Answer:
When the ball goes to first base it will be 4.23 m high.
Explanation:
Horizontal velocity = 30 cos17.3 = 28.64 m/s
Horizontal displacement = 40.5 m
Time
[tex]t=\frac{40.5}{28.64}=1.41s[/tex]
Time to reach the goal posts 40.5 m away = 1.41 seconds
Vertical velocity = 30 sin17.3 = 8.92 m/s
Time to reach the goal posts 40.5 m away = 1.41 seconds
Acceleration = -9.81m/s²
Substituting in s = ut + 0.5at²
s = 8.92 x 1.41 - 0.5 x 9.81 x 1.41²= 2.83 m
Height of throw = 1.4 m
Height traveled by ball = 2.83 m
Total height = 2.83 + 1.4 = 4.23 m
When the ball goes to first base it will be 4.23 m high.
What is the minimum index of refraction of a clear material if a minimum thickness of 121 nm , when laid on glass, is needed to reduce reflection to nearly zero when light of 675 nm is incident normally upon it? Assume that the film has an index less than that of the glass.
Answer:
[tex]\mu = 1.39[/tex]
Explanation:
Since the reflection is nearly zero intensity
so here we will say that the reflected light must show destructive interference
so here we have
[tex]path \: difference = \frac{\lambda}{2}[/tex]
[tex]2 \mu t = \frac{\lambda}{2}[/tex]
[tex]t = \frac{\lambda}{4\mu}[/tex]
here we have
[tex]\lambda = 675 nm[/tex]
t = 121 nm
now from above equation we have
[tex]\mu = \frac{675 nm}{4(121 nm)}[/tex]
[tex]\mu = 1.39[/tex]
Find the x-value of all points where the function below has any relative extrema. Find the value(s) of any relative extrema. G(x)equalsx cubed minus 3 x squared minus 24 x plus 2
Answer:
[tex]x_{1}=-2,x_{2}=4[/tex]
value of g(x) at these points are as follows
g(-2)=30
g(4)=-78
Explanation:
Given
g(x)=[tex]x^{3}-3x^{2}-24x+2[/tex]
Differentiating with respect to x we get
[tex]g'(x)=3x^{2}-6x-24[/tex]
to obtain point of extrema we equate g'(x) to zero
[tex]g'(x)=3x^{2}-6x-24\\\\\therefore 3x^{2}-6x-24=0\\\\\Rightarrow x^{2}-2x-8=0\\\\x^{2}-4x+2x-8=0\\x(x-4)+2(x-4)=0\\(x+2)(x-4)=0[/tex]
Thus the critical points are obtained as [tex]x_{1}=-2,x_{2}=4[/tex]
The values at these points are as
[tex]g(-2)=(-2)^{3}-3(-2)^{2}-24(-2)+2=30\\\\g(4)=(4)^{3}-3(4)^{2}-24(4)+2=-78[/tex]
In a certain cyclotron a proton moves in a circle of radius 0.740 m. The magnitude of the magnetic field is 0.960 T. (a) What is the oscillator frequency? (b) What is the kinetic energy of the proton?
Answer:
Part a)
[tex]f = 1.46 \times 10^7 Hz[/tex]
Part b)
[tex]KE = 3.87 \times 10^{-12} J[/tex]
Explanation:
Part a)
As we know that radius of circular path of a charge moving in constant magnetic field is given as
[tex]R = \frac{mv}{qB}[/tex]
now we have
[tex]v = \frac{qBR}{m}[/tex]
now the frequency of oscillator is given as
[tex]f = \frac{v}{2\pi R}[/tex]
[tex]f = \frac{qB}{2\pi m}[/tex]
[tex]f = \frac{(1.6 \times 10^{-19})(0.960)}{2\pi(1.67\times 10^{-27})}[/tex]
[tex]f = 1.46 \times 10^7 Hz[/tex]
PART b)
now for kinetic energy of proton we will have
[tex]KE = \frac{1}{2}mv^2[/tex]
[tex]KE = \frac{1}{2}m(\frac{qBR}{m})^2[/tex]
[tex]KE = \frac{q^2B^2R^2}{2m}[/tex]
[tex]KE = \frac{(1.6 \times 10^{-19})^2(0.960)^2(0.740)^2}{2(1.67\times 10^{-27})}[/tex]
[tex]KE = 3.87 \times 10^{-12} J[/tex]
Calculate your weight on the International Space Station (ISS), which orbits at roughly 400 km about the surface of Earth. What is "g at the ISS?
Answer:
517.6 N
8.63 m/s²
Explanation:
M = mass of earth = 5.98 x 10²⁴ kg
m = mass of the person = 60 kg
W = weight of the person
R = radius of earth = 6.4 x 10⁶ m
d = distance of ISS above the surface of earth = 400 km = 4 x 10⁵ m
Weight of person on the ISS is given as
[tex]W = \frac{GMm}{(R+d)^{2}}[/tex]
[tex]W = \frac{(6.67\times 10^{-11})(5.98\times 10^{24})(60)}{((6.4\times 10^{6})+(4\times 10^{5}))^{2}}[/tex]
W = 517.6 N
Acceleration due to gravity is given as
[tex]g = \frac{GM}{(R+d)^{2}}[/tex]
[tex]g = \frac{(6.67\times 10^{-11})(5.98\times 10^{24})}{((6.4\times 10^{6})+(4\times 10^{5}))^{2}}[/tex]
g = 8.63 m/s²
A body is projected downward at an angle of 30° with the horizontal from the top of a building 170 m high. Its initial speed is 40 m/s. 2.49 (c) At what angle with the horizontal will it strike? (c) 60
Answer:
The body will strike at angle 60.46°
Explanation:
Vertical motion of body:
Initial speed, u = 40sin30 = 20m/s
Acceleration, a = 9.81 m/s²
Displacement, s = 170 m
We have equation of motion, v² = u² + 2as
Substituting
v² = 20² + 2 x 9.81 x 170
v = 61.12 m/s
Final vertical speed = 61.12 m/s
Final horizontal speed = initial horizontal speed = 40cos30= 34.64m/s
Final velocity = 34.64 i - 61.12 j m/s
Magnitude
[tex]v=\sqrt{34.64^2+(-61.12)^2}=70.25m/s[/tex]
Direction
[tex]\theta =tan^{-1}\left ( \frac{-61.12}{34.64}\right )=-60.46^0[/tex]
The body will strike at angle 60.46°
. The force exerted by the wind on the sails of a sailboat is 390 N north. The water exerts a force of 180 N east. If the boat (including its crew) has a mass of 270 kg, what are the magnitude and direction of its acceleration?
Answer:
1.59 m/s^2, 65.2°
Explanation:
F1 = 390 N North
F2 = 180 N east
m = 270 kg
Net force is the vector sum of both the forces.
[tex]F = \sqrt{F_{1}^{2}+F_{2}^{2}}[/tex]
[tex]F = \sqrt{390^{2}+180^{2}}[/tex]
F = 429.53 N
Direction of force
tan∅ = F1 / F2 = 390 / 180 = 2.1667
∅ = 65.2°
The direction of acceleration is same as the direction of net force.
The magnitude of acceleration is
a = F / m = 429.53 / 270 = 1.59 m/s^2
Final answer:
The magnitude of the sailboat's acceleration is 1.59 m/s², and the direction is 25 degrees east of north. This is calculated using Newton's second law and vector addition of the orthogonal forces exerted by the wind and water.
Explanation:
To calculate the magnitude and direction of the sailboat's acceleration, we need to use Newton's second law, which states that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. Here, we combine the forces exerted by the wind and the water to obtain the net force. The net force can be calculated using vector addition, where the force due to the wind (390 N north) and the force due to the water (180 N east) are treated as orthogonal vectors.
The net force (Fnet) is the vector sum of the two individual forces. We calculate the net force using the Pythagorean theorem:
Fnet = √Fwind² + Fwater² = √390² + 180² = √152100 + 32400 = √184500 N
So, the magnitude of the net force is approximately 429.53 N. To find the acceleration (a), we use the formula:
a = Fnet / m
Substituting the mass of the sailboat (270 kg) and the net force, we get:
a = 429.53 N / 270 kg = 1.59 m/s²
To determine the direction of the acceleration, we take the arctangent of the ratio of the forces. Since arctan(180/390) equals approximately 25 degrees, this is the angle east of north.
The sailboat's acceleration has a magnitude of 1.59 m/s² and a direction of 25 degrees east of north.
A wave has a frequency of 0.5 kHz and two particles with a phase difference of \pi /3 are 1.5 cm apart. Calculate: the time period of the wave.
Answer:
The time period of the wave is 0.002 sec.
Explanation:
Given that,
Frequency = 0.5 kHz
Phase difference [tex]\phi=\dfrac{\pi}{3}[/tex]
Path difference = 1.5 cm
We need to calculate the time period
Using formula of time period
The frequency is the reciprocal of time period.
[tex]f =\dfrac{1}{T}[/tex]
[tex]T=\dfrac{1}{f}[/tex]
Where, f = frequency
Put the value into the formula
[tex]T=\dfrac{1}{0.5\times10^{3}}[/tex]
[tex]T=0.002\ sec[/tex]
Hence, The time period of the wave is 0.002 sec.
A gas expands from an initial volume of 0.040 m^3 and an initial pressure of 210 kPa to a final volume of 0.065 m^3 while its temperature is kept constant. How much work is done by the system?
Answer:
286
Explanation:
p1v1/T1=p2v2/T2
then subtitude your values
T2=760*0.65*273/210*0.040
T2=135/8.4=16+273=289
The pressure-volume work done by the system during the expansion is 5.25 x 10³J.
What is done by pressure-volume work?The work that is done when a fluid is compressed or expanded is known as pressure-volume work. Pressure-volume work occurs whenever there is a change in volume but the outside pressure stays the same.
Here,
The pressure of the system of gas, P = 210 kPa
Initial volume of the system of gas, V₁ = 0.04 m³
Final volume of the system of gas, V₂ = 0.065 m³
The expression for the pressure-volume work done by a system of gas is given by,
Work done,
W = PΔV
W = P(V₂ - V₁)
Applying the values of P, V₁ and V₂,
W = 210 x 10³(0.065 - 0.04)
W = 210 x 10³x 0.025
W = 5.25 x 10³J
Hence,
The pressure-volume work done by the system during the expansion is 5.25 x 10³J.
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A juggler tosses balls vertically to height H. To what height must they be tossed if they are to spend twice as much time in the air?
Answer:
The ball must be tossed to a height of 4 times the initial height H
Explanation:
We have equation of motion S = ut + 0.5at²
A juggler tosses balls vertically to height H.
That is
H = 0 x t + 0.5 x a x t²
H = 0.5at²
To what height must they be tossed if they are to spend twice as much time in the air.
H' = 0 x 2t + 0.5 x a x (2t)²
H' = 2at² = 4 H
So the ball must be tossed to a height of 4 times the initial height H
skateboarder, starting from rest, rolls down a 13.5 m ramp. When she arrives at the bottom of the ramp her speed is 7.37 m/s. If the ramp is inclined at 29.9 o with respect to the ground, what is the component of her acceleration that is parallel to the ground?
Answer:
1.7 m/s²
Explanation:
d = length of the ramp = 13.5 m
v₀ = initial speed of the skateboarder = 0 m/s
v = final speed of the skateboarder = 7.37 m/s
a = acceleration
Using the equation
v² = v₀² + 2 a d
7.37² = 0² + 2 a (13.5)
a = 2.01 m/s²
θ = angle of the incline relative to ground = 29.9
a' = Component of acceleration parallel to the ground
Component of acceleration parallel to the ground is given as
a' = a Cosθ
a' = 2.01 Cos29.9
a' = 1.7 m/s²
Two point masses a 6 Kg mass and an 18 kg mass are connected by a mass less rod 6 meters long. Calculate the distance of the center of mass from the 18 kg mass. Calculate the moment of inertial about an axis located at the center of mass that is perpendicular to the rod.
The center of mass is given by:
∑mx/∑m
m is the mass of each object
x is the position of each object
We will assign x = 0m to the 18kg mass, therefore x = 6m for the 6kg mass.
∑mx/∑m = (18×0+6×6)/(18+6) = 1.5m
The center of mass is located 1.5m away from the 18kg mass.
The total moment of inertia of the system about the center of mass is given by:
I = ∑mr²
I is the moment of inertia
m is the mass of each object
r is the distance of each object from the center of mass
We know r = 1.5m for the 18kg mass and the rod is 6m long, therefore the 6kg mass must be r = 4.5m from the center of mass.
I = 18(1.5)² + 6(4.5)²
I = 162kg×m²
A thin coil has 16 rectangular turns of wire. When a current of 3 A runs through the coil, there is a total flux of 4 × 10-3 T·m2 enclosed by one turn of the coil (note that , and you can calculate the proportionality constant ). Determine the inductance in henries.
Answer:
The inductance is 0.021 H.
Explanation:
Given that,
Number of turns = 16
Current = 3 A
Total flux [tex]\phi=4\times10^{-3}\ Tm^2[/tex]
We need to calculate the inductance
Using formula of total flux
[tex]N\phi = Li[/tex]
[tex]L=\dfrac{N\phi}{i}[/tex]
Where, i = current
N = number of turns
[tex]\phi[/tex] = flux
Put the value into the formula
[tex]L=\dfrac{16\times4\times10^{-3}}{3}[/tex]
[tex]L=0.021\ H[/tex]
Hence, The inductance is 0.021 H.
A 1.0-kg ball has a velocity of 12 m/s downward just before it strikes the ground and bounces up with a velocity of 12 m/s upward. What is the change in momentum of the ball?
The change in momentum of the ball is 24 kg·m/s.
Explanation:The change in momentum of the ball can be found by subtracting the initial momentum from the final momentum. The initial momentum is given by the formula p1 = m * v1, where m is the mass of the ball and v1 is its initial velocity. In this case, the initial velocity is 12 m/s downward, so the initial momentum is -12 kg·m/s. The final momentum is given by the formula p2 = m * v2, where v2 is the velocity of the ball after the bounce. Since the motion after the bounce is the mirror image of the motion before the bounce, the final velocity is 12 m/s upward. Thus, the final momentum is 12 kg·m/s. Subtracting the initial momentum from the final momentum, we get the change in momentum to be 24 kg·m/s.
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The change in momentum of a 1.0-kg ball that bounces with a velocity of 12 m/s upward after striking the ground with a velocity of 12 m/s downward is 24 kg·m/s. This is calculated using the formula for momentum, p = mv, and finding the difference between initial and final momentum.
Momentum (old{p}old{) is calculated by the formula:
p = mv
Step-by-Step Explanation:
The initial momentum just before striking the ground (p₁) is:
p₁ = mass × velocity
p₁ = 1.0 kg × (-12 m/s)
p₁ = -12 kg·m/s
The final momentum just after bouncing up (p₂) is:
p₂ = mass × velocity
p₂ = 1.0 kg × 12 m/s
p₂ = 12 kg·m/s
The change in momentum (Δp) is:
Δp = p₂ - p₁
Δp = 12 kg·m/s - (-12 kg·m/s)
Δp = 12 kg·m/s + 12 kg·m/s
Δp = 24 kg·m/s
So, the change in momentum of the ball is 24 kg·m/s.
Two circular rods, one steel and the other copper, are both 0.780 m long and 1.50 cm in diameter. Each is subjected to a force with magnitude 4350 N that compresses the rod. What is the difference in the length of the two rods when compressed?
Answer:
The difference in the length of the two rods when compressed is [tex]5.4\times10^{-5}\ m[/tex].
Explanation:
Given that,
Length = 0.780 m
Diameter = 1.50 cm
Force = 4350 N
(a). For steel rod
We know ,
The young modulus for steel rod
[tex]Y=2\times10^{11}[/tex]
Using formula of young modulus
[tex]e_{s}=\dfrac{Fl}{AY}[/tex]
[tex]e_{s}=\dfrac{4350\times0.780}{3.14\times(0.75\times10^{-2})^2\times2\times10^{11}}[/tex]
[tex]e_{s}=9.6\times10^{-5}\ m[/tex]
(b). For copper rod
We know ,
The young modulus for steel rod
[tex]Y=1.1\times10^{11}[/tex]
Using formula of young modulus
[tex]e_{c}=\dfrac{Fl}{AY}[/tex]
[tex]e_{c}=\dfrac{4350\times0.780}{3.14\times(0.75\times10^{-2})^2\times1.1\times10^{11}}[/tex]
[tex]e_{c}=1.5\times10^{-4}\ m[/tex]
The difference in the length of the two rods when compressed is
[tex]difference\ in\ length=e_{c}-e_{s}[/tex]
[tex]difference\ in\ length=1.5\times10^{-4}-9.6\times10^{-5}[/tex]
[tex]difference\ in\ length =5.4\times10^{-5}\ m[/tex]
Hence, The difference in the length of the two rods when compressed is [tex]5.4\times10^{-5}\ m[/tex].
An electron is released from rest in a uniform electric field. The electron accelerates, travelling 5.50 m in 4.00 µs after it is released. What is the magnitude of the electric field in N/C?
Answer:
3.91 N/C
Explanation:
u = 0, s = 5.50 m, t = 4 us = 4 x 10^-6 s
Let a be the acceleration.
Use second equation of motion
s = u t + 1/2 a t^2
5.5 = 0 + 1/2 a (4 x 10^-6)^2
a = 6.875 x 10^11 m/s^2
F = m a
The electrostatic force, Fe = q E
Where E be the strength of electric field.
So, q E = m a
E = m a / q
E = (9.1 x 10^-31 x 6.875 x 10^11) / ( 1.6 x 10^-19)
E = 3.91 N/C
Two ocean liners, each with a mass of 42,000 metric tons, are moving on parallel courses, 93 m apart. What is the magnitude of the acceleration of one of the liners toward the other due to their mutual gravitational attraction? Treat the ships as particles.
Answer:
0.324×10⁻³ m/s²
Explanation:
G=Gravitational constant=6.67408×10⁻⁸ m³/kg s
m₁=m₂=Mass of the ships=42000×10³ kg
r=Distance between the ships=93 m
The ships are considered particles
From Newtons Law of Universal Gravitation
[tex]F=G\frac {m_1m_2}{r^2}\\Here\ m_1=m_2\ hence\ m_1\times m_2=m_1^2\\\Rightarrow F=G\frac {m_1^2}{r^2}\\\Rightarrow F=\frac{6.67408\times 10^{-8}\times (42000\times 10^3)^2}{93^2}\\\Rightarrow F=13612.0674\ N\\[/tex]
[tex]F=ma\\\Rightarrow 13612.0674=42000\times 10^3a\\\Rightarrow a=0.324\times 10^{-3}\ m/s^2[/tex]
∴Acceleration of one of the liners toward the other due to their mutual gravitational attraction is 0.324×10⁻³
Given that a fluid at 260°F has a kinematic viscosity of 145 mm^2/s, determine its kinematic viscosity in SUS at 260°F.
Answer:
kinematic viscosity in SUS is = 671.64 SUS
Explanation:
given data
kinetic viscosity = 145 mm^2/s
we know
1 mm = 0.1 cm
so kinetic viscosity in cm is [tex]\nu =145 (0.1)^{2} =1.45 cm^{2}/s[/tex]
other unit of kinetic viscosity is centistokes
[tex]1 cm^{2}/s = 100 cst[/tex]
so 1.45 cm^2/s will be 145 cst
if the temperature is 260°f , then cst value should be multiplied by 4.632. therefore kinematic viscosity in SUS is = 4.362 *145 = 671.64 SUS
Mario, a hockey player, is skating due south at a speed of 4.79 m/s relative to the ice. A teammate passes the puck to him. The puck has a speed of 9.13 m/s and is moving in a direction of 17.8 ° west of south, relative to the ice. What are (a) the magnitude and (b) direction (relative to due south) of the puck's velocity, as observed by Mario
Mario's velocity vector relative to the ice is
[tex]\vec v_{M/I}=-4.79\,\vec\jmath[/tex]
(note that all velocities mentioned here are given in m/s)
The puck is moving in a direction of 17.8 degrees west of south, or 252.2 degrees counterclockwise relative to east. Its velocity vector relative to the ice is then
[tex]\vec v_{P/I}=9.13(\cos252.2^\circ\,\vec\imath+\sin252.2^\circ\,\vec\jmath)=-2.79\,\vec\imath-8.69\,\vec\jmath[/tex]
The velocity of the puch relative to Mario is
[tex]\vec v_{P/M}=\vec v_{P/I}+\vec v_{I/M}=\vec v_{P/I}-\vec v_{M/I}[/tex]
[tex]\vec v_{P/M}=-2.79\,\vec\imath-3.90\,\vec\jmath[/tex]
Then, relative to Mario,
a. the puck is traveling at a speed of [tex]\boxed{\|\vec v_{P/M}\|=4.80}[/tex], and
b. is moving in a direction [tex]\theta[/tex] such that
[tex]\tan\theta=\dfrac{-3.90}{4.80}\implies\theta=-126^\circ[/tex]
which is about [tex]\boxed{35.6^\circ}[/tex] west of south.
The magnitude of the puck's velocity as observed by Mario is 4.9 m/s and the direction is 35.6 degrees west of south.
Explanation:When addressing this problem, we're dealing with relative velocity in two dimensions. Since we need to find the puck's velocity as seen by Mario, we should use the principle that the relative velocity of the puck with respect to Mario is the vector difference between the velocity of the puck and Mario.
Given that Mario is travelling south at 4.79 m/s and the puck is travelling at an angle of 17.8 degrees west of south at a speed of 9.13 m/s, we can break the puck's velocity into southward and westward components. The southward component is 9.13 m/s × cos(17.8°) = 8.74 m/s and the westward component is 9.13 m/s × sin(17.8°) = 2.84 m/s.
The relative southward velocity of the puck with respect to Mario is 8.74 m/s - 4.79 m/s = 3.95 m/s. The westward component is unchanged since Mario has no westward velocity. So, the puck appears to Mario to be moving southwest at an angle of arctan(2.84 m/s / 3.95 m/s) = 35.6 degrees west of south.
The magnitude is given by the Pythagorean theorem, sqrt((2.84 m/s)² + (3.95 m/s)²) = 4.9 m/s.
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The angular position of a point on the rim of a rotating wheel is given by θ(t) = 4.0t - 3.0t2 + t3, where θ is in radians and t is in seconds. (a) What is θ(0)? What are the angular velocities at (b) t = 2.0 s and (c) t = 4.0 s? (d) What is the average angular acceleration for the time interval that begins at t = 2.0 s and ends at t = 4.0 s? What are the instantaneous angular accelerations at (e) the beginning and (f) the end of this time interval?
Answer:
(a) 0 rad
(b) 4 rad/s
(c) 28 rad/s
(d) 12 rad/s^2
(e) 6 rad/s^2
(f) 18 rad/s^2
Explanation:
[tex]\theta (t)=4t-3t^{2}+t^{3}[/tex] .... (1)
(a) here, we need to find angular displacement when t = 0 s
Put t = 0 in equation (1), we get
[tex]\theta (t=0)=0[/tex]
(b) Angular velocity is defined as the rate of change of angular displacement.
ω = dθ / dt
So, differentiate equation (1) with respect to t.
[tex]\omega =\frac{d\theta }{dt}=4-6t+3t^{2}[/tex] .... (2)
Angular velocity at t = 2 s
Put t = 2 s in equation (2), we get
ω = 4 - 6 x 2 + 3 x 4 = 4 rad/s
(c) Angular velocity at t = 4 s
Put t = 4 s in equation (2), we get
ω = 4 - 6 x 4 + 3 x 16 = 4 - 24 + 48 = 28 rad/s
(d) Average angular acceleration,
[tex]\alpha =\frac{\omega (t=4s)-\omega (t=2s)}{4-2}[/tex]
α = (28 - 4) / 2 = 12 rad/s^2
(e) The rate of change of angular velocity is called angular acceleration.
α = dω / dt
α = - 6 + 6 t
At t = 2 s
α = - 6 + 12 = 6 rad/s^2
(f) At t = 4 s
α = - 6 + 24 = 18 rad/s^2
When an electron enters a magnetic field, it will accelerate up the field. True OR False
Answer:
True
Explanation:
The force on the electron when it enters in a magnetic field is given by
F = q ( v x B)
F = -e x V x B x Sin∅
here, F is the force vector, B be the magnetic field vector and v be the velocity vector.
If the angle between the velocity vector and the magnetic field vector is 0 degree, then force is zero.
When the electrons enters in the magnetic field at any arbitrary angle, it experiences a force and hence it accelerate up.
A 1400 kg wrecking ball hangs from a 20-m-long cable. The ball is pulled back until the cable makes an angle of 23.0 ∘ with the vertical. Part A By how much has the gravitational potential energy of the ball changed?
Answer:
21813.46 J
Explanation:
The Potential energy stored at the time it is hanging freely is mgl.
the potential energy stored at the time when it makes a angle 23 degree from the vertical is mglCos 23
Change in gravitational potential energy
U = mgl = mgl cos 23
U = mgl (1 - cos23)
U = 1400 x 9.8 x 20 ( 1 - Cos 23)
U = 21813.46 J
(a) Neil A. Armstrong was the first person to walk on the moon. The distance between the earth and the moon is 3.85 × 108 m. Find the time it took for his voice to reach the earth via radio waves. (b) Someday a person will walk on Mars, which is 5.60 × 1010 m from the earth at the point of closest approach. Determine the minimum time that will be required for a message from Mars to reach the earth via radio waves.
Answer:
(a):The time what it takes his voice reach the earth via radio waves is t1= 1.28 seconds.
(b): The time what it takes his voice reach from Mars to the earth via radio waves is t2= 186.66 seconds.
Explanation:
d1= 3.85 *10⁸ m
d2= 5.6 *10¹⁰ m
C= 300,000,000 m/s
t1= d1/C
t1= 1.28 s
t2= d2/C
t2= 186.66 s
In softball, the pitcher throws with the arm fully extended (straight at the elbow). In a fast pitch the ball leaves the hand with a speed of 139 km/h. Find the rotational kinetic energy of the pitcher’s arm given its moment of inertia is 0.720 kg m2 and the ball leaves the hand at a distance of 0.600 m from the pivot at the shoulder.
The rotational kinetic energy of the pitcher's arm when throwing a softball at a speed of 139 km/h is calculated using the equation for rotational kinetic energy, and taking moment of inertia and angular velocity into account. The angular velocity is inferred from the linear speed of the ball and the distance it leaves the pitcher's hand from the pivot at the shoulder. The rotational kinetic energy is found to be approximately 1491 Joules.
Explanation:This question pertains to the rotational kinetic energy of the pitcher's arm during the act of throwing a softball. By definition, the kinetic energy associated with rotational motion (rotational kinetic energy) can be given by the equation: K_rot = 0.5 * I * ω² where 'I' is the moment of inertia and 'ω' is the angular velocity.
To calculate the angular velocity 'ω', we can infer it from the linear speed of the ball when it leaves the pitcher's hand, as 'ω = v/r', v = 139 km/h = 38.6 m/s, and r = 0.600 m (distance ball leaves hand from pivot at shoulder). Therefore, 'ω' is approximately 64.4 rad/s.
Substituting 'I' and 'ω' into the equation above, we get K_rot = 0.5 * 0.720 kg*m² * (64.4 rad/s)² = 1491 Joules, which is the rotational kinetic energy of the pitcher's arm.
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A 3.00 kg object is moving in the XY plane, with its x and y coordinates given by x = 5t³ !1 and y = 3t ² + 2, where x and y are in meters and t is in seconds. Find the magnitude of the net force acting on this object at t = 2.00 s.
Answer:
The net force acting on this object is 180.89 N.
Explanation:
Given that,
Mass = 3.00 kg
Coordinate of position of [tex]x= 5t^3+1[/tex]
Coordinate of position of [tex]y=3t^2+2[/tex]
Time = 2.00 s
We need to calculate the acceleration
[tex]a = \dfrac{d^2x}{dt^2}[/tex]
For x coordinates
[tex]x=5t^3+1[/tex]
On differentiate w.r.to t
[tex]\dfrac{dx}{dt}=15t^2+0[/tex]
On differentiate again w.r.to t
[tex]\dfrac{d^2x}{dt^2}=30t[/tex]
The acceleration in x axis at 2 sec
[tex]a = 60i[/tex]
For y coordinates
[tex]y=3t^2+2[/tex]
On differentiate w.r.to t
[tex]\dfrac{dy}{dt}=6t+0[/tex]
On differentiate again w.r.to t
[tex]\dfrac{d^2y}{dt^2}=6[/tex]
The acceleration in y axis at 2 sec
[tex]a = 6j[/tex]
The acceleration is
[tex]a=60i+6j[/tex]
We need to calculate the net force
[tex]F = ma[/tex]
[tex]F = 3.00\times(60i+6j)[/tex]
[tex]F=180i+18j[/tex]
The magnitude of the force
[tex]|F|=\sqrt{(180)^2+(18)^2}[/tex]
[tex]|F|=180.89\ N[/tex]
Hence, The net force acting on this object is 180.89 N.
To find the magnitude of the net force acting on the object at t = 2.00 s, we need to calculate the x and y components of the force separately. Once we have both components, we can use the Pythagorean theorem to find the magnitude of the net force.
Explanation:To find the magnitude of the net force acting on the object at t = 2.00 s, we need to calculate the x and y components of the force separately. The x-component of the force can be found by taking the derivative of the x-coordinate equation with respect to time and multiplying it by the mass of the object. The same process can be applied to find the y-component of the force. Once we have both components, we can use the Pythagorean theorem to find the magnitude of the net force.
Starting with the x-coordinate equation, x = 5t³ - 1, taking its derivative gives dx/dt = 15t². Multiplying the derivative by the mass of the object (3.00 kg) gives the x-component of the force: Fx = 3.00 kg * 15t².
Similarly, for the y-coordinate equation y = 3t² + 2, taking its derivative gives dy/dt = 6t. Multiplying the derivative by the mass of the object (3.00 kg) gives the y-component of the force: Fy = 3.00 kg * 6t.
Using the Pythagorean theorem, the magnitude of the net force Fnet can be calculated as: Fnet = sqrt(Fx² + Fy²). Substituting the values into the equation will give you the magnitude of the net force at t = 2.00 s.
A 1200 W microwave oven transforms 1.8 x10^5 J of energy while reheating some food. Calculate how long the food was in the microwave. Answer in minutes.
Answer:
2.5 min
Explanation:
Hello
by definition the power is the amount of work done per unit of time. in this case, we know the total power and the work developed
[tex]1 Watt= \frac{Joule }{sec}[/tex]
Let
[tex]P=1200 W\\E=1.8 *10^{5}\\X=\frac{1.8 *10^{5} j }{1200 W}\\X= unknown\ time\ (sec) \\\\we\ need\ the\ answer\ in\ minutes,\\ 1\ min=60\ sec\\\\x=150 sec \\150\ sec*(\frac{1 min}{ 60 sec})=2.5 min\\[/tex]
hence , the data is not altered, we did it like this to eliminate the sec units.
Answer 2.5 min
I hope it helps
The food was in the 1200 W microwave for 150 seconds, which is equivalent to 2.5 minutes.
To calculate how long the food was in the microwave, we use the formula for power, which is the rate at which work is done or energy is transferred:
Power (P) = Energy (E) / Time (t)
First, rearrange this formula to solve for time (t):
t = E / P
Plugging in the given values:
So:
t = (1.8 times 10⁵J) / (1200 W)
Calculate this to find the time in seconds, and then convert it to minutes as follows:
t = 150 seconds
Divide by 60 to get minutes:
t = 2.5 minutes
Aircraft sometimes acquire small static charges. Suppose a supersonic jet has a 0.55 μC charge and flies due west at a speed of 685 m/s over the Earth's south magnetic pole, where the 8.00 x 10^-5 T magnetic field points straight up. What is the magnitude of the magnetic force on the plane, in newtons?
Answer:
3.014 x 10⁻⁸ N
Explanation:
q = magnitude of charge on the supersonic jet = 0.55 μC = 0.55 x 10⁻⁶ C
v = speed of the jet = 685 m/s
B = magnitude of magnetic field in the region = 8 x 10⁻⁵ T
θ = angle between the magnetic field and direction of motion = 90
magnitude of the magnetic force is given as
F = q v B Sinθ
F = (0.55 x 10⁻⁶) (685) (8 x 10⁻⁵) Sin90
F = 3.014 x 10⁻⁸ N
A0.350 kg iron horseshoe that is initially at 600°C is dropped into a bucket containing 21.9 kg of water at 21.8°C. What is the final equilibrium temperature (in °C)? Neglect any heat transfer to or from the surroundings. Do not enter units.
Answer: [tex]22.8^0C[/tex]
Explanation:-
[tex]Q_{absorbed}=Q_{released}[/tex]
As we know that,
[tex]Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})[/tex]
[tex]m_1\times c\times (T_{final}-T_1)=-[m_2\times c\times (T_{final}-T_2)][/tex]
where,
[tex]m_1[/tex] = mass of iron horseshoe = 0.35 kg = 350 g (1kg=1000g[/tex]
[tex]m_2[/tex] = mass of water = 21.9 kg = 21900 g
[tex]T_{final}[/tex] = final temperature = ?
[tex]T_1[/tex] = temperature of iron horseshoe = [tex]600^oC[/tex]
[tex]T_2[/tex] = temperature of water = [tex]21.8^oC[/tex]
[tex]c_1[/tex] = specific heat of iron horseshoe = [tex]0.450J/g^0C[/tex]
[tex]c_2[/tex] = specific heat of water = [tex]4.184J/g^0C[/tex]
Now put all the given values in equation (1), we get
[tex]m_1\times c_1\times (T_{final}-T_1)=-[m_2\times c_2\times (T_{final}-T_2)][/tex]
[tex]350\times 0.450\times (T_{final}-600)^0C=-[21900g\times 4.184\times (T_{final}-21.8)][/tex]
[tex]T_{final}=22.8^0C[/tex]
Therefore, the final equilibrium temperature is [tex]22.8^0C[/tex].
A positive point charge Q1 = 2.5 x 10-5 C is fixed at the origin of coordinates, and a negative point charge Q2 = -5.0 x 10-6 C is fixed to the x axis at x = +2.0 m. Find the location of the place(s) along the x axis where the electric field due to these two charges is zero.
Answer:
3.62 m and - 1.4 m
Explanation:
Consider a location towards the positive side of x-axis beyond the location of charge Q₂
x = distance of the location from charge Q₂
d = distance between the two charges = 2 m
For the electric field to be zero at the location
E₁ = Electric field by charge Q₁ at the location = E₂ = Electric field by charge Q₂ at the location
[tex]\frac{kQ_{1}}{(2 + x)^{2}}= \frac{kQ_{2}}{x^{2}}[/tex]
[tex]\frac{2.5\times 10^{-5}}{(2 + x)^{2}}= \frac{5 \times 10^{-6}}{x^{2}}[/tex]
x = 1.62 m
So location is 2 + 1.62 = 3.62 m
Consider a location towards the negative side of x-axis beyond the location of charge Q₁
x = distance of the location from charge Q₁
d = distance between the two charges = 2 m
For the electric field to be zero at the location
E₁ = Electric field by charge Q₁ at the location = E₂ = Electric field by charge Q₂ at the location
[tex]\frac{kQ_{1}}{(x)^{2}}= \frac{kQ_{2}}{ (2 + x)^{2}}[/tex]
[tex]\frac{2.5\times 10^{-5}}{(x)^{2}}= \frac{5 \times 10^{-6}}{(2+x)^{2}}[/tex]
x = - 1.4 m
Electric field is zero due to positive point charge Q1 and negative Q2 along the x axis is at the location of 3.62 meters.
What is electric field?
The electric field is the field, which is surrounded by the electric charged. The electric field is the electric force per unit charge.
Given information-
The charge of the point 1 is [tex]2.5\times10^{-5} \rm C[/tex].
The charge of the point 2 is [tex]-5.0\times10^{-6} \rm C[/tex].
The distance between the point 1 and point 2 is 2 meters away from the x axis.
Let the position of the point 1 is at x.
Thus the electric force on point Q1 is,
[tex]F_1=\dfrac{kQ1}{x^2}[/tex]
As the distance between the point Q1 and point Q2 is 2 meters away from the x axis. Thus the position of it should be at (x+2). The electric force on point 2
[tex]F_2=\dfrac{kQ1}{(x+2)^2}[/tex]
As the force of two is equal and opposite thus,
[tex]\dfrac{kQ1}{x^2}=\dfrac{kQ2}{(x+2)^2}\\\dfrac{2.5\times10^{-5}}{x^2}=\dfrac{-5\times10^{-6}}{(x+2)^2}\\x=1.62[/tex]
Thus the position of point 2 is,
[tex]p_2=2+1.62\\p_2=3.62\rm m[/tex]
Thus, the location of the place along the x axis where the electric field due to these two charges is zero is 3.62 meters.
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What is log2 8 ? [watch out for the different base]
Answer:
[tex]log_28=3[/tex]
Explanation:
To calculate the value of [tex]log_28[/tex]
Also,Some relation for log values:
[tex]loga^m= mloga[/tex] -----------------------------------------------------1
[tex]log_aa= 1[/tex]---------------------------------------------------------------2
Also, [tex]2^3= 8[/tex]
Applying in the question as:
[tex]log_28=log_2(2^3)[/tex]
Using Relation-1 mentioned above:
[tex]log_28=3\times log_2(2)[/tex]
Using Relation-2 mentioned above:
[tex]log_28=3\times 1[/tex]
Thus,
[tex]log_28=3[/tex]
The resistivity of gold is 2.44×10-8 Ω•m at room temperature. A gold wire that is 1.8 mm in diameter and 40 cm long carries a current of 860 mA. What is the electric field in the wire?
To find the electric field in the wire, we first find the cross-sectional area of the wire and then substitute the given values and the calculated area into the formula for electric field. The electric field in the gold wire is approximately 8.24 kN/C.
Explanation:The question asks for the electric field in a gold wire of given dimensions and current. We'll solve this using the formula for electric field (E) in terms of resistivity (ρ),current (I) and area (A), which is given by E = ρI/A.
First, we need to find the cross-sectional area (A) of the wire using the formula for the area of a circle, since the wire is cylindrical. The area of a circle is given by A = π*(d/2)^2, where d is the diameter. Substituting d = 1.8 mm or 1.8 * 10^-3 m, we find A ≈ 2.54 * 10^-6 m^2.
Next, we need to substitute the known values into the formula E = ρI/A. Using the given values ρ = 2.44 *10^-8 Ω•m, I = 860 mA or 860 * 10^-3 A, and the calculated A ≈ 2.54 * 10^-6 m^2, we find E ≈ 8.24 * 10^3 N/C or 8.24 kN/C.
So, the electric field in the wire is approximately 8.24 kN/C.
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The electric field within the gold wire can be calculated utilizing resistivity formula, determining the cross-sectional area of the wire, and then computing the current density. After these intermediate calculations, the ultimate electric field in the wire is found to be approximately 72 mV/m.
Explanation:
The electric field in the gold wire can be calculated by using the formula for
resistivity: ρ=E/J where
ρ (rho) is the resistivity of the material,
E is the electric field, and
J is the current density.
Before we apply this formula, let's calculate the cross-sectional area of the wire (A) using the formula A=πr², where r is the radius of the wire. The diameter is given as 1.8mm, therefore
r=0.9mm = 0.9x10^-3 m.
Substituting this into the formula gives us
A=π(0.9x10^-3)²≈2.54x10^-6 m².
Now we calculate current density: J=I/A. where I is current, given as
860mA=860x10^-3 A.
Substituting values of I and A into the formula gives
J=860x10^-3/2.54x10^-6≈3.39x10^8 A/m².
Finally, the electric field E=ρ/J=2.44x10^-8/3.39x10^8 ≈ 7.2x10^-2 V/m or 72mV/m.
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