Answer:
C
Explanation:
100%
A 1.5 kg rock is dropped from a height of 2.0 meters onto a spring that
compresses and brings the rock to rest. (Assume no losses to thermal
energy.) How much energy is in the system before the drop.
Answer:
29.4 J
Explanation:
Before the drop, the system has only the gravitational potential energy, and this energy us given by mass×gravity×height:
1.5•9.8•2 = 29.4 J
A box rests on a frozen pond, which serves as a frictionless horizontal surfaceIf a fisherman applies a horizontal force with magnitude 51.0 N to the box and produces an acceleration of magnitude 3.00 m/s2 , what is the mass of the box?
Answer:
the mass of the box is 17 kg
Explanation:
Given;
magnitude of the force applied by the fisherman, F = 51 N
magnitude of acceleration of the box, a = 3 m/s²
the mass of the box is calculated using Newton's law of motion;
F = ma
where;
m is the mass of the box
m = F / a
m = (51) / (3)
m = 17 kg
Therefore, the mass of the box is 17 kg
A professional quarterback throws a 0.40 kg football. what is the force of weight?
Answer:
3.92N
Explanation:
Force= mass×accelerarion due gravity
But mass= 0.40kg
acceleration due to gravity = 9.8 m/s^2
Force = 0.40×9.8
Force=3.92N
1. What are Earth's natural climate cycles?
Find the ratio of speeds of an electron and a negative hydrogen ion (one having an extra electron) accelerated through the same voltage, assuming non-relativistic final speeds. Take the mass of the hydrogen ion to be 6.7 X 10 -27 Kg.
Answer:
v₂ /v₁ = 2.3 10⁺²
Explanation:
The energy is conserved so the total potential energy must be transformed into kinetic energy
K = U
½ m v² = q ΔV
v = [tex]\sqrt{\frac{2q \Delta V}{m} }[/tex]
a) Let's find the speed of the electron
m = 9.1 10⁻³¹ kg
as they do not indicate the value of the power difference, we will assume that ΔV = 1 V is worth one
v = [tex]\sqrt{ \frac{2 \ 1.6 \ 10^{-19} \ 1}{9.1 \ 10^{-31}} }[/tex]
v = [tex]\sqrt {0.3516 \ 10^{12}}[/tex]
v1 = 0.593 10⁶ m / s
b) the velocity of a hydrogen ion
M = M_H + m
M = 1.673 10⁻²⁷ + 9.1 10⁻³¹
M = 1.67391 10⁻²⁷ kg
M = 1.67 10⁻²⁷ kg
v = [tex]\sqrt{ \frac { 2 \ 1.6 \ 10^{-19} \ 1}{1.67 \ 10^{-27}} }[/tex]
v = [tex]\sqrt{ 1.916167 \ 10^8 }[/tex]
v₂ = 1.38 10⁴ m / s
the relationship between these speeds is
v₂ / v₁ = 1.38 10⁴ / 0.593 10⁶
v₂ /v₁ = 2.3 10⁺²
what happened on march 21 every year in the northern hemisphere
Answer:
B. The Spring equinox
Explanation:
The vernal equinox marks the moment the sun crosses the celestial equator. The vernal equinox happens on March 19, 20, or 21 every year in the Northern Hemisphere. In the Southern Hemisphere, this same event marks the beginning of fall. (Source: What Exactly Is The Spring Equinox? - Dictionary.com)
Hopefully this helps.
A light-emitting diode (LED) connected to a 3.0 V power supply emits 440 nm blue light. The current in the LED is 11 mA , and the LED is 51 % efficient at converting electric power input into light power output. How many photons per second does the LED emit?
Answer:
3.73 * 10^16 photons/sec
Explanation:
power supply = 3.0 V
Emits 440 nm blue light
current in LED = 11 mA
efficiency of LED = 51%
Calculate the number of photons per second the LED will emit
first step : calculate the energy of the Photon
E = hc / λ
=( 6.62 * 10^-34 * 3 * 10^8 ) / 440 * 10^-9
= 0.0451 * 10^-17 J
Next :
Number of Photon =( power supply * efficiency * current ) / energy of photon
= ( 3 * 0.51 * 11 * 10^-3 ) / 0.0451 * 10^-17
= 3.73 * 10^16 photons/sec
Compare the freezing point of water in the aquanaut’s apartment to its value at the surface. Is it higher, lower, or the same?
Answer:
Freezing Point - Lower
Boiling Point - Higher
Solid- liquid transition line in the phase diagram has a negative slope, but the liquid-gas transition line has a positive slope. Since there is more air pressure at 100m it will take less to freeze the water but more to boil it since it requires a larger temperature under larger pressures
If we double the mass of an object without a change in volume, its density would be
A) half.
B) double.
C) unchanged.
20 points, im begging for help‼️
How much capacitance is needed to
store 0.00100 J of energy when the
charge on the capacitor is
4.86 x 10-5 C?
[?] x 10?!F
Answer:
Capacitance= 1.18×10^-6
Answer: 1.18*10^-6
Explanation:
The drag force Fd, imposed by the surrounding air on a vehicle moving with velocity V is given by:
Fd =C dA 2 pV2
where Cd is a constant called the drag coefficient, A is the projected frontal area of the vehicle, and \rhorho is the air density.
Determine the power, in hp, required to overcome aerodynamic drag for an automobile moving at
(a) 25 miles per hour,
(b) 70 miles per hour.
Assume Cd=0.28,
A= 25ft2
and p=0.075Ib/ft2
Answer:
Explanation:
a)
Given that:
V = 25 mi/hr
To ft/sec, we have:
[tex]V = 25 \times \dfrac{5280}{3600} ft/s[/tex]
[tex]V = \dfrac{110}{3} ft/s[/tex]
[tex]\rho = 0.075 \ lb/ft^3[/tex]
[tex]\rho = 0.075 \times \dfrac{1 \ lbf s^2/ft}{32.174 \ lbm}[/tex]
[tex]\rho = \dfrac{0.075}{32.174 } lbf.s^2/ft^4[/tex]
[tex]C_d = 0.28[/tex]
A = 25ft²
Recall that:
The drag force [tex]F_d =\dfrac{C_dA \rho V^2}{2}[/tex]
[tex]F_d =\dfrac{1}{2}\times 0.28 \times 25\times \dfrac{0.075}{32.174}\times (\dfrac{110}{3})^2[/tex]
[tex]F_d =10.967 \ lbf[/tex]
[tex]P = F_dV \\ \\ P = 10.97 \times (\dfrac{110}{3}} \\ \\ P = 402.3 \ hp[/tex]
For 70 miles per hour, we have:
[tex]V = 70 \times \dfrac{5280}{3600} ft/s[/tex]
[tex]V = \dfrac{308}{3} ft/s[/tex]
The drag force [tex]F_d =\dfrac{C_dA \rho V^2}{2}[/tex]
[tex]F_d =\dfrac{1}{2}\times 0.28 \times 25\times \dfrac{0.075}{32.174}\times (\dfrac{308}{3})^2[/tex]
[tex]F_d =85.99 \ lbf[/tex]
[tex]P = F_dV \\ \\ P = 85.99 \times (\dfrac{308}{3}}) \\ \\ P = 8828.2 \ hp[/tex]
A vessel at rest at the origin of an xy coordinate system explodes into three pieces. Just after the explosion, one piece, of mass m, moves with velocity (-21 m/s) and a second piece, also of mass m, moves with velocity (-21 m/s) . The third piece has mass 3m. Just after the explosion, what are the (a) magnitude and (b) direction (as an angle relative to the x axis) of the velocity of the third piece
Answer:
25
Explanation:
magnitude and (b) direction (as an angle relative to the x axis) of the velocity
A motorist travels due North at 90 km/h for 2 hours. She changes direction and travels West at 60 km/for 1 hour.
a) Calculate the average speed of the motorist [4]
b) Calculate the average velocity of the motorist.
Answer:
a) S = 63.2 km/h
b) V = 63.2 km/h*(-0.316 , 0.949)
Explanation:
Let's define:
North as the positive y-axis
East as the positive x-axis.
Also, remember the relation:
Distance = Time*Speed
Let's assume that she starts at the position (0km, 0km)
Then she travels due North at 90km/h for two hours, then the displacement is
90km/h*2h = 180km to the north
Then the new position is:
(0km, 180km)
Then she travels West at 60km/h for one hour.
Then the distance traveled to the West (negative x-axis) is:
60km/h*1h = 60km to the west
Then the new position is:
(-60km, 180km).
a) The average speed is defined as the quotient between the displacement and the time.
We know that the total time traveled is 3 hours.
And the displacement is the difference between the final position and the initial position.
this is:
D = √( -60km - 0km)^2 + (180km - 0km)^2)=
D = √( (60km)^2 + (180km)^2) = 189.7 km
Then the average speed is:
S = (189.7 km)/(3 h) = 63.2 km/h
b) Now we want to find the average velocity, this will be equal to the average speed times a versor that points from the origin to the direction of the final position.
So, if the final position is (-60km, 180km)
We need to find a vector that represents the same angle, but that is on the unit circle.
Then, if the module of the final position is 189.7 km (as we found above), then the versor is just given by:
(-60km/ 189.7 km, 180km/ 189.7 km)
(-60/189.7 , 180/189.7)
We can just check that the module of the above versor is 1.
[tex]module = \sqrt{(\frac{-60}{189.7} )^2 + (\frac{180}{189.7} )^2} = \frac{1}{189.7}* \sqrt{(-60 )^2 + (180 )^2} = 1[/tex]
Then the average velocity is:
V = 63.2 km/h*(-60/189.7 , 180/189.7)
We can simplify our versor so the velocity equation is easier to read:
V = 63.2 km/h*(-0.316 , 0.949)
A dog accelerates at 1.50 m/s2 to reach a velocity of 13.5 m/s while covering a distance of 49.3 m. What was his initial velocity?
Let v be the dog's initial velocity. Then
(13.5 m/s)^2 - v ^2 = 2 (1.50 m/s^2) (49.3 m)
==> v ^2 = (13.5 m/s)^2 - 2 (1.50 m/s^2) (49.3 m)
==> v = √((13.5 m/s)^2 - 2 (1.50 m/s^2) (49.3 m))
==> v ≈ 5.86 m/s
The momentum of a falling rock is found to be 200 kg m/s. What is the mass of the rock if it falls with a velocity of 5.0 m/s
Answer:
[tex]\boxed {\boxed {\sf 40 \ kilograms}}[/tex]
Explanation:
Momentum is the product of velocity and mass. The formula is:
[tex]p=m*v[/tex]
We know the rock is falling. Its momentum is 200 kilograms meters per second and its velocity is 5 meters per second. Substitute the values into the formula.
[tex]200 \ kg \ m/s = m * 5.0 \ m/s[/tex]
We are solving for m, the mass. We must isolate the variable. It is being multiplied by 5 meters per second. The inverse of multiplication is division, so we divided both sides by 5.0 m/s.
[tex]\frac{200 \ kg \ m/s}{5.0 \ m/s}=\frac{ m* 5.0 \ m/s }{5.0 \ m/s}[/tex]
[tex]\frac{200 \ kg \ m/s}{5.0 \ m/s}=m[/tex]
The units of meters per second (m/s) cancel.
[tex]\frac{200 \ kg}{5.0 } =m[/tex]
[tex]40 \ kg = m[/tex]
The falling rock has a mass of 40 kilograms.
the magnitude of the electrical force acting between a +2.4x10-8c charge and 1+1.8x10-6 charge that are separated by 1.008m is
Answer:
3.83×10¯⁴ N
Explanation:
From the question given above, the following data were obtained:
Charge 1 (q₁) = +2.4x10¯⁸ C
Charge 2 (q₂) = +1.8x10¯⁶ C
Distance apart (r) = 1.008 m
Electrical constant (K) = 9×10⁹ Nm²/C²
Force (F) =?
The magnitude of the electrical force acting between the two charges can be obtained as follow:
F = Kq₁q₂ / r²
F = 9×10⁹ × 2.4x10¯⁸ × 1.8x10¯⁶ / (1.008)²
F = 0.0003888 / 1.016064
F = 3.83×10¯⁴ N
Thus the magnitude of the electrical force acting between the two charges is 3.83×10¯⁴ N
Which statement describes an action-reaction pair?
O A. You push on a car, and the car pushes back on you.
B. A book pushes down on a table, and the table pushes down on the
Earth.
C. The Moon pulls on Earth, and Earth pulls on the Sun.
D. You push down on your shoe, and Earth's gravity pulls down on the
shoe.
Answer:
A
Explanation:
a pex
Vectors 퐴, 퐵and 퐶are added together. 퐴has a magnitude of 20.0 units and makes an angle of 60.0° counterclockwise from the negativex-axis. 퐵has a magnitude of 40.0 units and makes an angle of 30.0° counterclockwise from the positive x-axis.퐶has a magnitude of 35.0 units and makes an angle of 60.0° clockwise from the negative y-axis. Determine the magnitude of the resultant vector 퐴+퐵+퐶and its direction as an angle measured counterclockwise from the positive x-axis.
Answer:
Magnitude = 15.86 units
direction = 69 degree below negative X axis
Explanation:
A = 20 units at 60.0° counterclockwise from the negative x - axis
B = 40 units at 30.0° counterclockwise from the positive x - axis
C = 35 units at 60.0° clockwise from the negative y - axis
Write the vectors in the vector form
[tex]\overrightarrow{A} =20 (- cos 60 \widehat{i} - sin 60 \widehat{j})=- 10\widehat{i} - 17.3 \widehat{j}\\\\\overrightarrow{B} =40 (cos 30 \widehat{i} + sin 30 \widehat{j})= 34.6\widehat{i} +20 \widehat{j}\\\\\overrightarrow{C} =35 (- sin 60 \widehat{i} - cos 60 \widehat{j})=- 30.3\widehat{i} - 17.5 \widehat{j}\\\\Now\\\\overrightarrow{A} + \overrightarrow{B} + \overrightarrow{C} = (- 10 + 34.6 - 30.3) \widehat{i} + (-17.3 + 20-17.5)\widehat{j}\\\\[/tex]
[tex]\\\overrightarrow{A} + \overrightarrow{B} + \overrightarrow{C} = - 5.7\widehat{i} -14.8\widehat{j}[/tex]
The magnitude is given by
[tex]= \sqrt{5.7^2 + 14.8^2} = 15.86 units[/tex]
The direction is given by
[tex]tan\theta = \frac{- 14.8}{- 5.7}\\\\\theta= 69^o[/tex]
below negative X axis.
Some runners train with parachutes that trail behind them to provide a large drag force. These parachutes are designed to have a large drag coefficient. One model expands to a square 1.8 mm on a side, with a drag coefficient of 1.4. A runner completes a 240 mm run at 6.0 m/s with this chute trailing behind.
Required:
How much thermal energy is added to the air by the drag force?
Answer:
by the drag force, 2.4004512 × 10⁻⁵ J is added to the air.
Explanation:
Given the data in the question;
drag coefficient of Cd = 1.4
speed v = 6.0 m/s
One model expands to a square 1.8 mm on a side
Area A = 1.8 × 1.8 = 3.24 mm² = 3.24 × 10⁻⁶ m²
distance travelled s = 240 mm = 0.24 m
we know that; density of air e = 1.225 kg/m³
Now,
Dragging force F[tex]_D[/tex] = ( Cd × e × v² × A ) / 2
thermal energy = F[tex]_D[/tex] × s
so
thermal energy = ( 1.4 × 1.225 × (6)² × (3.24 × 10⁻⁶) × 0.24 ) / 2
thermal energy = ( 4.8009024 × 10⁻⁵ ) / 2
thermal energy = 2.4004512 × 10⁻⁵ J
Therefore, by the drag force, 2.4004512 × 10⁻⁵ J is added to the air.
. How many meters away is a cliff if an echo is heard 0.5 s after the original sound? ( Assume that sound travels at 343 m/s on that day
Answer:
171.5 m
Explanation:
To find the distance, speed x time
342 x 0.5
171.5 m
Hope this helped!
The current through a 3.0 Ω resistor is 0.30 A. The resistor is wired in series with a 9.0 V battery and an unknown resistor. What is the value of the unknown resistor?
Answer:
answer is 37 ohm since current e
remains same in series
At 20 ◦C a copper wire has a resistance of 4×10−3 Ω and a temperature coefficient of resistivity of 3.9×10−3 (C◦)−1, its resistance at 100 ◦C is
A.
52.5 × 10-3 Ω
B.
5.25 × 10-3 Ω
C.
5.25 × 10-4 Ω
D.
5.25 × 10-2 Ω
E.
25.5 × 10-3 Ω
Answer:
[tex]R _{t} = R _{0}( \alpha t + 1) \\ = 4 \times {10}^{ - 3} (3.9 \times {10}^{ - 3} \times 20 + 1) \\ = 4 \times {10}^{ - 3} (1.078) \\ = 4.312 \times {10}^{ - 3} \: Ω[/tex]
Select the correct answer.
If you increase the frequency of a sound wave four times, what will happen to its speed?
A.
The speed will increase four times.
B.
The speed will decrease four times.
C.
The speed will remain the same.
D.
The speed will increase twice.
E.
The speed will decrease twice.
Answer:
A. The speed of the wave increases four times.
Answer:
A. The speed will increase four times.
Explanation:
If you increase the frequency of a sound wave four times, the speed will increase four times. So, option (A) is correct.
Which statements describe using genetic factors to influence the growth of organisms? Select the three (3) that apply.
-increasing use of hybrid crops
-altering genes in DNA to create new plants
-increasing human population
-increasing climate change
-developing disease or pest resistant crops
Answer:
- increasing use of hybrid crops
- altering genes in DNA to create new plants
- developing disease or pest resistant crops
Explanation:
The use of genetic factors to influence the growth of a plant encompasses manipulating the genetic constituent (gene) of such plant.
For example,
- Increasing use of hybrid crops entails mating two pure bred plants based on a gene of interest responsible for a particular trait, to form a hybrid.
- Altering genes in DNA to create new plants is also a genetic factor as it has to with gene modification.
- developing disease or pest resistant crops means that the genetic make up of such plant has been modified to be resistant to pest/disease.
. A car increases velocity from 20 m/s to 60 m/s in a time of 10 seconds. What was the acceleration of the car?
Answer:
0.3333
Explanation:
Acceleration = change in velocity/time
a = 20 m/s / 60 m/s
a = 0.3333 m/s^2
The viscid silk produced by the European garden spider (Araneus diadematus) has a resilience of 0.35. If 10.0 J of work are done on the silk to stretch it out, how many Joules of work are released as thermal energy as it relaxes?
Answer: The energy released as thermal energy is 6.5 J
Explanation:
Energy stored by the spider when it relaxes is given by:
[tex]E_o=\text{Resilience}\times \text{Work}[/tex]
We are given:
Resilience = 0.35
Work done = 10.0 J
Putting values in above equation, we get:
[tex]E_o=0.35\times 10\\\\E_o=3.5J[/tex]
Energy released at thermal energy is the difference between the work done and the energy it takes to relaxes, which is given by the equation:
[tex]E_T=\text{Work done}-E_o[/tex]
Putting values in above equation, we get:
[tex]E_T=(10-3.5)=6.5J[/tex]
Hence, the energy released as thermal energy is 6.5 J
The energy released as thermal energy when 10 J of work is done to stretch silk will be 6.5 J
What is thermal energy?Thermal energy refers to the energy contained within a system that is responsible for its temperature. Heat is the flow of thermal energy.
Energy stored by the spider when it relaxes is given by:
[tex]\rm E_o=Resilience \ \times Work[/tex]
We are given:
Resilience = 0.35
Work done = 10.0 J
Putting values in above equation, we get:
[tex]\rm E_o=0.35\times 10[/tex]
[tex]E_o=3.5\ J[/tex]
Energy released at thermal energy is the difference between the work done and the energy it takes to relaxes, which is given by the equation:
[tex]E_T=\rm Work done -E_o[/tex]
Putting values in above equation, we get:
[tex]E_T=(10-3.5)=6.5\ J[/tex]
Hence, the energy released as thermal energy is 6.5 J
To know more about thermal energy follow
https://brainly.com/question/19666326
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
What are the practical systems under continuos time system?
Answer:
REVIEWS
Wing-Kuen Ling, in Nonlinear Digital Filters, 2007
Controllability and observability
For a continuous time system, assume that x(0) = 0. ∀x1, if ∃t1 > 0 and u(t) such that x(t1) = x1, then the continuous time system is said to be reachable. Similarly, for a discrete time system, assume that x(0) = 0. ∀x1, if ∃n1 > 0 and u(n) such that x(n1) = x1, then the discrete time system is said to be reachable. For a continuous time system, if ∀x0, x1, ∃t1 > 0 and u(t) such that x(0) = x0 and x(t1) = x1, then the continuous time system is said to be controllable. Similarly, for a discrete time system, if ∀x0, x1, ∃n1 > 0 and u(n) such that x(0) = x0 and x(n1 = x1, then the discrete time system is said to be controllable. For LTI systems, the set of reachable state is R(|BAB… AnB|), where R(A) is defined as the range of A, that is R(A) = {y : y = Ax}. Also, the LTI systems are controllable if and only if R(A) = Rn Or in other words, rank(|BAB… AnB|) = n.
g is incident on 3 successive sheets of polarizing material. The transmission axis of the first sheet is vertical. The transmission axis of the second sheet is at 30 degrees from vertical. The transmission axis of the third is horizontal. What is the intensity of the light emerging from the third sheet
Answer:
The intensity of light passing from the third polarizer is 3Io/16.
Explanation:
The law of Malus is given by
[tex]I=I_o cos^2\theta[/tex]
Let the incident intensity of light is Io.
The intensity of light passing from the first polarizer is
[tex]I' = \frac{I_o}{2}[/tex]
The intensity of light passing from the second polarizer is
[tex]I''=\frac{I_o}{2}\times cos^230 =\frac{3I_o}{8}[/tex]
The intensity of light passing from the third polarizer is
[tex]I''' = \frac{3I_o}{8}\times cos^2 60\\\\\\I''' = \frac{3I_o}{16}[/tex]
A spring, with a spring constant of 4000 N/m, is oriented horizontally, and compressed by 10cm. When released, the spring launches a block of mass 1.0 kg along a 5.0-m horizontal section of track, where the coefficient of friction between the block and track is 0.20. The block then goes up a frictionless ramp angled at 60o with the horizontal. How high up the ramp does the block go before it starts to slide back down
Answer:
[tex]d=1.2m[/tex]
Explanation:
From the question we are told that:
Spring constant [tex]k= 4000 N/m[/tex]
Compressed [tex]l_d= 10cm=>0.10[/tex]
Mass [tex]m=1.0kg[/tex]
Length of horizontal section [tex]l=5.0-m[/tex]
Coefficient of friction [tex]\mu=0.20[/tex]
Angle [tex]\theta=60 \textdegree[/tex]
Generally the equation for Kinetic Energy K.E is mathematically given by
[tex]K.E=\mu mgL+mgdsin\theta[/tex]
[tex]\frac{1}{2}k*l_d^2=\mu mgL+mgdsin\theta[/tex]
[tex]\frac{1}{2}(4000)*0.1^2=0.2*1*9.8*5+1*9.8*d*sin60[/tex]
[tex]d=1.2m[/tex]