Answer:
The electric flux through the car's bottom is [tex]1.75\times 10^{4} Wb[/tex]
Solution:
As per the question:
Magnitude of vertical Electric field, [tex]E_{v} = 1.85\times 10^{4} N/C[/tex]
The area of the rectangular surface of the car, [tex]A_{bottom} = 6\times 3 = 18 m^{2}[/tex]
Downward slope at an angle, [tex]\angle\theta = 19.0^{\circ}[/tex]
Now, the electric flux, [tex]\phi_{E}[/tex] is given by:
[tex]\phi_{E} = \vec{E_{v}}.\vec{A_{bottom}} = E_{v}A_{bottom}cos\theta[/tex]
Now, substituting the appropriate values in the above formula:
[tex]\phi_{E} = 1.85\times 10^{4}\times 18cos19.0^{\circ}[/tex]
[tex]\phi_{E} = 1.75\times 10^{4} Wb[/tex]
What is the velocity of a 30-kg box with a kinetic energy of 6,000 J? 64 m/s
6.0 m/s
18 m/s
20 m/s
36 m/s
Answer: 20 m/s
Explanation: To solve this problem we have to consider the expression of the kinetic energy given by:
Ekinetic=(1/2)*(m*v^2)
then E=0.5*30Kg*(20 m/s)^2=15*400=6000J
A 0.676 m long section of cable carrying current to a car starter motor makes an angle of 57.7º with the Earth’s 5.29 x 10^−5T field. What is the current when the wire experiences a force of 6.53 x 10^−3 N?
Answer:
The current in the wire under the influence of the force is 216.033 A
Solution:
According to the question:
Length of the wire, l = 0.676 m
[tex]\theta = 57.7^{\circ}[/tex]
Magnetic field of the Earth, [tex]B_{E} = 5.29\times 10^{- 5} T[/tex]
Forces experienced by the wire, [tex]F_{m} = 6.53\times 10^{-3} N[/tex]
Also, we know that the force in a magnetic field is given by:
[tex]F_{m} = IB_{E}lsin\theta[/tex]
[tex]I = \frac{F_{m}}{B_{E}lsin\theta}[/tex]
[tex]I = \frac{6.53\times 10^{-3}}{5.29\times 10^{- 5}\times 0.676sin57.7^{\circ}[/tex]
I = 216.033 A
N women, each of mass m, stand on a railway flatcar of mass M. They jump off one end of the flatcar with velocity u relative to the car. The car rolls in the opposite direction without friction. (a) What is the final velocity of the flatcar if all the women jump at the same time? (b) What is the final velocity of the flatcar if the women jump off one at a time? (c) Which answer, either (a) or (b), is greater? Try to make physical and/or intuitive sense of this answer and comment on your thoughts.
Answer:
Explanation:
Given
Velocity of women relative to the car is u
Let v be the velocity of car after women jump off
Therefore women velocity relative to the ground is v-u
Conserving momentum
Mv+Nm(v-u)=0
Mv+Nmv=Nmu
[tex]v=\frac{Nm}{M+Nm}[/tex]
(b)Let [tex]v_r[/tex] be the velocity of car after r women has jumped and [tex]v_N[/tex] be the final velocity of the car after N women have jumped.
After r women jumped and N-r on cart momentum is given by
[tex]P_r=\left [ M+\left ( N-r\right )m\right ]v_r[/tex]
After next woman jumps car has velocity of [tex]v_{r+1}[/tex] while the woman has velocity is [tex]v_{r+1}-u[/tex](relative to the ground)
Total momentum
[tex]P_{r+1}=\left [ M+\left ( N-r-1\right )m\right ]v_{r+1}+m\left ( v_{r+1}-u\right )[/tex]
Since total momentum in horizontal direction is conserved then
[tex]P_{r+1}=P_r[/tex]
[tex]v_{r+1}-v_r=\frac{mu}{\left ( M+\left ( N-r\right )m\right )}[/tex]
Summing the above expression from r=0 to r=N-1 we get
[tex]\sum_{j=1}^{j=N}\frac{mu}{\left ( M+jm\right )}[/tex]
(c)Answer in part b is greater because j\leq N[/tex]
Thus Velocity in part b is greater.
Intuitively if you jump after an another person you will impart extra momentum to the cart compared to when all persons jumped off simultaneously.
When all women jump off the flatcar at the same time, the final velocity of the flatcar is zero. However, when the women jump off one at a time, the flatcar gains momentum and moves in the opposite direction, resulting in a non-zero final velocity.
Explanation:To answer the given question, we can apply the principle of conservation of momentum. When all the women jump off the flatcar at the same time, the total momentum of the system must remain constant. Since the women are jumping off the flatcar with equal and opposite velocities, the total momentum of the system before and after their jumps will be zero. Therefore, the final velocity of the flatcar will also be zero.
On the other hand, when the women jump off one at a time, the total momentum of the system does not remain constant. Initially, the flatcar and the first woman have a total momentum of zero. But as the first woman jumps off with a velocity u, the flatcar gains an equal and opposite momentum. When the second woman jumps off, the flatcar gains momentum again, and this process continues until all women have jumped off. As a result, the flatcar gains momentum with each jump and moves in the opposite direction of the women. Therefore, the final velocity of the flatcar when the women jump off one at a time will be greater than zero.
From a physical and intuitive perspective, the answer to (b) is greater because as each woman jumps off, the flatcar gains momentum in the opposite direction. This momentum accumulates with each jump, resulting in a higher final velocity for the flatcar compared to when all the women jump off simultaneously.
Learn more about Conservation of momentum here:https://brainly.com/question/33316833
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The pilot of an airplane reads the altitude 6400 m and the absolute pressure 46 kPa when flying over a city. Calculate the local atmospheric pressure in that city in kPa and in mmHg. Take the densities of air and mercury to be 0.828 kg/m3 and 13,600 kg/m3, respectively.
The local atmospheric pressure in the city in kPa is ?
The local atmospheric pressure in the city in mmHg is ?
Answer:
1. 6.672 kPa
2. 49.05 mm of mercury
Explanation:
h = 6400 m
Absolute pressure, p = 46 kPa = 46000 Pa
density of air, d = 0.823 kg/m^3
density of mercury, D = 13600 kg/m^3
(a) Absolute pressure = Atmospheric pressure + pressure due to height
46000 = Atmospheric pressure + h x d x g
Atmospheric pressure = 46000 - 6400 x 0.823 x 10 = 6672 Pa = 6.672 kPa
(b) To convert the pressure into mercury pressure
Atmospheric pressure = H x D x g
Where, H is the height of mercury, D be the density of mercury, g be the acceleration due to gravity
6672 = H x 13600 x 10
H = 0.04905 m
H = 49.05 mm of mercury
To calculate the local atmospheric pressure in the city, use the formula P = P0 * e^(-Mgh/RT), where P is the pressure at the current location, P0 is the pressure at sea level, M is the molar mass of air, g is the acceleration due to gravity, h is the altitude, R is the ideal gas constant, and T is the temperature. Plug in the given values into the formula and solve for P0 to find the local atmospheric pressure in kPa. To convert this pressure to mmHg, use the conversion factor that 101.325 kPa is equal to 760 mmHg.
Explanation:To calculate the local atmospheric pressure in the city, we can use the relationship between altitude, pressure, and density of air. Altitude affects atmospheric pressure because as we go higher in the atmosphere, the air becomes less dense and therefore exerts less pressure. The formula to calculate pressure is given by:
P = P0 * e^(-Mgh/RT)
Where P is the local atmospheric pressure, P0 is the pressure at sea level, M is the molar mass of air, g is the acceleration due to gravity, h is the altitude, R is the ideal gas constant, and T is the temperature. Since we know the altitude and pressure at the current location, we can rearrange the formula to solve for P0:
P0 = P / e^(-Mgh/RT)
Using the given values, we can substitute them into the formula. The density of air is 0.828 kg/m³, the altitude is 6400 m, the absolute pressure is 46 kPa, the molar mass of air is the average molar mass of air (28.97 g/mol), the acceleration due to gravity is 9.8 m/s², the ideal gas constant is 8.314 J/(mol·K), and the temperature can be assumed to be the average temperature at that altitude (around 0°C or 273 K).
Plugging in these values into the formula, we get:
Calculating this expression will give us the local atmospheric pressure in kPa.
To convert this pressure to mmHg, we can use the conversion factor that 101.325 kPa is equal to 760 mmHg. So, to find the pressure in mmHg, we can multiply the pressure in kPa by the conversion factor.
What is the beat frequency between a note at 350 Hz and a note at 353 Hz?
Answer:
[tex]f_{beat}=f_2-f_1=353Hz -350Hz=3Hz[/tex]
Explanation:
When there are two waves with similar frequency, the superposition of these waves create a new wave with a particular beat. The beat frequency from this constructive superposition is equal to the value of the difference in frequency of the two waves.
[tex]f_{beat}=f_2-f_1=353Hz -350Hz=3Hz[/tex]
A charge of -3.35 nC is placed at the origin of an xy-coordinate system, and a charge of 1.75 nC is placed on the y axis at y = 3.90 cm . A. If a third charge, of 5.00 nC , is now placed at the point x = 2.70 cm , y = 3.90 cm find the x and y components of the total force exerted on this charge by the other two charges.
B. Find the magnitude of this force.
C. Find the direction of this force. ( ° below the +x axis )
Answer:
Explanation:
Force due to charges 1.75 and 5 nC is given below
F =K Q₁Q₂ / d²
F₁ = [tex]\frac{9\times10^9\times1.75\times10^{-9}\times5\times 10^{-9}}{(2.7\times10^{-2})^2}[/tex]
F₁ = 10.8 X 10⁻⁵ N . It will at in x direction.
Force due to other charge placed at origin
F₂ = [tex]\frac{9\times10^9\times1.75\times10^{-9}\times5\times 10^{-9}}{22.5\times10^{-4}}[/tex]
F₂ = 3.5 x 10⁻⁵ N.
Its x component
= F₂ Cos θ
= 3.5 x 10⁻⁵ x 3.9/ 4.74
= 2.88 x 10⁻⁵ N
Its y component
F₂ sin θ
= 3.5 x 10⁻⁵ x 2.7/4.743
= 1.99 x 10⁻⁵ N
Total x component
= 10.8 X 10⁻⁵ +2.88 x 10⁻⁵
= 13.68 x 10⁻⁵ N.
Magnitude of total force F
F² = (13.68 x 10⁻⁵)² + (1.99 x 10⁻⁵ )²
F = 13.82 X 10⁻⁵ N
Direction θ with x axis .
Tanθ = 1.99/ 13.68
θ = 8 °
A 3.50 m -tall, 50.0 cm -diameter concrete column supports a 5.00 x 10^5 kg load. Part A : By how much is the column compressed?
Answer:
The column compressed is 57.142 Pa.
Explanation:
Given that,
Height = 3.50 m
Diameter = 50.0 cm
Radius = 25.0 cm
Load [tex]F=5.00\times10^{5}\ kg[/tex]
We need to calculate the column compressed
Using formula of compressed
[tex]P = \dfrac{F}{A}[/tex]
[tex]P=\dfrac{F}{l\times r}[/tex]
Where, F = load
l = length
r = radius
Put the value into the formula
[tex]P=\dfrac{5.00\times10^{5}}{3.50\times25\times10^{2}}[/tex]
[tex]P=57.142\ Pa[/tex]
Hence, The column compressed is 57.142 Pa.
What fraction of all the electrons in a 25 mg water
dropletmust be removed in order to obtain a net charge of 40
nC?
Answer:
[tex]9.11\times 10^{-15}.[/tex]
Explanation:
The water droplet is initially neutral, it will obtain a 40 nC of charge when a charge of -40 nC is removed from the water droplet.
The charge on one electron, [tex]\rm e=-1.6\times 10^{-19}\ C.[/tex]
Let the N number of electrons have charge -40 nC, such that,
[tex]\rm Ne=-40\ nC\\\Rightarrow N=\dfrac{-40\ nC}{e}=\dfrac{-40\times 10^{-9}\ C}{-1.6\times 10^{-19}\ C}=2.5\times 10^{11}.[/tex]
Now, mass of one electron = [tex]\rm 9.11\times 10^{-31}\ kg.[/tex]
Therefore, mass of N electrons = [tex]\rm N\times 9.11\times 10^{-31}=2.5\times 10^{11}\times 9.11\times 10^{-31}=2.2775\times 10^{-19}\ kg.[/tex]
It is the mass of the of the water droplet that must be removed in order to obtain a charge of 40 nC.
Let it is m times the total mass of the droplet which is [tex]25\ \rm mg = 25\times 10^{-6}\ kg.[/tex]
Then,
[tex]\rm m\times (25\times 10^{-3}\ kg) = 2.2775\times 10^{-19}\ kg.\\m=\dfrac{2.2775\times 10^{-19}\ kg}{25\times 10^{-3}\ kg}=9.11\times 10^{-15}.[/tex]
It is the required fraction of mass of the droplet.
If an astronaut has a weight of 450 N on the earth, what would be her weight on the moon? A. 101 Ib В. 46 N C. 77 N D. 7.8 N
Explanation:
Given that,
Weight of the astronaut on the surface of Earth, W = 450 N
We know that the acceleration due to gravity on the earth is, [tex]g=9.8\ m/s^2[/tex]
Weight of an object is given by, W = mg
[tex]m=\dfrac{W}{g}[/tex]
[tex]m=\dfrac{450\ N}{9.8\ m/s^2}[/tex]
m = 45.91 kg
Also, the acceleration due to gravity on the surface of moon is one -sixth of the acceleration due to gravity on the surface of Earth, [tex]g'=1.62\ m/s^2[/tex]
As mass remains constant. So, the weight on the moon is :
W' = mg'
[tex]W'=45.91\times 1.62[/tex]
W' = 74.37 N
Hence, this is the required solution.
For a satellite already in perfect orbit around the Earth, what would happen if: - the satellite's speed is reduced? - the satellite's mass is reduced?
Answer:
Answer
Explanation:
If the satellite is in perfect orbit around the earth then gravitational pull of the earth on the satellite and moving forward satellite is in perfect balance i.e gravitational pull does not effect the orbit as the velocity of the satellite is good enough to counter the gravitational pull of the earth and inertia makes the satellite move forward with greater velocity, but when its velocity is suddenly reduced, the gravitational pull will counter on the orbiting satellite and it will crash onto the earth.
This is same as in well of death when motorist moves in circular motion inside the well with centripetal and centrifugal forces balanced, suddenly crashes on the ground when velocity of the motorist is not enough to balance forces.
the satellite's mass has no affect on its orbit around the earth rather the mass of the earth is only affecting factor .
A +1.0μC charge sits at the origin, another +1.0μC sits at x = +0.01m, and another -2.0μC sits at x = +0.02m. Calculate the magnitude and direction of the electric field at the point x = +0.2m. (1 μC = 1.0x10^-6C). Draw a picture first.
Answer:
A)Ep=-81.3N/C :Electric field at the point x = +0.2
Ep Magnitude =81.3N/C
Direction of the electric field ( Ep): -x
B)Graphic attached
Explanation:
Conceptual analysis
The electric field at a point P due to a point charge is calculated as follows:
E = k*q/d²
E: Electric field in N/C
q: charge in Newtons (N)
k: electric constant in N*m²/C²
d: distance from charge q to point P in meters (m
Graphic attached
The attached graph shows the field due to the charges:
The electric field at a point P due to several point charges is the vector sum of the electric field due to individual charges.
E₁: Electric Field at point Xp=0.2 m due to charge q₁. As the charge q1 is positive (q₁+) ,the field leaves the charge.
E₁: Electric Field at point Xp=0.2 m due to charge q₂. As the charge q1 is positive (q₂+) ,the field leaves the charge
E₃: Electric Field at point Xp=0.2 m due to charge q₃. As the charge q₃ is negative (q₃-), the field enters the charge.
Equivalence
+1.0μC=1* 10⁶C
Data
q₁=+1.0μC=1* 10⁶C
q₂=+1.0μC=1* 10⁶C
q₃=-2.0μC=-2* 10⁶C
Xp=0.2m
x₁=0
x₂=0.01 m
x₃=0.02m
Calculation of the distances of the charges to the point P
d= Xp-x
d₁=Xp-x₁= 0.2-0= 0.2m
d₂=Xp-x₂=0.2-0.01= 0.19m
d₃=Xp-x₃=0.2-002= 0.18m
Calculation of electric fields due to charges q1, q2 and q3 at point P
E₁=k*q₁/d₁²=9*10⁹*1*10⁻⁶/0.2²=225*10³N/C
E₂=k*q₂/d₂²=9*10⁹*1*10⁻⁶/0.19²=249.3*10³N/C
E₃=-k*q₃/d₃²=9*10⁹*2*10⁻⁶/0.18²=-555.6*10³N/C
Calculation of electric field at point P due to charges q₁, q₂ and q₃
To calculate Ep, the electric fields E₁,E₂ and E₃ are added algebraically:
Ep=E₁+E₂ +E₃
Ep=(225*10³+249.3*10³ -555.6*10³)N/C
Ep=-81.3N/C
Ep Magnitude =81.3N/C in -x direction
A proton is accelerated down a uniform electric field of 450 N/C. Calculate the acceleration of this proton.
Answer:
The acceleration of proton will be [tex]431.137\times 10^8m/sec^2[/tex]
Explanation:
We have given electric field E = 450 N/C
Charge on proton [tex]=1.6\times 10^{-19}C[/tex]
Force on electron due to electric field is given by [tex]F=qE=1.6\times 10^{-19}\times 450=720\times 10^{-19}N[/tex]
Mass of electron [tex]m=1.67\times 10^{-27}kg[/tex]
Now according to second law of motion [tex]F=ma[/tex]
So [tex]720\times10^{-19}=1.67\times 10^{-27}a[/tex]
[tex]a=431.137\times 10^8m/sec^2[/tex]
So the acceleration of proton will be [tex]431.137\times 10^8m/sec^2[/tex]
How would you find the total energy stored in the
capacitorsif.....
A 2.0 microF capacitor and a 4.0 microF capacitor are connected
inPARALLEL across a 300V potential difference.
Answer:
The energy of the capacitors connected in parallel is 0.27 J
Given:
C = [tex]2.0\micro F = 2.0\times 10^{- 6} F[/tex]
C' = [tex]4.0\micro F = 4.0\times 10^{- 6} F[/tex]
Potential difference, V = 300 V
Solution:
Now, we know that the equivalent capacitance of the two parallel connected capacitors is given by:
[tex]C_{eq} = C + C' = 2.0 + 4.0 = 6.0\micro F = 6.0\times 10^{- 6} F[/tex]
The energy of the capacitor, E is given by;
[tex]E = \frac{1}{2}C_{eq}V^{2}[/tex]
[tex]E = \frac{1}{2}\times 6.0\times 10^{- 6}\times 300^{2} = 0.27 J[/tex]
A 34 kg block slides with an initial speed of 9 m/s up a
rampinclined at an angle of 10o with the horizontal.
Thecoefficient of kinetic friction between the block and the ramp
is0.6. Use energy conservation to find the distance the block
slidesbefore coming to rest.
Answer:
5.4 m
Explanation:
mass, m = 34 kg
initial velocity, u = 9 m/s
final velocity, v = 0 m/s
coefficient of friction, μ = 0.6
Angle of inclination, θ = 10°
Let teh distance traveled before to come into rest is d.
According to the work energy theorem, the work done by all the forces is equal to the change in kinetic energy of the block.
Work done by the gravitational force = W1 = - mg Sinθ x d
Work done by the frictional force = W2 = - μ N = - μ mg Cosθ x d
negative sign shows that the direction of force and the direction of displacement is opposite to each other.
Total work done W = W1 + w2
W = - 34 x 9.8 x Sin 10 x d - 0.6 x 34 x 9.8 x cos 10 x d
W = - 254.86 d
Change in kinetic energy = 0.5 x m (v^2 - u^2)
= 0.5 x 34 (0 - 81) = - 1377
So, W = change in KE
- 254.86 d = - 1377
d = 5.4 m
A block is placed at the top of a frictionless inclined plane with angle 30 degrees and released. The incline has a height of 15 m and the block has mass 2 kg. In the above problem the speed of the block when it reaches the bottom of the incline is 8.3 m/s 17 m/s 25 m/s 30 m/s
Answer:
The speed of the block when it reaches the bottom is 17 m/s.
(b) is correct option.
Explanation:
Given that,
Height = 15 m
Mass = 2 kg
We need to calculate the speed of the block when it reaches the bottom
Using conservation of energy
[tex]mgh=\dfrac{1}{2}mv^2[/tex]
[tex]v^2=2gh[/tex]
[tex]v=\sqrt{2gh}[/tex]
Where, g = acceleration due to gravity
h = height
Put the value into the formula
[tex]v=\sqrt{2\times9.8\times15}[/tex]
[tex]v=17\ m/s[/tex]
Hence, The speed of the block when it reaches the bottom is 17 m/s.
Kathy tests her new sports car by racing with Stan, an experienced racer. Both start from rest, but Kathy leaves the starting line 1.00 s after Stan does. Stan moves with a constant acceleration of 3.9 m/s^2 while Kathy maintains an acceleration of 4.73 m/s^2. (a) Find the time at which Kathy overtakes (b) Find the distance she travels before she catches him (c) Find the speeds of both cars at the instant she overtakes him.
Answer:
(a) 10 s
(b) 236.5 m
(c) Kathy's speed = 47.3 m/s
Stan's speed = 42.9 m/s
Explanation:
Given:
[tex]u_k[/tex] = initial speed of Kathy = 0 m/s[tex]u_s[/tex] = initial speed of Stan = 0 m/s[tex]a_k[/tex] = acceleration of Kathy = [tex]4.73\ m/s^2[/tex][tex]a_s[/tex] = acceleration of Stan = [tex]3.9\ m/s^2[/tex]Assumptions:
[tex]v_k[/tex] = final speed of Kathy when see catches Stan[tex]v_s[/tex] = final speed of Stan when Kathy catches him[tex]s_k[/tex] = distance traveled by Kathy to catch Stan[tex]s_s[/tex] = distance traveled by Stan when Kathy catches him[tex]t_k[/tex] = time taken by Kathy to catch Stan = [tex]t[/tex][tex]t_s[/tex] = time interval in which Kathy catches Stan = [tex]t+1[/tex]Part (a):
Kathy will catch Stan only if the distances traveled by each of them are equal at the same instant.
[tex]\therefore s_s=s_k\\\Rightarrow u_st_s+\dfrac{1}{2}a_st_s^2=u_kt_k+\dfrac{1}{2}a_kt_k^2\\ \Rightarrow (0)(t+1)+\dfrac{1}{2}(3.9)(t+1)^2=(0)(t)+\dfrac{1}{2}(4.73)t^2\\ \Rightarrow \dfrac{1}{2}(3.9)(t+1)^2=\dfrac{1}{2}(4.73)t^2\\\Rightarrow (3.9)(t+1)^2=(4.73)t^2\\\Rightarrow \dfrac{(t+1)^2}{t^2}=\dfrac{4.73}{3.9}\\\textrm{Taking square root in both sides}\\\dfrac{t+1}{t}= 1.1\\\Rightarrow t+1=1.1t\\\Rightarrow 0.1t = 1\\\Rightarrow t = 10\\[/tex]
Hence, Kathy catches Stan after 11 s from the Stan's starting times.
Part (b):
Distance traveled by Kathy to catch Stan will be distance the distance traveled by her in 10 s.
[tex]s_s = u_kt_k+\dfrac{1}{2}a_kt_k^2\\\Rightarrow s_s= (0)(t)+\dfrac{1}{2}(4.73)t^2\\\Rightarrow s_s= \dfrac{1}{2}(4.73)(10)^2\\\Rightarrow s_s= 236.5[/tex]
Hence, Kathy traveled a distance of 236.5 m to overtake Stan.
Part (c):
[tex]v_k = u_k+a_kt_k\\\Rightarrow v_k = 0+(4.73)(t)\\\Rightarrow v_k = (4.73)(10)\\\Rightarrow v_k =47.3[/tex]
The speed of Kathy at the instant she catches Stan is 47.3 m/s.
[tex]v_s = u_s+a_st_s\\\Rightarrow v_s = 0+(3.9)(t+1)\\\Rightarrow v_s = (3.9)(10+1)\\\Rightarrow v_s =42.9[/tex]
The speed of Stan at the instant Kathy catches him is 42.9 m/s.
If a truck is travelling east on a straight road and travels 100 meters in 25s what is the truck's velocity?
Answer:
Velocity of the truck, v = 4 m/s
Explanation:
It is given that,
Distance covered by the truck, d = 100 m
Time taken by the truck, t = 25 s
Let v is the velocity of the truck. We know that the velocity is a vector quantity. Mathematically, it is given by :
[tex]v=\dfrac{d}{t}[/tex]
[tex]v=\dfrac{100\ m}{25\ s}[/tex]
v = 4 m/s
So, the velocity of the truck is 4 m/s. Hence, this is the required solution.
A sports car accelerates uniformly from rest to a speed of 87 mi/hr in 8s. Determine: a.The acceleration of the car
b.The distance the car travels in the first 8 s
c.The velocity of the car after the first 10 s
Answer:
Part a)
[tex]a = 4.86 m/s^2[/tex]
Part b)
[tex]d = 155.52 m[/tex]
Part c)
[tex]v_f = 48.6 m/s[/tex]
Explanation:
As we know that car start from rest and reach to final speed of 87 mph
so we have
[tex]v_f = 87 mph = 38.88 m/s[/tex]
now we have
Part a)
acceleration is rate of change in velocity
[tex]a = \frac{v_f - v_i}{t}[/tex]
[tex]a = \frac{38.88 - 0}{8}[/tex]
[tex]a = 4.86 m/s^2[/tex]
Part b)
distance moved by car with uniform acceleration is given as
[tex]d = \frac{v_f + v_i}{2} t[/tex]
[tex]d = \frac{38.88 + 0}{2} 8[/tex]
[tex]d = 155.52 m[/tex]
Part c)
As we know that the car start from rest
so final speed after t = 10 s
[tex]v_f = v_i + at[/tex]
[tex]v_f = 0 + (4.86)10[/tex]
[tex]v_f = 48.6 m/s[/tex]
The acceleration of automobiles is often given in terms of the time it takes to go from 0 mi/h to 60 mi/h. One of the fastest street legal cars, the Bugatti Veyron Super Sport, goes from 0 to 60 in 2.70 s. To the nearest integer, what is its acceleration in m/s^2? Please show work.
Answer:
9.934 m/s²
Explanation:
Given:
Initial speed of the Bugatti Veyron Super Sport = 0 mi/h
Final speed of the Bugatti Veyron Super Sport = 60 mi/h
Now,
1 mi/h = 0.44704 m / s
thus,
60 mi/h = 0.44704 × 60 = 26.8224 m/s
Time = 2.70 m/s
Now,
The acceleration (a) is given as:
[tex]a=\frac{\textup{Change in speed}}{\textup{Time}}[/tex]
thus,
[tex]a=\frac{26.8224 - 0 }{2.70}[/tex]
or
a = 9.934 m/s²
Two small spherical insulators separated by 2.5 cm, which is much greater than either of their diameters. Both carry positive charge, one +60.0 microCoulombs and the other +6.66 microCoulombs. A third positive charge remains at rest between the two spheres and along the line joining them. What is the position of this charged sphere?
Answer:
1.875 cm from 60 microcoulomb charge.
Explanation:
Let the third charge be Q. Let it be put at x distance from 60 micro coulomb charge for balance.
Force on this charge due to first charge
= [tex]\frac{k\times60\times10^{-6}Q}{x^2}[/tex]
Force on this charge due to second charge
= [tex]\frac{k\times6.66\times10^{-6}Q}{(2.5-x)^2}[/tex]
Since both these forces are equal [tex]\frac{k\times60\times10^{-6}Q}{x^2}=\frac{k\times6.66\times10^{-6}Q}{(2.5-x)^2}[/tex]
[tex]\frac{60}{6.66} = \frac{x^2}{(2.5-x)^2}[/tex][tex]\frac{x}{2.5 -x} = \frac{3}{1}[/tex]
x = 1.875
1.875 cm from 60 microcoulomb charge.
Riding in a car, you suddenly put on the brakes. As you experience it inside the car, do Newton's law apply? Do they apply as seen by someone outside the car? Please explain reasoning, thank you!
Answer with Explanation:
Newton's laws are applicable for inertial frames of reference which is a frame which is not accelerating when seen from the observer standing on earth.
For the person as he presses the brakes his frame is a decelerating frame of reference hence he cannot apply the newtons laws of motion as they are in their original form but if he analyses the motion he has to apply a correction known as pseudo-force on the object he is analyzing. Pseudo Force has no basis in newton's laws but are a correction that needs to be applied if he wishes to analyse the motion from non inertial frame of reference
While as a person standing on earth outside the car since his frame is an inertial frame of reference he can apply newton's laws of motion without any correction.
At some airports there are speed ramps to help passengers get from one place to another. A speed ramp is a moving conveyor belt on which you can either stand or walk. Suppose a speed ramp has a length of 121 m and is moving at a speed of 2.2 m/s relative to the ground. In addition, suppose you can cover this distance in 78 s when walking on the ground. If you walk at the same rate with respect to the speed ramp that you walk on the ground, how long does it take for you to travel the 121 m using the speed ramp?
Answer:
It takes you 32.27 seconds to travel 121 m using the speed ramp
Explanation:
Lets explain how to solve the problem
- The speed ramp has a length of 121 m and is moving at a speed of
2.2 m/s relative to the ground
- That means the speed of the ramp is 2.2 m/s
- You can cover the same distance in 78 seconds when walking on
the ground
Lets find your speed on the ground
Speed = Distance ÷ Time
The distance is 121 meters
The time is 78 seconds
Your speed on the ground = 121 ÷ 78 = 1.55 m/s
If you walk at the same rate with respect to the speed ramp that
you walk on the ground
That means you walk with speed 1.55 m/s and the ramp moves by
speed 2.2 m/s
So your speed using the ramp = 2.2 + 1.55 = 3.75 m/s
Now we want to find the time you will take to travel 121 meters using
the speed ramp
Time = Distance ÷ speed
Distance = 121 meters
Speed 3.75 m/s
Time = 121 ÷ 3.75 = 32.27 seconds
It takes you 32.27 seconds to travel 121 m using the speed ramp
The vertical velocity of a projectile launch at 35 degree : O continuously decreases O continuously increases O sometimes decreases and sometimes increases O is zero O stays constant.
Answer:Sometime decreases and sometime increases.
Explanation:
Given
Projectile is launched at [tex]35 ^{\circ}[/tex]
so horizontal velocity=ucos35
vertical velocity=usin35
since gravity acts in the vertical direction, therefore, it affects only the vertical component of the projectile.
So Vertical velocity first decrease up to maximum height and then starts to increase.
An aircraft carrier is sailing at a constant speed of 15.0m/s. it is suddenly spotted by an enemy bomber aircraft flying in the same direction as the ship is moving. the bomber is flying at a constant speed of 100 m/s at an altitude of 1.00 km above the carrier when it decides to attack the carrier. if the bomber drops its bomb at a horizontal distance pf 1.3 km from the aircraft carrier, will it score a direct hit?
Answer:
Explanation:
Relative velocity = 100 - 15 = 85 m /s
Distance to be covered up with this relative speed = 1.3km
= 1300m
If t be the time taken
85 x t = 1300
t = 1300/ 85
= 15.29
Time of fall of the bomb t₁ by 1 km
1000 = 1/2 g t₁²
t₁² = 2000/9.8
t₁ = 14.28 s
Since t₁≠ t₂
It will not score a direct hit.
A uniformly accelerated car passes three equally spaced traffic signs. The signs are separated by a distance d = 25 m. The car passes the first sign at t = 1.3 s, the second sign at t = 3.9 s, and the third sign at t = 5.5 s.
(a) What is the magnitude of the average velocity of the car during the time that it is moving between the first two signs?
(b) What is the magnitude of the average velocity of the car during the time that it is moving between the second and third signs?
(c) What is the magnitude of the acceleration of the car?
The average velocity between the first two signs is 9.1 m/s, between the second and third signs is 10.4 m/s, and the car's acceleration is approximately 0.26 m/s².
To solve for the average velocity and acceleration of the car passing three equally spaced traffic signs, we can use the kinematic equations for uniformly accelerated motion.
Part a: average velocity between the first two signs
The average velocity is calculated by dividing the displacement (distance between the signs) by the time interval: vavg1,2 = d /
[tex](t_2 - t_1)[/tex]. Substituting the given values, we have vavg1,2 = 25 m / (3.9 s - 1.3 s), resulting in an average velocity of 9.1 m/s.
Part b: average velocity between the second and third signs
Similarly, for the average velocity between the second and third signs: vavg2,3 = d / ([tex]t_3 - t_2[/tex]). Using the given times, we calculate vavg2,3 = 25 m / (5.5 s - 3.9 s), giving an average velocity of 10.4 m/s.
Part c: magnitude of the acceleration
To find the acceleration, we can use the velocities we just calculated, and the time between the signs. We can set up an equation for motion for the time intervals. As initial velocity v0 for the second interval is the final velocity of the first interval, we get v0 = vavg1,2 and final velocity v = vavg2,3. Then we can use the equation v = v0 + a[tex](t_3 - t_2)[/tex] to find the acceleration. Rearranging for a, we have a = [tex](v - v_0) / (t_3 - t_2)[/tex]. After calculating, we find that the acceleration is approximately 0.26 m/s².
Ropes 3 m and 5 m in length are fastened to a holiday decoration that is suspended over a town square. The decoration has a mass of 3 kg. The ropes, fastened at different heights, make angles of 52° and 40° with the horizontal. Find the tension in each wire and the magnitude of each tension. (Use g
Answer:
T₁ = 19.39N with an angle of 52° with the horizontal.
T₂ = 15.64N with an angle of 40° with the horizontal.
Explanation:
1)We calculate the weight of holiday decoration
W = m×g
W: holiday decoration weight in Newtons (N)
m: holiday decoration weight = 3 kg
g: acceleration due to gravity = 9.8 m/s²
W = m×g = 3 kg × 9.8 m/s² = 29.4 N
2)We apply Newton's first law to the system in equilibrium:
ΣFx = 0
T₁Cos52° - T₂Cos40°=0
T₁Cos52° = T₂Cos40°
T₁ = T₂(Cos40° / T₁Cos52°)
T₁ = 1,24×T₂ Equation (1)
ΣFy = 0
T₁Sin52° + T₂Sin40° - 29.4 = 0 Equation (2)
We replace T₁ of Equation (1) in the Equation (2)
1.24×T₂×Sin52° + T₂Sin40° - 29.4 = 0
1.88*T₂ = 29.4
T₂ = 29.4/1.88
T₂ = 15.64N
We replace T₂ = 15.64N in the Equation (1) to calculate T₁:
T₁ = 1.24×15.64N
T₁ = 19.39N
A rocket sled used in old Air Force test could accelerate a volunteer at rates of 90m/s^2. With this acceleration being constant, how long would it take for one of these rocket sleds to travel the length of a one kilometer test track if it starts from rest? A) 11.1sec B) 22.2sec C)4.71sec D) 32.2sec
Answer:
option (c) 4.71 seconds
Explanation:
Given:
Distance to be covered = 1 km = 1000 m
Acceleration, a = 90 m/s²
From Newton's equation of motion
[tex]s=ut+\frac{1}{2}at^2[/tex]
where,
s is the distance
u is the initial speed = 0 m/s ( since it starts from rest )
a is the acceleration
t is the time
on substituting the respective values, we get
[tex]1000=0\times t+\frac{1}{2}\times90\times t^2[/tex]
or
45t² = 1000
or
t² = 22.22
or
t = 4.71 seconds
Hence, the correct answer is option (c) 4.71
Since vectors always have positive magnitudes, the sum of two vectors must have a magnitude greater than the magnitude of either one of them. O True O False
Answer:
False.
Explanation:
Yes the magnitude of a vector is always positive , but a vector consists of
magnitude directionwhen two vectors are added their direction may be opposite to each other For example-
[tex]a=3i+3j[/tex]
[tex]b=3i-6j[/tex] ,
then their resultant
[tex]r=b+a\\r=3i+3j+3i-6j\\r=6i-3j[/tex]
This resultant vector's x and y component equal to y and x component of vector b so its magnitude will be equal to magnitude of vector b.
Therefore, the resultant magnitude not necessary equal to the magnitude of either vector.
A 2.00- kg ball has zero potential and kinetic energy.
Mariadrops the ball into a 10.0- m deep well. After the ball comes
to astop in the mud, the sum of its potential and kinetic
energyis?
Answer:
- 196.2 J
Explanation:
Given:
Mass of the ball = 2.00 kg
Height through which ball drops = 10.0 m
Since it is mentioned that the ball comes to a stop in the mud, therefore the velocity of the ball will be zero i.e v = 0 m/s
Thus.
the kinetic energy = [tex]\frac{1}{2}mv^2[/tex]
= [tex]\frac{1}{2}m\times0^2[/tex] = 0 J
Now,
The potential energy = - mgh
here, g is the acceleration due to gravity
and, the negative sign depicts that the ball is falling down
thus,
The potential energy = - 2 × 9.81 × 10 = - 196.2 J
Hence, the total energy = - 196.2 + 0 = - 196.2 J
Final answer:
The sum of the gravitational potential and kinetic energy of the 2.00-kg ball after it comes to a stop in the mud at the bottom of a 10.0-m deep well is 0 Joules, since both potential and kinetic energy are zero at rest.
Explanation:
When the 2.00-kg ball is dropped into a 10.0-m deep well, it initially has a certain amount of gravitational potential energy which is converted entirely to kinetic energy just before it hits the bottom, assuming there is no air resistance. The sum of its potential and kinetic energy at the top is equal to the gravitational potential energy it has due to its position. This can be calculated using the formula for gravitational potential energy (GPE), which is GPE = mgh, where m is the mass, g is the acceleration due to gravity (approximately 9.81 m/s² on Earth), and h is the height above the reference point. In this case, when the ball comes to a stop in the mud at the bottom of the well, all its potential energy has been converted to kinetic energy during the fall, and then to other forms of energy (like heat or sound) upon impact. The ball's potential energy at the bottom of the well is zero and the kinetic energy is also zero since the ball is at rest, so the sum of its potential and kinetic energy after coming to a stop is:
mgh = 2*9.81*10 = -196.2 J (as the ball is falling down)
Sum = -196.2J
According to the conservation of energy principle, energy is neither created nor destroyed, only transformed from one form to another. Therefore, initially, the ball had potential energy which got converted to kinetic energy during the fall and finally to other forms of energy when it came to rest.
Two charges q1 and q2 have a total charge of 11 C. When they are separated by 4.5 m, the force exerted by one charge on the other has a magnitude of 8 mN. Find q1 and q2 if both are positive so that they repel each other, and q1 is the smaller of the two. (For the universal constant k use the value 8.99 109 N m2/C2.)
If[tex]\( q_1 = 0 \), then \( q_2 = 11 \)[/tex] which is positive.
Thus, [tex]\( q_1 = 0 \, \text{C} \) and \( q_2 = 11 \, \text{C} \)[/tex] are the valid solutions.
To find the charges [tex]\( q_1 \) and \( q_2 \),[/tex] we can use Coulomb's law, which states that the magnitude of the force between two point charges is given by:
[tex]\[ F = \frac{k \cdot |q_1 \cdot q_2|}{r^2} \][/tex]
Given that the force exerted by one charge on the other has a magnitude of 8 mN (milli-Newtons) when they are separated by 4.5 m, we can set up the equation as follows:
[tex]\[ 8 \times 10^{-3} \, \text{N} = \frac{(8.99 \times 10^9 \, \text{N} \cdot \text{m}^2/\text{C}^2) \cdot |q_1 \cdot q_2|}{(4.5 \, \text{m})^2} \][/tex]
Solving for[tex]\( |q_1 \cdot q_2| \[/tex]), we get:
[tex]\[ |q_1 \cdot q_2| = \frac{8 \times 10^{-3} \, \text{N} \cdot (4.5 \, \text{m})^2}{8.99 \times 10^9 \, \text{N} \cdot \text{m}^2/\text{C}^2} \][/tex]
[tex]\[ |q_1 \cdot q_2| = \frac{8 \times 10^{-3} \times 20.25}{8.99 \times 10^9} \, \text{C}^2 \][/tex]
[tex]\[ |q_1 \cdot q_2| = \frac{0.162}{8.99 \times 10^9} \, \text{C}^2 \][/tex]
[tex]\[ |q_1 \cdot q_2| = 1.802 \times 10^{-11} \, \text{C}^2 \][/tex]
Now, since both charges are positive and repel each other, we know that [tex]\( q_1 \)[/tex]is smaller than [tex]\( q_2 \)[/tex]. So, let's assume [tex]\( q_1 = x \) and \( q_2 = 11 - x \)[/tex]. Substituting these values into the equation for [tex]\( |q_1 \cdot q_2| \)[/tex], we get:
[tex]\[ x \cdot (11 - x) = 1.802 \times 10^{-11} \, \text{C}^2 \][/tex]
[tex]\[ 11x - x^2 = 1.802 \times 10^{-11} \, \text{C}^2 \][/tex]
Now, solve this quadratic equation to find x , which represents [tex]\( q_1 \).[/tex] Once we find x , we can find [tex]\( q_2 = 11 - x \).[/tex] Then, we verify that both charges are positive and satisfy the given conditions.
To solve the quadratic equation [tex]\( 11x - x^2 = 1.802 \times 10^{-11} \, \text{C}^2 \)[/tex], we first rearrange it to the standard form:
[tex]\[ x^2 - 11x + 1.802 \times 10^{-11} = 0 \][/tex]
We can use the quadratic formula to find[tex]\( x \):[/tex]
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
where [tex]\( a = 1 \), \( b = -11 \), and \( c = 1.802 \times 10^{-11} \).[/tex]
Plugging in these values:
[tex]\[ x = \frac{-(-11) \pm \sqrt{(-11)^2 - 4(1)(1.802 \times 10^{-11})}}{2(1)} \][/tex]
[tex]\[ x = \frac{11 \pm \sqrt{121 - 7.208 \times 10^{-11}}}{2} \][/tex]
[tex]\[ x = \frac{11 \pm \sqrt{121 - 7.208 \times 10^{-11}}}{2} \][/tex]
[tex]\[ x = \frac{11 \pm \sqrt{121 - 7.208 \times 10^{-11}}}{2} \][/tex]
[tex]\[ x = \frac{11 \pm \sqrt{121 - 7.208 \times 10^{-11}}}{2} \][/tex]
[tex]\[ x = \frac{11 \pm \sqrt{121 - 7.208 \times 10^{-11}}}{2} \][/tex]
[tex]\[ x \approx \frac{11 \pm \sqrt{121}}{2} \][/tex]
[tex]\[ x \approx \frac{11 \pm 11}{2} \][/tex]
[tex]\[ x_1 \approx \frac{11 + 11}{2} \][/tex]
[tex]\[ x_1 \approx \frac{22}{2} \][/tex]
[tex]\[ x_1 \approx 11 \][/tex]
[tex]\[ x_2 \approx \frac{11 - 11}{2} \][/tex]
[tex]\[ x_2 \approx \frac{0}{2} \][/tex]
[tex]\[ x_2 \approx 0 \][/tex]
So, we have [tex]\( x_1 = 11 \) and \( x_2 = 0 \).[/tex]
Since [tex]\( q_1 \)[/tex] must be smaller than [tex]\( q_2 \),[/tex] we take [tex]\( x_1 = 11 \) as \( q_1 \) and \( x_2 = 0 \) as \( q_2 \).[/tex]
Therefore,[tex]\( q_1 = 11 \, \text{C} \) and \( q_2 = 0 \, \text{C} \)[/tex]. But we need both charges to be positive, so this solution is not valid.
Let's check the other root [tex]\( x_2 = 0 \):[/tex]
If[tex]\( q_1 = 0 \), then \( q_2 = 11 \)[/tex] which is positive.
Thus, [tex]\( q_1 = 0 \, \text{C} \) and \( q_2 = 11 \, \text{C} \)[/tex] are the valid solutions.