Answer:
5.10×10⁻³ s
Explanation:
At terminal velocity, the acceleration is 0.
Sum of forces in the y direction:
∑F = ma
bv − mg = 0
mg = bv
m = bv / g
m/b = v/g
τ = v/g
τ = (5.00 cm/s) / (980 cm/s²)
τ = 5.10×10⁻³ s
What is the magnitude of the gravitational force between the earth and a 1 kg object on its surface? (Mass of the earth is 6 × 10 24 kg and radius of the earth is 6.4 × 10 6 m.)
Answer:
Explanation:
just use the gravational force equation which is G x m of earth x m of object divided by r squared (which is radius of earth)
Una tractomula se desplaza con rapidez de 69 km/h. Cuando el conductor ve una vaca atravesada enmedio de la carretera, acciona los frenos y se detiene 4 s después. Si la vaca estaba a 25 m de la tractomulacuando el conductor pisó el freno. ¿atropelló la vaca? Justifique su respuesta.
Answer:
Los datos que tenemos:
Rapidez: 69km/h
Tiempo que tarda en frenar = 4s.
Distancia inicial entre la tracto-mula y la vaca = 25m
Ok, la ecuación de desaceleración es:
D = (sf - si)/t
sf = velocidad final = 0m/s
si = velocidad inicial = 69km/h
t = tiempo = 4s
D = -69km/h/4s
ok, 1h = 3600s
D = (-69km/s)*1/(4*3600s) = -0.0048 km/s^2
Entonces la ecuación de aceleración es:
a(t) = -0.0048 km/s^2
Para la velocidad, integramos sobre el tiempo
v(t) = (-0.0048 km/s^2)*t + v0
donde v0 es la velocidad inicial, en este caso v0 = 69km/3600s = 0.0191km/s
v(t) = (-0.0048 km/s^2)*t + 0.0191km/s
Para la posición volvemos a integrar sobre el tiempo, esta vez suponemos la posición inicial igual a cero.
p(t) = (1/2)*(-0.0048 km/s^2)*t^2 + 0.0191m/s*t
Ahora, si p(t=4s) < 25m, esto implica que la tracto-mula no impacto con la vaca.
p(4s) = (1/2)*(-0.0048 km/s^2)*(4s)^2 + 0.0191km/s*4s = 0.038km
y 1km = 1000m
0.038km = 0.038*1000m = 38m
Entonces si, atropello a la vaca.
Please help asap. A soccer player can kick a 0.370 kg football at 55 km/h. How much work does the soccer player have to do on the ball in order to give it that much kinetic energy?
Answer: 43.2 J
Explanation:
Work = change in KE
initial KE = 0
final KE = 1/2mv^2 = 1/2(0.370 kg)(15.2778 m/s)^2 = 43.2 J
i'm not sure about sig figs though
A disk-shaped merry-go-round of radius 3.03 mand mass 145 kg rotates freely with an angular speed of 0.681 rev/s . A 65.4 kg person running tangential to the rim of the merry-go-round at 3.41 m/s jumps onto its rim and holds on. Before jumping on the merry-go-round, the person was moving in the same direction as the merry-go-round's rim. What is the final angular speed of the merry-go-round?
Answer:
[tex]\omega_2=0.891\ rev/s[/tex]
Explanation:
Given that
Radius , r= 3.03 m
Mass of disk , M= 145 kg
Initial angular velocity
ω=0.681 rev/s
Mass of person , m= 65.4 kg
Velocity of person , V= 3.41 m/s
Initial mass moment of inertia
[tex]I_1= \dfrac{M\times R^2}{2}[/tex]
[tex]I_1= \dfrac{145\times 3.03^2}{2}=665.61\ kg.m^2[/tex]
Final mass moment of inertia
[tex]I_2= \dfrac{M\times R^2}{2}+m\times R^2[/tex]
[tex]I_2= \dfrac{145\times 3.03^2}{2}+65.4\times 3.03^2=1266.04\ kg.m^2[/tex]
[tex]Final\ angular\ velocity =\omega_2[/tex]
By using angular momentum equation
[tex]I_1\times \omega+m\times V\times R=I_2\times \omega_2[/tex]
[tex]665.61\times 0.681+65.4\times 3.41\times 3.03=1266.04\times \omega_2[/tex]
[tex]1129.01= 1266.04\times \omega_2[/tex]
[tex]\omega_2=\dfrac{1129.01}{1266.04}[/tex]
[tex]\omega_2=0.891\ rev/s[/tex]
Thus the angular velocity will be 0.891 rev/s
The following passage has not been edited. There is an error in each line. Write the
incorrect word and the correction in your answer sheet against the correct question
number. The first one has been done as an example. ( 1 x 4 = 4 )
Community service sensitize people to Error: sensitize ; Correction: sensitizes
other‟s needs and supports inclusive (a) Error: _______ ; Correction: ______
development to the underprivileged (b) Error: _______ ; Correction: ______
sections with society. Courses about social (c) Error: ______ ; Correction: _______
work prepares frontline workers to (d) Error: _______ ; Correction: ______
Answer:
(a) Error: Other's ; Correction: Others'
(b) Error: to ; Correction: for
(c) Error: with ; Correction: of
(d) Error: prepares ; Correction: prepare
Explanation:
a) The error is in the word "other's" as the position of apostrophe is wrong, so the correct word will be "others'", it shows plural nouns.
b) The error is in the word "to", so the correct word will be "for" as for is use to talk about a purpose.
c) The error is in the word "with" and the correct word will be "of" as of indicates relationships between other words including things that made of other things.
d) The error is in the word "preapres" and the correct word will be "prepare".
The density of water is 1000 kg m^3. What is the value expressed in gcm^-3 units? please help me..
(1) 1000 (2) 100 (3) 1 (4) 0.1 (5) 0.01
Here's the neat, cool way to convert units like this:
-- 1 kilogram = 1,000 grams
-- 1 meter = 100 centimeters
So . . . . .
(1000 kg/m³) x (1000 g/kg) x (1 m/100 cm)³ =
(1,000 kg/m³) x (1,000 g/kg) x (1 m³/1,000,000 cm³) =
(1,000 x 1,000 x 1 / 1,000,000) (kg-g-m³ / m³-kg-cm³) = 1 g/cm³
Ratan took a thin, solid piece of material. When he put it in water, it rose and floated. He took it out of the water and cut out holes in it as shown below. When Ratan puts the material with holes back in the water, what will happen?
Answer:
The material will still remain afloat.
Explanation:
The fact that the material floated on water as a solid piece means that the material is less dense than water. When a material is less dense than another liquid material, the material floats in the denser material. Cutting holes in the material will not reduce the buoyancy of the material. If the material had not floated as a single solid piece, then it will be said to be denser than the liquid material, and in such a case, if a strategic hole is cut on it to reduce its weight while maintaining its immediate volume, then it can be made to float. This is the principle behind metal floating ships.
The specific latent heat of vaporisation of water is 2200 kJ kg, and the specific
heat capacity of water is 4200 J/ kg.K. A heating element is immersed in an
insulated cup of water. It takes five minutes to boil the water completely from an
initial temperature of 30°C. Assuming no heat is lost to the surroundings,
determine the mass of the water if the heating element is operating at 1000 W.
Answer:
0.12 kg
Explanation:
The amount of energy added is:
1000 W × (5 min × 60 s/min) = 300,000 J = 300 kJ
Heat to boil the water is:
q = mCΔT + mL
q = m (CΔT + L)
300 kJ = m (4.2 kJ/kg/K × (100°C − 30°C) + 2200 kJ/kg)
300 kJ = m (2494 kJ/kg)
m = 0.12 kg
how much heat is required to raise the temperature of 5kg of iron from 50°c to250°c
Answer:
462000J
Explanation:
Quantity of heat= mass x specific heat capacity of iron x change in temp
specific heat capacity of iron is 462J/Kg/K
change in temp = 250-50= 200°C
200°C is equivalent to 200K since 1°C is 1K
Q= mct
= 5x462x200
= 462000J
A 140-Hz sound travels through pure carbon dioxide. The wavelength of the sound is measured to be 1.92 m. What is the speed of sound in carbon dioxide?
Answer:
V = 268.8 m/s
Explanation:
The speed of a wave in general is given by the following formula:
V = fλ
where,
V = Speed of that wave
f = Frequency of the wave
λ = wavelength of the wave
In this case we have a sound wave, travelling across carbon dioxide. The properties of sound wave are as follows:
V = Speed of Sound in Carbon dioxide = ?
f = frequency of sound wave = 140 Hz
λ = wavelength of sound wave = 1.92 m
Therefore,
V = (140 Hz)(1.92 m)
V = 268.8 m/s
formula of minimmum pressure
Answer:
pressure=force/area
PLEASEEE HELP!!!!! I HAVE BEEN STRUGGLING FOR 2 DAYS
If i workout 90 minutes on earth, if I am on a rocket traveling 0.80c, according to the timer on the rocket, how long should I exercise?
Answer:
You should still workout 90 min.
The proper time is measured by a single clock in a single place.
The proper time on earth is 90 min.
The clock on the rocket is also in a single place in the frame of the rocket so you still need to workout for 90 min.
One star has a temperature of 30,000 K and another star has a temperature of 6,000 K. Compared to the cooler star, how much more energy per second will the hotter star radiate from each square meter of its surface?
Answer:
The hotter star radiates 625 times more energy per second from each square meter of its surface
Explanation:
Temperature of the hotter star is 30000 K
temperature of the cooler star = 6000 K
From Stefan-Boltzmann radiation laws, for a non black body
P = εσA[tex]T^{4}[/tex]
where
P is the energy per second or power of radiation
ε is the emissivity of the body
σ is the Stefan-Boltzmann constant of proportionality
A is the area of the sun
T is the temperature of the sun
The sun can be approximated as a black body, and the equation reduces to
P = σA[tex]T^{4}[/tex]
For the hotter body,
P = σA([tex]30000^{4}[/tex]) = 8.1 x 10^17σA J/s
For the cooler body,
P = σA([tex]6000^{4}[/tex]) = 1.296 x 10^15σA J/s
comparing the two stars energy
==> (8.1 x 10^17)/(1.296 x 10^15) = 625
This means that the hotter star radiates 625 times more energy per second from each square meter of its surface
The region of magnetic influence around either pole of a magnet is called the magnetic field. The magnetic field line points out from the south magnetic pole and in from the north magnetic pole. This statement is:
Answer:
This statement is not true
Explanation:
Because The normal magnetic field line points out from the north magnetic pole and in from the south magnetic pole.
Answer:
false
Explanation:
hope this helps :)
Which of the following object is in dynamic equilibrium?
Answer:
A car driving in a straight line 20 m/s
Explanation:
ayepecks silly
If y=5sin (3x -40)
Calculate the frequency and period
Answer:
0.477 Hz
2.09 s
Explanation:
y = A sin(ωx − φ)
A is the amplitude, ω is the angular frequency, and φ is the phase shift.
ω = 3 rad/s
f = ω / 2π ≈ 0.477 Hz
T = 1/f ≈ 2.09 s
pls help me with this question
Answer:
16 ms2 is the answer for this question
Experimenting with free fall, Mariana observes that her baseball takes 1.5 s to travel the last 30m before hitting the ground. From what height above the ground did Mariana drop the ball?
Answer:
37.8 m
Explanation:
At point 0, the ball is at height y₀.
At point 1, the ball is at height 30 m.
At point 2, the ball is at height 0 m.
Given:
y₁ = 30 m
y₂ = 0 m
v₀ = 0 m/s
a = -10 m/s²
t₂ − t₁ = 1.5 s
Find: y₀
Use constant acceleration equation.
y = y₀ + v₀ t + ½ at²
Evaluate at point 1.
y₁ = y₀ + v₀ t₁ + ½ at₁²
30 m = y₀ + (0 m/s) t₁ + ½ (-10 m/s²) t₁²
30 = y₀ − 5t₁²
Evaluate at point 2.
y₂ = y₀ + v₀ t₂ + ½ at₂²
0 m = y₀ + (0 m/s) t₂ + ½ (-10 m/s²) t₂²
0 = y₀ − 5t₂²
y₀ = 5t₂²
Substitute:
y₀ = 5 (1.5 + t₁)²
y₀ = 5 (2.25 + 3t₁ + t₁²)
y₀ = 11.25 + 15t₁ + 5t₁²
30 = 11.25 + 15t₁ + 5t₁² − 5t₁²
30 = 11.25 + 15t₁
t₁ = 1.25
30 = y₀ − 5t₁²
30 = y₀ − 5(1.25)²
y₀ ≈ 37.8
Which may result from an increase in friction?
A: decreased traction
B: increased speed
C: reduced wear and tear
D: generation of heat
Answer:
Its Generation of Heat. or "D"Explanation:
Friction causes generation of heat and causes increased wear and tear.
The friction is a resistive force and is related to heat energy, so an increase in friction results in the generation of heat, so option D is correct.
What is friction?Two solid objects cannot roll or slide over one another due to the force of friction. Although frictional forces can be useful, such as the traction needed to walk without slipping, they can also present a large degree of resistance to motion. Automobile engines need about 20% of their power to overcome frictional forces in moving parts.
The fundamental source of friction between metals appears to be the adhesion forces between the contact zones of the surfaces, which are always microscopically uneven. Friction is produced by shearing, these "welded" seams, and the rubbing action of the rougher, tougher surface against the softer, smoother surface.
The friction is a resistive force trying to oppose the force applied As friction is related to heat when we increase the heat, the friction increases, and vice versa.
To know more about Friction:
https://brainly.com/question/28356847
#SPJ6
A plane drops a package for delivery. The plane is flying horizontally at a speed of 120m/s,and the package travels 255 m horizontally during the drop. We can ignore air resistance.What is the package's vertical displacement during the drop?
Answer:
Package's vertical displacement(s) = 22.12 meter
Explanation:
Given:
Speed of plane = 120 m/s
Total distance = 255 m
Find:
Package's vertical displacement(s)
Computation:
Time taken = Distance / Speed
Time taken = Total distance / Speed of plane
Time taken = 255 / 120
Time taken = 2.125 s
Acceleration due to gravity(g) = 9.8 m/s²
Initial velocity (u) = 0
So,
Package's vertical displacement(s) = ut + (1/2)gt²
Package's vertical displacement(s) = (0)(2.125) + (1/2)(9.8)(2.125)²
Package's vertical displacement(s) = 22.12 meter
Answer: -22.1
Explanation:
I just did the Khan Academy and that was the answer, not the one provided by that one person. :)))
Two small, identical conducting spheres repel each other with a force of 0.035 N when they are 0.35 m apart. After a conducting wire is connected between the spheres and then removed, they repel each other with a force of 0.055 N. What is the original charge on each sphere? (Enter the magnitudes in C.)
Explanation:
Given:
distance between two sphere =0.35 m
Electrical repel force =0.035 N
Electrical repel force after connecting wire =0.055 N.
The electrical force between the two spheres:
[tex]F=k \frac{q_{1} q_{2}}{r^{2}}[/tex]
The electrical force between the two spheres after the wire is attached and removed:
[tex]F=k \frac{q^{2}}{r^{2}}[/tex]
[tex]q^{2}=\frac{F r^{2}}{k}[/tex]
[tex]q=r \sqrt{\frac{F}{k}}=0.35 \times \sqrt{\frac{0.055}{8.99 \times 10^{9}}}=6.46 \times 10^{-7} \mathrm{C}[/tex]
So the total charge of the two spheres [tex]=2 q=2 \times 6.46 \times 10^{-7}=1.29 \times 10^{-6} \mathrm{C}[/tex]
Then before connecting the wire, one sphere charge was [tex]q[tex] and the charge of the other sphere was [tex]\left(1.29 \times 10^{-6}-q\right)[/tex]
The electrical force between the two spheres before connecting the wire:
[tex]F=k \frac{q\left(1.29 \times 10^{-6}-q\right)}{r^{2}}[/tex]
[tex]q\left(1.29 \times 10^{-6}-q\right)=\frac{F r^{2}}{k}=\frac{0.035 \times(0.35)^{2}}{8.99 \times 10^{9}}=0.348 \times 10^{-12}[/tex]
[tex]-q^{2}+\left(1.29 \times 10^{-6}\right) q-\left(0.348 \times 10^{-12}\right)=0[/tex]
[tex]-q^{2}+\left(1.29 \times 10^{-6}\right) q-\left(0.348 \times 10^{-12}\right)=0[/tex]
A mass of 1.15 kg of air at 111 kPa and 27 C is contained in a gas-tight, frictionless piston cylinder device. The air is now compressed to a final pressure of 0.84 MPa. During this process, heat is transferred from the air such that the temperature inside the cylinder remains constant. Calculate the work input (in kJ) during this process.
Answer:
Work input W = -200.39 KJ
Explanation:
From the question, we are given;
m = 1.15 kg
Constant temperature T1 = T2 = 27 + 273 = 300k
Since the temperature is constant, we can say that the process is isothermal
P1 = 111 KPa
P2 = 0.84 MPa = 0.84 * 1000 KPa = 840 KPa
Now what we want to calculate is W1-2
Mathematically, for isothermal process;
W1-2 = mRTlnP1/P2
where R can be obtained from table and it is equal to 0.287 KJ/kg.k
Hence;
W1-2 = (1.15)(0.287)(300)(ln 111/840)
W1-2 = 99.015 * -2.023871690525 = -200.39 KJ
Kindly note that the value of the work is negative because work is done on the system and not by the system
Two students are trying to measure how high a ball bounces when it is dropped from different heights. They dropped a ball from P and it bounced up till Q. They now have to record two measurements as shown below. Which measurements should they take?A.A B.B C.C D.D
Answer:
A.A
Explanation:
The balls usually bounce 60% of the original height because it stores 60% of the energy it had before the bounce. When a ball is dropped from a great height it has kinetic energy before it hits the ground which is the result of the bounce of ball. The size of ball does matter in this case, Large balls will bounce higher.
Squids propel themselves by expelling water. They do this by keeping water in a cavity and then suddenly contracting the cavity to force out the water through an opening. A 9 kg squid (including the water in the cavity) at rest suddenly sees a dangerous predator. If the squid expels 2 kg of water out of its body with a speed of 8 m/s, what would be its own escape speed
Answer:
v_squid = - 2,286 m / s
Explanation:
This exercise can be solved using conservation of the moment, the system is made up of the squid plus the water inside, therefore the force to expel the water is an internal force and the moment is conserved.
Initial moment. Before expelling the water
p₀ = 0
the squid is at rest
Final moment. After expelling the water
[tex]p_{f}[/tex] = M V_squid + m v_water
p₀ = p_{f}
0 = M V_squid + m v_water
c_squid = -m v_water / M
The mass of the squid without water is
M = 9 -2 = 7 kg
let's calculate
v_squid = 2 8/7
v_squid = - 2,286 m / s
The negative sign indicates that the squid is moving in the opposite direction of the water
A ball is thrown vertically upwards. It returns 6s later. Calculate : (1) the greatest height reached by the ball, and (2) the initial velocity of the ball. (Take g=9.8m/s2)
Answer:
greatest displacement = 44.1m
initial velocity= 29.4m/s
Explanation:
Greatest displacement
s=1/2at^2
= (9.8/2 ×9)m
= 44.1m
initial velocity
s=ut-1/2at^2
44.1= 3u -(1/2×9.8×9)
44.1=3u-44.1
3u=88.2
u=29.4m/s
the resistor of values 6 ohm,6 ohm are connected in series and 12 ohm are connected in parallel. the equivalent resistance of the circuit is
Answer:
The equivalent or total resistance of the circuit is 6
Explanation:
6 &6 are in series
6+6=r
r= 12
1/Rtotal= 1/12+1/2
1/Rt=2/12=1/6
Rt=6
help me helpppp the last words are experience a total
Answer:
False
Explanation:
Total internal reflection is a condition that occurs when light travels from dense medium to less dense medium.
From the table given in the question above,
Refractive Index of Whale oil = 1.460
Refractive Index of vacuum = 1
From the above, we can see that the whale oil has a higher Refractive Index than vacuum. This implies that the whale oil is denser than the vacuum.
Therefore, total internal reflection will occur when light travels from whale oil to vacuum because the whale oil is denser than the vacuum.
The transfer of charge from clouds to the earth or cloud to cloud is called
That's called "lightning".
Answer:
The lightning itself is the transfer of charge from one region of a cloud to another or between the cloud and Earth. The narrow channel within which the flash of lightning occurs is heated suddenly to ~ 30,000 K, with essentially no time to expand.
Explanation:
Hope it helps
which discribes what a velocity/time graph would look like with no accelaration
A 15 kg object is pulled by a force of 68 N. If the surface exerts a friction force of 23 N, with what acceleration does the object move?
Answer:
3 m/s²
Explanation:
Sum of forces in the x direction:
∑F = ma
68 N − 23 N = (15 kg) a
a = 3 m/s²