it is approximately 4.24 light-years away.
Keep in mind that these distances are still minuscule compared to the vastness of the Universe, which contains billions of galaxies, each with billions of stars.
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_____ can be thrown, caught, and used to provide resistance for a variety of movements, in a variety of planes of motion, and at a variety of velocities.
Answer:
b. Medicine balls
Explanation:
A medicine ball (also called an exercise ball, or a health ball) is a weighted ball about the shoulder diameter (approx. 13.7 inches), often used for recovery and strength training.... In 1705 similar large balls had been used in Persia.So you can use your medicine ball as a projectile to boost the strength of your throws.
How much work is needed to assemble an atomic nucleus containing three protons (such as Li) if we model it as an equilateral triangle of side 2.00×10−15m2.00×10−15m with a proton at each vertex? Assume the protons started from very far away.
Answer:
2.1576 MeV
Explanation:
r = Distance = [tex]2\times 10^{-15}\ m[/tex]
k = Coulomb constant = [tex]8.99\times 10^{9}\ Nm^2/C^2[/tex]
q = Charge of electron = [tex]1.6\times 10^{-19}\ C[/tex]
The resulting energy of the system is
[tex]U=\dfrac{k3q^2}{r}\\\Rightarrow U=\dfrac{8.99\times 10^9\times 3(1.6\times 10^{-19})^2}{2\times 10^{-15}}\\\Rightarrow U=3.45216\times 10^{-13}\ J[/tex]
Converting to MeV
[tex]\dfrac{3.45216\times 10^{-13}}{1.6\times 10^{-13}}=2.1576\ MeV[/tex]
The work needed to assemble an atomic nucleus is 2.1576 MeV
All things in the universe are made up of elements. There are different types of elements in our surroundings. These elements are as follows:-
CarbonHydrogenOxygen and etcAccording to the question, the formula used in the question is [tex]U= \frac{k3q^{2} }{r}[/tex]
As the data is given in the question the solution is:-
[tex]U =\frac{8.99*10^{9} * )3(1.6 *10^{-19})^{2} }{2* 10^{-15} }[/tex]
After solving the question, the value of U we get is [tex]U =3.45216 \ X \ 10^{-13} J[/tex]
The joule must be converted into Mev.Hence, the solution is -[tex]\frac{3.45216 \ X \ 10^{-13} J}{1.6*10^{-13} } = 2.1576 Mev[/tex]
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The electric field strength 4.0 cm from the surface of a 10-cm-diameter metal ball is 60,000 N/C. What is the charge (in nC) on the ball?
Answer:
54 n C
Explanation:
given,
Electric field strength = 60,000 N/m
diameter of the sphere = 10 cm
distance of the electric field = 4 cm
distance of the point form the center of ball
r = 10/2 + 4
r = 9 cm = 0.09 m
the electric field strength
[tex]E = \dfrac{kq}{r^2}[/tex]
[tex]60000 = \dfrac{9\times 10^9\times q}{0.09^2}[/tex]
[tex]q = \dfrac{60000\times 0.09^2}{9\times 10^9}[/tex]
q = 54 x 10⁻⁹ C
q = 54 n C
the charge on the ball is equal to 54 n C
The magnitude of charge in the ball is 54 nC.
Given data:
The strength of Electric field is, E = 60,000 N/C.
The diameter of ball is, d = 10 cm = 0.1 m.
The distance is, r = 4.0 cm = 0.04 cm.
The region where the electric force has its significance is known as electric field strength. And the expression for electric field strength is given as,
[tex]E = \dfrac{kq}{(r+d/2)^{2}}[/tex]
Here, k is the Coulomb's constant and q is the magnitude of charge.
Solving as,
[tex]60,000 = \dfrac{9 \times 10^{9} \times q}{(0.04+0.1/2)^{2}}\\\\q =54 \times 10^{-9} \;\rm C\\\\q = 54 \;\rm nC[/tex]
Thus, we can conclude that the magnitude of charge in the ball is 54 nC.
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Your neighbor Paul has rented a truck with a loading ramp. The ramp is tilted upward at 35 ∘, and Paul is pulling a large crate up the ramp with a rope that angles 20 ∘ above the ramp. Paul pulls with a force of 350 N. (Force is measured in newtons, abbreviated N.)
Completed Question:
Your neighbor Paul has rented a truck with a loading ramp. The ramp is tilted upward at 35 ∘,and Paul is pulling a large crate up the ramp with a rope that angles 10 ∘ above the ramp. Paul pulls with a force of 350 N. (Force is measured innewtons, abbreviated N.)
What is the magnitude of the horizontal component of his force?
What is the magnitude of the vertical component of his force?
Answer:
Both are 175√2 N.
Explanation:
The ramp does an angle of 35° if the soil and the rope that he's pulling does an angle of 10° with the ramp. So, the total angle that the rope does with the soil is 45° (10° + 35°).
The force is a vector, it means that it has a module, direction, and sense. So, if the force acts at an angle of 45° with the horizontal (the soil), the vector can be decomposed to vertical and horizontal vectors.
The decomposition helps to study the movement because an inclined force acts both horizontally and vertically. By the sin and cos of a triangle, the horizontal (x) and vertical (y) forces are:
Fx = F*cosα
Fy = F*sinα
Where α is the angle with the horizontal, and sin45° = cos45° = √2/2
Fx = 350*cos45° =350*√2/2
Fx = 175√2 N
Fy = 350*sin45° = 350*√2/2
Fy = 175√2 N
An astronaut is in an all-metal chamber outside the space station when a solar storm results in the deposit of a large positive charge on the station. Which statement is correct?
a. The astronaut must abandon the chamber immediately to avoid being electrocuted.
b. The astronaut will be safe only if she is wearing a spacesuit made of non-conducting materials.
c. The astronaut does not need to worry: the charge will remain on the outside surface.
d. The astronaut must abandon the chamber if the electric field on the outside surface becomes greater than the breakdown field of air.
e. The astronaut must abandon the chamber immediately because the electric field inside the chamber is non- uniform.
Answer:
c. The astronaut does not need to worry: the charge will remain on the outside surface.
Explanation:
The all-metal chamber (conductor) receives an external electromagnetic field, so it is positively charged in the direction the electromagnetic field is going, and negatively charged in the opposite direction. Since the chamber is polarized, it generates an electric field equal in magnitude, but opposite the external electromagnetic field, then the net electric field inside the conductor is 0.
When all boxes (conductors) receive environmental electromagnetic frequencies, they become electrostatically attracted inside the plane of the magnetic wave and charged negatively in a reverse way.
Since the chambers are polarized, they generate an electric field of similar size and opposite to the outside magnetic wave, hence the net electric field within the wire is 0.Whenever a solar flare deposits large amounts of positive energy on the spaceship, one astronaut is in an all-metal room outside the station. An astronaut does not need to be concerned because the charge should remain from the outside surface.Therefore, the answer is " Option c".
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A 4.3 kg steel ball and 6.5 m cord of negligible mass make up a simple pendulum that can pivot without friction about the point O. This pendulum is released from rest in a horizontal position and when the ball is at its lowest point it strikes a 4.3 kg block sitting at rest on a shelf. Assume that the collision is perfectly elastic and take the coefficient of friction between the block and shelf to be 0.9. The acceleration of gravity is 9.81 m/s².
Answer
given,
mass of steel ball, M = 4.3 kg
length of the chord, L = 6.5 m
mass of the block, m = 4.3 Kg
coefficient of friction, μ = 0.9
acceleration due to gravity, g = 9.81 m/s²
here the potential energy of the bob is converted into kinetic energy
[tex]m g L = \dfrac{1}{2} mv^2[/tex]
[tex]v= \sqrt{2gL}[/tex]
[tex]v= \sqrt{2\times 9.8\times 6.5}[/tex]
v = 11.29 m/s
As the collision is elastic the velocity of the block is same as that of bob.
now,
work done by the friction force = kinetic energy of the block
[tex]f . d = \dfrac{1}{2} mv^2[/tex]
[tex]\mu m g. d = \dfrac{1}{2} mv^2[/tex]
[tex]d=\dfrac{v^2}{2\mu g}[/tex]
[tex]d=\dfrac{11.29^2}{2\times 0.9 \times 9.8}[/tex]
d = 7.23 m
the distance traveled by the block will be equal to 7.23 m.
Final answer:
The collision between the steel ball and the block is perfectly elastic. By using the equation µN = ff, we can calculate the frictional force and the acceleration of the block.
Explanation:
The collision between the steel ball and the block is a perfectly elastic collision, meaning that both momentum and kinetic energy are conserved.
When the ball strikes the block, it transfers its momentum to the block, causing the block to move with the same velocity as the ball had before the collision.
Because the coefficient of friction between the block and the shelf is given as 0.9, we can use the equation µN = ff to find the frictional force and calculate the acceleration of the block.
How many electrons must be removed from each of the two spheres so that the force of electrostatic repulsion exactly balances the gravitational attraction?
Answer / Explanation:
For proper clarity, let us recall Coulomb's Law,
Where, m = mass of an electron = 9.1 x 10⁻³¹
q = The electric charge of the electron = 1.6 x 10 ⁻¹⁹
r = The distance between the two electrons
G = Universal gravitational constant = 6.67 x 10⁻¹¹Nm²/Kg²
K = 8.9 x 10 ⁹ Nm²/C²
Now, considering the fact that the number of electron removed from the spheres was not given,
We assume this to be = "n"
Where the electric charge of the electron = 1.6 x 10 ⁻¹⁹ . n (where n = number of electron removed from each sphere)
Since we were not given the mass of the sphere, we try to calculate it from the volume using the formula:
V = 4/3πr³,
However, from coulombs law, mass of the electron = 9.1 x 10⁻³¹
Consequentially, where electrostatic repulsion = gravitational attraction
Therefore, recalling the formula,
Kq₁ q₂ / R² = Gмm / R²
Now inserting the value from the constant stated initially,
we have,
(8.9 x 10 ⁹)(1.6 x 10 ⁻¹⁹n)(1.6 x 10 ⁻¹⁹n)/R² = (6.67 x 10⁻¹¹)(9.1 x 10⁻³¹)/R₂
Doing a proper calculation of the above,
we should get n = 6.805838 × 10⁶
The answer is 6.805838 × 10⁶
A shopper pushes a grocery cart 20.0 m at constant speed on level ground, against a 35.0 N frictional force. He pushes in a direction 25.0° below the horizontal. (A) What is the work done on the cart by the friction? (B) What is the work done on the cart by the gravitational force? (C) What is the work done on the cart by the shopper? (D) Find the force the shopper exerts, using energy considerations. (E) What is the total work done on the cart?
Answer:
(A). The work done on the cart by the friction is -700 J.
(B). The work done on the cart by the gravitational force is 0 J.
(C). The work done on the cart by the shopper is 700 J.
(D). The force the shopper exerts 38.61 N.
(E). The total work done on the cart is zero.
Explanation:
Given that,
Distance = 20.0 m
Frictional force = 35.0 N
Angle = 25.0°
(A). We need to calculate the work done on the cart by the friction
Using formula of work done
[tex]W_{fr} = -F\cdot d[/tex]
Where, F = force
d = distance
Put the value into the formula
[tex]W_{fr}=-35.0\times20[/tex]
[tex]W_{fr}=−700\ J[/tex]
(B). The work done by the gravity is perpendicular to the direction of the motion
We need to calculate the work done on the cart by the gravitational force
Using formula of work done
[tex]W=fd\cos\theta[/tex]
Put the value into the formula
[tex]W=35.0\times20\cos90[/tex]
[tex]W=0\ J[/tex]
(C). We need to calculate the work done on the cart by the shopper
Using formula of work done
[tex]W_{sh}=W_{net}-W_{fr}[/tex]
Put the value into the formula
[tex]W_{sh}=0-(-700)[/tex]
[tex]W_{sh}=700\ J[/tex]
(D). We need to calculate the force the shopper exerts
Using formula of force
[tex]F_{sh}=\dfrac{W_{fr}}{d\cos\theta}[/tex]
Put the value into the formula
[tex]F_{sh}=\dfrac{700}{20\cos25}[/tex]
[tex]F_{sh}=38.61\ N[/tex]
(E). We need to calculate the total work done on the cart
Using formula of work done
[tex]W_{cart}=W_{fr}+W_{sh}[/tex]
Put the value into the formula
[tex]W_{cart}=700-(-700)[/tex]
[tex]W_{cart}=0\ J[/tex]
Hence, (A). The work done on the cart by the friction is -700 J.
(B). The work done on the cart by the gravitational force is 0 J.
(C). The work done on the cart by the shopper is 700 J.
(D). The force the shopper exerts 38.61 N.
(E). The total work done on the cart is zero.
The work done on a grocery cart by friction is -700 J, by gravitation is 0, by the shopper is designed to balance the friction, and the total net work considering only these factors tends to be 0, assuming the system is ideal.
(a) The work done on the cart by friction is given by the formula Work = force × distance × cos(θ). Since the force of friction is 35.0 N acting against the direction of the displacement (180° or π radians), and the distance is 20.0 m, the work done by friction = 35 × 20 × cos(180°) = -700 J.
(b) The work done on the cart by the gravitational force is zero because gravitational force acts vertically downwards and the displacement of the cart is horizontal, making the angle between the force and displacement 90° which leads to no work being done as cos(90°) = 0.
(c) The work done on the cart by the shopper can be found by calculating the component of the shopper's force in the direction of the displacement. Without the exact force the shopper applies, we directly cannot calculate the work done; however, it is designed to counteract friction and maintain constant speed.
(d) To find the force the shopper exerts using energy considerations, we equate the work done against friction to the work done by the shopper. Hence, 700 J of work is done by the shopper as well to overcome friction.
(e) The total work done on the cart is the sum of all works done by individual forces. However, without calculating the exact value of the shopper's force, we consider only the friction work, which is -700 J. If considering only friction, the external work (by the shopper) matches this in magnitude but is positive, suggesting a net work of 0 when only considering these two forces.
Three identical capacitors are connected in series across a potential source (battery). If a charge of Q flows into this combination of capacitors, how much charge does each capacitor carry? Explain.
A) 3Q
B) Q
C) Q/3
D) Q/9
Answer:
C.Q/3
Explanation:
The total capacitances in series
1/C=1/C1+1/C2+1/C3
=1 /C+1/C+1/C
3/C
Ctotal=C/3
Charge in each capacitances
1/3*Q
Q/3
The electric field 1.5 cm from a very small charged object points toward the object with a magnitude of 180,000 N/C. What is the charge on the object?
The charge on the object is 0.45 nanocoulombs (nC).
Explanation:The electric field 1.5 cm from a very small charged object can be calculated using the equation: E = kQ/r², where E is the electric field, k is the electrostatic constant (8.99 x 10⁹ Nm²/C²), Q is the charge on the object, and r is the distance from the object. In this case, we know that E = 180,000 N/C and r = 1.5 cm = 0.015 m.
Substituting the given values into the equation, we can solve for Q:
E = kQ/r²
180,000 N/C = (8.99 x 10⁹ Nm²/C²)(Q)/(0.015 m)²
Q = (180,000 N/C)(0.015 m)^2 / (8.99 x 10⁹ Nm²/C²)
Q = 0.45 x 10⁻⁹C
Therefore, the charge on the object is 0.45 nanocoulombs (nC).
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As you jump across a small stream, does a horizontal force keep you moving forward? If so, what is that force? [Note—this question seems to be deliberately designed to trick you! Is there really a horizontal force being applied once you’ve left the ground?]
There is no horizontal force
Explanation:
When you jump across the stream, your motion is the same of the motion of a projectile, which consists of two independent motions:
A uniform motion (constant velocity) along the horizontal directionA uniformly accelerated motion (with acceleration equal to the acceleration of gravity) in the vertical directionWe notice that as you are in the air, there is only one force acting on your body: the force of gravity, whose direction is downward, and causes your body to accelerate downward with an acceleration equal to
[tex]g=9.8 m/s^2[/tex]
However, there is no force acting on you in the horizontal direction: therefore, your acceleration in this direction is zero, and so your horizontal velocity is constant (that's why you keep moving forward).
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A heavy red ball is released from rest 2.0 m above a flat, horizontal surface. At exactly the same instant, a yellow ball with the same mass is fired horizontally at 3.0 m/s. Which ball hits the ground first?
Answer:
Both ball hit the ground about at the same time.
Explanation:
given information:
h = 2 m
the speed of red ball, v = 3 m/s
the time for red ball to reach the ground
h = [tex]\frac{1}{2}gt^{2}[/tex]
t = √2h/g
= √2(2)/9.8
= 0.64s
the time for yellow ball to reach the ground, it's considered as a vertical motion. thus
h = √2h/g
= √2(2)/9.8
= 0.64s
so, both ball hit the ground about at the same time.
The horizontal velocity does not influence how quickly an object falls to the ground. Thus, both the yellow and the red ball will hit the ground at the same time.
Explanation:In the context of gravity and projectile motion, the horizontal velocity of an object does not influence the time it takes for that object to fall to the ground. The yellow ball and the red ball will both hit the ground at the same time. When you drop an object, it accelerates towards the ground due to gravity. The same process happens to an object moving horizontally; gravity also pulls it downwards. Therefore, the time each ball takes to fall to the ground is only dependent on their initial height and the acceleration due to gravity, not any horizontal velocity. Both balls start from the same height, therefore, they will hit the ground at the same time.
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What is the moment of inertia of a 2.0 kg, 20-cm-diameter disk for rotation about an axis (a) through the center, and (b) through the edge of the disk?
Answer:
(a) I=0.01 kg.m²
(b) I=0.03 kg.m²
Explanation:
Given data
Mass of disk M=2.0 kg
Diameter of disk d=20 cm=0.20 m
To Find
(a) Moment of inertia through the center of disk
(b) Moment of inertia through the edge of disk
Solution
For (a) Moment of inertia through the center of disk
Using the equation of moment of Inertia
[tex]I=\frac{1}{2}MR^{2}\\ I=\frac{1}{2}(2.0kg)(0.20m/2)^{2}\\ I=0.01 kg m^{2}[/tex]
For (b) Moment of inertia through the edge of disk
We can apply parallel axis theorem for calculating moment of inertia
[tex]I=(1/2)MR^{2}+MD\\ Here\\D=R\\I=(1/2)(2.0kg)(0.20m/2)^{2}+(2.0kg)(0.20m/2)^{2}\\ I=0.03kgm^{2}[/tex]
The moment of inertia of the disk for rotation about the center is 0.01 kg•m^2, and for rotation about the edge is 0.02 kg•m^2.
Explanation:The moment of inertia of a solid disk about an axis through its center is given by the formula 1/2 * MR^2. In this case, the mass of the disk is 2.0 kg and the radius is 10 cm (half of the diameter). So, substituting the values into the formula, the moment of inertia about the center axis is:
1/2 * 2.0 kg * (0.1 m)^2 = 0.01 kg•m^2.
For rotation about an axis through the edge of the disk, the moment of inertia is:
MR^2 = 2.0 kg * (0.1 m)^2 = 0.02 kg•m^2.
For Aristotle, virtue refers to a type of a. good consequence. b. motive. c. excellence. d. none of these choices.
Answer:
The Correct Answer is C
Explanation:
According to Aristotle Virtue ethics
Virtue exist between a mean that has overabundance and inadequacy at either end.We required to select along with this mean path of action, towards the correct conclusion.people have different ability to be virtuous. some have great ability while some have a lack of ability.
Answer:
d.none of these choices.
Explanation:
Aristotle describes virtue as the average of abundance and lack, or the ' normal. ' The concept of virtue is simply, he suggests, ' ' all things in moderation. '' Human beings can value life, but not be greedy. We should escape pain and disappointment, but they should not expect a life completely devoid of them.
A lawyer drives from her home, located 88 milesmiles east and 1818 milesmiles north of the town courthouse, to her office, located 1313 milesmiles west and 5454 milesmiles south of the courthouse. Find the distance between the lawyer's home and her office.
Answer:
the distance between the lawyer's home and her office is 124 miles
Explanation:
given information:
first lets assume that
x-axis (west = positive, east = negative)
y-axis (north = positive, south = negative)
thus,
distance of the house = (-88,18)
distance of the office = (13, -54)
thus, the distance between the lawyer's home and her office
R = √(x₂ - x₁)² + (y₂ - y₁)²
= √(13 - (-88))² + (-54 -18)²
= 124 miles
Regarding the history of the universe, which of the following is true? Regarding the history of the universe, which of the following is true? Key elements of which Earth and life are made, including carbon, oxygen, and iron, did not exist when the universe was born and were created later in stars. All the chemical elements were created during the Big Bang, but some have been modified since that time as a result of radioactive decay. The Earth formed quickly after the Big Bang. All the current stars in our galaxy formed about the same time billions of years ago.
Explanation:
1. The universe consisted of hydrogen and helium initially. This statement is true.
2. It is important to understand radioactive decay in order to understand this question, here's a good analogy:
A snake will shed it's skin, just as an atom will shoot off different parts of itself. It would be very difficult to force that snake to reenter it's skin once it sheds, just as it takes a lot of energy to force fusion of atoms and the parts mentioned. In normal circumstances, nuclear decay is one-way.
3. The earth is a giant floating rock in space. It took many many years to gather a bunch of asteroids and dust to make this planet.
4. I shouldn't have to explain this one, it doesn't make much sense.
Key elements of which Earth and life are made, including carbon, oxygen, and iron, did not exist when the universe was born and were created later in stars through nucleosynthesis.
The statement that is true regarding the history of the universe is that key elements of which Earth and life are made, including carbon, oxygen, and iron, did not exist when the universe was born and were created later in stars.
Initially, after the Big Bang, all matter consisted mainly of hydrogen and helium.
As stars formed and evolved, they produced other elements through nuclear reactions, including the elements necessary for Earth and life as we know it.
This process, known as nucleosynthesis, enriched the universe with heavier elements over time.
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A 11.6-kg monkey is hanging by one arm from a branch and swinging on a vertical circle. As an approximation, assume a radial distance of 65.3 cm is between the branch and the point where the monkey's mass is located. As the monkey swings through the lowest point on the circle, it has a speed of 3.50 m/s. Find (a) the magnitude of the centripetal force acting on the monkey and (b) the magnitude of the tension in the monkey's arm.
Answer:
217.611 N
331.407 N
Explanation:
m = Mass of monkey = 11.6 kg
v = Velocity of the monkey = 3.5 m/s
r = Radius = 65.3 cm
g = Acceleration due to gravity = 9.81 m/s²
Centripetal force is given by
[tex]F_c=\dfrac{mv^2}{r}\\\Rightarrow F_c=\dfrac{11.6\times 3.5^2}{0.653}\\\Rightarrow F_c=217.611\ N[/tex]
The magnitude of the centripetal force is 217.611 N
Balancing the forces we get
[tex]F_c=T-mg\\\Rightarrow T=F_c+mg\\\Rightarrow T=217.611+11.6\times 9.81\\\Rightarrow T=331.407\ N[/tex]
The tension in the monkey's arm is 331.407 N
A car is driving around a circular track at constant speed. At one instant, the car is driving northward and sometime later the car is driving westward. What is the direction of the car’s average change in velocity during this time interval?
Answer:
Southwest
Explanation:
when the car is moving around a circular track at constant speed, velocity changes due to change in the direction. The direction at any instant is given by tangent drawn at the point on the circular path.
At instant 1: direction is northwards
At instant 2: direction is westwards
Let the constant speed be v. Then, at instant 1, the velocity would be:
[tex]v_1=v \hat j[/tex]
At instant 2, the velocity would be:
[tex]v_2 = v (-\hat i)[/tex]
Change in velocity = [tex]v_2-v_1 = v(-\hat i - \hat j)[/tex]
The direction of car's average change would be southwest.
The car's average change in velocity has a northwest direction.
Explanation:The direction of the car's average change in velocity is determined by the direction of the velocity vectors at the two instances. In this case, when the car is driving northward, its velocity vector points towards the north. Later, when the car is driving westward, its velocity vector points towards the west. Therefore, the car's average change in velocity will have a direction that is a combination of north and west, which is the northwest direction.
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You are climbing in the High Sierra when you suddenly find yourself at the edge of a fog-shrouded cliff. To find the height of this cliff, you drop a rock from the top; 7.00 s later you hear the sound of the rock hitting the ground at the foot of the cliff.(a) Ignoring air resistance, how high is the cliff if the speed of sound is 330 m/s? (b) Suppose you had ignored the time it takes the sound to reach you. In that case, would you have overestimated or underestimated the height of the cliff? Explain your reasoning.
The problem involves calculating the height of a cliff by considering both the rock's fall time and the sound's travel time back up. The actual height is found by equating the distance sound travels to the free fall distance, leading to an equation that, when solved, gives the cliff's height. Ignoring the sound's travel time results in overestimating the cliff's height.
Explanation:To solve this problem, we must first calculate the time it took for the rock to reach the ground and then use this to find the height of the cliff. The total time of 7.00 seconds includes both the time the rock is falling and the time the sound of the impact takes to travel back up to the top of the cliff.
Part A: Calculating the height of the cliff
Let's denote the time it takes for the rock to fall as t, and the remaining time for the sound to travel up as (7 - t) seconds. Knowing the speed of sound is 330 m/s, the distance the sound travels (which equals the height of the cliff) can be given by Distance = Speed × Time, hence 330 × (7 - t).
To find the height using the falling time, we use the formula for an object in free fall: Height = 0.5 × g × t^2 where g is the acceleration due to gravity (9.8 m/s^2). As both expressions describe the height of the cliff, they can be set equal to each other giving us 0.5 × 9.8 × t^2 = 330 × (7 - t). Solving for t and then substituting back to find the height will give us the answer.
Part B: Effect of Ignoring the Sound Travel Time
If you ignored the time for the sound to reach you, you would be overestimating the time it took for the rock to hit the ground. Since a part of the 7.00 seconds is actually the sound traveling back up, the actual falling time is less than 7.00 seconds. Consequently, the calculated height of the cliff would be *overestimated* because you would be applying the entire 7.00 seconds to the falling distance calculation.
Essentially, ignoring the sound's travel time makes it seem like the rock was in free fall for longer, leading to a larger computed height, which is inaccurate.
Your laundry basket weighs 22 N and your room is 3.0 m above you on the second floor. It takes you 6.0 seconds to carry the laundry basket up. What is your power?
Explanation:
Power = Work ÷ Time
Work = Force x Displacement
Force = 22 N
Displacement = 3 m
Time = 6 seconds
Substituting
Work = Force x Displacement
Work = 22 x 3 = 66 J
Power = Work ÷ Time
Power = 66 ÷ 6
Power = 11 W
Power is 11 W
Final answer:
To calculate power, divide the work done by the time taken. In this case, the power while carrying the laundry basket is 11 Watts.
Explanation:
To calculate power, we use the formula:
Power (P) = Work (W) / Time (t)
In this case, the work done is equal to the weight of the laundry basket multiplied by the distance it is lifted, which is
W = 22 N * 3.0 m = 66 J
Substituting the values into the formula:
P = W / t = 66 J / 6.0 s = 11 W
Therefore, your power while carrying the laundry basket is 11 Watts.
When the Darwin/Wallace theory of natural selection is summarized, four central postulates emerge. Which of the following is NOT one of these four natural selection postulates
Answer:
Four central postulates that emerges from Charles Darwin and Russel Wallace are detailed below. As there is no option provided in the question hence any option beyond the meaning of these four points will not be from these four postulates.
Explanation:
The four postulates were:
More individuals are produced each generation than can survive. Phenotypic variation exists among individuals and the variation is heritable. Those individuals with heritable traits better suited to the environment will survive. When reproductive isolation occurs new species will form.1st:
The number of individuals produced in a generation of a specie is more than that of the ones who survive. Some of them cannot withstand the environmental conditions and die.
2nd:
Observable characteristics of an individual can vary and this variation is inherited by parental cells/genes.
3rd:
The individuals who have characteristics which are suitable to the environment exists and survives while other ones do not bear the environment and die soon.
4th:
Reproductive isolation is the evolutionary mechanism which prevents members of different species from producing offspring but whenever this occurs, most of the times, new specie is formed. The survival of this new specie again depends upon suitability of its characteristics with environment.
An elevator packed with people has a mass of 1800 kg.
A) The elevator accelerates upward (in the positive direction) from rest at a rate of 1.5 m/s2 for 2.2 s. Calculate the tension in the cable supporting the elevator in newtons.
B) The elevator continues upward at constant velocity for 9 s. What is the tension in the cable during this time in newtons?
C) The elevator experiences a negative acceleration at a rate of 0.65 m/s2 for 2 s. What is the tension in the cable, in Newtons, during this period of negative accleration?
D) How far has the elevator moved above its original starting point in meters?
Answer:
a) 20.34 N
b)17.64 N
c)16.47 N
d)38.63m
Explanation:
a) F=m*a. Earth gravity is 9.8m/s2. If the elevator goes upwards, tension on the rope will be higher. To find the tension we need to all accelerates. Total accelerates effects the elevator is a=9.8+1.5=11.3 m/s2. Then;
[tex]F=1.8*11.3=20.34[/tex]
the tension is 20.34N
b) There is no accelerate in this situation therefore the tension is:
F=1.8*9.8=17.64 N
c) In this situation elevator goes down. We need to subtract gravity from elevator acceleration. a=9.8-0.65=9.15m/s2
F=1.8*9.15=16.47 N
d) Total distance is:
[tex]x=0.5*a*t^2+v*t+v*t-0.5*a*t^2\\=0.5*1.5*2.2^2+9*(1.5*2.2)+2*(1.5*2.2)-0.5*0.65*2^2\\=38.63 m[/tex]
A planet's moon travels in an approximately circular orbit of radius 7.0 ✕ 10⁷ m with a period of 6 h 38 min. Calculate the mass of the planet from this information. ___ kg
Answer:
3.56×10²⁶ Kg.
Explanation:
Note: The gravitational force is acting as the centripetal force.
Fg = Fc........................... Equation 1
Where Fg = gravitational Force, Fc = centripetal force.
Recall,
Fg = GMm/r²......................... Equation 2
Fc = mv²/r............................. Equation 3
Where M = mass of the planet, m = mass of the moon, r = radius of the orbit and G = Universal gravitational constant.
Substituting equation 2 and 3 into equation 1
GMm/r² = mv²/r
Simplifying the equation above,
M = v²r/G .............................. Equation 4.
The period of the moon in the orbit
T = 2πr/v
Making v the subject of the equation,
v = 2πr/T............................. Equation 5
where r = 7.0×10⁷ m, T = 6 h 38 min = (6×3600 + 38×60) s = (21600+2280) s
T = 23880 s, π = 3.14
v = (2×3.14×7.0×10⁷ )/23880
v = 18409 m/s
Also Given: G = 6.67×10⁻¹¹ Nm²/kg²
Also substituting into equation 4
M = 18409²×7.0×10⁷ /(6.67×10⁻¹¹)
M = 3.56×10²⁶ Kg.
Thus the mass of the planet = 3.56×10²⁶ Kg.
The mass of the planet is required.
The mass of the planet is [tex]3.56\times 10^{26}\ \text{kg}[/tex]
M = Mass of planet
r = Radius of orbit = [tex]7\times 10^7\ \text{m}[/tex]
t = Time period = [tex]6\times 60\times 60+38\times 60=23880\ \text{s}[/tex]
G = Gravitational constant = [tex]6.674\times 10^{-11}\ \text{Nm}^2/\text{kg}^2[/tex]
Mass is given by
[tex]M=\dfrac{4\pi^2r^3}{t^2G}\\\Rightarrow M=\dfrac{4\pi^2\times (7\times 10^7)^3}{23880^2\times 6.674\times 10^{-11}}\\\Rightarrow M=3.56\times 10^{26}\ \text{kg}[/tex]
The mass of the planet is [tex]3.56\times 10^{26}\ \text{kg}[/tex]
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