Answer:
C. 5730 years
Explanation:
N(t) = N(0)e^-kt
The half-life is T = 5730 years,
e^-kT = 1/2
→ k = - ln(1/2) / T
→ - ln(1/2) / 5730
→ 1.209681 x 10^-4 years^-1
The amount present dropped to 50%.
Then one half-life has elapsed, so the age is 5730 years.
The engine of a locomotive exerts a constant force of 6.8 105 N to accelerate a train to 80 km/h. Determine the time (in min) taken for the train of mass 1.1 107 kg to reach this speed from rest.
Answer:
t = 6 minutes
Explanation:
Given that,
Force,[tex]F=6.8\times 10^5\ N[/tex]
Initial speed of the train, u = 0
Final speed of the train, v = 80 km/h = 22.22 m/s
The mass of the train, [tex]m=1.1\times 10^7\ kg[/tex]
We need to find the time taken by the train to come to rest. We know that,
F = ma
[tex]F=\dfrac{m(v-u)}{t}\\\\t=\dfrac{m(v-u)}{F}\\\\t=\dfrac{1.1\times10^7\times (22.22-0)}{6.8\times 10^5}\\\\t=359.44\ s[/tex]
or
t = 6 minutes (approx)
So, the required time is equal to 6 minutes.
An electric drill starts from rest and rotates with a constant angular acceleration. After the drill has rotated through a certain angle, the magnitude of the centripetal acceleration of a point on the drill is 8.2 times the magnitude of the tangential acceleration. What is the angle?
Answer:
The angle is 4.1 rad.
Explanation:
The centripetal acceleration (α) is given by:
[tex] \alpha = \omega^{2} r [/tex] (1)
Where:
ω: is the angular velocity
r: is the radius
And the tangential acceleration (a) is:
[tex] a = \alpha r [/tex] (2)
Since the magnitude of "α" is 8.2 times the magnitude of "a" (equating (2) and (1)) we have:
[tex] \omega^{2} r = 8.2\alpha r [/tex]
[tex] \omega^{2} = 8.2\alpha [/tex] (3)
Now, we can find the angle with the following equation:
[tex] \omega_{f}^{2} = \omega_{0}^{2} + 2\alpha \Delta \theta [/tex]
Where:
[tex] \omega_{f}[/tex]: is the final angular velocity [tex] \omega_{0}[/tex]: is the initial angular velocity = 0 (it starts from rest)
[tex]\Delta \theta[/tex]: is the angle
[tex] \omega^{2} = 2\alpha \Delta \theta [/tex] (4)
By entering equation (3) into (4) we can calculate the angle:
[tex] 8.2\alpha = 2\alpha \Delta \theta [/tex]
[tex] \Delta \theta = 4.1 rad [/tex]
Therefore, the angle is 4.1 rad.
I hope it helps you!
In air an object weighs 15N, when immersed in water it weighs 12N, when immersed in another liquid, it weighs 13N, Calculate the density of the object and that of the other liquid?
M1 = 15/g = 15/9.8 = 1.53 kg = mass of object in air. M2 = 12/9.8 = 1.22 kg = mass of object immersed. M1-M2 = 1.53-1.22 = 0.31 kg lost by object = mass of water displaced. ... Do = 4.94 g/cm^3 = density of object.
A rock is thrown from the top of a building 146 m high, with a speed of 14 m/s at an angle 43 degrees above the horizontal. When it hits the ground, what is the magnitude of its velocity (i.e. its speed).
Answer:
time is 32 s and speed is 304.3 m/s
Explanation:
Height, h = 146 m
speed, u = 14 m/s
Angle, A = 43 degree
Let it hits the ground after time t.
Use second equation of motion
[tex]h = u t +0.5 at^2\\\\- 146 =14 sin 43 t - 4.9 t^2\\\\4.9 t^2 - 9.5 t - 146 =0 \\\\t =\frac{9.5\pm\sqrt {90.25 + 2861.6}}{9.8}\\\\t=\frac{9.5\pm 54.3}{9.8}\\\\t = 32.05 s, - 22.4 s[/tex]
Time cannot be negative so the time is t = 32 s .
The vertical velocity at the time of strike is
v' = u sin A - g t
v' = 14 sin 43 - 9.8 x 32 = 9.5 - 313.6 = - 304.1 m/s
horizontal velocity
v'' = 14 cos 43 =10.3 m/s
The resultant velocity at the time of strike is
[tex]v=\sqrt{v'^2 + v''^2}\\\\v = \sqrt{304.1^2 +10.3^2 }\\\\v = 304.3 m/s[/tex]
L Pretest Unit 7
Question 13 of 20
Andrew is trying to identify an unknown element. The element is shiny, but it
shatters when hammered and cannot be hammered into different shapes.
Where on the periodic table is this element most likely found?
A. On the left side
B. In one of the series below the main body of the table
C. On the right side
D. Along the metalloid stairstep line
SURMIT
Answer:
C
Explanation:
I think it would be there, it sounds like silicone and thats on the right side
A 7.5-kg rock and a 8.9 × 10-4-kg pebble are held near the surface of the earth. (a) Determine the magnitude of the gravitational force exerted on each by the earth. (b) Calculate the magnitude of the acceleration of each object when released.
Answer:
F' = 73.7 N
F = 8.749×10⁻³ N
a' = a = 9.83 m/s²
Explanation:
(a)
For the rock
Applying
F' = Gm'm/r²................... Equation 1
Where F = magnitude of the gravitational force on the rock, G = Gravitational constant, m' = mass of the rock, m = mass of the earth, r = radius of the earth.
From the question,
Given: m' = 7.5 kg
Constant: m = 5.98×10²⁴ kg, G = 6.67×10⁻¹¹ Nm²/kg², r = 6.37×10⁶ m
Substitute these values into equation 1
F' = 6.67×10⁻¹¹ (7.5)(5.98×10²⁴)/(6.37×10⁶)²
F' = 7.37×10¹ N
F' = 73.7 N
Also, For the pebble,
F = GMm/r².............. Equation 2
Where M = mass of the pebble, F = Gravitational force exerted on the pebble by the earth
Given: M = 8.9×10⁻⁴ kg,
Substitute into equation 2
F = 6.67×10⁻¹¹(8.9×10⁻⁴)(5.98×10²⁴)/(6.37×10⁶)²
F = 8.749×10⁻³ N
(b)
For the rock,
a' = F'/m'
Where a' = magnitude of the acceleration of the rock
a' = 73.7/7.5
a' = 9.83 m/s²
For the pebble,
a = F/M
Where a = acceleration of the pebble
a = (8.749×10⁻³)/(8.9×10⁻⁴)
a = 9.83 m/s²
Newton's law of cooling states that the rate of change of temperature of an object in a surrounding medium is proportional to the difference of the temperature of the medium and the temperature of the object. Suppose a metal bar, initially at temperature 50 degrees Celsius, is placed in a room which is held at the constant temperature of 40 degrees Celsius. One minute later the bar has cooled to 40.18316 degrees . Write the differential equation that models the temperature in the bar (in degrees Celsius) as a function of time (in minutes). Hint: You will need to find the constant of proportionality. Start by calling the constant k and solving the initial value problem to obtain the temperature as a function of k and t . Then use the observed temperature after one minute to solve for k .
Answer:
Newton's law of cooling says that the temperature of a body changes at a rate proportional to the difference between its temperature and that of the surrounding medium (the ambient temperature); dT/dt = -K(T - Tₐ) where T = the temperature of the body (°C), t = time (min), k = the proportionality constant (per minute),
Explanation:
A ball is thrown horizontally from the top of a building 59 m high. The ball strikes the ground at a point 65 m horizontally away from and below the point of release. What is the speed of the ball just before it strikes the ground
Answer:
Explanation:
We are looking for final velocity. Since the ball is thrown horizontally, there is no upwards velocity, so the y dimension here is only useful to us for finding how long the ball was in the air. In the y dimension, here's what we know:
a = -9.8 m/s/s
Δx = -59 m
[tex]v_0=0[/tex] (again, initial upwards velocity is 0 because the ball was thrown horizontally)
We can put all that together in the equation:
Δx = [tex]v_0t+\frac{1}{2}at^2[/tex] and filling in:
[tex]-59=0t+\frac{1}{2}(-9.8)t^2[/tex] which simplifies to
[tex]-59=\frac{1}{2}(-9.8)t^2[/tex] and solving for t:
[tex]t=\sqrt{\frac{2(-59)}{-9.8} }[/tex] and
t = 3.5 sec
Now we can use that time in the d = rt equation, which is all we need for the horizontal dimension (I'll show you why in just a second). In the horizontal dimension, here's what we know:
a = 0 m/s/s
Δx = 65 m
t = 3.5 sec
Putting that all together in our one-dimensional equation for displacement:
Δx = [tex]v_0t+\frac{1}{2}at^2[/tex] and acceleration is 0, we can simplify that down to
Δx = [tex]v_0t[/tex] which is the exact same thing as d = rt where r is the velocity we are looking for. Filling in:
65 = v(3.5) so
v = 18.6 m/s
That's the velocity with which the ball strikes the ground.
Which shows the formula for converting from degrees Celsius to degrees Fahrenheit?
°F = (9/5 × °C) +32
°F = 5/9 × (°C – 32)
°F = °C – 273
°F = °C + 273
Answer:
the first answer
Explanation:
(32°F − 32) × 5/9 = 0°C
Answer:
Answer: A
Explanation:
Who Knows?
Thank You
Answer:
Use below table to read the color codes.
When a condenser discharges electricity, the instantaneous rate of change of the voltage is proportional to the voltage in the condenser. Suppose you have a discharging condenser and the instantaneous rate of change of the voltage is -0.01 of the voltage (in volts per second). How many seconds does it take for the voltage to decrease by 90 %?
Answer:
460.52 s
Explanation:
Since the instantaneous rate of change of the voltage is proportional to the voltage in the condenser, we have that
dV/dt ∝ V
dV/dt = kV
separating the variables, we have
dV/V = kdt
integrating both sides, we have
∫dV/V = ∫kdt
㏑(V/V₀) = kt
V/V₀ = [tex]e^{kt}[/tex]
Since the instantaneous rate of change of the voltage is -0.01 of the voltage dV/dt = -0.01V
Since dV/dt = kV
-0.01V = kV
k = -0.01
So, V/V₀ = [tex]e^{-0.01t}[/tex]
V = V₀[tex]e^{-0.01t}[/tex]
Given that the voltage decreases by 90 %, we have that the remaining voltage (100 % - 90%)V₀ = 10%V₀ = 0.1V₀
So, V = 0.1V₀
Thus
V = V₀[tex]e^{-0.01t}[/tex]
0.1V₀ = V₀[tex]e^{-0.01t}[/tex]
0.1V₀/V₀ = [tex]e^{-0.01t}[/tex]
0.1 = [tex]e^{-0.01t}[/tex]
to find the time, t it takes the voltage to decrease by 90%, we taking natural logarithm of both sides, we have
㏑(0.01) = -0.01t
So, t = ㏑(0.01)/-0.01
t = -4.6052/-0.01
t = 460.52 s
Ion how to do this at all
URGENT
The component of solid-state physics that works with and/or heats metals and alloys to give them certain desired
shapes or properties is..
Answer:
the is metallurgy .....
Which of the statements below are TRUE! Group of answer choices The carbon rod in batteries react to form a carbon cation. A good car battery gives you a little bit of power for a long period of time. A good car battery gives you a lot of power in a short period of time. The carbon rod in batteries is used as an inert electrode.
Answer:
The carbon rod in batteries is used as an inert electrode
Explanation:
A battery is considered as a power source that consists of one or more electrochemical cells having an external connections to provide power to the electrical devices such as the lights, bulbs, fans, mobile phones, etc.
It contains a positive terminal and a negative terminal.
The carbon rod in the battery does not help in the electrochemical reactions. It acts as an inert electrode and helps to flow the electrons only.
Thus the true statement is :
The carbon rod in batteries is used as an inert electrode.
Does the same battery always deliver the same amount of flow to any circuit? Mention two observations of any circuits in this lab that support your answer. Explain.
Answer:
Yes
Explanation:
Given that the battery is the same the PD ( potential difference ) in the circuit will also be the same likewise the flow of charge in the circuit,
Hence the same amount of charge flow is delivered to any circuit.
attached below are examples
galileo was a contemporary of
A wave has a frequency of 87.00 Hz and has a wavelength of 74.62 m. What is its
velocity?
Answer:
v = 6491.94 m/s
Explanation:
We are given;
Frequency; f = 87 Hz
Wavelength;λ = 74.62 m
Formula for velocity(v) of waves from the wave equation is;
v = fλ
Thus;
v = 87 × 74.62
v = 6491.94 m/s
in which states of matter will a substance have a fixed volume
Answer:
Solid is the state in which Matter maintains a fixed volume
Answer:
The state of matter that has a fixed volume is Solid.
Explanation:
Solid substances will maintain a fixed volume and shape.
Speeding up
Slowing down
Standing still
Holding at a constant non-zero velocity
Answer:
speeding up is the answer
Explanation:
from the graph it can be seen that as the time (horizontal axis) increases the speed of vehicle (vertical axis) increases
You are using a constant force to speed up a toy car from an initial speed of 6.5 m/s
to a final speed of 22.9 m/s. If the toy car has a mass of 340 g, what is the work
needed to speed this car up?
By the work-energy theorem, the total work done on the car is equal to the change in its kinetic energy:
W = ∆K
W = 1/2 (0.34 kg) (22.9 m/s)² - 1/2 (0.34 kg) (6.5 m/s)²
W ≈ 82 J
The elastic energy stored in your tendons can contribute up to 35 % of your energy needs when running. Sports scientists have studied the change in length of the knee extensor tendon in sprinters and nonathletes. They find (on average) that the sprinters' tendons stretch 40 mm, while nonathletes' stretch only 32 mm. The spring constant for the tendon is the same for both groups, 32 N/mm. What is the difference in maximum stored energy between the sprinters and the nonathlethes?
Answer:
Explanation:
From the given information:
The difference in the maximum energy stored is can be determined by finding the difference in the maximum stored energy in the sprinters and that of the non-athlete:
[tex]\Delta U = \dfrac{1}{2}k(x_2^2 - x_1^2)[/tex]
[tex]\Delta U = \dfrac{1}{2} (32 \ N/mm) (\dfrac{1 \ mm}{10^{-3} \ m}) ((40\times 10^{-3})^2 - (32\times 10^{-3})^2)[/tex]
[tex]\Delta U =16000 \times (5.76\times 10^{-4})[/tex]
[tex]\mathbf{\Delta U =9.216\ J}[/tex]
A car starts from rest and accelerates uniformly in a straight line in the positive x direction. After 25 seconds, its speed is 90 km/h.
a) Determine the acceleration of the object. [5]
b) How far does the object travel during the first 25 seconds? [3]
c) What is the average velocity of the object during the first 25 seconds?
Answer:
A. 1 m/s²
B. 312.5 m
C. 12.5 m/s
Explanation:
We'll begin by converting the velocity i.e 90 Km/h to m/s. This can be obtained as follow:
Velocity (Km/h) = 90 Km/h
Velocity (m/s) =?
Velocity (m/s) = Velocity (Km/h) × 1000 / 3600
Velocity (m/s) = 90 × 1000 / 3600
Velocity (m/s) = 90000 / 3600
Velocity (m/s) = 25 m/s
A. Determination of the acceleration.
Initial velocity (u) = 0 m/s
Final velocity (v) = 25 m/s
Time (t) = 25 s
Acceleration (a) =?
v = u + at
25 = 0 + (a × 25)
25 = 0 + 25a
25 = 25a
Divide both side by 25
a = 25/25
a = 1 m/s²
B. Determination of the distance travelled.
Initial velocity (u) = 0 m/s
Final velocity (v) = 25 m/s
Acceleration (a) = 1 m/s²
Distance travelled (s) =?
v² = u² + 2as
25² = 0 + (2 × 1 × s)
625 = 0 + 2s
625 = 2s
Divide both side by 2
s = 625 / 2
s = 312.5 m
C. Determination of the average velocity.
Total distance travelled = 312.5 m
Total time = 25 s
Average velocity =?
Average velocity = Total distance / total time
Average velocity = 312.5 / 25
Average velocity = 12.5 m/s
what is the suitable way of using social media
Answer:
not using it too much and getting addicted
Explanation:
Give the number of protons and the number of neutrons in the nucleus of each of the following isotopes Aluminum 25 :13 protons and 12 neutrons
Answer:
No of proton is 13 and nucleus is 13
An insulated tank contains 50 kg of water, which is stirred by a paddle wheel at 300 rpm while transmitting a torque of 0.1 kN-m. At the same time, an electric resistance heater inside the tank operates at 110 V, drawing a current of 2 A. Determine the rate of heat transfer after the system achieves steady state.
Answer:
the rate of heat transfer after the system achieves steady state is -3.36 kW
Explanation:
Given the data in the question;
mass of water m = 50 kg
N = 300 rpm
Torque T = 0.1 kNm
V = 110 V
I = 2 A
Electric work supplied W₁ = PV = 2 × 110 = 220 W = 0.22 kW
Now, work supplied by paddle wheel W₂ is;
W₂ = 2πNT/60
W₂ = (2π × 0.1 × 300) / 60
W₂ = 188.495559 / 60
W₂ = 3.14 kW
So the total work will be;
W = 0.22 + 3.14
W = 3.36 kW
Hence total work done on the system is 3.36 kW.
At steady state, the properties of the system does not change so the heat transfer will be 3.36 KW.
The heat will be rejected by the system so the sign of heat will be negative.
i.e Q = -3.36 kW
Therefore, the rate of heat transfer after the system achieves steady state is -3.36 kW
The diagram shows the molecular structure of ethane. What is the chemical
formula for ethane?
Ethane
H H
H-C-C-H
| |
H H
Bola bermassa 200 gram dilempar
ke bawah dari ketinggian 20 m
dengan kecepatan 2 m/s. Jika
percepatan gravitasi bumi 10
m/s2 energi kinetik pada
ketinggian 8 m adalah ......
Answer:
0.4
Explanation:
[tex] \frac{1}{2} mv ^{2} [/tex]
kinetic energy formula , potential energy is not considered
0.5×0.2×2×2
The two most prominent wavelengths in the light emitted by a hydrogen discharge lamp are 656 nm (red) and 486 nm (blue). Light from a hydrogen lamp illuminates a diffraction grating with 550 lines/mm , and the light is observed on a screen 1.7 m behind the grating.
What is the distance between the first-order red and blue fringes?
Express your answer to two significant figures and include the appropriate units.
Answer:
Δd = 7.22 10⁻² m
Explanation:
For this exercise we must use the dispersion relationship of a diffraction grating
d sin θ = m λ
let's use trigonometry
tan θ = y / L
how the angles are small
tant θ = sinθ /cos θ = sin θ
we substitute
sin θ = y / L
d y / L = m λ
y = m λ L / d
let's use direct ruler rule to find the distance between two slits
If there are 500 lines in 1 me, what distance is there between two lines
d = 2/500
d = 0.004 me = 4 10⁻⁶ m
diffraction gratings are built so that most of the energy is in the first order of diffraction m = 1
let's calculate for each wavelength
λ = 656 nm = 656 10⁻⁹ m
d₁ = 1 656 10⁻⁹ 1.7 / 4 10⁻⁶
d₁ = 2.788 10⁻¹ m
λ = 486 nm = 486 10⁻⁹ m
d₂ = 1 486 10⁻⁹ 1.7 / 4 10⁻⁶
d₂ = 2.066 10⁻¹ m
the distance between the two lines is
Δd = d1 -d2
Δd = (2,788 - 2,066) 10⁻¹
Δd = 7.22 10⁻² m
A Michelson interferometer operating at a 600nm wavelength has a 2.02-cm-long glass cell in one arm. To begin, the air is pumped out of the cell and mirror M2 is adjusted to produce a bright spot at the center of the interference pattern. Then a valve is opened and air is slowly admitted into the cell. The index of refraction of air at 1.00 atm pressure is 1.00028.
How many bright-dark-bright fringe shifts are observed as the cell fills with air?
Answer:
19
Explanation:
Given that:
wavelength = 600 nm
Distance (d) = 2.02 cm = 2.02 × 10⁻² m
refraction index of air (n) = 1.00028
Pressure = 1.00 atm
∴
The number of bright-dark-bright fringe shifts can be determined by using the formula:
[tex]\Delta m = \dfrac{2d}{\lambda} (n -1 ) \\ \\ \Delta m = \dfrac{2\times2.02 \times 10^{-2}}{600\times 10^{-9}} (1.00028 -1 ) \\ \\ \Delta m = 67333.33 \times 10^{-5}(1.00028 -1) \\ \\ \Delta m = 67333.33 \times 10^{-5}(2.8\times 10^{-4}) \\ \\ \Delta m = 18.853 \\ \\ \mathbf{\Delta m = 19}[/tex]
A toy car of mass 600g moves through 6m in 2 seconds. The average kinetic energy of the toy car is
Answer:
12
Explanation:
I'm a beginner so am not sureeeeee
