Answer:
a) If both are charged
b) the first team has a load and the second half of it
c) The first team has a charge contrary to the rod, it is charged by induction and the second team has a charge equal to the rod, it is charged by contact
Explanation:
For this exercise let's analyze the charge on each electroscope separately
electroscope 1
Suppose the student with her finger on the device is grounded,
When bringing the rod closer to the electroscope a charge is indexed, on the side of the rod on the opposite side there is the same charge but with the opposite sign, as the student is grounded, this charge passes to the ground, this side remains neutral.
When removing the finger, loads can no longer pass, and when removing the rod the load from this side distributes throughout the equipment, therefore it has a load of the same magnitude as the rod
Electroscope 2
In this case, the equipment is touched, so the load redistributes in the two, when separating it remains with half the load of the cheek.
After analyzing each device separately we can answer the questions.
a) If both are charged
b) the first team has a load and the second half of it
c) The first team has a charge contrary to the rod, it is charged by induction and the second team has a charge equal to the rod, it is charged by contact
You are at the park with your little brother, when you notice a small merry-go-round with a radius that looks to be about 1.5 m. Your brother climbs on, and you give him a spin. From rest, you speed up smoothly, completing one full lap in 3.0 seconds. You wonder what sort of acceleration (magnitude and direction) your brother experiences at the very start of the motion, and at the very end of the first lap. He is quite young and hasn't studied any physics, so he does not know how to answer your question. You decide to give him your smartphone (which has a built-in accelerometer), and repeat the experiment. What will the smartphone record
Answer:
Explanation:
angle covered in one rotation = 2π radian
θ = ωt + 1/2 αt²
θ is angle rotated in time t with initial angular velocity of ω and angular acceleration α .
Putting the values
2π = 0 + 1/2 x α x 3²
α = 1. 4 radian / s²
linear acceleration = α x r = 1.4 x 1.5 = 2.1 m / s².
Initial acceleration = 2.1 m /s²
final angular velocity = α t = 1.4 x 3 = 4.2 radian / s
linear velocity = 4.2 x 1.5 = 6.3 m /s
centripetal acceleration = v² / R = 6.3² / 1.5 = 26.46 m /s²
radian acceleration = 26.46 m /s
tangential acceleration = 2.1 m /s²
Total final acceleration = √ ( 26.46² + 2.1² )
= √ ( 700.13 + 4.41)
Final acceleration = 26.53 m / s²
The kid will undergo a circular motion and the smartphone which has a built-in accelerometer will record the acceleration [tex]26.54 \,m/s^2[/tex], which is the magnitude of the vector sum of both radial acceleration and tangential acceleration.
Circular MotionThe angle covered in one full rotation, i.e.; the angular displacement is,
[tex]\theta = 2\pi\,\, radian[/tex]
The kinematics equation in the case of rotational motion is given by;
[tex]\theta = \omega_i \,t + \frac{1}{2} \alpha t^2[/tex]
But the initial angular velocity, [tex]\omega_i = 0[/tex]
Time taken to complete one lap is, [tex]t = 3\,s[/tex]
Therefore, substituting the given values in the kinematics equation;
[tex]2\pi \;rad=0 + \frac{1}{2} \alpha\times (3)^2[/tex]
[tex]\implies \alpha = \frac{4 \pi\,\,rad}{9\,s^2} =1.395\,rad/s^2 \approx 1.4\,rad/s^2[/tex]
We know that linear acceleration (here tangential direction) is given by,
[tex]a_T=a = r \alpha = 1.5\,m \times 1.4\; rad/s^2=2.1\,m/s^2[/tex]
Also, the final angular velocity after 3 s is given by;
[tex]\omega_f = \alpha t=1.4\,rad/s^2 \times 3\,s = 4.2\,rad/s[/tex]
Therefore, the final linear velocity is;
[tex]v_f = r \omega = 1.5\,m \times 4.2\,rad/s = 6.3\,m/s[/tex]
We know that the centripetal or the radial acceleration is given by;
[tex]a_r = \frac{v^2}{r} = \frac{(6.3\,m/s)^2}{1.5m}=26.46\,m/s^2[/tex]
Therefore, the total acceleration is given by;
[tex]a_{total} = \sqrt{(a_T)^2 + (a_r)^2\,} =\sqrt{(2.1)^2 + (26.46)^2} =\sqrt{704.54} =26.54 \,m/s^2[/tex]
Learn more about circular motion here:
https://brainly.com/question/2562955
a piano dropped from a plane in the air. As it falls, upward force of air resistance gets greater as the piano picks up speed eventually, the resistance forces equal to the downward force of gravity. the piano now
Answer:
When the piano is dropped, the first force acting on it will be the gravitational force, that accelerates the piano at 9.8m/s^2 downwards.
As the piano accelerates, the velocity increases, now appears other force, the air resistance, that opposes to the motion of the piano.
As the velocity of the piano increases, also does the force that the air applies on the piano.
There is a point where the velocity of the piano is such that the air resistance is equal in magnitude, but in the opposite direction, to the force of gravity.
Then the net force on the piano is zero, which means that there is no acceleration, so the piano will keep falling down at constant velocity after this point.
Using a set of observations to test a hypothesis.
2. The visible region of the hydrogen spectrum results from relaxation of electrons from excited states to energy level 2 (n1). Use the Rydberg equation and your measured wavelengths to determine the energy transitions associated with each of your observed wavelengths for hydrogen. In other words, calculate the excited state energy level (n2) for each of your observed wavelengths for hydrogen. n has integer values; so, calculate it first with appropriate significant digits, then round it to an integer. Use the key to show your work for at least one calculation. Must show energy levels for each hydrogen wavelength.
Answer:
E = 1.89 eV , E = 2.56 eV , E = 2.86 eV
Explanation:
The emission of light in the visible range, explained by the Balmer series with expression
1 /λ= [tex]R_{H}[/tex] (1/2² - 1 / n²)
n = 3, 4, 5 ...
the constant R_{H} called Rydberg's constant and is equal to 1,097 10⁷ m⁻¹
These transitions are clearly explained by Bohr's atomic model, where the empirical series of Balmer and Rydberg are deduced from a theoretical model of the hydrogen atom in natural form.
Let's calculate the wavelengths for each transition
State
initial final λ (10⁻⁷ m)
3 2 6.5634
4 2 4.8617
5 2 4.3408
Let's calculate the energy of each of these wavelengths using Planck's equation
E = h f = h c /λ
λ = 6.5634 10⁻⁷ m
E = 6.63 10⁻³⁴ 3 10⁸ / 6.5634 10⁻⁷
E = 3.03 10⁻¹⁹ J
we reduce to eV
E = 3.03 10⁻¹⁹ J (1 eV / 1.6 10⁻¹⁹ J)
E = 1.89 eV
λ = 4.8617 10⁻⁷m
E = 6.63 10⁻³⁴ 3 10⁸ / 4.8617 10⁻⁷
E = 4.09 10⁻¹⁹ J
E = 2.56 eV
λ= 4.3408 10⁻⁷ m
E = 6.63 10⁻³⁴ 3 10⁸ / 4.3408 10⁻⁷
E = 4.582 10⁻¹⁹J
E = 2.86 eV
A car is traveling north with a velocity of 18.1 m/s. Find the velocity of the car after 7.50 seconds if the acceleration is 2.4 m/s^2. *
A cloud mass moving across the ocean at an altitude of 2000 m encounters a coastal mountain range. As it rises to a height of 3500 m to pass over the mountains, it undergoes an adiabatic expansion. The pressure at 2000 m is 0.802 atm and at 3500 is 0.602 atm. If the initial temperature of the cloud mass is 288 K, what is the cloud temperature as it passes over the mountains? Assume that Cp,m for air is 28.86 J K-1 mol-1 and that the air obeys the ideal gas law. If you are on the mountain, should you expect rain or snow?
Answer:
snow
Explanation:
Since the process undergoes adiabatic expansion, hence q = 0 and ΔU = w.
We can sole this problem using the following derivation:
[tex]ln(\frac{T_2}{T_1} )=-(\gamma -1)ln(\frac{V_f}{V_i} )=-(\gamma -1)ln(\frac{T_2}{T_1}\frac{P_i}{P_f} )\\=-(\gamma -1)ln(\frac{T_2}{T_1})-(\gamma -1)ln(\frac{P_i}{P_f})\\=-(\frac{\gamma -1}{\gamma})ln(\frac{P_i}{P_f})\\=-(\frac{\frac{C_{p,m}}{C_{p,m}-R} -1}{\frac{C_{p,m}}{C_{p,m}-R}})ln(\frac{P_i}{P_f})\\\\ln(\frac{T_2}{T_1} )==-(\frac{\frac{C_{p,m}}{C_{p,m}-R} -1}{\frac{C_{p,m}}{C_{p,m}-R}})ln(\frac{P_i}{P_f})\\\\Substituting\ values:\\\\[/tex]
[tex]ln(\frac{T_2}{T_1} )=-(\frac{\frac{28.86}{28.86-8.314} -1}{\frac{28.86}{28.86-8.314}})ln(\frac{0.802\ atm}{0.602\ atm})=-0.0826\\\\ln(\frac{T_2}{T_1} )=-0.0826\\\\Taking\ exponential\ of\ both \ sides:\\\\\frac{T_2}{T_1} =e^{-0.0826}\\\\T_2=0.9207T_1\\\\T_2=0.9207*288\\\\T_2=265\ K\\[/tex]
Since T2 = 265 K, we should expect a snow
A .What is the difference between fossils fuels and nuclear fuel ?
B .Explain why it is important for us to find alternatives to fossil fuels to meet pur energy?
C .What are renewable energy sources?give three examples.
Answer:
A. Nuclear and fossil fuel-burning power plants differ mainly in where their energy comes from; a nuclear reactor produces heat from radioactive metals, and a fossil-fuel plant burns coal, oil or natural gas.
B. Environmental and economic benefits of using renewable energy include: Generating energy that produces no greenhouse gas emissions from fossil fuels and reduces some types of air pollution. Diversifying energy supply and reducing dependence on imported fuels
C. These are energy sources that are constantly being replenished, such as sunlight, wind, and water.
The most popular renewable energy sources currently are:
Solar energy.
Wind energy.
Hydro energy.
Tidal energy.
Geothermal energy.
Biomass energy.
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The items in a mixture can be returned to their original form.
True
False
Answer:
I believe it is true
If not, pls let me know. :)
The distance, x, covered by a particle in time, t, is given as x=a +bc+ct^2 +dt^3
.find the dimension of the constants a, b, c and d
Answer:
[tex]a[/tex] has units of distance
[tex]b[/tex] has units of distance over time
[tex]c[/tex] has units of distance over [tex]time^2[/tex]
[tex]d[/tex] has units of distance over [tex]time^3[/tex]
Explanation:
Since the expression for the distance is:
[tex]x = a+b\,t+c\,t^2+d\,t^3[/tex]
then:
[tex]a[/tex] has units of distance
[tex]b[/tex] has units of distance over time
[tex]c[/tex] has units of distance over [tex]time^2[/tex]
[tex]d[/tex] has units of distance over [tex]time^3[/tex]
because we are supposed to be able to add all of the terms and get a distance. So the products on each term that contains factors of time (t) should be cancelling those time units with units in the denominator of the multiplicative constant s that accompany them.
A car in a roller coaster moves along a track that consists of a sequence of ups and downs. Let the
x axis be parallel to the ground and +y axis point upward. In the time interval from t=0 to t=4 s the trajectory of the caralong a certain section of the track is given by
→r=A(1 m/s)t^i+A((1 m/s3)t3−6(1 m/s2)t2)^j, where A is positive dimensionless constant.a. At t=2.0s is the roller coaster car ascending or descending?
b. Derive a general expression for the speed v of the car. Make sure that your expression would give the correct value for the speed (in m/s), but you don't need to put in anything explicitly about units (unlike what you see in the original expressionfor →r. Express your answer in meters per second in terms of A and t.
Answer:
Explanation:
r=A(1 m/s)t^i+A((1 m/s3)t3−6(1 m/s2)t2)^j,
dr / dt = A i + (3 A t² - 6 x 2 t ) j
At t = 2
dr / dt = A i + (12A - 24 A ) j = A i - 12 A j .
Since slope is negative so velocity is downwards , hence it is descending
b ) velocity = dr / dr = . A i + (3 A t² - 6 x 2 t ) j
Vx = A
Vy = 3A t² - 12 t
A 25.0 kg bag of peat moss sits in the back of a flatbed truck, driving up a hill. The bag experiences a 225N normal force. The maximum acceleration the truck can have so the bag does not slip is 2.40 m/s2 . Calculate the (a) angle of the hill relative to horizontal and (b) coefficient of static friction between the bag and the truck. (c) The truck is now travelling on level ground at constant speed. The sand bag is tossed forward sliding along the truck bed with a horizontal speed of 2.55 m/s. If the coefficient of kinetic friction is 0.350, how far does the bag slide before coming to rest
Answer:
a
[tex]\theta = 23.32^o [/tex]
b
[tex] \mu_s = 0.27 [/tex]
c
[tex] s = 0.948 \ m [/tex]
Explanation:
From the question we are told that
The mass of the bag is [tex]m_b = 25.0 \ kg[/tex]
The normal force experienced is [tex]F_n = 225 \ N[/tex]
The maximum acceleration of the bag is [tex]a = 2.40 \ m/s^2[/tex]
Generally this normal force experience by the bag is mathematically represented as
[tex]F_n = mg cos \theta[/tex]
=> [tex]225 = (25 * 9.8) cos \theta[/tex]
=> [tex] 0.9183 = cos \theta[/tex]
=> [tex]\theta = cos^{-1}[0.9183][/tex]
=> [tex]\theta = 23.32^o [/tex]
Generally for the bag not to slip , it means that the frictional force is equal to the sliding force
[tex]F_f = F_s[/tex]
Hence [tex]F_f [/tex] is mathematically represented as
[tex]F_f = \mu_s * F_n [/tex]
While [tex]F_s [/tex] is mathematically represented as
[tex]F_s = m * a [/tex]
So
[tex] \mu_s * F_n = m * a [/tex]
=> [tex] \mu_s * 225 = 25 * 2.40 [/tex]
=> [tex] \mu_s = 0.27 [/tex]
Generally from the workdone equation we have that
[tex]KE_f - KE_i = W_f[/tex]
Here [tex]W_f[/tex] is the work done by friction which is mathematically represented as
[tex]W_f = m * g * \mu_k * s[/tex]
Here s is the distance covered by the bag
[tex]KE_f[/tex] is zero given that velocity at rest is zero
and
[tex]KE_i = \frac{1}{2} * m* v_i^2[/tex]
so
[tex] \frac{1}{2} * m* v_i^2 = m * g * \mu_k * s [/tex]
=> [tex] \frac{1}{2} * v_i^2 = g * \mu_k * s [/tex]
substituting 2.55 m/s for v_i and 0.350 for \mu_k we have that
[tex] \frac{1}{2} * 2.55^2 = 9.8 * 0.350 * s [/tex]
=> [tex] s = 0.948 \ m [/tex]
When a ball is thrown straight up with no air Resistance, the acceleration is in what direction ?
Answer:
The ball has a velocity upwards but the acceleration is downwards. Gravity is giving the ball a downwards acceleration from the moment the ball leaves the hands.
give five characteristics of a metal
Answer:
are good conductors of heat,are you used to make doors utensils e.t.c.
Q3. What is the symbol for a :-
(a) millimeter
(b) micrometer
(c) centimeter
(d) kilometer
(e) metre
(f) nanometer
Explanation:
a. mm
b. μm
c. cm
d. km
e. m
f. nm
Help please it’s due by 11:59p
Answer: 6 m/s squared
Explanation:
6m/s squared. Hope this helped
Plz give brainliest
There are two containers of equal volumes, each filled with a different gas. Both containers have the same number of moles of gas and are at the same temperature. The molecules of gas in container 1 are four times more massive than the molecules of gas in container 2. The number of moles in container 2 is increased until it is a factor of 4 larger than the number of moles in container
1. The volume and temperature of container 2 remain unchanged. After increasing the number of moles in container
2: Which container, 1 or 2, has a higher pressure, or are they the same? Which container, 1 or 2, has a higher average (rms) speed of gas molecules, or are they the same? Which container, 1 or 2, has a higher average kinetic energy of gas molecules, or are they the same? Which container, 1 or 2, has a higher thermal energy, or are they the same?
Answer:
Molecules in container 2 has a higher pressure.
Molecules in container 2 has a greater rms speed.
Molecules in container 2 has a greater kinetic energy.
Molecules in container 2 has a greater thermal energy.
Explanation:
At constant volume and temperature, The number of moles of a gas is proportional to its pressure.
The lower the pressure, the lesser the rms speed of molecules, kinetic energy of the molecules and thermal energy of the molecules, hence the answer above.
The coordinate of a particle in meters is given by x(t)=1 6t- 3.0t , where the time tis in seconds. The
particle is momentarily at rest at t is:
Select one:
a. 9.3s
b. 1.3s
C. 0.75s
d.5.3s
e. 7.3s
Complete question:
The coordinate of a particle in meters is given by x(t)=1 6t- 3.0t³ , where the time tis in seconds. The
particle is momentarily at rest at t is:
Select one:
a. 9.3s
b. 1.3s
C. 0.75s
d.5.3s
e. 7.3s
Answer:
b. 1.3 s
Explanation:
Given;
position of the particle, x(t)=1 6t- 3.0t³
when the particle is at rest, the velocity is zero.
velocity = dx/dt
dx /dt = 16 - 9t²
16 - 9t² = 0
9t² = 16
t² = 16 /9
t = √(16 / 9)
t = 4/3
t = 1.3 s
Therefore, the particle is momentarily at rest at t = 1.3 s
What the density of pure water?
a)1.0kg/l
b) 0.1 kg/l
c)100kg/l
d)10kg/l
Answer:
a) 1.0(kg/l) = 1 [kg/Lt]
Explanation:
The density of pure water is taken as 1000 [kg/m3], now we must convert the units to (kg/liters)
[tex]1000[\frac{kg}{m^{3}}]*\frac{1m^{3} }{1000Lt} = 1000[\frac{kg}{lt} ][/tex]
What is the acceleration at the apex of a vertical up and down problem?
Answer:
The correct option is +9.81 m/s²
Explanation:
The acceleration of a vertical (up and down) plane depends on if the object is going up or down. Acceleration due to gravity is 9.81 m/s/s or 9.81 m/s². When an object falls/comes down (vertically) without interference, the acceleration of such an object is same as acceleration due to gravity (+9.81 m/s²). However, when an object is thrown/goes up, the acceleration of such objects goes against the gravity (of earth) hence the acceleration is -9.81 m/s².
At the top/apex of a vertical up and down problem, the object will be pulled back down (because of gravity) and hence, it's acceleration becomes +9.81 m/s² (changing from negative while coming up to positive).
Imagine you are standing in the hallway. If the gravitational field strength were to suddenly double, what would change?
Answer:
If the gravitational field strength were to suddenly double while I was standing in the hallway, I'd suddenly feel like heavier.
Explanation:
The weight of a matter is given as the production between it's mass and the acceleration of gravity.
Mathematically, this is expressed as:
m x g that is m.g or mg.
Where m = mass and
g = gravity.
If the value of gravity was to double then weight (w) will become
w = m×2g
Assume for a moment that before the increase m was 5 and g was 5, it means that w would be 25.
But if g doubled, then we would have
w = 5 x 10
which is equals 50.
Thus, with the increase in gravitational field strenght, I'd find myself exerting more effort to stand upright, jump or run.
Cheers!
Metal reactivity
a. increases
b. decreases
from left to right in the periodic table.
C. stays the same
d. can increase or decrease depending on the
element
Answer:
i think its A
Explanation:
Sandra shot a rocket so that it moved with an initial velocity of 9.81 m/s straight upward. The rocket leaves and returns to ground level. What is the total time the rocket was in the air before it strikes the ground? (Ignore air resistance.
Answer:
2.00 seconds.
Explanation:
The following data were obtained from the question:
Initial velocity (u) = 9.81 m/s
Total time in air (T) =..?
Next, we shall determine the time taken for the rocket to reach it's maximum height. This can be obtained as follow:
Note: At maximum height, the final velocity is zero.
Initial velocity (u) = 9.81 m/s
Final velocity (v) = 0 m/s
Acceleration due to gravity (g) = 9.8 m/s²
Time to reach the maximum height (t) =.?
v = u – gt (since the rocket is going against gravity)
0 = 9.81 – 9.8t
Rearrange
9.8t = 9.81
Divide both side by 9.8
t = 9.81/9.8
t = 1.00 s
Therefore the time taken to reach the maximum height is 1.00 second.
Finally, we shall determine the total time spent by the rocket in the air as follow:
Time to reach the maximum height (t) = 1.00 s
Total time in air (T) =..?
T = 2t
T = 2 × 1.00
T = 2.00 s
Therefore, the total time spent by the rocket in the air is 2.00 seconds.
A train running along a straight track at 32.90 m/s is slowed at a rate of 0.783 m/s/s to a stop how far did the train travel before coming to rest ?
Answer:
The distance traveled by the train is 691.2 m
Explanation:
Given;
initial velocity of the train, u = 32.9 m/s
deceleration of the train, a = - 0.783 m/s²
final velocity of the train when brought to a stop, v = 0
The distance traveled by the train is given by;
v² = u² + 2as
0 = (32.9)² + 2(-0.783)s
0 = 1082.41 - 1.566s
1.566s = 1082.41
s = 1082.41 / 1.566
s = 691.2 m
Therefore, the distance traveled by the train is 691.2 m
2. A fish swimming at a constant speed of 18 m/s spots a sta-
tionary barracuda. Just as the fish passes the barracuda, the
predator begins swimming with a constant acceleration of
magnitude 2.2 m/s². The fish and barracuda are moving in the
same direction.
(a) How far does the barracuda swim before catching the fish?
(b) At what time will this occur? (Hint: Graphing may help you
visualize this problem.)
пип
Answer:
a) 294.55 meters
b) 16.4 seconds
Explanation:
The suggested graph shows ...
(a) The barracuda swims 294.55 meters before the it catches the fish.
__
(b) This occurs after about 16.4 seconds.
A truck heading east has an initial velocity of 6/ms. It accelerates at 2/ms2 for 12 seconds. What distance does the truck travel in the given time
Answer:
216m
Explanation:
As we know S=ut+1/2at^2
S=6×12+1/2×2×12×12
=72+144
=216 m
The electric field everywhere on the surface of a thin, spherical shell of radius 0.710 m is of magnitude 936 N/C and points radially toward the center of the sphere. (a) What is the net charge within the sphere's surface? nC (b) What is the distribution of the charge inside the spherical shell? The positive charge has an asymmetric charge distribution. The negative charge has an asymmetric charge distribution. The positive charge has a spherically symmetric charge distribution. The negative charge has a spherically symmetric charge distribution.
Explanation:
Given that,
Radius of a spherical shell, r = 0.71 m
Electric field that points radially toward the center of the sphere is 936 N/C
(a) Let q is the net charge within the sphere's surface. Using Gauss's law, we can find it :
[tex]\dfrac{1}{q}\times \epsilon_0=EA\\\\q=\dfrac{\epsilon_0}{EA}\\\\q=\dfrac{8.85\times 10^{-12}}{936\times \pi \times (0.71)^2}\\\\q=5.97\times 10^{-15}\ C[/tex]
(b) Gauss's law is used to find the amount of charge enclosed within a surface. It doesn't say anything about the distribution of the charge inside the spherical shell.
Calculate the distance moved by a runner who runs with a speed of 5 km/h for a period of 1.5 hours.
7.5 km
10 km
2.5 km
5 km
Answer:
7.5 km
Explanation:
h5 per hour means that he traveled 5 km in 1 our. And then half of the hour, which means half an hour 5 km which is 2.5.
5 + 2.5 = 7.5
or just 1.5 x 5 = 7.5
Let A be the last two digits, and let B be the last three digits, and the C be the sum of the last 4 digits of your 8-digit student ID. (Example: For 20245347, A = 47, B = 347, and C = 19) A train moves at an average speed of (23.0 + A) m/s for (250.0 + B) seconds and then at an average speed of (45.0 + C) m/s for (800.0 + B) seconds. Determine the average speed for the entire time in meters per second (m/s). Round your final answer to 3 significant figures.
Answer:
66.053m/s
Explanation:
A = 47
B = 347
C = 19
Train moves at
(23 + A)m/s
= 23 + 47 = 60m/s
At (250.0+B) seconds
250.0+347 =
547 seconds
Distance d,
= 70 x 597
= 41790
It also moves at
(45.0 + c)
= 45 + 19
= 64m/s
Time = 800 + B
= 800 + 347
= 1147
Distance,
= 64 x 1147
= 73408m
Total distance,
= 73408 + 41790
= 115,198
Total time,
= 597 + 1147
= 1744
Average speed,
= Total distance / total time
= 115198/1174
= 66.053m/s
A car is coasting to a stop with a constant acceleration. The car was going 35 m/s when the driver started coasting. The car moves 50. m forward while slowing down.
Show this on a vt graph.
Determine the acceleration of the car.
Answer: Plot each graph on the same coordinate system.
35 ( m/ s)
50 m
a v t.
Answer:
I think the car is accelerating at 85 m/s because the car is moving at 35/m plus when it was coasting it went 50.
Explanation:
Sorry I dont have a graph but here is what I think about on the acceleration.
What must be the units for the gravitational constant G in order for gravitational force to have units of newtons?
m3/(kg⋅s)
m3/(kg2⋅s2)
m3/(kg2⋅s)
m3/(kg⋅s2)
Answer:
m³/(kg⋅s²)
Explanation:
Hello.
In this case, since the involved formula is:
[tex]F=G*\frac{m_1m_2}{r^2}[/tex]
By writing a dimensional analysis with the proper algebra handling, we obtain:
[tex]N[=]G*\frac{kg*kg}{m^2}\\ \\kg*\frac{m}{s^2}[=]G *\frac{kg*kg}{m^2}\\\\G[=]\frac{kg*m*m^2}{kg^2*s^2}\\ \\G[=]\frac{m^3}{kg*s^2}[/tex]
Thus, answer is:
m³/(kg⋅s²)
Note that the [=] is used to indicate the units of G.
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