Answer:
m = 876.71 kg
Explanation:
This is an exercise of Archimedes' principle, which states that the thrust on a body is equal to the weight of the dislodged liquid
B = ρ g V
therefore the load that the balloon can lift is
B - W_structure - w_load = 0
w_load = B - W_structure
The volume of the balloon is
v = 4/3 π r³
let's substitute
w_carga = rho g 4/3 π r³ - m_structure g
the air density at T = 25ºc is ρ = 1.18 kg / m³
let's calculate
w_load = 1.18 9.8 4/3 π 7.15³ - 930 9.8
w_load = 17705,77 - 9114
w_ load = 8591.77 N
this corresponds to a mass of
w_load = m g
m = w_load / g
m = 8591.77 / 9.8
m = 876.71 kg
1. An object in motion tends to stay in
motion because it has ___?____. (inertia or
terminal velocity)
Answer: intertia
Explanation:
A piano of mass 852 kg is lifted to a height of 3.5 m. How much gravitational potential energy is added to the piano? Acceleration due to gravity is g = 9.8 m/s2 O A. 102,282 J O B. 29,224J O C. 2982 J O D. 304.3J SUSM
Answer:
29223.6J
Explanation:
Given parameters:
Mass of Piano = 852kg
Height of lifting = 3.5m
Unknown:
Gravitational potential energy = ?
Solution:
The gravitational potential energy of a body can be expressed as the energy due to the position of a body;
G.P.E = mgh
m is the mass
g is the acceleration due to gravity
h is the height
Now insert the given parameters and solve;
G.P.E = 852 x 9.8 x 3.5 = 29223.6J
True or False. A projectile is an object that once set in motion, continues in motion by its own inertia.
3. Sarah drops a 10.0 kg objed from a 200 m high bridge over a river. a) What work is done by the objec as a result of its fall?
Block A is also connected to a horizontally-mounted spring with a spring constant of 281 J/m2. What is the angular frequency (in rad/s) of simple harmonic oscillations of this system?
Answer:
This question is incomplete
Explanation:
This question is incomplete. However, the formula to be used here is
ω = 2π/T
Where ω is the angular frequency (in rad/s)
T is the period - the time taken for Block A to complete one oscillation and return to it's original position.
To solve for this period T, the formula below should be used
T = 2π√m/k
where m is the mass of the object (Block A) and k is the spring constant (281 J/m²)
Starting from rest, a coin and a ring roll down a ramp without slipping. Which of the following are true:
A. The ring reaches the bottom first
B. The coin reaches the bottom first
C. The coin and the ring arrive at the same time
D. The one the reaches the bottom first is the one with the largest mass
E. The one that reaches the bottom first is the one with the largest diameter.
Answer:
The answer is D
Explanation:
The object with the heavy mass will reach first because the cause it is heaver so it will go faster
22. What happens to the volume of a 1 kg of water when it is heated from 4oC to 6oC? A. Increases B. Decreases C. Stays the same
Answer:
a
Explanation:
..................
Squid use jet propulsion for rapid escapes. A squid pulls water into its body and then rapidly ejects the water backward to propel itself forward. A 1.5 kg squid (not including water mass) can accelerate at 20 m/s2 by ejecting 0.15 kg of water. Part A What is the magnitude of the thrust force on the squid
Answer: see attachment
Explanation:
Which statement best describes energy and matter in a closed system? (2 po
O Energy and matter flow into and out of the system.
Energy can flow into and out of the system but matter cannot.
Energy and matter are contained within a closed system.
O There is no energy in a closed system; there is matter.
Answer:
Energy can flow into and out of the system but matter cannot.
Explanation:
In a closed system, energy can flow in and out of the system but matter cannot.
A closed system prevents double way flow of matter. A closed system conserves matter.For an isolated system, energy and matter cannot flow out of the system.
For open systems, energy and matter can flow out of the system.
Such systems are used for certain thermodynamics experiment.
Two tiny conducting spheres are identical and carry charges of -20μC and +50μC. They are separeted by a distance of 2.50cm. (a) what is the magnitude of the force that each sphere each sphere experience, and is the force attractive or repulsive ? (b) The spheres are brought into contact and then separated toa distance of 2.50cm. Determine the magnitude of the force that each sphere now experiences, and state whether the force is attractive or repulsive.
Answer:
[tex]14400\ \text{N}[/tex], Attractive
[tex]3240\ \text{N}[/tex], Repulsive
Explanation:
[tex]q_1[/tex] = -20 μC
[tex]q_2[/tex] = 50 μC
r = Distance between the charges = 2.5 cm
k = Coulomb constant = [tex]9\times 10^9\ \text{Nm}^2/\text{C}^2[/tex]
Electrical force is given by
[tex]F=\dfrac{kq_1q_2}{r^2}\\\Rightarrow F=\dfrac{9\times 10^9\times (-20\times 10^{-6})\times (50\times 10^{-6})}{(2.5\times10^{-2})^2}\\\Rightarrow F=-14400\ \text{N}[/tex]
The magnitude of force each sphere will experience is [tex]14400\ \text{N}[/tex]
Since the charges have opposite charges they will attract each other.
Now the charges are brought into contact with each other so the resultant charge will be
[tex]q=\dfrac{q_1+q_2}{2}\\\Rightarrow q=\dfrac{-20+50}{2}\\\Rightarrow q=15\ \mu\text{C}[/tex]
[tex]F=\dfrac{kq^2}{r^2}\\\Rightarrow F=\dfrac{9\times 10^9\times (15\times 10^{-6})^2}{(2.5\times 10^{-2})^2}\\\Rightarrow F=3240\ \text{N}[/tex]
The magntude of the force the spheres experience will be [tex]3240\ \text{N}[/tex]
The spheres have the same charge now so they will repel each other.
when is thermal equilibrium achived between two identical objects
need help ASAP
Answer: When two objects in contact with each other are at different temperatures, they are said to be in thermal equilibrium.
Explanation: . When two objects not in contact with each other are at the same pressure, they are said to be in thermal equilibrium.
Why are the coral reefs suffering? (site 2) explain
Answer:
bcuz ov yo fat mamma
Explanation:
Answer:
Water pollution
Explanation:
A displacement vector with a magnitude of 20. meters could have perpendicular components with magnitudes of A. 10. m and 10. m B. 12 m and 8.0 m 12 m and 16 m D. 16 m and 8.0 m
Answer:10.m and 10. M
Explanation:
A displacement vector with a magnitude of 20. m could have perpendicular components with magnitudes of C. 12 m and 16 m.
A displacement vector with a magnitude of 20. meters can be decomposed in 2 perpendicular components.
They would form a right triangle, in which the displacement vector would be the hypotenuse (a) and the components would be the legs (b, c).
Given the magnitude of the legs, we can calculate the magnitude of the hypotenuse using the Pythagorean theorem.
[tex]c = \sqrt{a^{2} + b^{2} }[/tex]
Let's use this formula to calculate the displacement vector for each pair of legs.
A. 10. m and 10. m[tex]c = \sqrt{a^{2} + b^{2} } = \sqrt{(10.m)^{2} + (10.m)^{2} }= 14.1m[/tex]
B. 12 m and 8.0 m[tex]c = \sqrt{a^{2} + b^{2} } = \sqrt{(12m)^{2} + (8.0m)^{2} }= 14.4m[/tex]
C. 12 m and 16 m[tex]c = \sqrt{a^{2} + b^{2} } = \sqrt{(12m)^{2} + (16m)^{2} }= 20m[/tex]
D. 16 m and 8.0 m[tex]c = \sqrt{a^{2} + b^{2} } = \sqrt{(16m)^{2} + (8.0m)^{2} }= 17.9m[/tex]
A displacement vector with a magnitude of 20. m could have perpendicular components with magnitudes of C. 12 m and 16 m.
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A wire carries a current of 11.4 A in a direction that makes an angle of 11.4° with a magnetic field of magnitude 11.4 à 10-3 T. The magnitude of the force on 11.4 cm of this wire is:____.a) 11.4 * 10^-3 N
b) 0.130 N
c) 1.48 * 10^-2 N
d) 2.93 * 10^-3 N
Answer:
(d) 2.93 x 10⁻³ N
Explanation:
Given;
current in the wire, I = 11.4 A
angle of inclination, θ = 11.4⁰
magnetic field on the wire, B = 11. 4 x 10⁻³
length of the wire, L = 11.4 cm = 0.114 m
The magnitude of magnetic force on the wire is given by;
F = BILsinθ
F = (11.4 x 10⁻³)(11.4)(0.114)(sin 11.4°)
F = 0.00293 N
F = 2.93 x 10⁻³ N
Therefore, the correct option is "D"
Your teacher placed a 3.5 kg block at the position marked with a “ + ” (horizontally, 0.5 m from the origin) on a large incline outlined on the graph below and let it slide, starting from rest. ***There are two images included!***
Answer:
x = 10.75 m
Explanation:
For this problem we will solve it in two parts, the first using energy and the second with kinematics
Let's use the energy work relationship to find the velocity of the block as it exits the ramp
W = [tex]Em_{f}[/tex] - Em₀
Starting point. Higher
Em₀ = U = m g h
the height from the edge of the ramp of the graph has a value
h = 9-3 = 6 m
Final point. At the bottom of the ramp
Em_{f} = K = ½ m v²
Friction force work
W = - fr d
The friction force has the formula
fr = μ N
On the ramp, we can use Newton's second law
N - W cos θ = 0
N = W cos θ
where the angle is obtained from the graph
tan θ = (9-3) / (0.5-4) = -6 / 3.5
θ = tan⁻¹ (-1,714)
θ = -59.7º
the distance d is
d = √ (Δx² + Δy²)
d = √ [(0.5-4)² + (9-3)²]
d = 6.95 m
for which the work is
W = - μ mg cos 59.7 d
we substitute
W = Em_{f} -Em₀
- μ mg cos 59.7 d = ½ m v² - m g h
In the graph o text the value of the friction coefficient is not observed, suppose that it is μvery = 0.2
- μ g cos 59.7 d = ½ v² - g h
v² = 2g (h - very d coss 59.7)
let's calculate
v² = 2 9.8 (6 - 0.2 6.95 cos 59.7)
v = √ 103.8546
v = 10.19 m / s
in the same direction as the ramp
in the second part we use projectile launch kinematics
let's look for the components of velocity
v₀ₓ = vo cos -59.7
[tex]v_{oy}[/tex] = vo sin (-59,7)
v₀ₓ = 10.19 cos (-59.7) = 5.14 m / s
v_{oy} = 10.19 if (-59.7) = -8.798 m / s
Let's find the time to get to the floor (y = o)
y = y₀ + v_{oy} t - ½ g t²
to de groph y₀=3 m
0 = 3 - 8.798 t - ½ 9.8 t²
t² - 1.796 t - 0.612 = 0
we solve the quadratic equation
t = [1.796 ±√(1.796² + 4 0.612)] / 2
t = [1,795 ± 2,382] / 2
t₁ = 2.09 s
t₂ = -0.29 s
since time must be a positive quantity the correct value is t = 2.09 s
we calculate the horizontal displacement
x = v₀ₓ t
x = 5.14 2.09
x = 10.75 m
The motion of the box, after it exits the incline is the motion and trajectory
of a projectile.
Horizontal distance from the right-hand edge of the incline to the point of
contact with the floor is approximately 1.24613 m.
Reasons:
Mass of the block, m = 3.5 kg
Coefficient of kinetic friction, μ = 1.2
Location of the = 0.5 m from the origin
Required:
Horizontal distance between the block's point of contact with the floor and
the bottom right-hand edge of the incline.
Solution:
Let θ represent the angle the incline make with the horizontal.
The normal reaction of the incline on the block, [tex]F_N[/tex] = m·g·cos(θ)
Work done on friction = [tex]F_N[/tex]×μ×Length of the incline, L
Rise of the incline = 10 - 3 = 7
Run of the incline = 4
L = √(6.125² + 3.5²) = [tex]\dfrac{7 \times \sqrt{65} }{8}[/tex]
Let ΔP.E.₁ represent the potential energy transferred to kinetic energy
and work along the incline, we have;
Energy of the block at the bottom of the incline, M.E.₂, is found as follows;
K.E.₂ = mgh - m·g·μ·cos(θ)·L
[tex]K.E. =\frac{1}{2} \times 3.5 \times v^2 = 3.5 \times 9.81 \times 6.125 - 3.5 \times 9.81 \times 1.2 \times \dfrac{4}{\sqrt{65} } \times \dfrac{7 \times \sqrt{65} }{8}[/tex]
v ≈ 6.1456 m/s
The vertical component of the velocity is therefore;
[tex]v_y = v \cdot sin(\theta)[/tex]
[tex]v_y = 6.1456 \times \dfrac{7}{\sqrt{65} } \approx 5.33588[/tex]
From the equation, h = u·t + 0.5·g·t² derived from Newton's Laws of motion, we have;
ΔP.E.₁ = 3.5×9.81×7
3 = 5.33588·t + 0.5×9.81·t²
Factorizing, the above quadratic equation, we get;
The time it takes the block to reach the floor, t ≈ 0.40869 seconds
Horizontal component of the velocity is [tex]v_x \approx 6.1456 \times \dfrac{4}{\sqrt{65} } \approx 3.04908[/tex]
The horizontal distance, x = vₓ × t
∴ x = 3.04908 × 0.40869 ≈ 1.08194
Horizontal distance from the right-hand edge of the incline to the point of
contact with the floor, x ≈ 1.24613 m.
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Question 1 of 5
In which way are electromagnetic waves different from mechanical waves?
Answer:
Electromagnetic waves differ from mechanical waves in that they do not require a medium to propagate. This means that electromagnetic waves can travel not only through air and solid materials, but also through the vacuum of space.
Explanation:
Answer:
In which way are electromagnetic waves different from mechanical waves?Electromagnetic waves can travel through empty space
Explanation:
I took A P E X Quiz.
Which of the following descriptions best describe a liquid
A
takes the shape and volume of its container
B
matter is made of atoms so tightly packed together that they cannot move around
C
has a definite volume, but takes the shape of its container
Answer:
C
Explanation:
A crossbow is fired horizontally off a cliff with an initial velocity of 15 m/s. If the arrow takes 4s to hit the ground, what is the range of the projectile?
Answer:
The range of the projectile is 60 m
Explanation:
Horizontal Motion
When an object is thrown horizontally with a speed vo from a height h, it describes a curved path ruled exclusively by gravity until it eventually hits the ground.
The horizontal component of the velocity is always constant because no acceleration acts in that direction, thus:
[tex]v_x=v_o[/tex]
The vertical component of the velocity changes in time because gravity makes the object fall at increasing speed given by:
[tex]v_y=g.t[/tex]
The horizontal distance is calculated as a constant speed motion:
[tex]x = v_x.t[/tex]
Knowing the crossbow is fired horizontally at vo=vx=15 m/s and it takes t=4 s to hit the ground, thus the range of the projectile is:
x = 15*4 = 60
The range of the projectile is 60 m
A 1200 N force acts on an object, resulting in an acceleration of 8.0 m/s2. What is the mass of the object?
Answer:
The answer is 150 kgExplanation:
The mass of the object can be found by using the formula
[tex]m = \frac{f}{a} \\ [/tex]
f is the force
a is the acceleration
From the question we have
[tex]m = \frac{1200}{8} \\ [/tex]
We have the final answer as
150 kgHope this helps you
The propeller of an aircraft accelerates from rest with an angular acceleration α = 7t + 8, where α is in rad/s2 and t is in seconds. What is the angle in radians through which the propeller rotates from t = 1.00 s to t = 6.10 s?
Answer:
The value is [tex]\theta =407.3 \ radian[/tex]
Explanation:
From the question we are told that
The angular acceleration is [tex]\alpha = (7t + 8) \ rad/ s^2[/tex]
The first time is [tex]t_1 = 1.00 \ s[/tex]
The second time [tex]t_2 = 6.10 \ s[/tex]
Generally the angular velocity is mathematically represented as
[tex]w = \int\limits {\alpha } \, dt[/tex]
=> [tex]w = \int\limits {7t + 8 } \, dt[/tex]
=> [tex]w =\frac{ 7t^2}{2} + 8 t[/tex]
Generally the angular displacement is mathematically represented as
[tex]\theta = \int\limits^{t_2}_{t_1} { w} \, dt[/tex]
=> [tex]\theta = \int\limits^{t_2}_{t_1} { \frac{7t^2}{2} + 8t } \, dt[/tex]
=> [tex]\theta = { \frac{7t^3}{6} + \frac{8t^2}{2} } | \left \ t_2} \atop {t_1}} \right.[/tex]
=> [tex]\theta = { \frac{7t^3}{6} + 4t^2} } | \left \ 6.10} \atop {1}} \right.[/tex]
=> [tex]\theta =[ { \frac{7}{6}[6.10 ]^3 + 4[6.10]^2} } ] -[ { \frac{7}{6}[1 ]^3 + 4[1]^2} } ][/tex]
=> [tex]\theta =407.3 \ radian[/tex]
A 6.0-kilogram cart initially traveling at 4.0 meters per second east accelerates uniformly at 0.50 meter per second squared east for 3.0 seconds. What is the speed of the cart at the end of this 3.0 second interval? A. 1.5 m/s B. 5.5 m/s G. 3.0 m/s D. 7.0 m/s
Answer: 5.5m/s
Explanation:
vf=vi+at
vf= 4.0m/s + (0.50m/s^2)(3.0s)
The speed of the cart at the end of this 3.0 second interval is 5.5 meter per seconds.
Given the following data:
Initial velocity = 4 m/sMass of cart = 6 KgAcceleration = 0.5 [tex]m/s^2[/tex]Time = 3 secondsTo find the speed of the cart at the end of this 3.0 second interval, we would use the first equation of motion;
[tex]V = U + at[/tex]
Where:
U is the initial velocity.V is the final velocity. a is the acceleration. t is the time measured in seconds.Substituting the given parameters into the formula, we have;
[tex]V = 4 + 0.5(3)\\\\V = 4 + 1.5[/tex]
Final velocity, V = 5.5 m/s.
Therefore, the speed of the cart at the end of this 3.0 second interval is 5.5 meter per seconds.
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You are lifting a 10 kg block straight up at a constant speed of 10 m/s. How much force are you exerting on the block?
Answer:
The force exerted is [tex]F = 100 \ N[/tex]
Explanation:
From the question we are told that
The mass of the block is [tex]m_b = 10 \ kg[/tex]
The speed is [tex]v = 10 \ m/s[/tex]
Generally the force exerted to lift the object at constant speed is equivalent to the wight of the ball, this is mathematically represented as
[tex]F = m * g[/tex] Here [tex]g = 10 \ m/s^2[/tex]
=> [tex]F = 10 * 10[/tex]
=> [tex]F = 100 \ N[/tex]
The force are you exerting on the block when the block is lifting straight up with constant speed is 98 N and this can be determined by using the given data.
Given :
You are lifting a 10 kg block straight up at a constant speed of 10 m/s.
The following steps can be used in order to determine the force are you exerting on the block:
Step 1 - According to the given data, the block is lifting straight up at a constant speed. So, the acceleration is zero.
Step 2 - So, the only force exerted on the block is the weight of the block.
Step 3 - So, the force are you exerting on the block is given by:
F = mg
F = 10 [tex]\times[/tex] 9.8
F = 98 N
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What specific changes in two climate variables are expected to lead to major decreases in soil moisture southern Africa and the Mediterranean region?
Answer:
Less precipitation, droughts9: How might agriculture in southern Europe change by the end of the century if conditions follow the RCP8.
Explanation:
Precipitation and droughts are the specific changes in two climate variables that are expected to lead to major decreases in soil moisture.
What is drought?
Drought is defined as a period of protracted water scarcity, whether it is due to atmospheric surface water, or groundwater constraints.
Droughts can last months or years, although they can be proclaimed in as little as 15 days.
It has the potential to have a significant influence on the afflicted region's ecology and agriculture as well as harm the local economy.
Precipitation and droughts are the specific changes in two climate variables that are expected to lead to major decreases in soil moisture in southern Africa and the Mediterranean region.
Hence Precipitation and droughts are the specific changes in two climate variables.
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An object that moves in uniform circular motion has a centripetal acceleration of 13 m/s^2 . If the radius of the motion is 0.02m, what is the frequency of motion?
Answer:
f = 3.97 Hz
Explanation:
Given that,
Centripetal acceleration, [tex]a=13\ m/s^2[/tex]
The radius of motion is 0.02 m
The formula for the centripetal acceleration is given by :
[tex]a=\dfrac{v^2}{r}\\\\v=\sqrt{ar} \\\\v=\sqrt{13\times 0.02} \\\\v=0.5\ m/s[/tex]
The speed of an object in a circular path is given by :
[tex]v=\dfrac{2\pi r}{t}[/tex]
t is time period
Also, f=1/t (f is frequency)
[tex]f=\dfrac{v}{2\pi r}\\\\f=\dfrac{0.5}{2\pi \times 0.02}\\\\f=3.97\ Hz[/tex]
Hence, the frequency of motion s 3.97 Hz.
The frequency of the motion is 4.1 Hz.
Linear velocity?The linear velocity of the of the object is calculated as follows;
[tex]a = \frac{v^2}{r} \\\\v^2 = ar\\\\v = \sqrt{ar} \\\\v = \sqrt{13 \times 0.02} \\\\v = 0.51 \ m/s[/tex]
Angular speed of the objectThe angular speed of the object is calculated as follows;
[tex]\omega =\frac{v}{r} \\\\\omega = \frac{0.51}{0.02} \\\\\omega = 25.5 \ rad/s[/tex]
Frequency of motionThe frequency of the motion is calculated as follows;
[tex]\omega = 2\pi f\\\\f = \frac{\omega }{2\pi} \\\\f = \frac{25.5}{2\pi } \\\\f = 4.1 \ Hz[/tex]
Thus, the frequency of the motion is 4.1 Hz.
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<
Question
An airplane flies eastward and always accelerates at a constant rate. At one position along its path, it has a velocity of
32.7 m/s. It then flies a further distance of 44500 m, and afterwards, its velocity is 50.3 m/s. Find the
airplane's acceleration
acceleration:
.016m/s2
Calculate how much time clapses while the airplane covers those 44500 m
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49 ENG
1:45 PM
9/17/2020
Answer:
Explanation:
initial velocity u = 32.7 m /s
final velocity v = 50.3 m /s
displacement s = 44500 m
acceleration a = ?
v² = u² + 2 a s
50.3² = 32.7² + 2 x a x 44500
2530.09 = 1069.29 + 89000a
a .016 m /s²
time taken t = ?
v = u + at
50.3 = 32.7 + .016 t
t = 1100 s
A rack of seven spherical bolwing balls (each 7.00 kg, radius of 9.50 cm) is positioned along a line located a distance =0.850 m from a point ,
Calculate the gravitational force
the bowling balls exert on a ping-pong ball of mass 2.70 g, centered at point .
Answer:
8.72*[tex]10^{-12}[/tex] N
Explanation:
force of attraction f = G m1m2/ r^2 = [tex]\frac{6.67*10^{-11}*7*5*2.70 }{.85*.85*1000}[/tex] = 8.72*[tex]10^{-12}[/tex] N
The gravitational force the bowling balls exert on a ping-pong is given as 8.72*[tex]10^{-12}[/tex] N.
What is force?A force is an effect that can alter an object's motion according to physics. An object with mass can change its velocity, or accelerate, as a result of a force. An obvious way to describe force is as a push or a pull. A force is a vector quantity since it has both magnitude and direction.
force of attraction
f = G [tex]m_1m_2[/tex]/r²
= [tex]6.67*10^{-11}[/tex]*7*5*2.70/.85²*1000
= 8.72*[tex]10^{-12}[/tex] N
The gravitational force the bowling balls exert on a ping-pong is given as 8.72*[tex]10^{-12}[/tex] N.
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#SPJ3
A car is moving at an average speed of 20 meters per second. This is equivalent to
Answer:
44.73 MP/H or 71.98 KM/H
Explanation:
I need emergency help we only have 3 minutes left
Answer:
Option A
Explanation:
Ast the force is equal and the diayance is equal the beam is also balanced
1. Opposite charges
O repel
attract
Answer:
attract
Explanation:
that is the answer
Answer:
Attract.
Explanation:
I took the quiz.
The speed of a space shuttle is 10 / express this in /�
Answer:
268.22m/s
Explanation:
Given;
10mile/min to m/s
We need to convert between the two units;
1 mile = 1609.34m
60s = 1min
Now;
10 x [tex]\frac{mile}{min}[/tex] x [tex]\frac{1min}{60s}[/tex] x [tex]\frac{1609.34m}{1mile}[/tex]
= 268.22m/s