Answer:
The smallest amount of lead that needs to be attached to John's boat in order to sink it has to have a mass slightly greater than the mass of the boat.
Explanation:
A body floats in a fluid when its density is less than the density of the fluid.
The body sinks when its density is more than the density of the fluid.
Density of the pond = density of pure water = 1 g/cm³
Specific gravity of an object = (density of object)/(density of water)
0.5 = (density of John's boat)/1
Density of the John's boat = 0.5 g/cm³
And density is given as
Density = (mass/volume)
0.5 = (mass of John's boat)/(volume of John's boat)
Let the volume of John's boat be v and the mass of John's boat be m
0.5 = (m/v)
v = (m/0.5) = 2m
To sink the boat, we need the total mass on the boat to increase the density to a value greater than 1.
Let the minimum mass of lead required for this to be M
The volume of the boat remains the same, but the total mass on the boat is now (m+M)
1 < (m + M)/v
M + m > v
Recall, v = 2m
M + m > 2m
M > 2m - m
M > m
Hence, the smallest amount of lead that needs to be attached to John's boat in order to sink it has to have a mass slightly greater than the mass of the boat.
Hope this Helps!!!!
You have two charges; Q1 and Q2, and you move Q1 such that the potential experienced by Q2 due to Q1 increases.
Gravity should be ignored.
Then, you must be:
a) Moving Q1 further away from Q2.
b) Moving in the opposite direction to that of the field due to Q1
c) Moving Q1 closer to Q2.
d) Moving in the same direction as the field due to Q1.
e) Any of the above
Given that,
First charge = Q₁
Second charge = Q₂
The potential experienced by Q2 due to Q1 increases
We know that,
The electrostatic force between two charges is defined as
[tex]F=\dfrac{kQ_{1}Q_{2}}{r^2}[/tex]
Where,
k = electrostatic constant
[tex]Q_{1}[/tex]= first charge
[tex]Q_{2}[/tex]= second charge
r = distance
According to given data,
The potential experienced by Q₂ due to Q₁ increases.
We know that,
The potential is defined from coulomb's law
[tex]V=\dfrac{Q_{1}}{4\pi\epsilon_{0}r}[/tex]
[tex]V\propto\dfrac{1}{r}[/tex]
If r decrease then V will be increases.
If V decrease then r will be increases.
Since, V is increases then r will decreases that is moving Q₁ closer to Q₂.
Hence, Moving Q₁ closer to Q₂.
(c) is correct option.
HELP ASAP!
There is a lever with 5 m long. The fulcrum is 2 m from the right end. Each end hangs a box. The whole system is in balance. If the box hung to the right end is 12 kg, then what is the mass of the box hung to the left end?
Answer:
8kg
Explanation:
For the box to be in equilibrium. The clockwise moment ensued by the box on the right should be same as that ensued by the one on the right. Hence :
M ×3 = 12 ×2
M = 24/3 = 8kg
Note mass is used because trying to compute the weight by multiplying by the acceleration of free fall due to gravity on both sides will cancel out.
How far does a roller coaster travel if it accelerates at 2.83 m/s2 from an initial
velocity of 3.19 m/s for 12.0 s?
Answer:
b
Explanation:
A spaceship travels toward the Earth at a speed of 0.97c. The occupants of the ship are standing with their torsos parallel to the direction of travel. According to Earth observers, they are about 0.50 m tall and 0.50 m wide. Calculate what the occupants’ height and width according to the others on the spaceship?
Answer:
Explanation:
We shall apply length contraction einstein's relativistic formula to calculate the length observed by observer on the earth . For the observer , increased length will be observed for an observer on the earth
[tex]L=\frac{.5}{\sqrt{1-(\frac{.97c}{c})^2 } }[/tex]
[tex]L=\frac{.5}{.24}[/tex]
L= 2.05
The length will appear to be 2.05 m . and width will appear to be .5 m to the observer on the spaceship. . It is so because it is length which is moving parallel to the direction of travel. Width will remain unchanged.
Please help in the 2nd question
Answer:
[tex]q=4\times 10^{-16}\ C[/tex]
Explanation:
It is given that,
The charge on an object is 2500 e.
We need to find how many coulombs in the object. The charge remains quantized. It says that :
q = ne
[tex]q=2500\times 1.6\times 10^{-19}\ C\\\\q=4\times 10^{-16}\ C[/tex]
So, the charge on the object is [tex]4\times 10^{-16}\ C[/tex].
An aluminium bar 600mm long, with diameter 40mm, has a hole drilled in the center of the bar. The hole is 30mm in diameter and is 100mm long. If the modulus of elasticity for the aluminium is 85GN/m2, calculate the total contraction on the bar due to a compressive load of 180kN.
Answer:
[tex]\delta = 0.385\,m[/tex] (Compression)
Explanation:
The aluminium bar is experimenting a compression due to an axial force, that is, a force exerted on the bar in its axial direction. (See attachment for further details) Under the assumption of small strain, the deformation experimented by the bar is equal to:
[tex]\delta = \frac{P\cdot L}{A \cdot E}[/tex]
Where:
[tex]P[/tex] - Load experimented by the bar, measured in newtons.
[tex]L[/tex] - Length of the bar, measured in meters.
[tex]A[/tex] - Cross section area of the bar, measured in square meters.
[tex]E[/tex] - Elasticity module, also known as Young's Module, measured in pascals, that is, newtons per square meter.
The cross section area of the bar is now computed: ([tex]D_{o} = 0.04\,m[/tex], [tex]D_{i} = 0.03\,m[/tex])
[tex]A = \frac{\pi}{4}\cdot (D_{o}^{2}-D_{i}^{2})[/tex]
Where:
[tex]D_{o}[/tex] - Outer diameter, measured in meters.
[tex]D_{i}[/tex] - Inner diameter, measured in meters.
[tex]A = \frac{\pi}{4}\cdot [(0.04\,m)^{2}-(0.03\,m)^{2}][/tex]
[tex]A = 5.498 \times 10^{-4}\,m^{2}[/tex]
The total contraction of the bar due to compresive load is: ([tex]P = -180\times 10^{3}\,N[/tex], [tex]L = 0.1\,m[/tex], [tex]E = 85\times 10^{9}\,Pa[/tex], [tex]A = 5.498 \times 10^{-4}\,m^{2}[/tex]) (Note: The negative sign in the load input means the existence of compressive load)
[tex]\delta = \frac{(-180\times 10^{3}\,N)\cdot (0.1\,m)}{(5.498\times 10^{-4}\,m^{2})\cdot (85\times 10^{9}\,Pa)}[/tex]
[tex]\delta = -3.852\times 10^{-4}\,m[/tex]
[tex]\delta = -0.385\,mm[/tex]
[tex]\delta = 0.385\,m[/tex] (Compression)
If the mass of the ladder is 12.0 kgkg, the mass of the painter is 55.0 kgkg, and the ladder begins to slip at its base when her feet are 70% of the way up the length of the ladder, what is the coefficient of static friction between the ladder and the floor
Answer:
μ = 0.336
Explanation:
We will work on this exercise with the expressions of transactional and rotational equilibrium.
Let's start with rotational balance, for this we set a reference system at the top of the ladder, where it touches the wall and we will assign as positive the anti-clockwise direction of rotation
fr L sin θ - W L / 2 cos θ - W_painter 0.3 L cos θ = 0
fr sin θ - cos θ (W / 2 + 0,3 W_painter) = 0
fr = cotan θ (W / 2 + 0,3 W_painter)
Now let's write the equilibrium translation equation
X axis
F1 - fr = 0
F1 = fr
the friction force has the expression
fr = μ N
Y Axis
N - W - W_painter = 0
N = W + W_painter
we substitute
fr = μ (W + W_painter)
we substitute in the endowment equilibrium equation
μ (W + W_painter) = cotan θ (W / 2 + 0,3 W_painter)
μ = cotan θ (W / 2 + 0,3 W_painter) / (W + W_painter)
we substitute the values they give
μ = cotan θ (12/2 + 0.3 55) / (12 + 55)
μ = cotan θ (22.5 / 67)
μ = cotan tea (0.336)
To finish the problem, we must indicate the angle of the staircase or catcher data to find the angle, if we assume that the angle is tea = 45
cotan 45 = 1 / tan 45 = 1
the result is
μ = 0.336
A 60-kg skier is stationary at the top of a hill. She then pushes off and heads down the hill with an initial speed of 4.0 m/s. Air resistance and the friction between the skis and the snow are both negligible. How fast will she be moving after she is at the bottom of the hill, which is 10 m in elevation lower than the hilltop
Answer:
The velocity is [tex]v = 8.85 m/s[/tex]
Explanation:
From the question we are told that
The mass of the skier is [tex]m_s = 60 \ kg[/tex]
The initial speed is [tex]u = 4.0 \ m/s[/tex]
The height is [tex]h = 10 \ m[/tex]
According to the law of energy conservation
[tex]PE_t + KE_t = KE_b + PE_b[/tex]
Where [tex]PE_t[/tex] is the potential energy at the top which is mathematically evaluated as
[tex]PE_t = mg h[/tex]
substituting values
[tex]PE_t = 60 * 4*9.8[/tex]
[tex]PE_t = 2352 \ J[/tex]
And [tex]KE_t[/tex] is the kinetic energy at the top which equal to zero due to the fact that velocity is zero at the top of the hill
And [tex]KE_b[/tex] is the kinetic energy at the bottom of the hill which is mathematically represented as
[tex]KE_b = 0.5 * m * v^2[/tex]
substituting values
[tex]KE_b = 0.5 * 60 * v^2[/tex]
=> [tex]KE_b = 30 v^2[/tex]
Where v is the velocity at the bottom
And [tex]PE_b[/tex] is the potential energy at the bottom which equal to zero due to the fact that height is zero at the bottom of the hill
So
[tex]30 v^2 = 2352[/tex]
=> [tex]v^2 = \frac{2352}{30}[/tex]
=> [tex]v = \sqrt{ \frac{2352}{30}}[/tex]
[tex]v = 8.85 m/s[/tex]
Answer:
The Skier's velocity at the bottom of the hill will be 18m/s
Explanation:
This is simply the case of energy conversion between potential and kinetic energy. Her potential energy at the top of the hill gets converted to the kinetic energy she experiences at the bottom.
That is
[tex]mgh = 0.5 mv^{2}[/tex]
solving for velocity, we will have
[tex]v= \sqrt{2gh}[/tex]
hence her velocity will be
[tex]v=\sqrt{2 \times 9.81 \times 10}=14.00m/s[/tex]
This is the velocity she gains from the slope.
Recall that she already has an initial velocity of 4m/s. It is important to note that since velocities are vector quantities, they can easily be added algebraically. Hence, her velocity at the bottom of the hill is 4 + 14 = 18m/s
The Skier's velocity at the bottom of the hill will be 18m/s
A body moves due north with velocity 40 m/s. A force is applied
on it and the body continues to move due north with velocity 35 m/s. W. .What is the direction of rate of change of momentum,if it takes
some time for that change and what is the direction of applied
external force?
Answer:
the direction of rate of change of the momentum is against the motion of the body, that is, downward.
The applied force is also against the direction of motion of the body, downward.
Explanation:
The change in the momentum of a body, if the mass of the body is constant, is given by the following formula:
[tex]\Delta p=\Delta (mv)\\\\\Delta p=m\Delta v[/tex]
p: momentum
m: mass
[tex]\Delta v[/tex]: change in the velocity
The sign of the change in the velocity determines the direction of rate of change. Then you have:
[tex]\Delta v=v_2-v_1[/tex]
v2: final velocity = 35m/s
v1: initial velocity = 40m/s
[tex]\Delta v =35m/s-40m/s=-5m/s[/tex]
Hence, the direction of rate of change of the momentum is against the motion of the body, that is, downward.
The applied force is also against the direction of motion of the body, downward.
Two forces are acting on an object as shown in Fig. on the right. What is the magnitude of the resultant force?
A) 47.5 N
B) 185 N
C) 198 N
D) 200 N
Answer:
185 N
Explanation:
Sum of forces in the x direction:
Fₓ = -(80 N cos 75°) + (120 N cos 60°)
Fₓ = 39.3 N
Sum of forces in the y direction:
Fᵧ = (80 N sin 75°) + (120 N sin 60°)
Fᵧ = 181.2 N
The magnitude of the net force is:
F = √(Fₓ² + Fᵧ²)
F = √((39.3 N)² + (181.2 N)²)
F = 185 N
We have that for the Question "Two forces are acting on an object as shown in Fig. on the right. What is the magnitude of the resultant force?" it can be said that the magnitude of the resultant force is
R=200N
From the question we are told
Two forces are acting on an object as shown in Fig. on the right. What is the magnitude of the resultant force?
A) 47.5 N
B) 185 N
C) 198 N
D) 200 N
Generally the equation for the Resultant force is mathematically given as
For x axis resolution
[tex]Fx=80cos75+120cos60\\\\Fx=80.7N[/tex]
For y axis resolution
[tex]Fx=80sin75+120sin60\\\\Fx=181.2N[/tex]
Therefore
[tex]R=\sqrt{80.7^2+181.2N^2}\\\\R=200N[/tex]
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https://brainly.com/question/23379286
(III) A baseball is seen to pass upward by a window with a vertical speed of If the ball was thrown by a person 18 m below on the street, (a) what was its initial speed, (b) what altitude does it reach, (c) when was it thrown, and (d) when does it reach the street again? Giancoli, Douglas C.. Physics (p. 45). Pearson Education. Kindle Edition.
Answer:
Assuming that the vertical speed of the ball is 14 m/s we found the given values:
a) V₀ = 23.4 m/s
b) h = 27.9 m
c) t = 0.96 s
d) t = 4.8 s
Explanation:
a) Assuming that the vertical speed is 14 m/s (founded in the book) the initial speed of the ball can be calculated as follows:
[tex] V_{f}^{2} = V_{0}^{2} - 2gh [/tex]
Where:
[tex]V_{f}[/tex]: is the final speed = 14 m/s
[tex]V_{0}[/tex]: is the initial speed =?
g: is the gravity = 9.81 m/s²
h: is the height = 18 m
[tex] V_{0} = \sqrt{V_{f}^{2} + 2gh} = \sqrt{(14 m/s)^{2} + 2*9.81 m/s^{2}*18 m} = 23.4 m/s [/tex]
b) The maximum height is:
[tex] V_{f}^{2} = V_{0}^{2} - 2gh [/tex]
[tex] h = \frac{V_{0}^{2}}{2g} = \frac{(23. 4 m/s)^{2}}{2*9.81 m/s^{2}} = 27.9 m [/tex]
c) The time can be found using the following equation:
[tex] V_{f} = V_{0} - gt [/tex]
[tex] t = \frac{V_{0} - V_{f}}{g} = \frac{23.4 m/s - 14 m/s}{9.81 m/s^{2}} = 0.96 s [/tex]
d) The flight time is given by:
[tex] t_{v} = \frac{2V_{0}}{g} = \frac{2*23.4 m/s}{9.81 m/s^{2}} = 4.8 s [/tex]
I hope it helps you!
How many times can a three-dimensional object that has a radius of 1,000 units fit something with a radius of 10 units inside of it? How many times can something with a radius of 2,000 units fit something with a radius of 1 unit?
Answer:
# _units = 1000
Explanation:
This exercise we can use a direct proportion rule.
If a volume of radius r = 1 is one unit, how many units can fit in a volume of radius 10?
# _units = V₁₀ / V₁
The volume of a body of radius 1 is
V₁ = 4/3 π r₁³
V₁ = 4/3π
the volume of a body of radius r = 10
V₁₀ = 4/3 π r₂³
V10 = 4/3 π 10³
the number of times this content is
#_units = 4/3 π 1000 / (4/3 π 1)
# _units = 1000
What is the gravitational potential energy of a ball of mass 2.00 kg which is tossed to a height of 13.0 m above the ground? Answer in J, taking the potential energy to be 0.00 J at the ground.
Answer:
I believe the answer is 254.8 J, or rounded 255 J.
Explanation:
The formula for potential energy is:
PE=m(h)g
This means the mass (m) times height (h) times gravity (m). Gravity is 9.8 m/s (meters per second). Putting all of the numbers into it would equal:
PE=2(13)9.8
This equals 254.8 exactly, or if the assignment calls for you to round, 255.