Answer:
Initial velocity, u = 60 km/h = 16.7 m/s
Final velocity, v = 72 km/h = 20 m/s
time, t = 2 sec
From first equation of motion:
[tex]{ \bf{v = u + at}}[/tex]
Substitute the variables:
[tex]{ \tt{20 = 16.7 + (a \times 2)}} \\ { \tt{2a = 3.3}} \\ { \tt{acceleration = 1.65 \: {ms}^{ - 2} }}[/tex]
A covalent bond is formet by of electrons..?
Answer:
The covalent bond is formed by pairs of electrons that are shared between two atom
Explanation:
The covalent bond is formed by pairs of electrons that are shared between two atoms, in general the electrons must have opposite spins to have a lower energy state.
In this bond, the electrons are between the two atoms and are shared between them in such a way that there is a configuration of eight electrons in the orbit.
1. Consider a 1000 kg car rounding a curve on a flat road of radius 50 m at a speed of
50 km/h (14 m/s).
a. Will the car make the turn if the pavement is dry and the coefficient of static
friction is 0.60?
Answer:
The car will make the turn perfectly
Explanation:
Given that the centripetal force= mv^2/r
M= mass of the car
v = speed of the car
r= radius
Hence;
F = 1000 × (14)^2/50
F= 3920 N
The frictional force = μmg
μ = coefficient of static friction
m= mass
g = acceleration due to gravity
Frictional force= 0.6 × 1000× 10
Frictional force = 6000 N
The car will not skid off the curve because the frictional force is greater than the centripetal force.
Object A has a mass m and a speed v, object B has a mass m/2 and a speed 4v, and object C has a mass 3m and a speed v/3. Rank the objects according to the magnitude of their momentum.
Required:
Rank from smallest to largest.
Answer:
Momentum of object A = Momentum of object C < momentum of B.
Explanation:
The momentum of an object is equal to the product of mass and velocity.
Object A has a mass m and a speed v. Its momentum is :
p = mv
Object B has a mass m/2 and a speed 4v. Its momentum is :
p = (m/2)×4v = 2mv
Object C has a mass 3m and a speed v/3. Its momentum is :
p = (3m)×(v/3) = mv
So,
Momentum of object A = Momentum of object C < momentum of B.
A force of 15 N toward the WEST is applied to a 4.0 kg box. Another force of 42 N toward the EAST is also applied to the 4.0 kg box. The net force on the 4.0 kg box
is
[tex]\implies F_1 < F_2[/tex]
[tex] \implies F_{net} = F_2 - F1[/tex]
[tex]\implies F_{net} = 42 -15[/tex]
[tex]\implies \underline{ \boxed{ F_{net} = 27 \: N}}[/tex]
The net force on the 4.0 kg box is 27 N towards EAST.
Can a conductor be given limitless charge
Answer:
No
Explanation:
You could try to give it enough to fill all valence electrons in all of the atoms in the conductor, but practically this could not be achieved.
A car starting at rest accelerates at 3m/seconds square How far has the car travelled after 4s?
Answer:
24 meters
Explanation:
Find the final velocity. 12m/s
d=[final-initial]/2×time
D=(6m/s)×4=24 m/s
The following are ways to properly manage your stress except
Answer:
you should provide the options for us to answer?
HEELLPPPPPpppppppppppppppp
Explanation:
Given:
[tex]A_1[/tex] = 4.5 cm[tex]^2[/tex]
[tex]v_1[/tex] = 40 cm/s
[tex]v_2[/tex] = 90 cm/s
[tex]A_2[/tex] = ?
a) The continuity equation is given by
[tex]A_1v_1 = A_2v_2[/tex]
Solving for [tex]A_2[/tex],
[tex]A_2 = \dfrac{v_1}{v_2}A1 = \left(\dfrac{40\:\text{cm/s}}{90\:\text{cm/s}}\right)(4.5\:\text{cm}^2)[/tex]
[tex]= 2\:\text{cm}^2[/tex]
b) If the cross-sectional area is reduced by 50%, its new area [tex]A_2'[/tex] now is only 1 cm^2, which gives us a radius of
[tex]r = \sqrt{\dfrac{A_2'}{\pi}} = 0.564\:\text{cm}[/tex]
Una bala de 10 g se dispara contra un bloque de madera de 102 g inicialmente en reposo sobre una superficie horizontal. Después del impacto el bloque se desliza 8 m antes de detenerse. Si el coeficiente de fricción entre el bloque y la superficie es 0,5, ¿Cuál es la velocidad de la bala inmediatamente antes del impacto?
Answer:
una ess abola cola sola answer
Two wires are made of the same material and have the same length but different radii. They are joined end-to- end and a potential difference is maintained across the combination. Of the following quantities that is same for both wires is
A. Potential difference
B. Electric current
C. Current density
D. Electric field
Answer:
Current
I think The choose (B)
B. Electric current
Which two factors does the power of a machine depend on? А. work and distance B.. force and distance C. work and time D. time and distance?
[tex]Hello[/tex] [tex]There![/tex]
[tex]AnimeVines[/tex] [tex]is[/tex] [tex]here![/tex]
The answer is...
C. Work and time.
[tex]HopeThisHelps!![/tex]
[tex]AnimeVines[/tex]
During a particular thunderstorm, the electric potential difference between a cloud and the ground is Vcloud - Vground = 4.20 108 V, with the cloud being at the higher potential. What is the change in an electron's electric potential energy when the electron moves from the ground to the cloud?
Answer:
The electric potential energy is 6.72 x 10^-11 J.
Explanation:
Potential difference, V = 4.2 x 10^8 V
charge of electron, q = - 1.6 x 10^-19 C
Let the potential energy is U.
U = q V
U = 1.6 x 10^-19 x 4.2 x 10^8
U = 6.72 x 10^-11 J
A 59.0 kg bungee jumper jumps off a bridge and undergoes simple harmonic motion. If the period of oscillation is 0.250 mins, what is the spring constant (in N/m) of the bungee cord, assuming it has negligible mass compared to that of the jumper
Answer:
The spring constant of the spring is 10.3 N/m.
Explanation:
Given that,
Mass of a bungee jumper, m = 59 kg
The period of oscillation, T = 0.25 min = 15 sec
We need to find the spring constant of the bungee cord. We know that the period of oscillation is given by :
[tex]T=2\pi\sqrt{\dfrac{m}{k}}[/tex]
Where
k is the spring constant
[tex]T^2=4\pi^2\times \dfrac{m}{k}\\\\k=4\pi^2\times \dfrac{m}{T^2}\\\\k=4\pi^2\times \dfrac{59}{(15)^2}\\\\k=10.3\ N/m[/tex]
So, the spring constant of the spring is 10.3 N/m.
TRUE or FALSE: The acceleration of projectile is 0 m/s/s at the peak of the trajectory. Identify the evidence which supports your answer.
The vertical acceleration of the projectile is at 0 m/s while the horizontal acceleration of the projectile at the peak of the trajectory = initial acceleration ( i.e. statement in the question is False )
Projectile motion follows a parabolic path with x and y components of its velocity and acceleration. also the acceleration of a projectile is subject only to the acceleration due to gravity unlike other kinds of motions.
In a parabolic motion an object ( projectile ) is thrown into the air and left to move through a parabolic path under the effect of acceleration due to gravity.
Hence we can conclude that the statement is false, because horizontal acceleration of the projectile at the peak of the trajectory = initial acceleration
Learn more : https://brainly.com/question/24658194
Wind instruments like trumpets and saxophones work on the same principle as the "tube closed on one end" that we examined in our last experiment. What effect would it have on the pitch of a saxophone if you take it from inside your house (at 76 degrees F) to the outside on a cold day when the outside temperature is 45 degrees F ?
Answer:
f = v / 4L
the frequency of the instruments is reduced by the decrease in the speed of the wave with the temperature.
Explanation:
In wind instruments the wave speed must meet
v = λ f
λ = v / f
from v is the speed of sound that depends on the temperature
v = v₀ [tex]\sqrt{1+ \frac{T [C]}{273} }[/tex]
where I saw the speed of sound at 0ºC v₀ = 331 m/s the temperature is in degrees centigrade, we can take the degrees Fahrenheit to centigrade with the relation
(F -32) 5/9 = C
76ºF = 24.4ºC
45ºF = 7.2ºC
With this relationship we can see that the speed of sound is significantly reduced when leaving the house to the outside
at T₁ = 24ºC v₁ = 342.9 m / s
at T₂ = 7ºC v₂ = 339.7 m / s
To satisfy this speed the wavelength of the sound must be reduced, so the resonant frequencies change
λ / 4 = L
λ= 4L
v / f = 4L
f = v / 4L
Therefore, the frequency of the instruments is reduced by the decrease in the speed of the wave with the temperature.
In a physics lab, light with a wavelength of 560 nm travels in air from a laser to a photocell in a time of 17.2 ns . When a slab of glass with a thickness of 0.810 m is placed in the light beam, with the beam incident along the normal to the parallel faces of the slab, it takes the light a time of 20.8 ns to travel from the laser to the photocell.
Required:
What is the wavelength of the light in the glass?
Answer:
Distance traveled = 3 * 10E8 * 17.2 * 10E-9 = 5.16 m
.81 / 3 * 10E8 = 2.7 * 10E-9 normal time thru glass
(20.8 - 17.2) * E10-9 = 3.6 * 10E-9 additional time due to glass
c tg = c n ta where tg and ta are the times spent in glass and air
(Note you can also write Va = n Vg or D / ta = n D / tg)
n = tg / ta = 3.6 / 2.7 = 1.33 the index of refraction of the glass
Wavelength (air) = Wavelength (glass) * n
Wavelenght = 560 nm / 1.33 = 421 nm
A small object with mass 0.20 kg swings as a pendulum on the end of a long light rope. For small amplitude of swing, the period of the motion is 3.0 s. If the object is replaced by one with mass 0.400 kg, what is the period for small amplitude of swing? (a) 1.5 s (b) 3.05 (c) 6.0 s (d) 12.0 s (e) none of the above answers
Answer:
The correct option is (e) "none of the above".
Explanation:
Given that,
A small object with mass 0.20 kg swings as a pendulum on the end of a long light rope. For small amplitude of swing, the period of the motion is 3.0 s.
If the object is replaced by one with mass 0.400 kg, then we have to find the period for small amplitude of the swing.
We know that the time period can be calculated as :
[tex]T=2\pi \sqrt{\dfrac{l}{g}}[/tex]
Where
l is the length
g is acceleration due to gravity
It means the time period is independent of the mass. So, if the mass is replaced by one with mass 0.400 kg, there is no effect on the time period.
A barge is hauled along a straight-line section of canal by two horses harnessed to tow ropes and walking along the tow paths on
either side of the canal. Each horse pulls with a force of 839 N at an angle of 15° with the centerline of the canal. Find the sum of these
two forces on the barge.
answer in ___kN
Answer:
1.621 kN
Explanation:
Since each horse pulls with a force of 839 N at an angle of 15° with the centerline of the canal, the horizontal component of the force due to the first horse along the canal is F= 839cos15° N and its vertical component is F' = 839sin15° N(it is positive since it is perpendicular to the centerline of the canal and points upwards).
The horizontal component of the force due to the second horse along the canal is f = 839cos15° N and its vertical component is f' = -839sin15° N (it is negative since it is perpendicular to the centerline of the canal and points downwards).
So, the resultant horizontal component of force R = F + f = 839cos15° N + 839cos15° N = 2(839cos15°) N = 2(839 × 0.9659) = 2 × 810.412 = 1620.82 N
So, the resultant vertical component of force R' = F' + f' = 839sin15° N + (-839sin15° N) = 839sin15° N - 839sin15° N = 0 N
The magnitude of the resultant force which is the sum of the two forces is R" = √(R² + R'²)
= √(R² + 0²) (since R' = 0)
= √R²
= R
= 1620.82 N
= 1.62082 kN
≅ 1.621 kN
So, the sum of these two forces on the barge is 1.621 kN
write a note on unity of ant
Answer: When a pathogen enters their colony, ants change their behavior to avoid the outbreak of disease. In this way, they protect the queen, brood and young workers from becoming ill. These results, from a study carried out in collaboration between the groups of Sylvia Cremer at the Institute of Science and Technology Austria (IST Austria) and of Laurent Keller at the University of Lausanne, are published today in the journal Science.
Explanation: search for it.
Consider two oppositely charged, parallel metal plates. The plates are square with sides L and carry charges Q and -Q. What is the magnitude of the electric field in the region between the plates
Answer:
E = [tex]\frac{Q}{L^2 \epsilon_o}[/tex]
Explanation:
For this exercise we use that the electric field is a vector, so the resulting field is
E_total = E₁ + E₂ (1)
since the field has the same direction in the space between the planes
Let's use Gauss's law for the electric field of each plate
Let's use a Gaussian surface that is a cylinder with the base parallel to the plate, therefore the normal to the surface and the field lines are parallel and the angle is zero so cos 0 = 1
Ф = ∫ .dA = [tex]q_{int}[/tex] /ε₀
if we assume that the charge is uniformly distributed on the plate we can define a charge density
σ = q_{int} A
as the field exists on both sides of the plate on the inside
E A = A σ / 2ε₀
E = σ / 2ε₀
we substitute in equation 1
E = σ /ε₀
for the complete plate
σ = Q / A = Q / L²
we substitute
E = [tex]\frac{Q}{L^2 \epsilon_o}[/tex]
The same constant force is used to accelerate two carts of the same mass, initially at rest, on horizontal frictionless tracks. The force is applied to cart A for twice as long a time as it is applied to cart B. The work the force does on A is WA; that on B is WB. Which statement is correct?
a. WA = WB
b. WA = 2WB.
c. WA=4WB
d. WB= 2WA
Answer:
Option (c).
Explanation:
Let the mass of each cart is m and the force is F.
Time for cart A is 2t and for cart B is t.
Work done is given by the
W= force x displacement
As the distance is given by
S= u t +0.5 at^2
So, when the time is doubled the distance is four times.
So, WA = F x 4 S
WB = F x S
WA= 4 WB
This percentage of all water on the planet is salt water . 97 % 95% 93% 91%
hurry please !
Answer:
none of those are right, its technically 96.5%. so i would say 97% is your best bet because thats closest and it just rounds up :)
Explanation:
A source of emf is connected by wires to a resistor and electrons flow in the circuit the wire diameter is teh same throughout teh circuit compared to the drift speed of the electrons before entering the source of emf, the drift speed ot eh electrons afte rleaving the source of emf is: ___________
a. faster.
b. slower.
c. the same.
d. either A or B depending on circumstances.
e. any of A, B , or C depending on circumstances.
A cylindrical water tank has a height of 20cm and a radius of 14cm. If it is filled to 2/5 of its capacity, calculate.
I. Quantity of water in the tank
II. Quantity of water left to fill the tank to its capacity.
Answer:
4.926 L Y 7.389 L
Explanation:
first you calculate the tank volume
V = π[tex](14 cm)^{2}[/tex](10 cm = [tex]12315 cm^{3}[/tex]
then you convert to liters
[tex]12315 cm^{3}[/tex] = 12.315 l
then you calculate the liters of water
2/5(12.35 l) = 4.926 l
finally we calculate the amount without water
12.315 l - 4.926 l = 7.389 l
HERE IS MORE INFORMATION ON THE SUBJECT. THEY REMOVED THE
ENGLISH SITE BUT YOU CAN USE TRANSLATOR
LINK: https://gscourses.thinkific.com/courses/fisicai
A baseball pitcher throws a fastball by spinning his arm at 27.7m/s. The ball has a mass of 0.700kg and experiences a net centripetal force of 625N. How long is the pitchers arm (the radius of the curve)?
In the historical sense, postmodern society is simply a society that occurs after the modern society. ... Many of the elements of a society like this are reactions to what the modern society stood for: industrialism, rapid urban expansion, and rejection of many past principles.
A hungry monkey is sitting at the top of a tree 69 m above ground level. A person standing on the ground wants to feed the monkey. He uses a tee-shirt cannon to launch bananas at the monkey. If the person knows that the monkey is going to drop from the tree at the same instant that the person launches the bananas, how should the person aim the banana cannon
Answer:
Well if you want to be sure you should just throw it to the ground so then when he lands he can catch it.
If the cannon throws the banana with the same force the monkey falls
(m.g=Fz <=> m.9,81N/kg=...N).
Then the throw will slow down because of the gravitational pull.
Because the banana cannon is selfmade you can choose what mass the bananas in question have, so let that be the same as the monkeys.
The monkey falls with the speed of 9,81m.s => so it takes the monkey 7,1s to land.
If the cannon can shoot the banana at the same speed the monkey falls then they would cross in the middle.
So to do so you need to throw the bananas with a speed of at least 9,81m.s
Soo ... throw them with a force of that is greater then the gravitational pull and things will work out.
I'm sorry I don't know why I wrote all of this irrelevant information it's 2:21 right now and I'm tired.
kind regards
The _______ principle encourages us to resolve a set of stimuli, such as trees across a ridgeline, into smoothly flowing patterns
A.) depth perception.
B.) perception.
C.) similarity.
D.) continuity.
Answer:
The answer is continuity ( D )
Explanation:
PLZ MARK AS BRAINLIEST
please help me .finish this paper
Solution-1:-
[tex]\boxed{\sf \dfrac{10\times 1000}{60\times 60}}[/tex]
Solution:-2
[tex]\boxed{\sf Sodium\:and\:Potassium}[/tex]
Solution:-3
[tex]\boxed{\sf 320m}[/tex]
Solution:-4
[tex]\boxed{\sf Rough\:tiles\:are\:used\:in\:bathroom}[/tex]
Solution:-5
[tex]\boxed{\sf Mg_3N_2}[/tex]
Solution:-6
[tex]\boxed{\sf Grapes\:and\:Rambutan}[/tex]
Solution:-7
[tex]\boxed{\sf {}^{}_{}N}[/tex]
Solution:-8
[tex]\boxed{\sf Galactuse}[/tex]
Solution:-9
[tex]\boxed{\sf Y-X}[/tex]
Solution:-10
[tex]\boxed{\sf Cell\:wall}[/tex]
After a successful experiment involving the Sun, an astrophysicist proposes a theory for how stars change during the course of their existence. Is the astrophysicist correct in doing this
Answer:
Yes
Explanation:
A scientific theory is an explanation of a natural phenomenon based on available facts. If an Astrophysicist (a scientist who studies the universe and its celestial constituents) performs several successful experiments about a natural body like the sun and tests the results of his experiment using the principles of maths and physics and they come out as true, then he can propose a scientific theory based on his findings.
Unlike a scientific law that simply states a fact, a scientific theory provides evidence and explanation to support the facts.
A compact disk with a 12 cm diameter is rotating at 5.24 rad/s.
a. What is the linear speed _______m/s
b. What is the centripetal acceleration of a point on its outer rim _______
c. Consider a point on the CD that is halfway between its center and its outer rim. Without repeating all of the calculations required for parts (a) and (b), determine the linear speed of this point. _______m/s
d. Determine the centripetal acceleration of this point. _______
Answer:
(a) 31.44 m/s (b) 164.74 m/s²
Explanation:
Given that,
The diameter of a disk, d = 12 cm
Radius, r = 6 cm
Angular speed = 5.24 rad/s
(a) Linear speed,
[tex]v=r\omega\\\\v=6\times 5.24\\\\v=31.44\ m/s[/tex]
(b) Centripetal acceleration,
[tex]a=\dfrac{v^2}{r}\\\\a=\dfrac{31.44^2}{6}\\\\a=164.74\ m/s^2[/tex]
