Answer:
a) t = 19.6 s, b) fr = 1.274 10⁴ N
Explanation:
This is a Newton's second law problem
Y Axis
for the cabin
N₁-W₁ = 0
N₁ = W₁
for the trailer
N₂- W₂ = 0
N₂ = W₂
X axis
for the cabin plus trailer, where friction is only in the cabin
fr = (m₁ + m₂) a
the friction force equation is
fr = μ N
we substitute
μ N₁ = (m₁ + m₂) a
μ m₁ g = (m₁ + m₂) a
a = μ g [tex]\frac{m_1}{m_1 + m_2}[/tex]
let's calculate
a = 0.65 9.8 [tex]\frac{2000}{2000+8000}[/tex]
a = 1,274 m / s²
a) to find the stopping distance we can use kinematics
Let's slow down the sI system
v₀ = 90 km / h (1000 m / 1km) (1h / 3600s) = 25 m / s
v = v₀ - a t
when it is stopped its speed is zero
0 = v₀ - at
t = v₀ / a
t = 25 / 1.274
t = 19.6 s
b) the friction force is
fr = 0.65 2000 9.8
fr = 1.274 10⁴ N
This is the braking force and also the forces that couple the cars.
Conductivity in aqueous solutions, is a measure of the ability of water to conduct an electric current.
a. True
b. False
Answer:
true
Explanation:
i think its true beacuse water can cause electricity
pls let me know if corrct
if yes pls mark brainlest
Key Stage 3 Science - Physics
Question 19 of 35
Calculate the force applied in newtons) if a pressure of 150Pa is acting on an area of 25m².
Answer:
F = 3750 N
Explanation:
Given that,
Pressure, P = 150 Pa
Area, a = 25m²
We need to find the force applied. We know that, pressure is equal to the force acting per unit area. It can be given by :
[tex]P=\dfrac{F}{A}\\\\F=P\times A\\\\F=150\ Pa\times 25\ m^2\\\\F=3750\ N[/tex]
So, the required force is 3750 N.
The speed of sound in air is around 330 m/s. If a bat emits a single high-pitched ‘click’ of sound in a cave that is 25m wide, calculate the time taken for the echo of the sound to return to the bat.
Answer:
0.15 s
Explanation:
From the question given above, the following data were obtained:
Speed of sound (v) = 330 m/s
Distance (x) = 25 m
Time (t) =?
The time taken for the echo of the sound to the bat can be obtained as follow:
v = 2x / t
330 = 2 × 25 / t
330 = 50 / t
Cross multiply
330 × t = 50
Divide both side by 330
t = 50 / 330
t = 0.15 s
Thus, it will take 0.15 s for the echo of the sound to the bat
elophase II is the final and the fourth stage in meiosis II when the chromosomes reach the opposite poles of the
Answer: The correct answer is formation of four haploid nuclei.
The event occurring during telophase II includes formation of four haploid daughter cells. Telophase II is the final and the fourth stage in meiosis II when the chromosomes reach the opposite poles of the nuclear spindle. During this stage both the daughter cells get divided forming four haploid cells. The development of the nuclear envelope around each set of the chromosome and cytokinesis also takes place during telophase II.
Explanation:
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how can you Make different objects using blocks
What are the differences between atoms, molecules, and compounds?
Answer:
Atoms are the thing that make up molecules and compounds. Molecule. Two or more atoms joined together with covalent bonds. Molecules contain two or more atoms and are held together by covalent bonds, whereas compounds are held together by ionic bonds. Compound. Two or more elements bonded together through ionic attraction.
I need help! thank you
Answer:
it is coice A and choice D
A proton moves at a speed of 0.12 x 10^7 m/s
at right angles to a magnetic field with a
magnitude of 0.58 T.
Find the magnitude of the acceleration of
the proton. The elemental charge is 1.60 x
10^-19 C.
Answer in units of m/s?.
Answer:
13/23 ydfffgggggggffttf
Why is manganin wire is thicker than copper wire?
Answer:
Manganin is an alloy of Cu with manganese and nickel. Since the latter two metals have resistivity greater than copper, the pure copper has lower resistivity and the manganin thus has to be thicker to have the same resistance.
I HOPE IT WILL HELP U.....#VERIFIED ANSWERWhat holds the moon in orbit around the earth?
A. The sun's gravity
B. The Earth's gravity
Answer:
Earth's gravity
Explanation:
hope this helps
Answer:
the earth gravity
Explanation:
Gravitational attraction provides the centripetal force needed to keep planets in orbit around the Sun and all types of satellite in orbit around the Earth. The Earth's gravity keeps the Moon orbiting us.
PLEASE HELP ASAPPPPP
Answer:
link
Explanation:
ww.comhwhelp
two identical springs, each with a spring force constant k, are attached end to end. If a weight is hung from a single spring, it stretches the spring by a distance d. When this same mass is hung from the end of the two springs, which, again, are connected end-to-end, the total stretch of these springs is
Answer:
Δx = 2*d
Explanation:
According to Hooke's Law, in order to the mass be in equilibrium, when attached to one spring, no net force must act on it, so the algebraic sum of the elastic force and gravity must be zero, as follows:[tex]k*\Delta x = m*g (1)[/tex]
If we hang the mass from the end of the two springs attached end to end, in order to be in equilibrium, the total elastic force must be equal to gravity, as we have already said.We can express this elastic force, as the product of a Keff times the distance stretched by the two springs combined, as follows:[tex]F = k_{eff} * \Delta x_{eff} = m*g (2)[/tex]
Due to F is a tension, it will be the same at any point of the chain of springs, so we can write the following expression, for the distance stretched by any of the springs:[tex]\Delta x_{1} = \frac{F}{\ k_{1} } (3)[/tex]
The total distance stretched will be the sum of the distances stretched by any spring individually:[tex]\Delta x_{} = \frac{F}{\ k_{1} } + \frac{F}{\ k_{2} } (4)[/tex]
Replacing (4) in (2) and rearranging, we have:[tex]\frac{F}{k_{eff} } = \frac{F}{\ k_{1} } + \frac{F}{\ k_{2} } (5)[/tex]
Since k₁ = k₂ = k, we can find keff, as follows:
[tex]k_{eff} = \frac{k^{2} }{2*k} = \frac{k}{2} (6)[/tex]
Replacing (6) in (2), and making (2) equal to (1) we finally get:[tex]F = \frac{k}{2} * \Delta x_{eff} = k*\Delta x = m*g (7)[/tex]
Solving for Δxeff:[tex]\Delta x_{eff} = \frac{2*k*\Delta x}{k} = 2* \Delta x = 2*d (8)[/tex]
PLZZZ HELP!!! 80PTS!!!
You are given a toy car and told it has a density of 0.54 g/mL and a volume of 38.2 mL. Calculate impulse, given that the toy car's acceleration is 10 m/s^2 and time is 10 s. WRITE THE NUMBER ONLY and DO NOT ROUND.
Answer:
Impulse = [tex]206.28 * 10 = 2062.8[/tex] g.m/s
Explanation:
Impulse is equal to the product of force and time
Here
Density of car = [tex]0.54[/tex] g/mL
Volume of car [tex]= 38.2[/tex] mL
Mass of the car is equal to the product of density and volume of car
Mass of the car [tex]= 0.54 * 38.2[/tex] [tex]= 20.628[/tex] grams
Acceleration of the car is [tex]10[/tex] m/s^2
Force is equal to product of mass and acceleration
F = [tex]20.628 * 10 = 206.28[/tex] g .m/s^2
Impulse = F * t
Impulse = [tex]206.28 * 10 = 2062.8[/tex] g.m/s
An important problem that all electric circuits have is the fact that since current is flowing through the traces connected the components, the circuit is generating a complex magnetic field during operation. You can shield electric fields pretty easily using a Faraday cage, but how do you shield your device from the magnetic field it is generating itself
Answer:
grid or metal sheets or metal foams
Explanation:
Electric and magnetic fields are generated in electrical circuits that often cause interference in the measurements made.
To avoid these interferences, shields are generally used for the electric fields, Farday boxes are used, which are metal grids connected to earth.
For the magnetic field there is also a grid or metal sheets or metal foams,
It should be noted that the gaps in the screening must be smaller than the wavelength of the radiation to be screened.
describe the relationship of the atoms shown above
choices:
Answer:
I don't know.
How does a parallel circuit compare to a series in circuit in terms of the paths through which
the current can flow in the circuit
that is thoramo in gach branch of a parallel circuit?
Answer:
yes
Explanation:
gl m8
A 1500 kg car traveling at 30 m/s hits a stationary 1200 kg car. If they stick together on collision, what is the final velocity of the two cars?
Answer:
16.67 m/s
Explanation:
Let that velocity be v.
Using conservation of momentum:
Initial momentum = final momentum
momentum of car1 = momentum of cars
mass1 x velocity1 = (m1 + m2)v
1500*30 = (1500 + 1200)v
45000/2700 = v
16.67 m/s = v
ANSWER THIS! Which of these statements about the Sun is the result of its closeness to Earth? A) It is the only star with a gravitational pull. B) It is the largest star in the Milky Way Galaxy. C) It appears (from Earth) to be the brightest star. D) It has the lowest temperature of any star seen from Earth.
Answer:
C) It appears (from Earth) to be the brightest star.
Explanation:
This is because due to sun's brightness we can predict its distance.
Calculate the displacement Vector for a particle moved From Thepoint
(4, 3, 2) to a point (8,3,6)
Answer:
first braliest me
Explanation:
HELP ITS DUE IN 4 MINUTES
Answer:
igneous = melted rocks formed by cooled magma
sedimentary is brken rocks, kayers with fossil...
metamophic is rocks formed by pressure and heat
Two point masses are held in place a distance d apart. Another point mass M is midway between them. M is then displaced a small distance x perpendicular to the line connecting the two fixed masses and released.
a. Show that the magnitude of the net gravitational force on M due to the fixed masses is guven approximately by Fnet = 16GmM/d^3x if x ≪ d.
b. What would the period be if m = 100kg and d = 25.0cm?
c. Will M oscillate if it is displaced from the center a small distance x toward either of the fixed masses?.
What is the mechanical advantage of a lever that has an input arm of 6 meters and an output arm of 2 meters
Answer:A
Explanation:I TOOK THE Test
Which statement best compares gamma rays to the other components of the electromagnetic spectrum?
Am I correct ???
Answer:
no
Explanation:
gamma rays do have the most frequency, but they also have the highest amount of energy because they happen the most often.
16. Two capacitors have an equivalent
capacitance of 30 pF, if connected in
parallel, and 7.2 pF, if connected in
series. Find C1 and C2.
Answer:
C1 + C2 = 30 parallel connection
C1 * C2 / (C1 + C2) = 7.2 series connection
C1 * C2 = 7.2 * (C1 + C2) = 216
C2 + 216 / C2 = 30 using first equation
C2^2 + 216 = 30 C2
C2^2 - 30 C2 + 216 = 0
C2 = 12 or 18 solving the quadratic
Then C1 = 18 or 12
A population pyramid is created by
graphing the distribution of ages in a population
Explanation:
Belive me if you dont then dont use the answer
Answer:
a graph that shows the distribution of ages across a population
18. Write conversion factors (as ratios) for the number of:
(a) yards in 1 meter
(b) liters in 1 liquid quart
(c) pounds in 1 kilogram
Answer:
Conversion tables show:
1 m = 1.09361 yds
1 Lit = .26418 gal = 1.05672 qt or 1 qt = .944632 Lit
1 lb = .45359 kg = 2.2046 Lbs / Kg
So X yds = X m * 1.09361 yds / m = 1.09361 * X yds
Likewise X Lit = X qt / 1.05672 qt/ Lit = X / 1.05672 Lit = .94632 X Lit
So X Lbs = X kg * 2.2046 Lbs / Kg = 2.2046 Lbs
____ is based on the acceptance of rules and laws.
Multiple Choice
1. Rational-legal authority
2. Traditional authority
3. Legal power
4. Political power
A vibrating mass of 300 kg mounted on a massless support by a spring of stiffness 40,000 N>m and a damper of unknown damping coefficient is observed to vibrate with a 10-mm amplitude while the support vibration has a maximum amplitude of only 2.5 mm (at resonance). Calculate the damping constant and the amplitude of the force on the base.
Answer:
400 N
Explanation:
[tex]\text { Given: } m=300 \mathrm{~kg}, k=40,000 \mathrm{~N} / \mathrm{m}, \omega_{b}=\omega_{n}(r=1), X=10 \mathrm{~mm}, Y=2.5 \mathrm{~mm}[/tex] .
Find damping constant
[tex] \frac{X}{Y}=\left[\frac{1+(2 \zeta r)^{2}}{\left[\left(1-r^{2}\right)^{2}+(2 \zeta r)^{2}\right.}\right]^{1 / 2} \\ \left.\frac{10}{2.5}=\frac{\left\lceil 1+4\zeta^{2}\right]^{1 / 2}}{4 \zeta^{2}}\right] \\ 16=\frac{1+4 \zeta^{-2}}{4 \zeta^{2}} \\ \zeta^{2}=\frac{1}{60}=\frac{c^{2}}{4 k m} \\ c=\sqrt{\frac{4(40,000)(300)}{60}} \\ c=894.4 \mathrm{~kg} / \mathrm{s}[/tex]
Amplitude of force on base:
[tex]F_{T}=k Y r^{2}\left[\frac{1+(2 \zeta r)^{2}}{\left(1-r^{2}\right)^{2}+(2 \zeta r)^{2}}\right]^{1 / 2}[/tex]
substituting the values in above formula we get
F_T = 400 N
According to Newton's 3rd Law of Motion, Doug, a baseball
player hits a ball with his bat with a force of 1,000N. The ball
exerts a reaction force equally against the bat of
A.less than 1,000N
B.more than 1,00N
C.1,000N
D.double 1,000N
While being thrown, a net force of 132 N acts on a baseball (mass = 140g) for a period of 4.5 x 10^-2 sec. what is the magnitude of the change in momentum of the ball?
Answer:
5.94 N·s
Explanation:
F = 132 N
t = 0.045 s
Impulse = Ft = (132 N)(0.045 s) = 5.94 N·s