Answer:
Area is a vector quantity.
Explanation:
A vector quantity has both magnitudes as well as direction. On the other hand a scalar quantity has only magnitude.
Out of given options,
Volume and mass is a scalar quantity because they don't have directions. While area can be taken as a vector quantity. The direction of area vector of a surface is along perpendicular to the surface.
Hence, the correct option is (d).
In the series circuit, if the potential difference across the battery is 20 V and the potential difference across R1 is 12 V, what is the potential difference across R2
Answer:
The correct answer is "8 V".
Explanation:
Given:
Potential difference across battery,
= 20 V
Potential difference across R1,
= 12 V
Now,
On applying the Kirchorff loop, we get
⇒ [tex]E-I_1R_1-I_2R_2=0[/tex]
⇒ [tex]E-V_1-V_2=0[/tex]
⇒ [tex]V_2=E-V_1[/tex]
On putting values, we get
⇒ [tex]=20-12[/tex]
⇒ [tex]= 8 \ V[/tex]
The potential difference across the resistance R2 will be "8 Volts.
What is Kirchoff;s law?According to the kirchoff's law in a loop of a circuit when there are number of the resistances so the sum of all the potential differences will be zero.
It is given that:
Potential difference across battery,= 20 V
Potential difference across R1,= 12 V
Now,
On applying the Kirchorff loop, we get
⇒ [tex]\rm E-I_1R_1-I_2R_2=0[/tex]
⇒ [tex]\rm E-V_1-V_2=0[/tex]
⇒ [tex]\rm V_2=E-V_1[/tex]
On putting values, we get
⇒ [tex]V_2=20-12=8\ volt[/tex]
Hence the potential difference across the resistance R2 will be "8 Volts.
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Question 24 of 33 Which of the following is an example of uniform circular motion? A. A car speeding up as it goes around a curve O B. A car slowing down as it goes around a curve 2 C. A car maintaining constant speed as it goes around a curve D. A car traveling along a straight road
Answer:
Uniform Circular Motion is the Movement or Rotation of an Object along a circular Path at constant speed.
OPTION C IS YOUR ANSWER!.
A student is comparing the speed of sound in air and water. She measures the frequency, f, and wavelength, λ, of sound waves in both air and water. The results are shown in the following table. Based on her data, how does the speed of sound in water compare to the speed of sound in air?
A.The speed of sound through water is 4.3 times faster than sound through air.
B.The speed of sound through water is 2.6 times slower than sound through air.
C.The speed of sound through water is 8.4 times faster than sound through air.
D.The speed of sound through air is approximately equal to the speed of sound through water.
Answer:
Option A. The speed of sound through water is 4.3 times faster than sound through air.
Explanation:
To answer the question correctly, we shall determine the speed of the wave in both cases. This is illustrated below:
For Air:
Frequency (fₐ) = 195 Hz
Wavelength (λₐ) = 1.76 m
Velocity (vₐ) =?
vₐ = λₐfₐ
vₐ = 1.76 × 195
vₐ = 343.2 m/s
For Water:
Frequency (fᵥᵥ) = 195 Hz
Wavelength (λᵥᵥ) = 7.6 m
Velocity (vᵥᵥ) =?
vᵥᵥ = λᵥᵥfᵥᵥ
vᵥᵥ = 7.6 × 195
vᵥᵥ = 1482 m/s
Finally, we shall compare the speed in water to that of air. This can be obtained as follow:
Velocity in air (vₐ) = 343.2 m/s
Velocity in water (vᵥᵥ) = 1482 m/s
Water : Air
vᵥᵥ : vₐ =>
vᵥᵥ / vₐ = 1482 / 343.2
vᵥᵥ / vₐ = 4.3
Cross multiply
vᵥᵥ = 4.3 × vₐ
Thus, the speed in water is 4.3 times the speed in air.
Option A gives the correct answer to the question.
pair of two hollow cups is term as
Answer:
Magdeburg hemispheres are a pair of large copper hemispheres, with mating rims.
Explanation:
They were used to demonstrate the power of atmospheric pressure. When the rims were sealed with grease and the air was pumped out, the sphere contained a vacuum and could not be pulled apart by teams of horses.
You need to produce a set of cylindrical copper wire 3.5 m long that will have a
resistance of 0.125 Ω each. What will be the mass of each of these wires?
(ρ = 1.72X10-8 Ωm, density of copper = 8.9X103 kg/m3)
Solution :
We know, resistance is given by :
[tex]R = \dfrac{\rho l}{A}[/tex]
[tex]A = \dfrac{\rho l }{R}\\\\A = \dfrac{1.72\times 10^{-8} \times 3.5 }{0.125}\\\\A = 4.816 \times 10^{-7} \ m^2[/tex]
Now, we know mass of wire is given by :
[tex]Mass = Density \times Volume\\\\\M = 8.9 \times 10^3 \times 4.816 \times 10^{-7} \times 3.5 kg\\\\M = 0.01500\ kg\\\\M = 15.00\ gram[/tex]
Hence, this is the required solution.
Given:
Length of wire, l = 3.5 mResistance, R = 0.125 ΩThe resistance will be:
→ [tex]R = \frac{\rho l}{A}[/tex]
or,
→ [tex]A = \frac{\rho l}{R}[/tex]
By substituting the values, we get
[tex]= \frac{1.72\times 10^{-8}\times 3.5}{0.125}[/tex]
[tex]= 4.816\times 10^{-7} \ m^2[/tex]
hence,
The mass will be:
→ [tex]Mass = Density\times Volume[/tex]
[tex]= 8.9\times 10^3\times 4.816\times 10^{-7}\times 3.5[/tex]
[tex]= 0.01500 \ kg[/tex]
[tex]= 15.00 \ g[/tex]
Thus the above answer is right.
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Appliances A, B, and C consume 250, 480 and 1450 watts of power, respectively. The system voltage is 120V, and the circuit breaker is rated at 15 amps. Which combinations of the three appliances can be on at the same time, and which combinations will trip the circuit breaker
Answer:
Appliance A and B can work together without tripping
Explanation:
We will calculate the amount of current consumed by each appliances.
Appliance A
P = VI
I = P/V
I = 250/120 = 2.08 A
Appliance B
I = 480 /120 = 4 A
Appliance C
I = 1450/120
I = 12.08 A
Hence, appliance C will trip the circuit as it consumes a lot of electricity.
If, for a given velocity, the maximum range is at a projection angle of 45, then there must be equal ranges for angles above and below this. Show this explicitly.
Explanation:
The range R of a projectile is given by
[tex]R = \frac{v_0^2}{g} \sin 2\theta[/tex]
The maximum range [tex]R_{max}[/tex] occurs when [tex]\sin 2\theta = 1\:\text{or}\:\theta = 45°[/tex]. Let [tex]\alpha[/tex] be the angle above or below 45°. Now let's look at the ranges brought about by these angle differences.
Case 1: Angle above 45°
We can write the range as
[tex]R_+ = \dfrac{v_0^2}{g} \sin 2(45° + \alpha)= \dfrac{v_0^2}{g} \sin (90° + 2\alpha)[/tex]
[tex]\:\:\:\:\:\:\:= \dfrac{v_0^2}{g} (\sin 90° \cos 2\alpha + \cos 90° \sin 2\alpha)[/tex]
[tex]\:\:\:\:\:\:\:= \dfrac{v_0^2}{g} \cos 2\alpha\:\:\:\:\:(1)[/tex]
Case 2: Angle below 45°
We can write the range as
[tex]R_- = \dfrac{v_0^2}{g} \sin 2(45° - \alpha)= \dfrac{v_0^2}{g} \sin (90° - 2\alpha)[/tex]
[tex]\:\:\:\:\:\:\:= \dfrac{v_0^2}{g} (\sin 90° \cos 2\alpha - \cos 90° \sin 2\alpha)[/tex]
[tex]\:\:\:\:\:\:\:= \dfrac{v_0^2}{g} \cos 2\alpha\:\:\:\:\:(2)[/tex]
Note that the equations (1) and (2) are identical. Therefore, the ranges are equal if they differ from 45° by the same amount.
What does the law of the conservation of energy state?
A 45.00 kg person in a 43.00 kg cart is coasting with a speed of 19 m/s before it goes up a hill.Assuming there is no friction, what is the maximum vertical height the person in the cart can reach?
Answer:
the maximum vertical height the person in the cart can reach is 18.42 m
Explanation:
Given;
mass of the person, m₁ = 45 kg
mass of the cart, m₂ = 43 kg
velocity of the system, v = 19 m/s
let the maximum vertical height reached = h
Apply the principle of conservation mechanical energy;
[tex]P.E = K.E\\\\mgh_{max} = \frac{1}{2} mv^2_{max}\\\\gh_{max} = \frac{1}{2} v^2_{max}\\\\h_{max} = \frac{v_{max}^2}{2g} \\\\h_{max} = \frac{19^2}{2\times 9.8} \\\\h_{max} = 18.42 \ m[/tex]
Therefore, the maximum vertical height the person in the cart can reach is 18.42 m
A gymnast is swinging on a high bar. The distance between his waist and the bar is 1.06 m, as the drawing shows. At the top of the swing his speed is momentarily zero. Ignoring friction and treating the gymnast as if all of his mass is located at his waist, find his speed at the bottom of the swing.
Answer: The speed of gymnast at the bottom of the swing is 6.44 m/s.
Explanation:
Given: Distance = 1.06 m
According to the law of conservation of energy, the speed is calculated as follows.
[tex]mgh = - mgh + \frac{1}{2}mv^{2}\\gh = - gh + \frac{1}{2}v^{2}\\v = \sqrt{4gh}[/tex]
where,
g = acceleration due to gravity = 9.8 [tex]m/s^{2}[/tex]
h = distance
v = speed
Substitute the values into above formula as follows.
[tex]v = \sqrt{4gh}\\= \sqrt{4 \times 9.8 m/s^{2} \times 1.06}\\= 6.44 m/s[/tex]
Thus, we can conclude that speed of gymnast at the bottom of the swing is 6.44 m/s.
a ball dropped from a height of 10 meters will bounce more times before coming to rest than a ball dropped from a height of 5 meters. Use evidence and scientific reasoning to explain this phenomenon.
Answer:
plz mark brainliest again lol :)
Explanation:
When you drop a ball from a greater height, it has more kinetic energy just before it hits the floor and stores more energy during the bounce—it dents farther as it comes to a stop.
Answer:
When you drop a ball from a greater height, there is more potential energy. When you release the ball, the potential energy turns into kinetic energy. When the ball bounces off the ground, the ball go upward and then it has more potential energy. Then when it goes down it has more kinetic energy. The ball keeps doing this until there is not enough potential energy left.
Explanation:
A family uses an electric frying pan with a power rating of 1.2 X 10^3 W. Although the pan is thermostatically controlled, its element was drawing power for 6.3 X 10^2 min in a period of one month. Calculate the electrical energy in kWh used by the pan during the month
Answer:
378 KWh
Explanation:
We'll begin by converting 1.2×10³ W to KW. This can be obtained as follow:
10³ W = 1 KW
Therefore,
1.2×10³ W = 1.2×10³ W × 1 KW / 10³ W
1.2×10³ W = 1.2 KW
Next, we shall convert 6.3×10² mins to hours (h). This can be obtained as follow:
60 mins = 1 h
Therefore,
6.3×10² mins = 6.3×10² mins × 1 h / 60 mins
6.3×10² mins = 10.5 h
Finally, we shall determine the electrical energy in KWh used for 1 month (i.e 30 days). This can be obtained as follow:
Power (P) = 1.2 KW
Time (t) for 1 month (30 days) = 10.5 h × 30
= 315 h
Energy (E) =?
E = Pt
E = 1.2 × 315
E = 378 KWh
Thus, the electrical energy used for 1 month (i.e 30 days) is 378 KWh.
What is the average speed (in km/h) of Zhana, who runs to the store that is 4.0 km away in 30.0 minutes?
0.13 km/ h
8.0 km/h
2.0 km/h
Answer:
2.0km/h.
Explanation:
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Which scenario is an example of a scientific way of thinking?
O A. A scientist collects different kinds of rocks and then takes them to
museums where the public can appreciate them.
B. A scientist makes a judgment that a rock sample is volcanic
because of the rock's texture and mineral composition.
O C. A scientist decides that taking rocks from national parks is wrong
because later park visitors will not be able to see them.
D. A scientist makes a judgment that rocks are more interesting to
study than living organisms because they tell more about Earth's
history
The scenario when a scientist makes a judgment that a rock sample is volcanic because of the rock's texture and mineral composition is an example of a scientific way of thinking, therefore the correct answer is option B.
What is the scientific claim?Scientific claims are statements made in science based on an experiment.
These Scientific claims are backed by experimental data and their true results obtained from scientific investigation and experimentation.
A good illustration of a scientific mode of thinking is when a scientist determines that a rock sample is volcanic based on its texture and mineral makeup.
Thus, The scenario when a scientist makes a judgment that a rock sample is volcanic because of the rock's texture and mineral composition is an example of a scientific way of thinking, therefore the correct answer is option B.
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Which image illustrates the absorption of a wave as it strikes a surface?
Answer: I think D
Explanation:
Answer:
Correct option is (C) . D
Explanation
In diagram D the ray incidents on the surface but it does not reflected.It is clear that surface absorbed the ray.Surface may be perfect black body that has absorbed incident ray.What is black body ?A surface that absorbs all the incident rays or radiation , is called black body . Platinum black can be taken as a perfect black body with the 98% absorption.
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a 2kg block of wood starts at rest and slides down a ramp. Its initail height is 12m. if the final velocity of the block is 13m/s, determine the energy of this system that has been turned into heat
Answer:
E = 66.44 J
Explanation:
From the law of conservation of energy:
Total Mechanical Energy at Start = Total Mechanical Energy at the End
Potential Energy at Start = Kinetic Energy at End + Energy Lost
[tex]mgh = \frac{1}{2} mv^2 + E\\\\E = mgh - \frac{1}{2} mv^2\\\\[/tex]
where,
E = Energy turned into heat = ?
m = mass of block = 2 kg
g = acceleration due to gravity = 9.81 m/s²
h = height = 12 m
v = final speed = 13 m/s
Therefore,
[tex]E = (2\ kg)(9.81\ m/s^2)(12\ m)-\frac{1}{2} (2\ kg)(13\ m/s)^2\\\\E = 235.44\ J - 169\ J\\[/tex]
E = 66.44 J
Model a hydrogen atom as a three-dimensional potential well with Uo = 0 in the region 0 < x
a. 283 eV
b. 339 eV
c. 113 eV
d. 226 eV
This question is incomplete, the complete question is;
Model a hydrogen atom as a three-dimensional potential well with U₀ = 0 in the region 0 < x < L, 0 < y < L and 0 < z < L, and infinite otherwise, with L = 1.0 × 10⁻¹⁰ m.
Which of the following is NOT one of the lowest three energy levels of an electron in this model?
a. 283 eV
b. 339 eV
c. 113 eV
d. 226 eV
Answer:
the lowest three energy are; 113 eV, 225 eV, and 339 eV.
Hence Option a) 283 eV is not among the three lowest energy
Explanation:
Given the data in the question;
Three dimension cube or particle in a cubic box
the energy value is given by;
[tex]E_{nx,ny,nz[/tex] = [tex]( n_x^2 + n_y^2 + n_z^2 )[/tex] × π²h"² / 2ml²
where h" = h/2π and h is Planck's constant ( 6.626 × 10⁻³⁴ m² kg / s )
m is mass of electron ( 9.1 × 10⁻³¹ kg )
l is length of side of box ( 1.0 × 10⁻¹⁰ m )
for ground level ( [tex]n_x = n_y = n_z = 1[/tex] )
so
[tex]( n_x^2 + n_y^2 + n_z^2 )[/tex] × π²h"² / 2ml²
since h" = h/2π
[tex]( n_x^2 + n_y^2 + n_z^2 )[/tex] × π²h² / (2π)²2ml²
so we substitute
[tex]E_{111[/tex] = ( 1² + 1² + 1² ) × [ π²( 6.626 × 10⁻³⁴ )² ] / [ (2π)² × 2 × 9.1 × 10⁻³¹ kg × ( 1.0 × 10⁻¹⁰)² ]
[tex]E_{111[/tex] = 3 × [ (4.333188779 × 10⁻⁶⁶) / ( 7.185072 × 10⁻⁴⁹ ) ]
[tex]E_{111[/tex] = 3 × [ 6.03082165 × 10⁻¹⁸ ]
Now, we know that electric charge = 1.602 x 10⁻¹⁹
so
[tex]E_{111[/tex] = 3 × [ (6.03082165 × 10⁻¹⁸) / (1.602 x 10⁻¹⁹) ]
[tex]E_{111[/tex] = 3 × [ 37.645578 ]
[tex]E_{111[/tex] = 112.9 ≈ 113 eV
[tex]E_{211[/tex] = [tex]( n_x^2 + n_y^2 + n_z^2 )[/tex] × π²h² / (2π)²2ml²
we substitute
[tex]E_{211[/tex] = ( 1² + 1² + 2² ) × [ 37.645578 ]
[tex]E_{211[/tex] = 6 × [ 37.645578 ]
[tex]E_{211[/tex] = 225.87 ≈ 226 eV
[tex]E_{221[/tex] = [tex]( n_x^2 + n_y^2 + n_z^2 )[/tex] × π²h² / (2π)²2ml²
we substitute
[tex]E_{221[/tex] = ( 2² + 2² + 1² ) × [ 37.645578 ]
[tex]E_{211[/tex] = 9 × [ 37.645578 ]
[tex]E_{211[/tex] = 338.8 ≈ 339 eV
Therefore, the lowest three energy are; 113 eV, 225 eV, and 339 eV.
Hence Option a) 283 eV is not among the three lowest energy
Genes are arranged on a twisted strands called.
Si un electrón recorre el acelerador lineal de Stanford de 2 millas de longitud a 99% de la velocidad de la luz, ¿Cuál es la longitud del acelerador según el electrón?
Q2 A source of frequency 500 Hz emits waves of
wavelength 0.2m. How long does it take the waves to
travel 400m?
Answer:
4 secs
Explanation:
The first step is to calculate the velocity
V= frequency × wavelength
= 500× 0.2
= 100
Therefore the time can be calculated as follows
= distance/velocity
= 400/100
= 4 secs
Assume the following vehicle are all moving at the same speed.it would be harder to change the velocity of which vehicle.a.bicycle.B.car.c.motorcycle.D.semi-tractor trailer.
Answer:im just guessing d but i think its d though
Explanation:
it pretty obvious
amount of pressure of liquid increases with ?
Answer: Pressure increases as the depth increases.
Answer:
depths cause it help it with a lot and that it the answer
An object moves 20 m east in 30 s and then returns to its starting point taking an additional 50 s. If west is chosen as the positive direction, what is the average speed of the object
Answer:
v = 0.5 m/s.
Explanation:
Total distance, d = 20 m + 20 m = 40 m
Total time taken, t = 30 s + 50 s = 80 s
The average speed of an object is the total distance divided by time taken. So,
[tex]v=\dfrac{40}{80}\\\\=0.5\ m/s[/tex]
So, the average speed of the object is 0.5 m/s.
As a 2.0-kg object moves from (4.4 i + 5j) m to ( 11.6 i - 2j) m, the constant resultant force
acting on it is equal to (41 - 9j) N. If the speed of the object at the initial position is 4.0 m/s,
what is its kinetic energy at its final position?
Answer:
Answer:
v_f = 10.38 m / s
Explanation:
For this exercise we can use the relationship between work and kinetic energy
W = ΔK
note that the two quantities are scalars
Work is defined by the relation
W = F. Δx
the bold are vectors. The displacement is
Δx = r_f -r₀
Δx = (11.6 i - 2j) - (4.4 i + 5j)
Δx = (7.2 i - 7 j) m
W = (4 i - 9j). (7.2 i - 7 j)
remember that the dot product
i.i = j.j = 1
i.j = 0
W = 4 7.2 + 9 7
W = 91.8 J
the initial kinetic energy is
Ko = ½ m vo²
Ko = ½ 2.0 4.0²
Ko = 16 J
we substitute in the initial equation
W = K_f - K₀
K_f = W + K₀
½ m v_f² = W + K₀
v_f² = 2 / m (W + K₀)
v_f² = 2/2 (91.8 + 16)
v_f = √107.8
v_f = 10.38 m / s
what is parallelogram law of vector addition ???
Answer:
According to the parallelogram law of vector addition if two vectors act along two adjacent sides of a parallelogram(having magnitude equal to the length of the sides) both pointing away from the common vertex, then the resultant is represented by the diagonal of the parallelogram passing through the same common vertex
Explanation:
A 17.6-kg block rests on a horizontal table and is attached to one end of a massless, horizontal spring. By pulling horizontally on the other end of the spring, someone causes the block to accelerate uniformly and reach a speed of 3.58 m/s in 1.77 s. In the process, the spring is stretched by 0.250 m. The block is then pulled at a constant speed of 3.58 m/s, during which time the spring is stretched by only 0.0544 m. Find (a) the spring constant of the spring and (b) the coefficient of kinetic friction between the block and the table.
In order get the block up to a speed of 3.58 m/s in 1.77 s, it must undergo an acceleration a of
a = (3.58 m/s) / (1.77 s) ≈ 2.02 m/s²
When the spring is getting pulled, Newton's second law tells us
• the net vertical force is
∑ F = n - mg = 0
where ∑ F is the net force, n is the magnitude of the normal force, and mg is the weight of the block - it follows that n = mg ; and
• the net horizontal force is
∑ F = F - f = ma
where F is the applied force, f is kinetic friction, m is the block's mass, and a is the acceleration found earlier. F stretches the spring by x = 0.250 m, so we have
F - f = kx - µn = kx - µmg = ma
where k is the spring constant and µ is the coefficient of kinetic friction.
When the block is being pulled at a constant speed, Newton's second law says
• the net vertical force is still
∑ F = n - mg = 0
so that n = mg again; and
• the net horizontal force is
∑ F = F - f = 0
This time, F stretches the spring by y = 0.0544 m, so we have
F - f = ky - µmg = 0
Solve the equations in boldface for k and µ :
kx - µmg = ma
ky - µmg = 0
==> k (x - y) = ma
==> k = ma / (x - y)
==> k = (17.6 kg) (2.02 m/s²) / (0.250 m - 0.0544 m) ≈ 182 N/m
Then
ky - µmg = 0
==> µ = ky / (mg)
==> µ = (182 N/m) (0.0544 m) / ((17.6 kg) g) ≈ 0.0574
DL: Activity 2.3 Resistance Complete the questions based on the Resistance presentation. a. All resistors ___________ or _____________ the flow of electrons. b. As resistance __________________ current _______________________
DL: Activity 2.3 Resistance
Complete the questions based on the Resistance presentation.
a. All resistors ___________ or _____________ the flow of electrons.
b. As resistance __________________ current _______________________
Answer:a) limits or opposes
b) increases, decreases
Explanation:Resistors are electrical devices used to resist, limit, oppose or hinder the flow of electrons in a circuit. This resistance causes a reduction in current and an increase in voltage in the circuit. In order words, as resistance increases, the current decreases and voltage increases.
This was further stated by Ohm's law that states that as long as the resistance in a wire/conductor remains constant, the voltage across it is directly proportional to the current flowing through a conductor. i.e
V = IR
Where;
R = constant called resistance
I = current flowing through the wire
V = voltage across the wire
g A 50 kg box is resting on a horizontal surface. Results for item 1. 1 1 / 1 point Determine the weight of the box in [N]. Correct answer: 490 Results for item 2. 2 1 / 1 point Determine the magnitude of the normal force acting on the box in [N]. Correct answer: 490 Results for item 3. 3 0 / 1 point Find the magnitude of the upward applied force, in [N], necessary to lift the box with an acceleration of 1 m/s2
Answer:
(a) 490 N
(b) 490 N
(c) 540 N
Explanation:
mass, m = 50 kg
acceleration, a = 1 m/s2
(a) The weight is given by
W = m g = 50 x 9.8 = 490 N
(b) The normal force is
N = m g = 490 N
(c) Let the force required is F.
Use Newton's second law
F - m g = m a
F = m(g + a)
F = 50(9.8 + 1)
F = 540 N
A steel playground slide is 5.25 m long and is raised 2.75 m on one end. A 45.0 kg child slides down from the top starting at rest. The final speed of the child at the bottom is 6.81 m/s. Find the average force of friction between the child and the slide.
Answer:
[tex]F=32.24N[/tex]
Explanation:
From the question we are told that:
Height [tex]h= 2.75 m[/tex]
Length[tex]l = 5.25 m[/tex]
Mass [tex]m=45kg[/tex]
Final speed [tex]v_f=6.81[/tex]
Generally the equation for Potential Energy P.E is mathematically given by
[tex]P.E=mgh[/tex]
Therefore
Initial potential energy
[tex]P.E_1=45*9.8*2.75 \\\\P.E_1= 1212.75 J[/tex]
Generally the equation for Kinetic Energy K.E is mathematically given by
[tex]K.E=0.5mv^2[/tex]
Therefore
Final kinetic energy
[tex]K.E_2= 1/2*45*6.81*6.81 \\\\K.E_2= 1043.46J[/tex]
Generally the equation for Work_done is mathematically given by
[tex]W=P.E_1-K.E_2\\\\W=169.3[/tex]
Therefore
[tex]F=\frac{W}{d}\\\\F=\frac{169.3}{5.25}[/tex]
[tex]F=32.24N[/tex]
A banked curve is designed for a roadway. The curve is designed so that a car traveling at 19.2 m/s can negotiate it without relying on the force of static friction. The radius of the curve is 51.2 in and the car has a mass of 2300 kg. a. What angle is the curve banked at (degrees)? b. What is the magnitude of the Normal force (N) between the road and the car when it is traveling at the designed speed? c. If a careless driver tries to negotiate the curve at a speed of 25 m/s ,will the direction of the frictional force be inward or outward?
