Find the angle θ made by the road. When rounding the curve at 15.0 m/s, the car has a radial acceleration of
a = (15.0 m/s)² / (75.0 m) = 3.00 m/s²
There are two forces acting on the car in this situation:
• the normal force of the road pushing upward on the car, perpendicular to the surface of the road, with magnitude n
• the car's weight, pointing directly downward; its magnitude is mg (where m is the mass of the car and g is the acceleration due to gravity), and hence its perpendicular and parallel components are, respectively, -mg cos(θ) and mg sin(θ)
By Newton's second law, the net forces in the perpendicular and parallel directions are
(perp.) ∑ F = n - mg cos(θ) = 0
(para.) ∑ F = mg sin(θ) = ma
==> sin(θ) = a/g ==> θ = arcsin(a/g) ≈ 17.8°
(Notice that in the paralell case, the positive direction points toward the center of the curve.)
When rounding the curve at 31.8 m/s, the car's radial acceleration changes to
a = (31.8 m/s)² / (75.0 m) ≈ 13.5 m/s²
and there is now static friction (mag. f = µn, where µ is the coefficient of static friction) acting on the car and keeping from sliding off the road, hence pointing toward the center of the curve and acting in the parallel direction. Newton's second law gives the same equations, with an additional term in the parallel case:
(perp.) ∑ F = n - mg cos(θ) = 0
(para.) ∑ F = mg sin(θ) + f = ma
The first equation gives
n = mg cos(θ)
and substituting into the second equation, we get
mg sin(θ) + µmg cos(θ) = ma
==> µ = (a - g sin(θ)) / (g cos(θ)) = a/g sec(θ) - tan(θ) ≈ 1.12
Answer:
Explanation:
You are in the chapter on Physics about uniform circular motion and gravity. This is a centripetal force problem in particular, and the equation for that is
[tex]F_c=\frac{mv^2}{r}[/tex] where
[tex]F_c[/tex] is the centripetal force needed to keep the car moving in its circular path,
m is the mass of the car,
v is the velocity with which the car is moving, and
r is the radius of the circle that the car is moving around.
For us, the centripetal force is supplied by the friction keeping the car on the road, altering the equation to become
[tex]f=\frac{mv^2}{r}[/tex] and friction is defined by
f = μ[tex]F_n[/tex] (the coefficient of friction multiplied by the weight of the car).
Going on and getting buried even deeper,
[tex]F_n=mg[/tex] which says that the weight of the car is equal to its mass times the pull of gravity. Putting all that together, finally, we have the equation we need to solve this problem:
μ·m·g = [tex]\frac{mv^2}{r}[/tex] and we solve this for μ:
μ = [tex]\frac{mv^2}{mgr}[/tex] and it just so happens that the mass of the car cancels out. (I'll tell you why the mass of the car doesn't matter at the end of this problem). Filling in and solving for the coefficient of friction:
μ = [tex]\frac{31.8^2}{(9.8)(75.0)}[/tex] to 2 significant figures is
μ = 1.4
The mass of the car doesn't affect whether or not the car can stay on the curve. Even though a car with a greater mass will have a greater frictional force, that doesn't mean that it's easier for that car to stay on the road; a larger mass only means that a larger centripetal force is needed to keep it moving in a circle. This makes the gain in friction become offset by the fact that a larger centripetal force is necessary. Thus,
On a flat curve, the mass of the object experiencing circular motion does not affect the velocity at which it can stay on the curve.
Parallel incident rays appear to bounce like they have all originated from the same point. What is this point called?
A. cross point
B. midpoint
C. bounce point
D. focal point
HELP, SCIENCE QUESTION I AM STUCK
6. Which of the following is NOT part of a circuit?
A. rim B. load C. power source D. conductor
a sphere of mass 5kg and volume 2×10-5completely immersed in water find the buoyant force exerted water
Answer:
Buoyant force exerted water = 0.196 Newton
Explanation:
Given:
Mass of sphere ball = 5 kg
Volume = 2 x 10⁻⁵
Find:
Buoyant force exerted water
Computation:
Buoyant force exerted water = Gravity due to acceleration x volume of object x density of given liquid
Buoyant force exerted water = 9.8 x 2 x 10⁻⁵ x 1000
Buoyant force exerted water = 0.196 Newton
Which type of wave causes particles of matter to vibrate in a direction
perpendicular to the direction of its motion?
O A. Sound
B. Transverse
C. Longitudinal
D. Compression
Answer:
C.) Longitudinal
A car is travelling at 60m/s. and brakes to a speed of 14m/s, in 2.7 seconds. What is the deceleration?
Answer:
by using v = u + at equation we can find "a"
14 = 60 - 2.7a
2.7a = 60 - 14
2.7a = 46
decceleration = 17.03
Explain: What happens to the velocity of a stream as the size of the sediment increases?
Answer:
Also, as stream depth increases, the hydraulic radius increases thereby making the stream more free flowing. Both of these factors lead to an increase in stream velocity. The increased velocity and the increased cross-sectional area mean that discharge increases.
A ship anchored at sea is rocked by waves that have crests Lim apart the waves travel at 70m/S, at what frequency do the waves reach the ship?
Question: A ship anchored at sea is rocked by waves that have crests 100 m apart the waves travel at 70m/S, at what frequency do the waves reach the ship?
Answer:
0.7 Hz
Explanation:
Applying,
v = λf............... Equation 1
Where v = velocity of the wave, f = frequency fo the wave, λ = wavelength of the wave
make f the subject of the equation
f = v/λ................. Equation 2
From the question,
Given: v = 70 m/s, λ = 100 m ( distance between successive crest)
Substitute these values into equation 2
f = 70/100
f = 0.7 Hz
Hence the frequency at which the wave reach the ship is 0.7 Hz
1. A vehicle of mass 1500 kg braked to a standstill from a
velocity of 24 m/s in 12 s.
i. Show that the deceleration of the vehicle was 2.0 m/s2.
ii. Calculate the resultant force on the vehicle.
Explanation:
i. Vi=24
Vf=0
t= -2
a=vf-vi/t =0-24/12 = -2m/S2
ii. F=ma = 1500×-2= -3000 N
What is an example of a series circuit
Answer:
Explanation
The most famous and common example is Christmas tree lights. You can't tell easily by looking at them whether they are in series or parallel. But you sure know the difference when one of them burns out. When that happens, the whole string goes dead. No matter what you do (other than find out which bulb burned out) will not fix the problem.
Another example is anything that is temperature controlled. For example a furnace is controlled by a thermostat. When the room temperature reaches a certain point, the thermostat is constructed in a certain way so that it forms an open circuit and no current can flow through it. The furnace motor turns off and the furnace stops pumping hot air into a room.
Use the universal law of gravitation to solve the following problem.
Hint: mass of the Earth is = 5.97 x 1024 kg
A scientific satellite of mass 1300 kg orbits Earth 200 km above its surface. If Earth has a radius of 6378 km, what is the force of gravity acting on the scientific satellite?
a. Write out the formula for this problem.
b. Plug in the values from this problem into the formula.
c. Solve the problem, writing out each step.
d. Correct answer
Answer:
a.
[tex]F=G\cdot\dfrac{M \cdot m}{r^{2}}[/tex]
b.
[tex]F=6.67430 \times 10^{-11} \dfrac{N \cdot m^2}{kg^2} \times \dfrac{5.97 \times 10^{24} \ kg \times 1,300 \ kg}{(6,578 \ m)^{2}}[/tex]
c.
[tex]F=6.67430 \times 10^{-11} \dfrac{N \cdot m^2}{kg^2} \times \dfrac{5.97 \times 10^{24} \ kg \times 1,300 \ kg}{(6,578 \ m)^{2}} \approx 1.144 \times 10^{13} \ N[/tex]
d. The force of gravity acting on the satellite is approximately 1.144 × 10¹³ N
Explanation:
a. The formula for finding the force of gravity, F, acting object on an object is given as follows;
[tex]F=G\cdot\dfrac{M \cdot m}{r^{2}}[/tex]
Where;
F = The force acting between the Earth and the object
G = The gravitational constant = 6.67430 × 10⁻¹¹ N·m²/kg²
M = The mass of the Earth = 5.97 × 10²⁴ kg
m = The mass of the object
r = The distance between the center of the Earth and the object
b. Finding the gravitational force, 'F', between the Earth and the scientific satellite, we have;
The given mass of the satellite, m = 1,300 kg
The distance between the center of the Earth and the center of the satellite, r = The length of the radius of the Earth + The height of orbit of the satellite
The given height of orbit of the satellite, h = 200 km
∴ r = R + h = 6,378 km + 200 km = 6,578 m
Therefore, by plugging in the values, we get;
[tex]F=6.67430 \times 10^{-11} \dfrac{N \cdot m^2}{kg^2} \times \dfrac{5.97 \times 10^{24} \ kg \times 1,300 \ kg}{(6,578 \ m)^{2}}[/tex]
c. Solving the above equation gives;
[tex]F=6.67430 \times 10^{-11} \dfrac{N \cdot m^2}{kg^2} \times \dfrac{5.97 \times 10^{24} \ kg \times 1,300 \ kg}{(6,578 \ m)^{2}} \approx 1.144 \times 10^{13} \ N[/tex]
d. The force of gravity acting on the satellite, F ≈ 1.144 × 10¹³ Newton
Please answer in your own words and I will mark brainlist, Compare all matters from the view of passing light through them (name 3 types and write if they pass the light or not.)
Answer:
Light passes through the gas
Light passes through the pure water
Light passes through some solids
Explanation:
In gasses, there are many spaces between the molecules. These spaces allow light to pass through them without any interruption.
In pure water, there are some spaces between particles. these particles allow some light rays to pass theough, some to move through the common boundary and reflec5 some of them.
in solids, some allow light to pass through as they are transparent or translucent
I need help with question 6
Answer:
0.16 h
Explanation:
Speed: 25km per hour
Speed=distance/time
25=4/t
t=4/25
t=0.16 hour
t=9.6 minuites
Brainliest please~~
Rewrite the false statements correctly
1.If an object sinks in one liquid and floats on another liquid,it implies that the density of second liquid is less than the first liquid.
2.The immersed volume of body in a liquid depends on density of the liquid.
3.Relative density of a body is usually expressed in kgm^-3
Explanation:
1. if an object sjnks in one liquid and floats on another liquid it implies that the density of second liquid is greater than the density of first liquid
which forces are capable of affecting particles or objects from large distance
Answer:
only long-range force that affects all particles is the gravitational force.
Explanation:
In nature there are four fundamental forces: nuclear, weak, gravitational and electrical.
The last two are long-range, that is, the forces are zero for infinite distances, the current gravitational on all the particles and the electric one acts on the charged particles, without the chosen charge it is zero, the forces is also zero.
Consequently the only long-range force that affects all particles is the gravitational force.
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Answer: The equations in column A is matched with gas laws in column B as follows:
21. PV = nRT : (g) Ideal gas law
22. [tex]V_{1}n_{2} = V_{2}n_{1}[/tex] : (f) Avogadro's law
23. [tex]P_{1}V_{1}T_{2} = P_{2}V_{2}T_{1}[/tex] : (e) Combined Gas Law
24. [tex]P_{1}T_{2} = P_{2}T_{1}[/tex] : (d) Gay-Lusaac's law
25. [tex]V_{1}T_{2} = V_{2}T_{1}[/tex] : (c) Charles' law
26. [tex]P_{1}V_{1} = P_{2}V_{2}[/tex] : (b) Boyle's law
27. [tex]\frac{v_{1}}{v_{2}} = \frac{\sqrt{MM_{1}}}{MM_{2}} = \frac{\sqrt{p_{1}}}{p_{2}}[/tex] : (a) Graham's Law of effusion
Explanation:
(A) Ideal gas law: It states that the product of pressure and volume is directly proportional to the product of number of moles and temperature.
So, PV = nRT
where,
P = pressure
V = volume
n = no. of moles
R = gas constant
T = temperature
Boyle's law: At constant temperature, the pressure of a gas is inversely proportional to volume.So, [tex]P_{1}V_{1} = P_{2}V_{2}[/tex]
Charles' law: At constant pressure, the volume of a gas is directly proportional to temperature. So,[tex]V \propto T\\\frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}}\\V_{1}T_{2} = V_{2}T_{1}[/tex]
Gay-Lussac's law: At constant volume, the pressure of a gas is directly proportional to temperature.So, [tex]P_{1}T_{2} = P_{2}T_{1}[/tex]
Avogadro's law: At same temperature and pressure, the volume of gas is directly proportional to moles of gas.So, [tex]V_{1}n_{2} = V_{2}n_{1}[/tex]
Combined gas law: When Boyle's law, Charles' law, and Gay-lussac's law are combined together then it is called combined gas law. So,[tex]\frac{P_{1}V_{1}}{T_{1}} = \frac{P_{2}V_{2}}{T_{2}}\\or, P_{1}V_{1}T_{2} = P_{2}V_{2}T_{1}[/tex]
Graham's law of effusion: It states that the rate of effusion of a gas is inversely proportional to the square root of mass of its particles.[tex]\frac{v_{1}}{v_{2}} = \frac{\sqrt{MM_{1}}}{MM_{2}} = \frac{\sqrt{p_{1}}}{p_{2}}[/tex]
Thus, we can conclude that equation in column A is matched with gas laws in column B as follows:
21. PV = nRT : (g) Ideal gas law
22. [tex]V_{1}n_{2} = V_{2}n_{1}[/tex] : (f) Avogadro's law
23. [tex]P_{1}V_{1}T_{2} = P_{2}V_{2}T_{1}[/tex] : (e) Combined Gas Law
24. [tex]P_{1}T_{2} = P_{2}T_{1}[/tex] : (d) Gay-Lusaac's law
25. [tex]V_{1}T_{2} = V_{2}T_{1}[/tex] : (c) Charles' law
26. [tex]P_{1}V_{1} = P_{2}V_{2}[/tex] : (b) Boyle's law
27. [tex]\frac{v_{1}}{v_{2}} = \frac{\sqrt{MM_{1}}}{MM_{2}} = \frac{\sqrt{p_{1}}}{p_{2}}[/tex] : (a) Graham's Law of effusion
What happens when a negatively charged object A is brought near a neutral object B?
A.
Object B gets a negative charge.
Ο
O o
B.Object B gets a positive charge.
O C.
Object B stays neutral but becomes polarized.
D.
Object A gets a positive charge.
O
E.
Object A loses all its charge.
Reset
Next
Answer:
A.
Explanation:Object b will get a negative charge .
help me please
only if you really know
Assume R is measured in meters (m) and M in kilograms (kg). Then
R ² / (GM) = [m]² / ([N•m²/kg²] [kg]) = m•kg / N = m•kg / (kg•m/s²) = s²
so t ² is indeed proportional to R ²/(GM).
used to measure temperature
used to measure force
prefix that means 1/100
prefix that means 1,000
prefix that means 1/1,000
Answer:
prefix that means 1/100 = Centi
prefix that means 1,000 = Kilo
prefix that means 1/1000 = Milli
Explanation:
Which energy store is increased when an object is heated?
Answer:
Kinetic Energy
Explanation:
A person skateboards at 3.25 m/s for 55.0 s. How far did he travel?
A 15 cm length of wire is moving perpendicularly
through a magnetic field of strength 1.4 T at the rate
of 0.12 m/s. What is the EMF induced in the wire?
A. OV
C. 0.025 v
B. 0.018 V
D. 2.5 V
Answer: C or B
Explanation:
The EMF induced in the wire moving perpendicularly through a magnetic field is 0.025V. The correct option is C.
What is EMF?The EMF is the electro motive force which causes the current to induce in the object moving in the magnetic field.
Given is the length of wire L =15cm =0.15m, magnetic field strength B = 1.4T and velocity of wire V =0.12 m/s
EMF is related to the length of wire, magnetic field strength and velocity of wire proportionally.
ε = B x L x V
Plug the values, we get
ε = 1.4 x 0.15 x 0.12
ε = 0.025 Volts
Thus, the correct option is C.
Learn more about EMF.
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The current in a resistor is 5 A and the voltage between its terminals is 40 V. Calculate the resistance.
An iron wire has a resistance of 24 Ω. If the voltage across its ends is 12 V, calculate the current in the wire.
Answer:
1=8 ohms 2=0.5 Amps
Explanation:
What type of electromagnetic waves do heat lamps give off?
A. infrared
B. ultraviolet
C. microwaves
D. radio waves
A educação física, enquanto componente curricular da educação básica. Qual a tarefa que educação física?? Alguem me ajuda por fvr ??
Answer:
como assim qual a tarefa que educação física? se você me explicar melhor eu consigo te responder !!
Explanation:
At the local grocery store, you push a 14.5-kg shopping cart. You stop for a moment to add a bag of dog food to your cart. With a force of 12.0 N you now accelerate the cart from rest through a distance of 2.29 m in 3.00 s. What was the mass of the dog food?
Answer:
The mass of the dog food added is 9.03 kg
Explanation:
Given;
mass of the shopping cart, m₁ = 14.5 kg
let the mass of the bag added = m₂
the force applied, F = 12 N
initial velocity of the cart-bag system, u = 0
distance traveled by the system, d = 2.29 m
time of motion of the system, t = 3.0 s
The acceleration of the system is calculated as;
[tex]d = ut + \frac{1}{2} at^2\\\\2.29 = 0 + (\frac{1}{2} \times 3^2)a\\\\2.29 = 4.5 a\\\\a = \frac{2.29}{4.5} \\\\a = 0.51 \ m/s^2[/tex]
The total mass of the system (M) is calculated as follows;
F = Ma
M = F/a
M = (12)/(0.51)
M = 23.53 kg
The mass of the dog food added is calculated as;
m₂ = M - m₁
m₂ = 23.53 kg - 14.5 kg
m₂ = 9.03 kg
The force of friction acting on a sliding crate is 223 N.
How much force must be applied to main- tain a constant velocity?
Answer:
Friction Opposes Motion of an Object.
Now
To get the Net force that Moves an Object and causes acceleration....You subtract the Frictional force
Net force = Pushing Force - Frictional Force
Recall
Net Force; F=Ma
Ma = P - Fr
Now the question asked for How Much force Must be applied to Maintain a Constant velocity.
In a Constant Velocity Motion... Acceleration do not change... Its Zero
So Putting this into the formula above
M(0) = P - Fr
0=P - Fr
Fr = P.
This means
That The force needed to keep this object Moving at Constant Velocity Must be equal to its Frictional Force
Since Frictional Force; Fr =223N
The Applied Force(Pushing Force) Must be equal to 223N too.
In the picture below, a car hits a wall. Using what you know about Newton’s Third Law, which is true?
The force of the wall on the car and the car on the wall are equal
The force of the wall on the car is greatest
The force of the car on the wall is greatest
There is not enough information to tell
Answer:
A...................................
The force of the wall on the car and the car on the wall are equal is true about Newton’s Third Law. Option A is the correct answer.
According to Newton's Third Law of Motion, for every action, there is an equal and opposite reaction. This means that if the car hits the wall, there will be a force exerted by the car on the wall, and an equal and opposite force exerted by the wall on the car. Option A is the correct answer.
The forces involved in the interaction between the car and the wall are equal in magnitude but opposite in direction, as dictated by Newton's Third Law. Newton's Third Law of Motion states that for every action, there is an equal and opposite reaction. This means that when an object exerts a force on another object, the second object exerts a force of equal magnitude but in the opposite direction on the first object.
Learn more about Newton here:
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The complete question is, "In the picture below, a car hits a wall. Using what you know about Newton’s Third Law, which is true?
a. The force of the wall on the car and the car on the wall are equal
b. The force of the wall on the car is greatest
c. The force of the car on the wall is greatest
d. There is not enough information to tell"
A girl travels 50m in 12s and then another 30m in 5s .Calculate her average speed?
Answer:
4.71m/s
Explanation:
Average speed = Total distance travelled ÷ Total time taken.
80/17=4.71
4.71m/s
Answer:
Average speed = Total distance travelled ÷ Total time taken.In this question,
Total distance travelled = 50m + 30m= 80m.
Total time taken = 12s + 5s= 17s
So, Average speed would be 80 ÷ 17
= [tex]\frac{80}{17}[/tex]
= 4.71 m/s. or 4.71 meter per second.
what is the speed of a wave with a wavelength of 3.0 m and a period of 0.40 s?
in a series circuit, how does the voltage supplied by the battery compare to the voltages on each load?
Answer:
In a series circuit, how does the voltage supplied by the battery compare to the voltage on each load? The voltage of the battery is equal to the voltage of each load added together. ... The voltage across the two resistors must both have the same voltage of the battery.
Explanation:
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Answer:
The voltage of the battery is equal to the voltage of each load added together. The voltage across the two resistors must both have the same voltage of the battery.
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