Answer:
Part a)
the two frequencies are in ratio of odd numbers so it must be closed at one end
Part b)
L = 34.3 cm
Explanation:
Part a)
Since the shortest frequency is known as fundamental frequency
It is given as
[tex]f_o = 250 Hz[/tex]
next higher frequency is given as
[tex]f_1 = 750 Hz[/tex]
since the two frequencies here are in ratio of
[tex]\frac{f_1}{f_o} = \frac{750}{250} = 3 : 1[/tex]
since the two frequencies are in ratio of odd numbers so it must be closed at one end
Part b)
For the length of the pipe we can say that fundamental frequency is given as
[tex]f_o = \frac{v}{4L}[/tex]
here we have
[tex]250 = \frac{343}{4(L)}[/tex]
now we will have
[tex]L = \frac{343}{4\times 250}[/tex]
[tex]L = 34.3 cm[/tex]
A 360.0 $g$ block is dropped onto a vertical spring with a spring constant k = 254.0 $N/m$. The block becomes attached to the spring, and the spring compresses 0.26 $m$ before momentarily stopping. While the spring is being compressed, what work is done by the block's weight?
Answer:
8.6 J
Explanation:
Work done by the block = change in energy in the spring
W = ½ kx²
W = ½ (254.0 N/m) (0.26 m)²
W = 8.6 J
Ball A (7kg) and ball B (3kg) hit each other in a perfectly elastic collision. If ball A has an initial velocity of 12 m/s and ball B has an initial velocity of -1 m/s. What are the final velocities of each ball?
Explanation:
Mass of ball A, [tex]m_A=7\ kg[/tex]
Mass of ball B, [tex]m_B=3\ kg[/tex]
Initial velocity of ball A, [tex]u_A=12\ m/s[/tex]
Initial velocity of ball B, [tex]u_B=-1\ m/s[/tex]
We need to find the final velocity of each ball. For a perfectly elastic collision, the coefficient of restitution is equal to 1. It is given by :
[tex]e=\dfrac{v_B-v_A}{u_A-u_B}[/tex]
[tex]v_A\ and\ v_B[/tex] are final velocities of ball A and B
[tex]1=\dfrac{v_B-v_A}{13}[/tex]
[tex]v_B-v_A=13[/tex]...........(1)
Using the conservation of linear momentum as :
[tex]m_Au_A+m_Bu_B=m_Av_A+m_Bu_B[/tex]
[tex]7(12)+3(-1)=7v_A+3u_B[/tex]
[tex]7v_A+3u_B=81[/tex]..............(2)
On solving equation (1) and (2) using calculator we get :
[tex]v_A=4.2\ m/s[/tex]
[tex]v_B=17.2\ m/s[/tex]
So, the final velocities of ball A and B are 4.2 m/s and 17.2 m/s. Hence, this is the required solution.
A ball dropped from the roof of a tall building will fall approximately how far in two seconds? (A) 10 m, (B) 20 m, (C) 30 m, (D) 40 m, (E) 50 m.
Answer:
The ball covered the distance in 2 second is 20 m.
(b) is correct option.
Explanation:
Given that,
Time = 2 sec
Using equation of motion
[tex]s = ut+\dfrac{1}{2}gt^2[/tex]
Where, s = height
u = initial velocity
g = acceleration due to gravity
t = time
Put the value in the equation
[tex]s = 0+\dfrac{1}{2}\times9.8\times2^2[/tex]
[tex]s = 19.6\ approx\ 20\ m[/tex]
Hence, The ball covered the distance in 2 second is 20 m.
A basket of negligible weight hangs from a vertical spring scale of force constant 1500 N/m. Part A If you suddenly put a 3.0-kg brick in the basket, find the maximum distance that the spring will stretch. (Express your answer in units of cm.)
Answer:
3.92 cm
Explanation:
k = spring constant of the spring scale = 1500 N/m
m = mass of the brick = 3 kg
x = stretch caused in the spring
h = height dropped by the brick = x
Using conservation of energy
Spring potential energy gained by the spring scale = Potential energy lost
(0.5) k x² = mgh
(0.5) k x² = mgx
(0.5) k x = mg
(0.5) (1500) x = (3) (9.8)
x = 0.0392 m
x = 3.92 cm
Suppose you take a trip that covers 1650 km and takes 35 hours to make it. What is your average speed in km/h? Please round your answer to the nearest whole number (integer).
Answer:
47.14 Km/h
Explanation:
distance = 1650 km
time = 35 hours
The average speed is defined as the ratio of total distance to the total time taken.
Average speed = total distance / total time
Average speed = 1650 / 35 = 47.14 Km/h
Given that the internal energy of water at 28 bar pressure is 988 kJ kg–1 and that the specific volume of water at this pressure is 0.121 × 10–2 m3 kg–1, calculate the specific enthalpy of water at 55 bar pressure.
Answer:
1184 kJ/kg
Explanation:
Given:
water pressure P= 28 bar
internal energy U= 988 kJ/kg
specific volume of water v= 0.121×10^-2 m^3/kg
Now from steam table at 28 bar pressure we can write
[tex]U= U_{f}= 987.6 kJ/Kg[/tex]
[tex]v_{f}=v=1.210\times 10^{-3}m^{^{3}}/kg[/tex]
therefore at saturated liquid we have specific enthalpy at 55 bar pressure.
that the specific enthalpy h = h at 50 bar +(55-50)/(60-50)*( h at 50 bar - h at 60 bar)
[tex]h= 1154.5 + \frac{5}{10}\times(1213-1154)[/tex]
h= 1184 kJ/kg
A tennis player swings her 1000 g racket with a speed of 12 m/s. she hits a 60 g tennis ball that was approaching her at a speed of 16 m/s. The ball rebounds at 42 m/s.
(a) How fast is her racket moving immediatley after the impact? You can ignore the interaction of the racket with her hand for the brief duration of the collision.
__________ m/s
(b) If the tennis ball and racket are in contatc for 10 ms, what is the average force that the racket exerts on the ball?
________ N
How does this compare to the gravitational force on the ball?
F avg / W ball=________
Answer:
should be the answer B
Explanation:
A wind turbine is initially spinning at a constant angular speed. As the wind's strength gradually increases, the turbine experiences a constant angular acceleration 0.155 rad/s2. After making 2870 revolutions, its angular speed is 127 rad/s. (a) What is the initial angular velocity of the turbine? (b) How much time elapses while the turbine is speeding up?
Answer:
a) Initial angular velocity of the turbine = 102.66 rad/s
b) Time elapsed while the turbine is speeding up = 157 s
Explanation:
a) Considering angular motion of turbine:-
Initial angular velocity, u = ?
Acceleration , a = 0.155 rad/s²
Final angular velocity, v = 127 rad/s
Angular displacement, s = 2π x 2870 = 18032.74 rad
We have equation of motion v² = u² + 2as
Substituting
v² = u² + 2as
127² = u² + 2 x 0.155 x 18032.74
u = 102.66 rad/s
Initial angular velocity of the turbine = 102.66 rad/s
b) We have equation of motion v = u + at
Initial angular velocity, u = 102.66 rad/s
Acceleration , a = 0.155 rad/s²
Final angular velocity, v = 127 rad/s
Substituting
v = u + at
127 = 102.66 + 0.155 x t = 157 s
Time elapsed while the turbine is speeding up = 157 s
A wave on a string is reflected from a fixed end. The reflected wave 1. Is in phase with the original wave at the end. 2. Has a larger speed than the original wave. 3. Is 180◦ out of phase with the original wave at the end. 4. Cannot be transverse. 5. Has a larger amplitude than the original wave.
Answer:
3. Is 180◦ out of phase with the original wave at the end.
Explanation:
Here when wave is reflected by the rigid boundary then due to the rigidly bounded particles at the end or boundary they have tendency not to move and remains fixed at their position.
Due to this fixed position we can say when wave reach at that end the particles will not move and they apply equal and opposite force at the particles of string
Due to this the reflected wave is transferred back into the string in opposite phase with respect to the initial wave
so here correct answer will be
3. Is 180◦ out of phase with the original wave at the end.
The pressure at constant velocity, flowing through a pipe measuring 8 inches in diameter with a flow rate of 2,000 gpmS 1.1 psi 32.2 psi 34.8 psi 28.9 psi
Answer:
The pressure at constant velocity is 1.1 psi.
(1). is correct option.
Explanation:
Given that,
Diameter = 8 inch = 20.32 cm = 0.2032 m
Flow rate =2000 g/m
We need to calculate the velocity
[tex]v = \dfrac{Flow\ rate}{Area}[/tex]
[tex]2000\ g/m=\dfrac{0.00378\times2000}{60}[/tex]
[tex]2000\ g/m=0.1262 m^3/s[/tex]
[tex]v=\dfrac{0.1262}{\pi (0.1016)^2}[/tex]
[tex]v=3.89\ m/s[/tex]
We need to calculate the pressure
Using formula of pressure
[tex]P = \dfrac{1}{2}\rho v^2[/tex]
[tex]P=\dfrac{1}{2}\times1000\times(3.89)^2[/tex]
[tex]P=7566\ Pa[/tex]
[tex]P=1.09\ psi=1.1\ psi[/tex]
Hence, The pressure at constant velocity is 1.1 psi.
13. A diver swims to a depth of 3.2 m in a freshwater lake. What is the increase in the force pushing in on her eardrum, compared to what it was at the lake surface? The area of the eardrum is 0.60 cm².
The increase in force exerted on a diver's eardrum as they descend to a depth of 3.2m in a freshwater lake is due to the increased underwater pressure from the weight of the water above them. By calculating the pressure increase, and then multiplying this by the area of the eardrum, we find the increase in force is approximately 1.88 N.
Explanation:The increase in pressure experienced by a diver as they go deeper underwater is due to the weight of the water above them; this weight results from the water's density, which is approximately 775 times greater than air. Hence, the force exerted increases with the increasing depth. To calculate the pressure increase, we use the formula: Pressure = fluid density * gravity * depth, and then the force on the eardrum is obtained by: Force = Pressure * Area.
In this case, the fluid density of freshwater is 1000 kg/m³, gravity is 9.81 m/s², the depth is 3.2 m, and the eardrum area is 0.60 cm² (or 0.00006 m² when converted to square meters).
So, the Pressure increase due to depth = 1000kg/m³ * 9.81m/s² * 3.2m = 31428 Pascal (or Pa); therefore, increase in force = 31428 Pa * 0.00006m² = 1.88 N, approximately. So, the increase in the force on the eardrum of the diver when she swims down to a depth of 3.2 m in the freshwater lake is around 1.88 Newtons.
Learn more about Underwater Pressure here:https://brainly.com/question/6444848
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Final answer:
The increase in force pushing in on the eardrum at a depth of 3.2 m compared to the surface is 1.8816 N.
Explanation:
The student has asked about the increase in force on a diver's eardrum when diving to a depth of 3.2 m in a freshwater lake. The increase in force is due to the increase in water pressure with depth. To find the increase in the force on the eardrum, we need to calculate the water pressure at the depth of 3.2 m and multiply it by the area of the eardrum.
Water pressure increases by approximately 9.8 kPa per meter of depth (this is due to the weight of the water above). Therefore, at a depth of 3.2 m, the pressure is 3.2 m * 9.8 kPa/m = 31.36 kPa. The area of the eardrum is given as 0.60 cm², which is 0.60 * 10^-4 m² in SI units.
The increase in force F = pressure * area = 31.36 kPa * 0.60 * 10^-4 m². To get the force in newtons, we convert kPa to Pa by multiplying by 1,000, giving F = 31.36 * 1,000 Pa * 0.60 * 10^-4 m² = 1.8816 N.
The increase in force pushing in on the eardrum at a depth of 3.2 m compared to the surface is 1.8816 N.
A solid nonconducting sphere of radius R carries a charge Q distributed uniformly throughout its volume. At a certain distance r1 (r1 < R) from the center of the sphere, the electric field has magnitude E. If the same charge Q were distributed uniformly throughout a sphere of radius 2R, the magnitude of the electric field at the same distance r1 from the center would be equal to:
Answer:
[tex]E' = \frac{E}{8}[/tex]
Explanation:
As we know that that electric field inside the solid non conducting sphere is given as
[tex]\int E.dA = \frac{q_{en}}{\epsilon_0}[/tex]
[tex]\int E.dA = \frac{\frac{Q}{R^3}r_1^3}{\epsilon_0}[/tex]
[tex]E(4\pi r_1^2) = \frac{Qr_1^3}{R^3 \epsilon_0}[/tex]
so electric field is given as
[tex]E = \frac{Qr_1}{4\pi \epsilon_0 R^3}[/tex]
now if another sphere has same charge but twice of radius then the electric field at same position is given as
[tex]E' = \frac{Qr_1}{4\pi \epsilon_0 (2R)^3}[/tex]
so here we have
[tex]E' = \frac{E}{8}[/tex]
Final answer:
The electric field at distance r1 from the center of a uniformly charged sphere with charge Q and radius 2R would be one-eighth of the field when the charge is within a sphere with radius R.
Explanation:
To find the electric field at a distance r1 from the center of a uniformly charged sphere, we can use Gauss's law. Inside a uniformly charged sphere, the electric field is proportional to the distance from the center because only the charge enclosed by a Gaussian surface contributes to the field at a point. The field inside the sphere can be described using the formula E = (Qr) / (4πε₀R³), where Q is the total charge, r is the distance from the center, R is the radius of the sphere, and ε0 is the permittivity of free space.
When the same total charge Q is distributed uniformly throughout a sphere of radius 2R, for a point at a distance r1 inside the sphere, which is still less than 2R, the electric field magnitude would be calculated with the new radius. Since the enclosed charge within the distance r1 would be the same and the volume of the larger sphere is greater, the electric field at r1 would be one-eighth of its original value, due to the volume of the sphere increasing by eight times when the radius is doubled (since the volume is proportional to the cube of the radius). Therefore, the new electric field magnitude at distance r1 would be E/8.
The position of an object is given by the equation x = 4.0t2 - 2.0 t - 4.5where x is in meters and t is in seconds. What is the instantaneous acceleration, velocity and position of the object at t = 3.0 s? At what time t is the velocity zero. At what time t is the position zero. At what time t is the velocity zero.
Answer:
j
Explanation:
x = 4 t^2 - 2 t - 4.5
Position at t = 3 s
x = 4 (3)^2 - 2 (3) - 4.5 = 25.5 m
Velocity at t = 3 s
v = dx / dt = 8 t - 2
v ( t = 3 s) = 8 x 3 - 2 = 22 m/s
Acceleration at t = 3 s
a = dv / dt = 8
a ( t = 3 s ) = 8 m/s^2
When is the velocity = 0
v = 0
8 t - 2 = 0
t = 0.25 second
When is the position = 0
x = 0
4 t^2 - 2 t - 4.5 = 0
[tex]t = \frac{2 \pm \sqrt{4 + 72}}{8}[/tex]
t = 1.4 second
An electric field with a magnitude of 6.0 × 104 N/C is directed parallel to the positive y axis. A particle with a charge q = 4.8 × 10–19 C is moving along the x axis with a speed v = 3.0 × 106 m/s. The force on the charge is approximately:
Answer:
Electric force, [tex]F=2.88\times 10^{-14}\ N[/tex]
Explanation:
It is given that,
Magnitude of electric field, [tex]E=6\times 10^4\ N/C[/tex]
Charge, [tex]q=4.8\times 10^{-19}\ C[/tex]
The electric field is directed parallel to the positive y axis. We need to find the force on the charge particle. It is given by :
[tex]F=q\times E[/tex]
[tex]F=4.8\times 10^{-19}\ C\times 6\times 10^4\ N/C[/tex]
[tex]F=2.88\times 10^{-14}\ N[/tex]
So, the electric force on the charge is [tex]2.88\times 10^{-14}\ N[/tex]. Hence, this is the required solution.
a baseball is hit 3 feet above ground level at 100 feet per second and at an angle of 45 with respect to the ground. (g=32 feet/sec^2) find the maximum height
Final answer:
To find the maximum height, we need to use the initial velocity and the angle of the baseball's trajectory. By breaking down the initial velocity into its horizontal and vertical components, we can then use the equation h = v_y^2 / (2g) to find the maximum height.
Explanation:
To find the maximum height, we first need to break down the initial velocity into its horizontal and vertical components. The initial velocity of the baseball is given as 100 ft/s at an angle of 45° with respect to the ground.
The horizontal component of velocity can be found using the equation: vx = v * cos(θ), where v is the initial velocity and θ is the angle.
The vertical component of velocity can be found using the equation: vy = v * sin(θ).
Once we have the vertical component of velocity, we can use the equation h = vy2 / (2g) to find the maximum height.
Substituting the given values, we have:
h = (100 * sin(45°))2 / (2 * 32)
Calculating this will give us the maximum height of the baseball above its initial position.
A ball is on the end of a rope that is 1.72 m in length. The ball and rope are attached to a pole and the entire apparatus, including the pole, rotates about the pole's symmetry axis. The rope makes an angle of 64.0° with respect to the vertical. What is the tangential speed of the ball
Answer:
Tangential Speed equals 5.57m/s
Explanation:
In the figure shown for equilibrium along y- axis we have
[tex]\sum F_{y}=0[/tex]
Resolving Forces along y axis we have
[tex]Tcos(\theta )=mg............(i)[/tex]
Similarly along x axis
[tex]\sum F_{x}=ma_{x}[/tex]
[tex]Tsin(\theta )=m[tex]\frac{v^{2} }{r}[/tex]............(ii)[/tex]
Dividing ii by i we have
[tex]tan(\theta )=\frac{v^{2}}{rg}[/tex]
In the figure below we have [tex]r=lsin(\theta )[/tex]
Thus solving for v we have
[tex]v=\sqrt{lgsin(\theta) tan(\theta )}[/tex]
Applying values we get
v=5.576m/s
Two point charges q1 and q2 are held in place 4.50 cm apart. Another point charge Q = -1.85 mC, of mass 5.50 g, is initially located 3.00 cm from both of these charges and released from rest. You observe that the initial acceleration of Q is 314 m>s2 upward, parallel to the line connecting the two point charges. Find q1 and q2.
Answer:
[tex]q_1 = 6.22 \times 10^{-11} C[/tex]
[tex]q_2 = -6.22 \times 10^{-11} [/tex]
Explanation:
Force on the charge Q = -1.85 mC is along the line joining the two charges
so here we can say that net force on it is given by
[tex]F = ma[/tex]
[tex]F = (0.00550)(314)[/tex]
[tex]F = 1.727 N[/tex]
Now this is the force due to two charges which are in same magnitude but opposite sign
so this force is given as
[tex]F = 2\frac{kq_1q}{r^2} cos\theta[/tex]
here we know that
[tex]F = 2\frac{(9\times 10^9)(1.85 \times 10^{-3})(q)}{0.03^2}cos\theta[/tex]
here we know that
[tex]cos\theta = \frac{2.25}{3} = 0.75[/tex]
now we have
[tex]1.727 = 2\frac{(9\times 10^9)(1.85 \times 10^{-3})(q)}{0.03^2}(0.75)[/tex]
[tex]1.727 = 2.775 \times 10^{10} q[/tex]
[tex]q = 6.22 \times 10^{-11} C[/tex]
A projectile launcher is loaded by providing an average force of 23 N to compress a spring 12 cm. If the projectile has a mass of 5.7 grams and is shot at an angle of 57 degrees at a height of 1.37 meters above the floor, what is the spring constant of the launcher, muzzle velocity of the ball, time in the air, maximum height and horizontal distance travelled?
Answer:
a) k = 191.67 N\m
b) V = 22 m/s
c) t = 3.83s
d) 17.36m
e) 45.89 m
Explanation:
given:
F = 23 N
x = 12 cm = 0.12 m
mass of projectile, m = 5.7g = 5.7×10⁻³ kg
a) For a spring
F = kx
where,
F = Applied force
k = spring constant
x = change in in spring length
thus, we have
23 = k×0.12
or
k = 23/0.12
⇒ k = 191.67 N\m
b) From the conservation of energy between the start and the point of interest we have
[tex]\frac{1}{2}\times kx^{2}=\frac{1}{2}mV^{2}[/tex]
where,
V = velocity of the projectile at 1.37 m above the floor
[tex]\frac{1}{2}\times 191.67\times 0.12^{2}=\frac{1}{2}5.7\times 10^{-3}V^{2}[/tex]
V = 22 m/s
c) time in air (t)
applying the Newtons's equaton of motion
[tex]h = Vsin\Theta\times t +\frac{1}{2}a_{y}t^{2}[/tex]
substituting the values in the above equation we get
[tex]-1.37 = 22sin57^{\circ} \times t +\frac{1}{2}(-9.8)t^{2}[/tex]
or
[tex]4.9t^{2}-18.45t-1.37 = 0[/tex]
solving the qudratic equation for 't', we get
t = 3.83s
d) For maximum height ([tex]H_{max}[/tex])
we have from the equations of projectile motion
[tex]H_{max} =\frac{V^{2}sin^{2}\theta }{2g}[/tex]
substituting the values in the above equation we get
[tex]H_{max} =\frac{22^{2}sin^{2}57^{\circ} }{2\times 9.8}[/tex]
or
[tex]H_{max} =17.36 m[/tex]
the height with respect to the ground surface will be = 17.36m + 1.37 m 18.73m
e)For horizontal distance traveled (R) we have the formula
[tex]R = Vcos\theta\times t[/tex]
substituting the values in the above equation we get
[tex]R =22\times cos57^{\circ} \times 3.83[/tex]
or
[tex]R =45.89 m[/tex]
The container is fitted with a piston so that the volume can change. When the gas is heated at constant pressure, it expands to a volume of 25 L. What is the temperature of the gas in kelvins?
Answer:
[tex]T_{2}[/tex] =[tex]\frac{25T_{1}}{V_{1}}[/tex]
where
[tex]V_{1}[/tex] is initial volume in liters
[tex]T_{1}[/tex] is initial temperature in kelvins
Explanation:
Let the initial volume be [tex]V_{1}[/tex] and the initial temperature be [tex]P_{1}[/tex]
Now by ideal gas law
[tex]\frac{P_{1}V_{1}}{T_{1}}=nR..............(i)[/tex]
Similarly let
[tex]V_{2}[/tex] be final volume
[tex]T_{2}[/tex] be the final volume
thus by ideal gas law we again have
[tex]\frac{P_{2}V_{2}}{T_{2}}=nR..............(ii)[/tex]
Equating i and ii we get
[tex]\frac{P_{1}V_{1}}{T_{1}}=\frac{P_{2}V_{2}}{T_{2}}[/tex]
For system at constant pressure the above expression reduces to
[tex]\frac{V_{1}}{T_{1}}=\frac{V_{2}}{T_{2}}[/tex]
Solving for [tex]T_{2}[/tex] we get
[tex]T_{2}[/tex] =[tex]\frac{25T_{1}}{V_{1}}[/tex]
where
[tex]V_{1}[/tex] is initial volume in liters
[tex]T_{1}[/tex] is initial temperature in kelvins
A car initially contains just its driver. The combined mass of car and driver is 869 kg. When the gas pedal is pushed down all the way, the car reaches its maximum acceleration of 5.6 m/s2.If the driver then picks up three of his friends, the combined mass is increased by 300 kg. What is the maximum acceleration possible in units m/s2 now if the force provided by the engine remains the same as before? Round to one decimal place.
Answer:
4.2 m/s²
Explanation:
m = combined mass of car and driver = 869 kg
a = maximum acceleration reached by the car with driver in it = 5.6 m/s²
Force provided by the engine is given as
[tex]F_{eng}[/tex] = ma
[tex]F_{eng}[/tex] = (869) (5.6)
[tex]F_{eng}[/tex] = 4866.4 N
M = Combined mass of car, driver and his three friends = 869 + 300 = 1169 kg
a' = maximum acceleration reached with friends in the car = ?
Force provided by the engine remains same and is then given as
[tex]F_{eng}[/tex] = M a'
4866.4 = (1169) a'
a' = 4.2 m/s²
A cyclist is coasting at 12 m/s when she starts down a 450-m-long slope that is 30 m high. The cyclist and her bicycle have a combined mass of 70 kg. A steady 12 N drag force due to air resistance acts on her as she coasts all the way to the bottom. What is her speed at the bottom of the slope?
Answer:
The speed of her at the bottom is 24.035 m/s.
Explanation:
Given that,
Speed of cyclist = 12 m/s
Height = 30 m
Distance d = 450 m
Mass of cyclist and bicycle =70 kg
Drag force = 12 N
We need to calculate the speed at the bottom
Using conservation of energy
K.E+P.E=drag force+K.E+P.E
Potential energy is zero at the bottom.
K.E+P.E=drag force+K.E
[tex]\dfrac{1}{2}mv^2+mgh=Fx+\dfrac{1}{2}mv^{2}[/tex]
[tex]\dfrac{1}{2}\times70\times12^2+70\times9.8\times30=12\times450+\dfrac{1}{2}\times70\timesv^2[/tex]
[tex]25620=5400+35v^2[/tex]
[tex]35v^2=25620-5400[/tex]
[tex]35v^2=20220[/tex]
[tex]v^2=\dfrac{20220}{35}[/tex]
[tex]v=\sqrt{\dfrac{20220}{35}}[/tex]
[tex]v=24.035\ m/s[/tex]
Hence, The speed of her at the bottom is 24.035 m/s.
Answer:
Speed at the bottom of the slope is 24.03 m /sec
Explanation:
We have given speed of the cyclist v = 12 m/sec
She starts down a 450 m long that is 30 m high
Si distance = 450 m and height h = 30 m
Combined mass of cyclist and bicycle m = 70 kg
Drag force = 12 N
We have to find the speed at the bottom
According to conservation theory
KE +PE = work done by drag force + KE + PE
As we know that at the bottom there will be no potential energy
So [tex]\frac{1}{2}\times 70\times 12^2+70\times 9.8\times 30=12\times 450+0+\frac{1}{2}\times 70\times v^2[/tex]
[tex]\frac{1}{2}\times 70\times v^2=20220[/tex]
[tex]v=24.03m/sec[/tex]
Points A [at (2, 3) m] and B [at (5, 7) m] are in a region where the electric field is uniform and given by E = (4i+3j)N/C. What is the potential difference VA - VB?
To calculate the potential difference VA - VB, first determine the displacement from point A to B. This results in a vector, which you dot product with the electric field vector to find the work done. This work done is the potential difference.
Explanation:To compute the potential difference VA - VB, we first need to determine the displacement from point A to point B. This is determined by subtracting the coordinates of point A from point B, so (5 - 2)i + (7 - 3)j = 3i + 4j (meters). The next step involves calculating the work done by the electric field in moving a unit positive charge from A to B, which is obtained by taking the dot product of the displacement and the electric field vectors. Hence, the work done is (3i + 4j) . (4i + 3j) = 24 Joules/Coulomb. Now, since potential difference is defined as the work done per unit charge in moving a positive charge from one point to another, the potential difference VA - VB in this case would just be equal to this work done. So VA - VB = 24 Volts.
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A girl with a mass of 40 kg is swinging from a rope with a length of 3.3 m. What is the frequency of her swinging?
Answer:
Frequency, f = 0.274 Hz
Explanation:
It is given that,
Mass of the girl, m = 40 kg
Length of the rope, L = 3.3 m
We need to find the frequency of her swinging. It is an example of simple harmonic motion whose time period is given by :
[tex]T=2\pi\sqrt{\dfrac{L}{g}}[/tex]
[tex]T=2\pi\sqrt{\dfrac{3.3\ m}{9.8\ m/s^2}}[/tex]
T = 3.64 seconds
The frequency, [tex]f=\dfrac{1}{T}[/tex]
[tex]f=\dfrac{1}{3.64\ s}[/tex]
f = 0.274 Hz
So, the frequency of her swinging is 0.274 Hz. Hence, this is the required solution.
A 4.4 kg mess kit sliding on a frictionless surface explodes into two 2.2 kg parts, one moving at 2.9 m/s, due north, and the other at 6.8 m/s, 35° north of east. What is the original speed of the mess kit?
Answer:
Original speed of the mess kit = 4.43 m/s at 50.67° north of east.
Explanation:
Let north represent positive y axis and east represent positive x axis.
Here momentum is conserved.
Let the initial velocity be v.
Initial momentum = 4.4 x v = 4.4v
Velocity of 2.2 kg moving at 2.9 m/s, due north = 2.9 j m/s
Velocity of 2.2 kg moving at 6.8 m/s, 35° north of east = 6.9 ( cos 35i + sin35 j ) = 5.62 i + 3.96 j m/s
Final momentum = 2.2 x 2.9 j + 2.2 x (5.62 i + 3.96 j) = 12.364 i + 15.092 j kgm/s
We have
Initial momentum = Final momentum
4.4v = 12.364 i + 15.092 j
v =2.81 i + 3.43 j
Magnitude
[tex]v=\sqrt{2.81^2+3.43^2}=4.43m/s[/tex]
Direction
[tex]\theta =tan^{-1}\left ( \frac{3.43}{2.81}\right )=50.67^0[/tex]
50.67° north of east.
Original speed of the mess kit = 4.43 m/s at 50.67° north of east.
For a brass alloy, the stress at which plastic deformation begins is 345 MPa (50,000 psi), and the modulus of elasticity is 103 GPa (15.0 ×106 psi). (a) What is the maximum load that may be applied to a specimen with a cross-sectional area of 130 mm2 (0.2 in2 ) without plastic deformation? (b) If the original specimen length is 76 mm (3.0 in.), what is the maximum length to which it can be stretched without causing plastic deformation?
Answer:
a) P = 44850 N
b) [tex]\delta l =0.254\ mm[/tex]
Explanation:
Given:
Cross-section area of the specimen, A = 130 mm² = 0.00013 m²
stress, σ = 345 MPa = 345 × 10⁶ Pa
Modulus of elasticity, E = 103 GPa = 103 × 10⁹ Pa
Initial length, L = 76 mm = 0.076 m
a) The stress is given as:
[tex]\sigma=\frac{\textup{Load}}{\textup{Area}}[/tex]
on substituting the values, we get
[tex]345\times10^6=\frac{\textup{Load}}{0.00013}[/tex]
or
Load, P = 44850 N
Hence the maximum load that can be applied is 44850 N = 44.85 KN
b)The deformation ([tex]\delta l[/tex]) due to an axial load is given as:
[tex]\delta l =\frac{PL}{AE}[/tex]
on substituting the values, we get
[tex]\delta l =\frac{44850\times0.076}{0.00013\times103\times 10^9}[/tex]
or
[tex]\delta l =0.254\ mm[/tex]
The maximum load that may be applied to the specimen without causing plastic deformation is 44850 N. The maximum length to which the specimen can be stretched without causing plastic deformation is 76.254 mm.
Explanation:To find the maximum load that may be applied to the specimen without causing plastic deformation, we need to calculate the stress.
Stress = Force / Area
Where the force can be calculated by multiplying the stress with the cross-sectional area of the specimen.
Hence, maximum load = Stress × Area = 345 MPa × 130 mm² = 44850 N
To find the maximum length to which the specimen can be stretched without causing plastic deformation, we need to calculate the strain.
Strain = Change in length / Original length
Maximum length = Original length + (Strain × Original length) = 76 mm + (345 MPa / 103 GPa × 76 mm) = 76 mm + 0.254 mm = 76.254 mm
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What is the maximum speed at which a car could safely negotiate a curve on a highway, if the radious of the curve is 250 m and the coefficient of friction is 0.4? Give the answer in meter/sec.
Answer:
31.3 m/s
Explanation:
The relation between the coefficient of friction and the velocity, radius of curve path is given by
μ = v^2 / r g
v^2 = μ r g
v^2 = 0.4 x 250 x 9.8 = 980
v = 31.3 m/s
Practice Exercises Name: : Billy-Joe stands on the Talahatchee Bridge kicking stones into the water below a) If Billy-Joe kicks a stone with a horizontal velocity of 3.50 m/s, and it lands in the water a horizontal distance of 5.40 m from where Billy-Joe is standing what is the height of the bridge? b) If the stone had been kicked harder, how would this affect the time it would take to fall
(a) The height is [tex]h = 11.66\ m[/tex]. (b) If the speed is greater, the time required will be longer.
The height can be computed from the second equation of motion. However, for height to be computed, time is required. Therefore, the required time can be computed from distance and speed.
Given:
Horizontal velocity, [tex]u = 3.5 m/s\\[/tex]
Horizontal distance, [tex]x = 5.4 m/s\\[/tex]
(a)
The time is computed as:
[tex]t = x/u\\t = 5.4/3/5\\t = 1.5428/ s[/tex]
The height is given as:
[tex]h = 1/2gt^2\\h = 1/2 \times9.8\times1.5428^2\\h = 11.66\ m[/tex]
Hence, the height is [tex]h = 11.66\ m[/tex].
(b)
The relation between time and velocity is given as:
[tex]t= x/u\\t \alpha1/u[/tex]
The time and speed are inversely proportional. Therefore, if the velocity is larger then the time will be shorter.
Hence, if the speed is greater, the time required will be longer.
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Organ pipe A, with both ends open, has a fundamental frequency of 310 Hz. The third harmonic of organ pipe B, with one end open, has the same frequency as the second harmonic of pipe A. How long are (a) pipe A and (b) pipe B? (Take the speed of sound to be 343 m/s.)
Answer:
Part a)
55.3 cm
Part b)
41.5 cm
Explanation:
Pipe A is open at both ends so the fundamental frequency of this pipe is given as
[tex]f_o = \frac{V}{2L}[/tex]
here we know that
V = 343 m/s
[tex]f_o = 310 Hz[/tex]
now we have
[tex]310 = \frac{343}{2L}[/tex]
[tex]L = 55.3 cm[/tex]
Now we also know that second harmonic of pipe A and third harmonic of pipe B has same frequency
so we will have
[tex]\frac{2V}{2L_a} = \frac{3V}{4L_b}[/tex]
[tex]L_b = \frac{3}{4}L_a[/tex]
[tex]L_b = \frac{3}{4}(55.3) = 41.5 cm[/tex]
At one instant, an electron (charge = –1.6 x 10–19 C) is moving in the xy plane, the components of its velocity being vx = 5.0 x 105 m/s and vy = 3.0 x 105 m/s. A magnetic field of 0.80 T is in the positive y direction. At that instant, what is the magnitude of the magnetic force on the electron?
A magnetic field of 0.80 T is there, then the magnitude of the magnetic force on the electron will be equal to - 6.4 x 10⁻¹⁴ k.
What is Charge?Charged material exerts a force when it is introduced to an electromagnetic field because of the chemical property of electric charge. You can have a negative or a positive electric charge. Unlike charge is created one another while like charges repel one another. Neutral refers to an item that carries no net charge.
q = - 1.6 x 10⁻¹⁹ c
v(x) = 5 x 10⁵ m/s,
v(y) = 3 x 10⁵ m/s,
B = 0.8 T along Y-axis
The velocity vector is given by
v = 5 x 10⁵ i + 3 x 10⁵ j
B = 0.8 j
Now calculate the force,
F = q (v x B)
F = -1.6 x 10⁻¹⁹ {(5 x 10⁵ i + 3 x 10⁵ j) x (0.8 j)}
F = - 1.6 x 10⁻¹⁹ (5 x 0.8 x 10^5 k)
F = - 6.4 x 10⁻¹⁴ k
The force's amplitude and direction are along the negative Z axis Is 6.4 x 10⁻¹⁴ N.
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The magnitude of the magnetic force on the electron, given its charge, velocity components, and a magnetic field of 0.80 T in the positive y direction, is 8.0 x 10^(-14) N.
What is the magnitude of the magnetic force on the electron?The magnetic force on a charged particle moving in a magnetic field can be calculated using the following formula:
F = q * v * B * sin(θ)
Where:
- F is the magnetic force.
- q is the charge of the particle.
- v is the velocity of the particle.
- B is the magnetic field strength.
- θ is the angle between the velocity vector and the magnetic field vector.
In this case, the charge of the electron, q, is -1.6 x 10^(-19) C, the velocity components are given as vx = 5.0 x 10^5 m/s and vy = 3.0 x 10^5 m/s, and the magnetic field strength is 0.80 T in the positive y direction.
The angle θ between the velocity and the magnetic field is 90 degrees (π/2 radians) because the electron is moving in the xy plane, and the magnetic field is in the positive y direction.
Now, let's calculate the magnitude of the magnetic force:
F = (-1.6 x 10^(-19) C) * (5.0 x 10^5 m/s) * (0.80 T) * sin(π/2)
F = (-8.0 x 10^(-14) N) * 1
F = -8.0 x 10^(-14) N
So, the magnitude of the magnetic force on the electron is 8.0 x 10^(-14) N.
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A star has a mass of 1.48 x 1030 kg and is moving in a circular orbit about the center of its galaxy. The radius of the orbit is 1.9 x 104 light-years (1 light-year = 9.5 x 1015 m), and the angular speed of the star is 1.8 x 10-15 rad/s. (a) Determine the tangential speed of the star. (b) What is the magnitude of the net force that acts on the star to keep it moving around the center of the galaxy?
Answer:
a)
3.25 x 10⁵ m/s
b)
8.7 x 10²⁰ N
Explanation:
(a)
w = angular speed of the star = 1.8 x 10⁻¹⁵ rad/s
r = radius of the orbit = 1.9 x 10⁴ ly = 1.9 x 10⁴ (9.5 x 10¹⁵) m = 18.05 x 10¹⁹ m
tangential speed of the star is given as
v = r w
v = (18.05 x 10¹⁹) (1.8 x 10⁻¹⁵)
v = 32.5 x 10⁴ m/s
v = 3.25 x 10⁵ m/s
b)
m = mass of the star = 1.48 x 10³⁰ kg
Net force on the star to keep it moving is given as
F = m r w²
F = (1.48 x 10³⁰) (18.05 x 10¹⁹) (1.8 x 10⁻¹⁵)²
F = 8.7 x 10²⁰ N