Answer:
1 month
Explanation:
In the following calculations, be sure to express the answer in standard scientific notation with the appropriate number of
significant figures.
3.88 x 1079 - 4.701 x 1059
x 10
g
Answer:
-45,597.07
Explanation:
if not in scientific calculator and yung answer nung sa scientific sa comment na lang dinadownload ko ka eh
The image of the object formed by the lens is real, enlarged and inverted. What is the kind of lens ?
Answer:
Converging (convex) lens.
Explanation:
A lens can be defined as a transparent optical instrument that refracts rays of light to produce a real image.
Basically, there are two (2) main types of lens and these includes;
I. Diverging (concave) lens.
II. Converging (convex) lens.
A converging (convex) lens refers to a type of lens that typically causes parallel rays of light with respect to its principal axis to come to a focus (converge) and form a real image. Thus, this type of lens is usually thin at the lower and upper edges and thick across the middle.
Basically, the image of the object formed by a converging (convex) lens. lens is real, enlarged and inverted.
A spacecraft on its way to Mars has small rocket engines mounted on its hull; one on its left surface and one on its back surface. At a certain time, both engines turn on. The one on the left gives the spacecraft an acceleration component in the x direction of
ax = 5.10 m/s2,
while the one on the back gives an acceleration component in the y direction of
ay = 7.30 m/s2.
The engines turn off after firing for 670 s, at which point the spacecraft has velocity components of
vx = 3670 m/s and vy = 4378 m/s.
What was the magnitude and the direction of the spacecraft's initial velocity before the engines were turned on? Express the magnitude as m/s and the direction as an angle measured counterclockwise from the +x axis.
magnitude m/s
direction ° counterclockwise from the +x-axis
Answer:
a) v = 517.99 m / s, b) θ = 296.3º
Explanation:
This is an exercise in kinematics, we are going to solve each axis independently
X axis
the acceleration is aₓ = 5.10 1 / S², they are on for t = 670 s and reaches a speed of vₓ= 3670 m / s, let's use the relation
vₓ = v₀ₓ + aₓ t
v₀ₓ = vₓ - aₓ t
v₀ₓ = 3670 - 5.10 670
v₀ₓ = 253 m / s
Y axis
the acceleration is ay = 7.30 m / s², with a velocity of 4378 m / s after
t = 670 s
v_y = v_{oy} + a_y t
v_{oy} = v_y - a_y t
v_oy} = 4378 - 7.30 670
v_{oy} = -513 m / s
to find the velocity modulus we use the Pythagorean theorem
v = [tex]\sqrt{v_o_x^2 + v_o_y^2}[/tex]
v = [tex]\sqrt{253^2 +513^2}[/tex]
v = 517.99 m / s
to find the direction we use trigonometry
tan θ ’= [tex]\frac{v_o_y}{v_o_x}[/tex]
θ'= tan⁻¹ [tex]\frac{voy}{voy}[/tex]
θ'= tan⁻¹ (-513/253)
tea '= -63.7
the negative sign indicates that it is below the ax axis, in the fourth quadrant
to give this angle from the positive side of the axis ax
θ = 360 - θ
θ = 360 - 63.7
θ = 296.3º
General Circulation Models (GCM) :_________
a) use data collected exclusively from high-resolution satellites.
b) use spectral models derived from energy released from the earth and clouds.
c) can be run on powerful home computers, allowing citizen scientists to run models.
d) use complicated two-dimensional grid systems that change temporally.
Answer:
b)
Explanation:
GCMs (general circulation models) are useful instruments for gaining a quantitative knowledge of climate processes. Physical processes in the atmosphere, cryosphere, and land surface are represented by them. They are used for modeling the global climate system's reaction to rising greenhouse gas concentrations available at the moment by utilizing spectral models based on the energy emitted by the biosphere and clouds.
You have two identical beakers A and B. Each beaker is filled with water to the same height. Beaker B has a rock floating at the surface (like a pumice stone). Which beaker, with all its contents, weighs more. Or are they equal?
Answer:
a) if we assume that the water does not spill, Beaker B weighs more than beaker S, or which in this case Beaker A weighs more
b) If it is spilled in water the weight of the two beakers is the same
Explanation:
The beaker weight is
beaker A
W_total = W_ empty + W_water
Beaker B
W_total = W_ empty + W_water + W_roca
a) if we assume that the water does not spill, Beaker B weighs more than beaker S, or which in this case Beaker A weighs more
b) If it is spilled in water, the weight of the two beakers is the same because the amount of liquid spilled and equal to the weight of the stone, therefore the two beakers weigh the same
You are helping your friend move a new refrigerator into his kitchen. You apply a horizontal force of 275 N in the positive x direction to try and move the 61 kg refrigerator. The coefficient of static friction is 0.58. (a) How much static frictional force does the floor exert on the refrigerator
Answer:
f = 347.08 N
Explanation:
The frictional force exerted by the floor on the refrigerator is given as follows:
[tex]f = \mu R = \mu W[/tex]
where,
f = frictional force = ?
μ = coefficient of static friction = 0.58
W = Weight of refrigerator = mg
m = mass of refrigerator = 61 kg
g = acceleration due to gravity = 9.81 m/s²
Therefore,
[tex]f = \mu mg\\f = (0.58)(61\ kg)(9.81\ m/s^2)\\[/tex]
f = 347.08 N
What is the value of the charge that experiences a force of 2.4×10^-3N in an electric field of 6.8×10^-5N/C
Hi there!
[tex]\large\boxed{\approx 35.29 C}[/tex]
Use the following formula:
E = F / C, where:
E = electric field (N/C)
F = force (N)
C = Charge (C)
Thus:
6.8 × 10⁻⁵ = 2.4 × 10⁻³ / C
Isolate for C:
C = 2.4 × 10⁻³ / 6.8 × 10⁻⁵
Solve:
≈ 35.29 C
Suppose oil spills from a ruptured tanker and spreads in a circular pattern. If the radius of the oil spill increases at a constant rate of 2 m/s, exactly how fast (in m2/s) is the area of the spill increasing when the radius is 39 m?
Explanation:
The area of a circle of radius r is given by
[tex]A = \pi r^2[/tex]
Taking the derivative of A with respect to time t, we get
[tex]\dfrac{dA}{dt} = 2\pi r \dfrac{dr}{dt}[/tex]
We also know that
[tex]\dfrac{dr}{dt} = 2\:\text{m/s}\:\text{at}\:r = 39\:\text{m}[/tex]
[tex]\dfrac{dA}{dt} = 2\pi (39\:\text{m})(2\:\text{m/s})= 490\:\text{m}^2\text{/s}[/tex]
A 10.0 L tank contains 0.329 kg of helium at 28.0 ∘C. The molar mass of helium is 4.00 g/mol . Part A How many moles of helium are in the tank? Express your answer in moles.
Answer:
82.25 moles of He
Explanation:
From the question given above, the following data were obtained:
Volume (V) = 10 L
Mass of He = 0.329 Kg
Temperature (T) = 28.0 °C
Molar mass of He = 4 g/mol
Mole of He =?
Next, we shall convert 0.329 Kg of He to g. This can be obtained as follow:
1 Kg = 1000 g
Therefore,
0.329 Kg = 0.329 Kg × 1000 g / 1 Kg
0.329 Kg = 329 g
Thus, 0.329 Kg is equivalent to 329 g.
Finally, we shall determine the number of mole of He in the tank. This can be obtained as illustrated below:
Mass of He = 329 g
Molar mass of He = 4 g/mol
Mole of He =?
Mole = mass / molar mass
Mole of He = 329 / 4
Mole of He = 82.25 moles
Therefore, there are 82.25 moles of He in the tank.
A beam of light has a wavelength of 549nm in a material of refractive index 1.50. In a different material of refractive index 1.07, its wavelength will be:_________.
Explanation:
someone to check if the answer is correct
An electrostatic paint sprayer has a 0.100 m diameter metal sphere at a potential of 30.0 kV that repels paint droplets onto a grounded object. (a) What charge (in C) is on the sphere?(b) What charge must a 0.100-mg drop of paint have to arrive at the object with a speed of 10.0 m/s?
Answer:
A) q = 1.67 × 10^(-7) C
B) q = 1.67 × 10^(-10) C
Explanation:
We are given;
Potential; V = 30 KV = 30000 V
Radius of sphere; r = diameter/2 = 0.1/2 = 0.05 m
A) To find the charge of the sphere, we will use the formula;
V = kq/r
Where;
q is the charge
k is electric force constant = 9 × 10^(9) N.m²/C²
Thus;
q = Vr/k
q = (30000 × 0.05)/(9 × 10^(9))
q = 1.67 × 10^(-7) C
B) Now, potential energy here is a formula; U = qV
However, for the drop of paint to move, the potential energy will be equal to the kinetic energy. Thus;
qV = ½mv²
q = mv²/2V
Where;
v is speed = 10 m/s
V = 30000 V
m = mass = 0.100 mg = 0.1 × 10^(-6) Thus;
q = (0.1 × 10^(-6) × 10²)/(2 × 30000)
q = 1.67 × 10^(-10) C
Which best describes the relationship between heat,intemal energy, and thermal energy?
Internal energy is heat that flows and heat is the part of thermal energy that can be transferred
Internal energy is thermal energy that flows, and thermal energy is the part of heat that can be transferred,
Thermal energy is heat that flows, and heat is the part of intemal energy that can be transferred
Heat is thermal energy that flows, and hennal energy is the part of internal energy that can be transferred.
Answer:
It is all a thermodynamic system that is highly related to each other.
Explanation:
Because they are in the physics of thermodynamics it is not wrong to say they follow the same thermodynamic rules and has highly the same properties of energy.
Where would the normal force exerted on the rover when it rests on the surface of the planet be greater
Answer:
Normal force exerted on the rover would be greater at a point on the surface of the planet where the weight of the rover is experienced to be greater.
Explanation:
Since weight is a vector quantity, it can vary with position. Weight is the amount of force the planet exerts on the rover centered towards the planet.
Such a force is the result of gravitational pull and is quantified as:
[tex]F=G\times \frac{M.m}{R^2}[/tex]
and [tex]M=\rho\times \frac{4\pi.r^3}{3}[/tex]
where:
R = distance between the center of mass of the two bodies (here planet & rover)
G = universal gravitational constant
M = mass of the planet
m = mass of the rover
This gravitational pull varies from place to place on the planet because the planet may not be perfectly spherical so the distance R varies from place to place and also the density of the planet may not be uniform hence there is variation in weight.
Weight is basically a force that a mass on the surface of the planet experiences.
According to Newton's third law the there is an equal and opposite reaction force on the body (here rover) which is the normal force.
g A mass of 2.0 kg traveling at 3.0 m/s along a smooth, horizontal plane hits a relaxed spring. The mass is slowed to zero velocity when the spring has been compressed by 0.15 m. What is the spring constant of the spring
By the work-energy theorem, the total work done on the mass by the spring is equal to the change in the mass's kinetic energy:
W = ∆K
and the work done by a spring with constant k as it gets compressed a distance x is -1/2 kx ²; the work it does is negative because the restoring force of the spring points opposite the direction in which it's getting compressed.
So we have
-1/2 k (0.15 m)² = 0 - 1/2 (2.0 kg) (3.0 m/s)²
Solve for k to get k = 800 N/m.
a nano second is what
Answer:
one thousand-millionth of a second.
A nanosecond is an SI unit of time equal to one billionth of a second, that is, ¹⁄₁ ₀₀₀ ₀₀₀ ₀₀₀ of a second, or 10⁻⁹ seconds. The term combines the prefix nano- with the basic unit for one-sixtieth of a minute. A nanosecond is equal to 1000 picoseconds or ¹⁄₁₀₀₀ microsecond.
Starting with the Ideal Gas Law, show that the relationship between volume and temperature in an adiabatic process is the one given by :
TfVf^γ^-1 = TiVi^γ-1 = Constant
Answer:
hope it helps
explanation:
del tema de fuerza centripeta
1.- Un chico va en bicicleta a 10m/s por una curva plana de 200m de radio.
a) ¿Cuál es la aceleración?
b) si el chico y la bicicleta tienen una masa total de 70kg, ¿Qué fuerza se necesita para producir esta aceleración?
Answer:
a. C = 0.5 m/s²
b. F = 35 Newton
Explanation:
Given the following data;
Radius, r = 200 m
Velocity, v = 10 m/s
Mass, m = 70 kg
a. To find the centripetal acceleration;
Mathematically, centripetal acceleration is given by the formula;
C = v²/r
Where:
C is the centripetal acceleration
v is the velocity
r is the radius
Substituting into the formula, we have;
C = 10²/200
C = 100/200
C = 0.5 m/s²
b. To find the force;
F = mv²/r
F = (70*10²)/200
F = (70 * 100)/200
F = 7000/200
F = 35 Newton
Larger animals have sturdier bones than smaller animals. A mouse's skeleton is only a few percent of its body weight, compared to 16% for an elephant. To see why this must be so, recall that the stress on the femur for a man standing on one leg is 1.4% of the bone's tensile strength.
Suppose we scale this man up by a factor of 10 in all dimensions, keeping the same body proportions. (Assume that a 70 kg person has a femur with a cross-section area (of the cortical bone) of 4.8 x 10−4 m2, a typical value.)
Both the inside and outside diameter of the femur, the region of cortical bone, will increase by a factor of 10. What will be the new cross-section area?
Answer:
[tex]a_s=4.8\times 10^{-2}~m^2[/tex]
Explanation:
Given:
cross sectional area of the bone, [tex]a=4.8 \times 10^{-4} ~m^2[/tex]
factor of up-scaling the dimensions, [tex]s=10[/tex]
Since we need to find the upscaled area having two degrees of the dimension therefore the scaling factor gets squared for the area being it in 2-dimensions.
The scaled up area is:
[tex]a_s=a\times s^2[/tex]
[tex]a_s=[4.8 \times 10^{-4}]\times 10^2[/tex]
[tex]a_s=4.8\times 10^{-2}~m^2[/tex]
The area is defined as the space covered by an object in 2 d dimension. For a rectangle, it is a product of length and breadth. The new cross-section area will be 4.8×10⁻² m².
What is the area?The area is defined as the space covered by an object in 2 d dimension. For a rectangle, it is a product of length and breadth. Its unit is m².
Given data in the problem
a is the crossectional area of conical bone = 4.8×10⁻⁴m².
s is the factor of up-scaling the dimensions =10
For two degrees of dimension, the upscaled area will be square of the given area.
The scaled-up area will be
[tex]\rm a_s=a\times s^2\\\\ a_s= 4.8\times10^{-4}\times {10}^2\\\\\ \rm a_s=4.8\times10^{-2}\;m^2[/tex]
Hence the new cross-section area will be 4.8×10⁻² m².
To learn more about the area refer to the link;
https://brainly.com/question/1631786
In Trial II, the same spring is used as in Trial I. Let us use this information to find the suspended mass in Trial II. Use 0.517 ss for the value of the period.
Trial 1 Spring constant is 117N/m, period of oscillations .37s, mass of the block is .400kg .
Trial 2 oscillation period is .52s
Answer:
[tex]M_2=0.79kg[/tex]
Explanation:
From the question we are told that:
Period [tex]T=0.517s[/tex]
Trial 1
Spring constant [tex]\mu=117N/m[/tex]
Period [tex]T_1=0.37[/tex]
Mass [tex]m=0.400kg[/tex]
Trial 2
Period [tex]T_2=0.52[/tex]
Generally the equation for Spring Constant is mathematically given by
\mu=\frac{4 \pi^2 M}{T^2}
Since
[tex]\mu _1=\mu_2[/tex]
Therefore
[tex]\frac{4 \pi^2 M_1}{T_1^2}=\frac{4 \pi^2 M_2}{T_2^2}[/tex]
[tex]M_2=M_1*(\frac{T_2}{T_1})^2[/tex]
[tex]M_2=0.400*(\frac{0.52}{0.37}})^2[/tex]
[tex]M_2=0.79kg[/tex]
a baseball is thrown vertically upward with an initial velocity of 20m/s.
A,what maximum height will it attain? B,what time will elapse before it strike the ground?
C,what is the velocity just before it strike the ground?
Answer:
Look at explanation
Explanation:
a)Only force acting on the object is gravity, so a=-g (consider up to be positive)
use: v^2=v0^2+2a(y-y0)
plug in givens, at max height v=0
0=400-19.6(H)
Solve for H
H= 20.41m
b) Use: y=y0+v0t+1/2at^2
Plug in givens
0=0+20t-4.9t^2
solve for t
t=4.08 seconds
c) v=v0+at
v=20-39.984= -19.984m/s
which vector best represents the net force acting on the +3 C charge
Which simple machine is shown in the diagram?
a wedge
a screw
an inclined plane
a wheel and axle
Answer:
Wheel and axle
Explanation:
Which simple machine is shown in the diagram?
a wheel and axle
From the given diagram, the machine shown is actually a wheel and axle
Description of wheel and axle
The wheel and axle is a machine consisting of a wheel attached to a smaller axle so that these two parts rotate together in which a force is transferred from one to the other.
Answer:
Wheel and axle
Explanation:
Drawing a shows a displacement vector (450.0 m along the y axis). In this x, y coordinate system the scalar components are Ax 0 m and Ay 450.0 m. Suppose that the coordinate system is rotated counterclockwise by 35.0, but the magnitude (450.0 m) and direction of vector remain unchanged, as in drawing b. What are the scalar components, Ax and Ay, of the vector in the rotated x, y coordinate system
Answer:
x ’= 368.61 m, y ’= 258.11 m
Explanation:
To solve this problem we must find the projections of the point on the new vectors of the rotated system θ = 35º
x’= R cos 35
y’= R sin 35
The modulus vector can be found using the Pythagorean theorem
R² = x² + y²
R = 450 m
we calculate
x ’= 450 cos 35
x ’= 368.61 m
y ’= 450 sin 35
y ’= 258.11 m
A body initially at rest travels a distance 100 m in 5 s with a constant acceleration. calculate
(i) Acceleration
(ii) Final velocity at the end of 5 s.
Answer:
(i)8m/s²(ii)40m/s
Explanation:
according to the formula
½at²=s.
then substituting the data
½a•5²=100
a=8m/s²
v=at=8•5=40m/s
Answer:
(I)
[tex]{ \bf{s = ut + \frac{1}{2} a {t}^{2} }} \\ 100 = (0 \times 5) + \frac{1}{2} \times a \times {5}^{2} \\ 200 = 25a \\ { \tt{acceleration = 8 \: m {s}^{ -2} }}[/tex]
(ii)
[tex]{ \bf{v = u + at}} \\ v = 0 + (8 \times 5) \\ { \tt{final \: velocity = 40 \: m {s}^{ - 1} }}[/tex]
PLZ help asap :-/
............................
Explanation:
[16][tex]\underline{\boxed{\large{\bf{Option \; A!! }}}} [/tex]
Here,
[tex]\rm { R_1} [/tex] = 2Ω[tex]\rm { R_2} [/tex] = 2Ω[tex]\rm { R_3} [/tex] = 2Ω[tex]\rm { R_4} [/tex] = 2ΩWe have to find the equivalent resistance of the circuit.
Here, [tex]\rm { R_1} [/tex] and [tex]\rm { R_2} [/tex] are connected in series, so their combined resistance will be given by,
[tex]\longrightarrow \rm { R_{(1,2)} = R_1 + R_2} \\ [/tex]
[tex]\longrightarrow \rm { R_{(1,2)} = (2 + 2) \; Omega} \\ [/tex]
[tex]\longrightarrow \rm { R_{(1,2)} = 4 \; Omega} \\ [/tex]
Now, the combined resistance of [tex]\rm { R_1} [/tex] and [tex]\rm { R_2} [/tex] is connected in parallel combination with [tex]\rm { R_3} [/tex], so their combined resistance will be given by,
[tex]\longrightarrow \rm {\dfrac{1}{ R_{(1,2,3)}} = \dfrac{1}{R_{(1,2)}} + \dfrac{1}{R_3} } \\ [/tex]
[tex]\longrightarrow \rm {\dfrac{1}{ R_{(1,2,3)}} = \Bigg ( \dfrac{1}{4} + \dfrac{1}{2} \Bigg ) \;\Omega} \\ [/tex]
[tex]\longrightarrow \rm {\dfrac{1}{ R_{(1,2,3)}} = \Bigg ( \dfrac{1 + 2}{4} \Bigg ) \;\Omega} \\ [/tex]
[tex]\longrightarrow \rm {\dfrac{1}{ R_{(1,2,3)}} = \Bigg ( \dfrac{3}{4} \Bigg ) \;\Omega} \\ [/tex]
Reciprocating both sides,
[tex]\longrightarrow \rm {R_{(1,2,3)}= \dfrac{4}{3} \;\Omega} \\ [/tex]
Now, the combined resistance of [tex]\rm { R_1} [/tex], [tex]\rm { R_2} [/tex] and [tex]\rm { R_3} [/tex] is connected in series combination with [tex]\rm { R_4} [/tex]. So, equivalent resistance will be given by,
[tex]\longrightarrow \rm {R_{(1,2,3,4)}= R_{(1,2,3)} + R_4} \\ [/tex]
[tex]\longrightarrow \rm {R_{(1,2,3,4)}= \Bigg ( \dfrac{4}{3} + 2 \Bigg ) \; \Omega} \\ [/tex]
[tex]\longrightarrow \rm {R_{(1,2,3,4)}= \Bigg ( \dfrac{4 + 6}{3} \Bigg ) \; \Omega} \\ [/tex]
[tex]\longrightarrow \rm {R_{(1,2,3,4)}= \Bigg ( \dfrac{10}{3} \Bigg ) \; \Omega} \\ [/tex]
[tex]\longrightarrow \bf {R_{(1,2,3,4)}= 3.33 \; \Omega} \\ [/tex]
Henceforth, Option A is correct.
_________________________________[17][tex]\underline{\boxed{\large{\bf{Option \; B!! }}}} [/tex]
Here, we have to find the amount of flow of current in the circuit. By using ohm's law,
[tex] \longrightarrow [/tex] V = IR
[tex] \longrightarrow [/tex] 3 = I × 3.33
[tex] \longrightarrow [/tex] 3 ÷ 3.33 = I
[tex] \longrightarrow [/tex] 0.90 Ampere = I
Henceforth, Option B is correct.
____________________________[tex] \tt \purple{Hope \; it \; helps \; you, Army! \heartsuit } \\ [/tex]
A 36.0 kg child slides down a long slide in a playground. She starts from rest at a height h1 of 24.00 m. When she is partway down the slide, at a height h2 of 11.00 m, she is moving at a speed of 7.80 m/s.
Calculate the mechanical energy lost due to friction (as heat, etc.).
Answer:
E = 3495.96 J
Explanation:
From the law of conservation of energy:
Total Mechanical Energy at h1 = Total Mechanical Energy at h2
Kinetic energy at h1 + potential energy at h1 = Kinetic energy at h2 + potential energy at h2 + Mechanical Energy Lost due to Friction
[tex]K.E_{h1}+P.E_{h1} = K.E_{h2}+P.E_{h2} + E\\\\\frac{1}{2}mv_1^2\ J + mgh_1 = \frac{1}{2}mv_2^2 + mgh_2 + E\\\\\frac{1}{2}(36\ kg)(0\ m/s)_1^2\ J + (36\ kg)(9.81\ m/s^2)(24\ m)_1 = \frac{1}{2}(36\ kg)(7.8\ m/s)_2^2 + (36\ kg)(9.81\ m/s^2)(11\ m)_2 + E\\\\0\ J + 8475.84\ J = 1095.12\ J + 3884.76\ J + E\\E = 8475.84\ J - 1095.12\ J - 3884.76\ J\\[/tex]
E = 3495.96 J
A uniform horizontal bar of mass m1 and length L is supported by two identical massless strings. String A Both strings are vertical. String A is attached at a distance d
Answer:
a) T_A = [tex]\frac{g}{d}\ ( m_2 x + m_1 \ \frac{L}{2} )[/tex] , b) T_B = g [m₂ ( [tex]\frac{x}{d} -1[/tex]) + m₁ ( [tex]\frac{L}{ 2d} -1[/tex]) ]
c) x = [tex]d - \frac{m_1}{m_2} \ \frac{L}{2d}[/tex], d) m₂ = m₁ ( [tex]\frac{ L}{2d} -1[/tex])
Explanation:
After carefully reading your long sentence, I understand your exercise. In the attachment is a diagram of the assembly described. This is a balancing act
a) The tension of string A is requested
The expression for the rotational equilibrium taking the ends of the bar as the turning point, the counterclockwise rotations are positive
∑ τ = 0
T_A d - W₂ x -W₁ L/2 = 0
T_A = [tex]\frac{g}{d}\ ( m_2 x + m_1 \ \frac{L}{2} )[/tex]
b) the tension in string B
we write the expression of the translational equilibrium
∑ F = 0
T_A - W₂ - W₁ - T_B = 0
T_B = T_A -W₂ - W₁
T_ B = [tex]\frac{g}{d}\ ( m_2 x + m_1 \ \frac{L}{2} )[/tex] - g m₂ - g m₁
T_B = g [m₂ ( [tex]\frac{x}{d} -1[/tex]) + m₁ ( [tex]\frac{L}{ 2d} -1[/tex]) ]
c) The minimum value of x for the system to remain stable, we use the expression for the endowment equilibrium, for this case the axis of rotation is the support point of the chord A, for which we will write the equation for this system
T_A 0 + W₂ (d-x) - W₁ (L / 2-d) - T_B d = 0
at the point that begins to rotate T_B = 0
g m₂ (d -x) - g m₁ (0.5 L -d) + 0 = 0
m₂ (d-x) = m₁ (0.5 L- d)
m₂ x = m₂ d - m₁ (0.5 L- d)
x = [tex]d - \frac{m_1}{m_2} \ \frac{L}{2d}[/tex]
d) The mass of the block for which it is always in equilibrium
this is the mass for which x = 0
0 = d - \frac{m_1}{m_2} \ \frac{L}{2d}
[tex]\frac{m_1}{m_2} \ (0.5L -d) = d[/tex]
[tex]\frac{m_1}{m_2} = \frac{ d}{0.5L-d}[/tex]
m₂ = m₁ [tex]\frac{0.5 L -d}{d}[/tex]
m₂ = m₁ ( [tex]\frac{ L}{2d} -1[/tex])
find the distance travelled by a moving body if it attained acceleration of 2m/s2 after starting from rest in 5min
Answer:
300 meters
Explanation:
a= 2m/s^2
t= 5 min
Convert into seconds, 5*60= 300seconds
v0= 0
x0=0
use x-x0= v0t + 1/2at^2
plug in values
x= 1/2*2*(300)
Solve
x= 300 meters
A car accelerates at 2 meters/s/s. Assuming the car starts from rest how far will it travel in 10 seconds
Answer:
Distance = velocity x time, so 10 m/s X 10 s = 100 m
Explanation:
If you accelerate at 2 m/s^2 for 10 seconds, at the end of the 10 seconds you are moving at a rate of 20 m/s.
V(f) = V(i) + a*t
Final velocity = initial velocity + acceleration x time
Your average velocity will be half of your final, because you accelerated at a constant rate. So your average velocity is 10 m/s.
Distance = velocity x time, so 10 m/s X 10 s = 100 m
Answer:
100 m
Explanation:
Given,
Initial velocity ( u ) = 0 m/s
Acceleration ( a ) = 2 m/s^2
Time ( t ) = 10 sec s
To find : Displacement ( s ) = ?
By 2nd equation of motion,
s = ut + at^2 / 2
= ( 0 ) ( 10 ) + ( 2 ) ( 10 )^2 / 2
= 0 + ( 2 ) ( 100 ) / 2
= 200 / 2
s = 100 m
g A spherical container of inner diameter 0.9 meters contains nuclear waste that generates heat at the rate of 872 W/m3. Estimate the total rate of heat transfer from the container to its surroudings ignoring radiation.
Answer: The total rate of heat transfer from the container to its surroundings ignoring radiation is 332.67 W.
Explanation:
Given: Inner diameter = 0.9 m
q = 872 [tex]W/m^{3}[/tex]
Now, radii is calculated as follows.
[tex]r = \frac{diameter}{2}\\= \frac{0.9}{2}\\= 0.45 m[/tex]
Hence, the rate of heat transfer is as follows.
[tex]Q = q \times V[/tex]
where,
V = volume of sphere = [tex]\frac{4}{3} \pi r^{3}[/tex]
Substitute the values into above formula as follows.
[tex]Q = q \times \frac{4}{3} \pi r^{3}\\= 872 W/m^{3} \times \frac{4}{3} \times 3.14 \times (0.45 m)^{3}\\= 332.67 W[/tex]
Thus, we can conclude that the total rate of heat transfer from the container to its surroundings ignoring radiation is 332.67 W.
