Answer:595.32 N
Explanation:
Given
Mechanic shop has a car of weight([tex]F_1[/tex]) 16,500 N
Area of lift([tex]A_1[/tex])=[tex]125 cm^2[/tex]
Area of small piston([tex]A_2[/tex])=[tex]4.51 cm^2[/tex]
According to pascal's law pressure transmit will be same in all direction in a closed container
[tex]P_1=P_2[/tex]
[tex]\frac{F_1}{A_1}=\frac{F_2}{A_2}[/tex]
[tex]F_2=F_1\frac{A_2}{A_1}[/tex]
[tex]F_2=16,500\times \frac{4.51}{125}=595.32 N[/tex]
The quartz crystal used in an electric watch vibrates with a frequency of 32,768 Hz. What is the period of the crystal's motion?
Answer:
Time period, [tex]T=3.05\times 10^{-5}\ s[/tex]
Explanation:
Given that,
The quartz crystal used in an electric watch vibrates with a frequency of 32,768 Hz, f = 32768 Hz
We need to find the period of the crystal's motion. The relationship between the frequency and the time period is given by :
[tex]T=\dfrac{1}{f}[/tex]
T is the time period of the crystal's motion.
Time period is given by :
[tex]T=\dfrac{1}{32768}[/tex]
[tex]T=3.05\times 10^{-5}\ s[/tex]
So, the time period of the crystal's motion is [tex]3.05\times 10^{-5}\ s[/tex]. Hence, this is the required solution.
A well hole having a diameter of 5 cm is to be cut into the earth to a depth of 75 m. Determine the total work (in joules) required to raise the earth material to the surface if the average mass of 1 m is 1830 kg. (Data: g = 9.81 m/s) (Hint: How much work is required to raise a volume of ad/4 x dx from a depth of x feet to the surface?
Answer:
total work is 99.138 kJ
Explanation:
given data
diameter = 5 cm
depth = 75 m
density = 1830 kg/m³
to find out
the total work
solution
we know mass of volume is
volume = [tex]\frac{\pi}{4} d^2 dx[/tex]
volume = [tex]\frac{\pi}{4} d^2 1830 dx[/tex]
so
work required to rise the mass to the height of x m
dw = [tex]\frac{\pi}{4} d^2 1830[/tex] gx dx
so total work is integrate it with 0 to 75
w = [tex]\int\limits^{75}_{0} {\frac{\pi}{4} d^2 1830 gx dx}[/tex]
w = [tex]\frac{\pi}{4}[/tex] × 0.05² × 1830× 9.81× [tex](\frac{x^2}{2})^{75}_0[/tex]
w = 99138.53 J
so total work is 99.138 kJ
If a 1kg English swallow is carrying a 1kg coconut and flies northward from rest to a final velocity of 10m/s in 7.5 seconds, what is its acceleration? A. What force was exerted by the bird during that acceleration? B. How far did the bird travel during that time?
Answer:
[tex]a=1.33 m/s^{2}[/tex]
A. [tex]F=2.67N[/tex]
B. [tex]x=37.50m[/tex]
Explanation:
From the exercise we know that the swallow:
[tex]m_{total} =2kg[/tex]
[tex]v_{o}=0\\v_{f}=10m/s\\t=7.5s[/tex]
To find its acceleration we must calculate:
[tex]a=\frac{v_{f}-v_{o} }{t_{f}-t_{o} }=\frac{(10-0)m/s}{(7.5-0)s}=1.33m/s^{2}[/tex]
A. Now, from Newton's second law we know that force is:
[tex]F=m*a[/tex]
[tex]F=(2kg)*(1.33m/s^{2})=2.67N[/tex]
B. To find the distance which the bird travels we need to find how long does it take
[tex]v_{f}=v_{o}+at[/tex]
Solving for t
[tex]t=\frac{v_{f} }{a}=\frac{10m/s}{1.33m/s^{2} }=7.51s[/tex]
Now, from the equation of position we know that
[tex]x=x_{o}+v_{o}t+\frac{1}{2}at^{2}[/tex]
[tex]x=\frac{1}{2}(1.33m/s^2)(7.51s)^2=37.50m[/tex]
What wavelength photon is required to excite a hydrogen from the n=1 state to the n=3 state? What is the lowest frequency photon that will be observed?
Explanation:
It is given that,
Initial state of electron, [tex]n_i=1[/tex]
Final state of electron, [tex]n_f=3[/tex]
The wavelength of the excited electron is given by :
[tex]\dfrac{1}{\lambda}=R(\dfrac{1}{n_f^2}-\dfrac{1}{n_i^2})[/tex]
Where
R is Rydberg's constant
[tex]\dfrac{1}{\lambda}=1.097\times 10^{7}\ J\times (\dfrac{1}{3^2}-\dfrac{1}{1^2})[/tex]
[tex]\lambda=-1.02\times 10^{-7}\ m[/tex]
or
[tex]\lambda=102\ nm[/tex]
Let f is the frequency of the observed photon. It is given by :
[tex]f=\dfrac{c}{\lambda}[/tex]
[tex]f=\dfrac{3\times 10^8\ m/s}{1.02\times 10^{-7}\ m}[/tex]
[tex]f=2.94\times 10^{15}\ Hz[/tex]
Hence, this is the required solution.
Answer:
1.032 x 10^-7 m, 2.9 x 10^15 Hz
Explanation:
n = 1 to n = 3
Rydberg's constant, R = 1.09 × 10^7 per metre
Use the formula for the wavelength
[tex]\frac{1}{\lambda }=R\left ( \frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}} \right )[/tex]
here, n1 = 1 and n2 = 3
[tex]\frac{1}{\lambda }=1.09\times 10^{7}\left ( \frac{1}{1^{2}}-\frac{1}{3^{2}} \right )[/tex]
[tex]\frac{1}{\lambda }=1.09\times 10^{7}\times \frac{8}{9}[/tex]
λ = 1.032 x 10^-7 m
Let the frequency is f.
Use the relation
v = f x λ
[tex]3 \times 10^8=f\times 1.032 \times 10^{-7}[/tex]
f = 2.9 x 10^15 Hz
Consider a gray squirrel falling out of a tree to the ground. (a) If we ignore air resistance in this case (only for the sake of this problem), determine a squirrel's velocity (in m/s) just before hitting the ground, assuming it fell from a height of 3.0 m
Answer: 7.66 8 m/s
Explanation:
Since the squirrel fell out and it is not thrown, we assume the initial velocity is zero ([tex]V_{o}=0[/tex]). On the other hand, the squirrel only experiences the acceleration due gravity, which is constant and in the downward direction [tex]g=-9.8 m/s^{2}[/tex]. So, the following equation will be useful to find the squirrel's final velocity [tex]V_{f}[/tex]:
[tex]{(V_{f})}^{2}={(V_{o})}^{2}-2gd[/tex]
Where [tex]d=3 m[/tex] is the height from which the squirrel fell
As [tex]V_{o}=0[/tex]:
[tex]{(V_{f})}^{2}=-2gd[/tex]
Then:
[tex]{(V_{f})}^{2}=-2(-9.8 m/s^{2})(3 m)[/tex]
[tex]V_{f}=\sqrt{58.8 m^{2}/s^{2}}[/tex]
Finally:
[tex]V_{f}=7.668 m/s[/tex]
Change a speed of 72.4 miles per hour to its equivalent in meters per second.
Final answer:
To convert 72.4 miles per hour to meters per second, multiply by 1609.34 to convert miles to meters, then divide by 3600 to convert hours to seconds. The result is approximately 32.34 meters per second.
Explanation:
To convert a speed of 72.4 miles per hour to its equivalent in meters per second, you can use the conversion factors that 1 mile is equal to 1609.34 meters and 1 hour is equal to 3600 seconds. Since the conversion from kilometers per hour to meters per second involves dividing by 3.6, and knowing from the reference material that 60 miles per hour is approximately equal to 27.8 meters per second, we can calculate the speed in meters per second as follows:
72.4 mi/h * (1609.34 m / 1 mi) * (1 h / 3600 s) = 72.4 * 1609.34 / 3600 m/s.Doing the math:
72.4 * 1609.34 / 3600 = 32.34 m/s (rounded to two decimal places).Check if the answer is reasonable: Since 72.4 mi/h is a little more than 60 mi/h, and we know that 60 mi/h is approximately 27.8 m/s, it seems reasonable that 72.4 mi/h would be slightly over 27.8 m/s, so our answer of 32.34 m/s seems plausible within the context.
In an experiment to measure the density of iron, the following data was obtained. Book value is 7.86 g/cm3. One of the two sets of data is more precise, while one is more accurate. Which of these two sets has the greater precision and which has the greater accuracy
(a) 7.72, 7.74. 7.73. 7.75, 7.74
(b) 7.86, 7.90, 7.78, 7.93, 7.83
Which is more precise?
Which is more accurate?
Answer:
a) is more precise
b) is more accurate
Explanation:
(a) 7.72, 7.74. 7.73. 7.75, 7.74
For this set the average is: (7.72+7.74+7.73+7.75+7.74)/5=7.736
(b) 7.86, 7.90, 7.78, 7.93, 7.83
For this set the average is: (7.86+7.90+7.78+7.93+7.83)/5=7.86
Accuracy refers to the closeness of a measured set of measures to a known value.In our case that the book value is 7.86g/cm3
We can see that the set (b) is more accurate.
Precision correspond to the closeness of the measurements to each other.
We can see that the set of measures a) is more precise, they are more close one of the each other than the set b)
Answer:
(a) is more precise
(b) is more accurate
Explanation:
When a given set of values are precise, this implies that all the values are close to each other but may not be accurate. But a set of values are accurate when they are compared and it can be observed that they are close to a given measured value. So we can infer that the values in option (a) are precise, while that in (b) are accurate when compared with the book value given.
A cheetah is walking at 5.0m/s when it sees a zebra 35m away. What acceleration would be required for the cheetah to reach 25.0 m/s in that distance?
Answer:
Acceleration, [tex]a=8.57\ m/s^2[/tex]
Explanation:
Given that,
Initial speed of the Cheetah, u = 5 m/s
Distance covered, s = 35 m
Final speed of the cheetah, v = 25 m/s
We need to find the acceleration required for the cheetah to reach its final speed. The formula is as follows :
[tex]a=\dfrac{v^2-u^2}{2s}[/tex]
[tex]a=\dfrac{(25)^2-(5)^2}{2\times 35}[/tex]
[tex]a=8.57\ m/s^2[/tex]
So, the acceleration of the cheetah is [tex]8.57\ m/s^2[/tex]. Hence, this is the required solution.
How does a capacitor store energy? Can a resistor store energy?
Answer:
A capacitor stores energy in the form of an electric field.A resistor can’t store energy.Explanation:
A capacitor stores energy in the form of an electric field formed between the plates of the capacitor. To establish this electric field ( or to move against it) the charges must do work, and this work its stored in the capacitor.
A resistor can’t store energy. There is a loss of energy in a current passing a resistor, but this is thanks to the heat dissipated by the resistor, and can’t be recovered.
Final answer:
A capacitor stores energy in its electric field between its plates, whereas a resistor dissipates energy as heat.
Explanation:
A capacitor stores energy in the electric field between its plates. When a voltage is applied across the capacitor, it charges up and stores electrical potential energy.
The energy stored in a capacitor can be calculated using the formula:
E = ½CV²
Where E is the energy stored in joules, C is the capacitance of the capacitor in farads, and V is the voltage across the capacitor in volts.
On the other hand, a resistor dissipates energy as heat when a current passes through it and cannot store energy like a capacitor.
A medical cyclotron used in the production of medical isotopes accelerates protons to 6.5 MeV. The magnetic field in the cyclotron is 1.2 T What is the diameter of the largest orbit, just before the protons exit the cyclotron?
Answer:
diameter of largest orbit is 0.60 m
Explanation:
given data
isotopes accelerates KE = 6.5 MeV
magnetic field B = 1.2 T
to find out
diameter
solution
first we find velocity from kinetic energy equation
KE = 1/2 × m×v² ........1
6.5 × 1.6 × [tex]10^{-19}[/tex] = 1/2 × 1.672 × [tex]10^{-27}[/tex] ×v²
v = 3.5 × [tex]10^{7}[/tex] m/s
so
radius will be
radius = [tex]\frac{m*v}{B*q}[/tex] ........2
radius = [tex]\frac{1.672*10^{-27}*3.5*10^{7}}{1.2*1.6*10^{-19}}[/tex]
radius = 0.30
so diameter = 2 × 0.30
so diameter of largest orbit is 0.60 m
The diameter of the largest orbit, just before the protons exit the cyclotron is equal to 0.614 meter.
Given the following data:
Potential difference = 6.5 MeVMagnetic field, B = 1.2 TScientific data:
Mass of proton = [tex]1.67 \times 10^{-27}\;kg[/tex]Charge of proton = [tex]1.6 \times 10^{-19}\;C[/tex]To calculate the diameter of the largest orbit, just before the protons exit the cyclotron:
First of all, we would determine the velocity of protons.
According to the law of conservation of energy, the work done in accelerating the proton is equal to the kinetic energy of the proton possesses. Mathematically, this is given by this expression:
[tex]qV_d = \frac{1}{2} MV^2[/tex]
Where:
q is the charge of proton.[tex]V_d[/tex] is the potential difference of proton.M is the mass of proton.V is the velocity of proton.Making V the subject of formula, we have:
[tex]V=\sqrt{\frac{2qV_d}{M} }[/tex]
Substituting the given parameters into the formula, we have;
[tex]V=\sqrt{\frac{2 \times 1.6 \times 10^{-19}\times \;6.5 \times 10^6 }{1.67 \times 10^{-27}} }\\\\V=\sqrt{\frac{2.08 \times 10^{-12} }{1.67 \times 10^{-27}}}\\\\V=\sqrt{1.25 \times 10^{15}}\\\\V=3.53 \times 10^7\;m/s[/tex]
In a magnetic field, diameter is given by this formula:
[tex]Diameter = \frac{2MV}{Bq} \\\\Diameter = \frac{2 \times 1.67 \times 10^{-27}\times 3.53 \times 10^7}{1.2 \times 1.6 \times 10^{-19}} \\\\Diameter = \frac{1.179 \times 10^{-19}}{1.92 \times 10^{-19}}[/tex]
Diameter = 0.614 meter
Read more on magnetic field here: https://brainly.com/question/12757739
A 6 kg object falls 10 m. The object is attached mechanically to a paddle-wheel which rotates as the object falls. The paddle-wheel is immersed in 600 g of water at 15°C. What is the maximum temperature that the water could be increased to? Is this a very efficient way of heating water? The specific heat of water is 4.186 J/(g* ˚C).
Answer:[tex]16.234^{\circ}C[/tex]
Explanation:
Given
Mass of object (m)=6 kg
falling height(h)=10 m
mass of water([tex]m_w[/tex])=600 gm
temperature of water =15
specific heat of water [tex]=4.186 j/g-^{\circ}C[/tex]
Let T be the Final Temperature of water
Here Object Potential Energy is converted into Heat energy which will be absorbed by water
Potential Energy(P.E.)[tex]=mgh=6\times 9.81\times 10=588.6 J[/tex]
Heat supplied[tex]=m_wc(\Delta T)[/tex]
[tex]H.E.=600\times 4.186\times (T-16) [/tex]
[tex]588.6=2511.6\times (T-16)[/tex]
T-16=0.234
[tex]T=16.234^{\circ}C[/tex]
This is not an efficient way of heating water as there is only[tex] 0.234^{\circ}C [/tex]increase in temperature.
what is the necessary condition on a force the result the conservation of angular momentum for a particle affected by that force?
Answer:
The necessary condition is that the force that is applied should not be able to produce any torque about the axis of rotation which is achieved only if the force passes through the axis of rotation.
Explanation:
Mathematically the angular momentum is related to force as
[tex]\overrightarrow{F}\times \overrightarrow{r}=\frac{d\overrightarrow{l}}{dt}[/tex]
where
'r' is arm of the force 'F'
For angular momentum to be conserved the right hand side of the above equation should be zero which can happen only if:
1) F = 0
2) r = 0
3) F and r are parallel.
Thus fulfilling any of the above condition shall conserve the angular momentum of a particle.
The car traveling 56.0 km/h is 25.0 m towards a haystack when the driver brakes. The car hits the haystack 2.11 s later. How fast is the car traveling at impact? Answer in m/s
Answer:
36.13 m/s
Explanation:
initial speed, u = 56 km/h = 15.56 m/s
distance, s = 25 m
t = 2.11 s
Let the speed of the car at the impact is v and the acceleration is a.
Use second equation of motion
[tex]s=ut+\frac{1}{2}at^{2}[/tex]
[tex]25=1.56 \times 2.11+\frac{1}{2}\times a \times 2.11^{2}[/tex]
a = 9.75 m/s^2
Use first equation of motion
v = u + at
v = 15.56 + 9.75 x 2.11
v = 36.13 m/s
Consider flipping N fair coins (assume N is even) (a) 1 point] What is the probability of getting half heads and half tails, in no particular order? 1, simplify your answer, and show the limiting behavior of the (b) [1.5 points] Assume that N probability as N -> co.
Answer with Explanation:
The problem can be simplified as follows
The number of possible outcomes after tossing a coin N times is [tex]2^N[/tex] since for each toss 2 outcomes are possible
Since we need equal heads and equal tails the no of cases amont the [tex]2^n[/tex] cases are
[tex]\binom{N}{N/2}=\frac{N!}{(N-N/2)!(N/2)!}=\frac{n!^2}{(n/2)!^2}[/tex]
Thus the required probability is
[tex]P(E)=\frac{\frac{n!}{(n/2)!^2}}{2^n}=\frac{n!}{(n/2)!^2\cdot 2^n}[/tex]
Part 2)
For the limit as N approaches infinity we have
[tex]P(E')=\lim_{n\rightarrow \infty }\frac{n!}{(n/2)!^2\cdot 2^n}[/tex]
Using Stirling's approximation and solving we get
[tex]\lim_{n\rightarrow \infty }\binom{n}{i}\approx \sqrt{\frac{n}{2\pi i(n-i)}}\times \frac{n^n}{i^i(n-i)^i}\\\\\lim_{n\rightarrow \infty }\binom{n}{n/2}\approx \sqrt{\frac{2n}{\pi n^2}}\times \frac{n^n}{(n/2)^{n/2}\cdot (n/2)^{n/2}}\\\\\lim_{n\rightarrow \infty }\binom{n}{i}\approx \sqrt{\frac{2n}{\pi n^2}}\times 2^n\\\\P(E')=\frac{\sqrt{\frac{2n}{\pi n^2}}\times 2^n}{2^n}=\sqrt{\frac{2}{\pi N}}[/tex]
During a plane showcase, a pilot makes circular "looping" with a speed equal to the sound speed (340 m/s). However, the pilot can fall when the acceleration is more than 8g, with g being the gravitational constant. Find the radius of the smallest circle that the pilot can take. (HINT: Think about the centripetal acceleration).
Answer:
1472.98 m
Explanation:
Data provided:
Speed of circular looping, v = 340 m/s
Acceleration, a = 8g
here,
g is the acceleration due to the gravity = 9.81 m/s²
Now,
the centripetal acceleration is given as,
[tex]a=\frac{v^2}{r}[/tex]
r is the radius of the loop
on substituting the respective values, we get
[tex]8\times9.81=\frac{340^2}{r}[/tex]
or
r = 1472.98 m
Please help asap!!
Assume that the cannon is fired at time t=0 and that the cannonball hits the ground at time tg. What is the y position of the cannonball at the time tg/2?
Answer numerically in units of meters.
Answer:
Explanation:
Let's use the equation Δy = [tex]v(t)-\frac{1}{2}(g)(t)^{2}[/tex]
That would mean -h = [tex]v(\frac{9.8}{2})-\frac{1}{2}(9.80)(\frac{9.8}{2})^{2}.
Since the ball has stopped at t = 9.8 = g, then that would mean that the final velocity v = 0.
-h = [tex](0)(\frac{9.8}{2})-(\frac{1}{2})(9.80)(\frac{9.8}{2})^{2}[/tex]
[tex]h = -(\frac{1}{2})(9.80)(\frac{9.8}{2})^{2}[/tex]
[tex]h = (\frac{1}{2})(9.80)(\frac{9.8}{2})^{2}[/tex]
[tex]h = (\frac{9.8}{2})(\frac{9.8}{2})(\frac{9.8}{2})[/tex]
[tex]h = \frac{9.8^{3}}{2^{3}}[/tex]
[tex]h = \frac{941.192}{8}[/tex]
[tex]h = 117.649[/tex]
The height of the cannonball at [tex]t = \frac{9.8}{2}[/tex] should be 117.649 m
Hope this helps!
Answer:
[tex]h = 67.5\,m[/tex]
Explanation:
The position of the cannonball is given by the following expressions:
Half position
[tex]h = H -\frac{1}{2}\cdot g \cdot \left(\frac{t_{g}}{2} \right)^{2}[/tex]
Final position
[tex]0 = H -\frac{1}{2}\cdot g \cdot t_{g}^{2}[/tex]
The instant when the cannonball hits the ground is:
[tex]t_{g} = \sqrt{\frac{2\cdot H}{g} }[/tex]
Lastly, this result is applied in the other equation, which simplified afterwards:
[tex]h = H -\frac{1}{8}\cdot g \cdot \left(\frac{2\cdot H}{g} \right)[/tex]
[tex]h = H -\frac{1}{4}\cdot H[/tex]
[tex]h = \frac{3}{4}H[/tex]
[tex]h = 67.5\,m[/tex]
A european sports car dealer claims that his car
willaccelerates ar a constant rate from rest to 100 km/hr in8.00 s.
If so, what is the acceleration?
Answer:
Acceleration of the car, [tex]a=3.47\ m/s^2[/tex]
Explanation:
Given that,
Initial speed of the car, u = 0 (at rest)
Final speed of the car, v = 100 km/h = 27.77 m/s
Time, t = 8 s
We need to find the acceleration of the car. Mathematically, it is given by :
[tex]a=\dfrac{v-u}{t}[/tex]
[tex]a=\dfrac{27.77\ m/s}{8\ s}[/tex]
[tex]a=3.47\ m/s^2[/tex]
So, the acceleration of the car is [tex]3.47\ m/s^2[/tex]. Hence, this is the required solution.
Answer:
acceleration is 3.47 m/s²
Explanation:
given data
speed = 100 km/hr = 27.78 m/s
time = 8 s
to find out
acceleration?
solution
we will apply here first equation of motion
v = u +at ..............1
here v is final speed and u is initial speed that is 0 and t is time that is 8 sec
so put all value in equation 1
v = u +at
27.78 = 0 +a(8)
a = [tex]\frac{27.78}{8}[/tex]
a = 3.47 m/s²
so acceleration is 3.47 m/s²
A plane wall of thickness 0.1 mm and thermal conductivity 25 W/m K having uniform volumetric heat generation of 0.3 MW/m3 is insulated on one side, while the other side is exposed to a fluid at 92 oC. The convection heat transfer coefficient between the wall and the fluid is 500 W/m2 K. Determine the maximum temperature in the wall.
Answer:
The maximum temperature is 90.06° C
Explanation:
Given that
t= 0.1 mm
Heat generation
[tex]q_g=0.3\ MW/m^3[/tex]
Heat transfer coefficient
[tex]h=500\ W/m^2K[/tex]
Here one side(left side) of the wall is insulated so the all heat will goes in to right side .
The maximum temperature will at the left side.
Lets take maximum temperature is T
Total heat flux ,q
[tex]q=q_g\times t[/tex]
[tex]q=0.3\times 1000000\times 0.1 \times 10^{-3}\ W/m^2[/tex]
[tex]q=30\ W/m^2[/tex]
So the total thermal resistance per unit area
[tex]R=\dfrac{t}{K}+\dfrac{1}{h}[/tex]
[tex]R=\dfrac{0.1\times 10^{-3}}{25}+\dfrac{1}{500}[/tex]
R=0.002 K/W
We know that
q=ΔT/R
30=(T-90)/0.002
T=90.06° C
The maximum temperature is 90.06° C
Solids in which electrons are able to move easily from one atom to another are: a) conductors b) electrically charged c) capacitors d) insulators
Answer:
Option a) conductors
Explanation:
Conductors are the solids which allows the movement of electrons from atom to atom.
In conductors, electrons are free to move in the conductor's vicinity thus these free electrons contributes to the electrical conductivity of the conductor. A body is said to be electrically charged when the electrons in the atoms of the body are not evenly shared.Capacitors are the elements for the storage of electric charge. Insulators are the materials with no free electrons and have a high energy gap which do not allow the conduction of electricity.Answer:
a) Conductors.
Explanation:
Conductors are solids that have free electrons. Some of the examples of solids that are conductors are Steel, Iron, Silver, Copper etc. How best a metallic solid can conduct depends on the number of conduction electrons present in it.These electrons are called conduction electrons.
Not all solids are conductors. For example, wood is a solid, but it is not a conductor. It is an insulator.
Insulators are those materials that do not have free electrons.
Semiconductors are the materials that behave as conductors under certain conditions
Calculate the gravitational force on a 6.50 x 10^2 kg that is 4.15 x 10^6 m above the surface of the Earth?
Answer:
1.5×10^4 N
Explanation:
mass of earth M= 6×10^(24) kg
mass of object m= 6.5×10^2 kg
distance between them R = 4.15×10^6 m
we know from the newtons law of gravitation
[tex]F=G\frac{Mm}{R^2}[/tex]
putting values in the above equation we get
[tex]F=6.67\times10^{-11}\frac{6\times10^{24}6.5\times10^2}{4.15^2\times10^{12}}[/tex]
F= 1.5×10^4 N
the gravitational force between earth and the object
= 1.5×10^4 N
The gravitational force exerted on a 6.50 x 10^2 kg object that is 4.15 x 10^6 m above the Earth's surface is approximately 3573 N, as calculated using Newton's Universal Law of Gravitation.
Explanation:The question asks us to calculate the gravitational force acting on a 6.50 x 10^2 kg mass that is 4.15 x 10^6 m above the Earth's surface. This calculation can be done using Newton's Universal Law of Gravitation, specifically the formula F = G * (M1*M2) / d². In this formula, F is the force, G is the gravitational constant, M1 and M2 are the masses of the two objects, and d is the distance between their centers.
Given the mass of the Earth as 5.98 x 10^24 kg and the radius of the Earth as 6.371 x 10^6 m, we can calculate the gravitational force. The distance from the center of the Earth to the location of our object is the radius of the Earth plus the height of the object above the Earth, which is 6.371 x 10^6 m + 4.15 x 10^6 m = 1.0521 x 10^7 m. Plugging these values into the equation, we get F = (6.67 x 10^-11 N(m²/kg²) * 6.50 x 10^2 kg * 5.98 x 10^24 kg) / (1.0521 x 10^7 m)² = 3573 N, approximately.
Learn more about Gravitational Force here:https://brainly.com/question/32609171
#SPJ3
A parallel-plate capacitor is formed from two 1.6 cm -diameter electrodes spaced 2.8 mm apart. The electric field strength inside the capacitor is 6.0 × 10^6 N/C . What is the charge (in nC) on each electrode?
Express your answer using two significant figures.
Answer:
12 nC
Explanation:
Capacity of the parallel plate capacitor
C = ε₀ A/d
ε₀ is constant having value of 8.85 x 10⁻¹² , A area of plate , d is distance between plate
Area of plate = π r²
= 3.14 x (0.8x 10⁻²)²
= 2 x 10⁻⁴
C = ( 8.85 X 10⁻¹² x 2 x 10⁻⁴ ) / 2.8 x 10⁻³
= 7.08 x 10⁻¹³
Potential difference between plate = field strength x distance between plate
= 6 x 10⁶ x 2.8 x 10⁻³
= 16.8 x 10³ V
Charge on plate = CV
=7.08 x 10⁻¹³ X 16.8 X 10³
11.9 X 10⁻⁹ C
12 nC .
The charge on each electrode of the parallel-plate capacitor is 534 nC.
Explanation:The charge on each electrode of a parallel-plate capacitor can be calculated using the formula Q = CV, where Q is the charge, C is the capacitance, and V is the potential difference.
First, we need to find the capacitance of the capacitor using the formula C = ε₀A/d, where ε₀ is the permittivity of free space, A is the area of the electrodes, and d is the separation between the electrodes.
Substituting the given values (diameter = 1.6 cm, separation = 2.8 mm) into the formula, we have:
C = (8.85 × 10⁻¹² F/m)(π(0.08 m)²)/0.0028 m = 8.90 × 10⁻¹⁰ F
Finally, we can calculate the charge on each electrode using Q = CV:
Q = (8.90 × 10⁻¹⁰ F)(6.0 × 10⁶ N/C) = 5.34 × 10⁻⁴ C = 534 nC
A 30.0 kg packing crate in a warehouse is pushed to the loading dock by a worker who applies a horizontal force. The coefficient of kinetic friction between the crate and the floor is 0.20. The loading dock is 15.0 m southwest of the initial position of the crate. (a) If the crate is pushed 10.6 m south and then 10.6 m west, what is the total work done on the crate by friction? (b) If the crate is pushed along a straight-line path to the dock, so that it travels 15.0 m southwest, what is the work done on the crate by friction?
The total work done on the crate by friction is -294.6 J for part (a) and -160.1 J for part (b).
Explanation:
The total work done on the crate by friction is determined by calculating the work done on each leg of the trip separately. For part (a), the crate is pushed 10.6 m south and then 10.6 m west. The work done on the crate by friction is the sum of the work done on each leg. The work done on the first leg is -219.1 J and the work done on the second leg is -75.5 J, so the total work done by friction is -294.6 J. For part (b), the crate is pushed along a straight-line path to the dock, traveling 15.0 m southwest. The work done on the crate by friction is -160.1 J.
Learn more about work done by friction here:https://brainly.com/question/30280752
#SPJ3
The total work done by friction on the crate when pushed south and then west (21.2 m total) is 1246.56 J, and when pushed directly southwest (15.0 m) is 882 J.
(a) Crate Pushed 10.6 m South and then 10.6 m West
In this case, the work done will be the sum of work done in pushing the crate 10.6 m south and in pushing the crate 10.6 m west.
Distance south: - 10.6 m = d₁
Distance west: - 10.6 m = d₂
force of friction = f = μ[tex]_k[/tex] N = 0.20 × m g = 0.20 × 30.0 kg × 9.8 m/s² = 58.8 N
∴ work done due south = W₁ = f d₁ = 58.8 N × 10.6 m = 623.28 N
and, work done due west = W₂ = f d₂ = 58.8 N × 10.6 m = 623.28 N
so total work done = W₁ + W₂ = 623.28 N + 623.28 N = 1246.56 N
(b) Crate Pushed Along a Straight-Line Path of 15.0 m Southwest
Let us analyze this problem with the vector approach:
Distance south: - 10.6 m
Distance west: - 10.6 m
The total displacement of the crate can be calculated using the Pythagoras Theorem:
d² = (10.6 m)² + (10.6 m)² = 224.72 m²
d = 14.99 m ≈ 15 m
the angle between the resultant vector and the negative x-axis can be given as :
θ = tan⁻¹ [tex]|\frac{10.6}{10.6} |[/tex] = 45°, the direction of the resultant vector be towards south-west (graphically in the [tex]3^r^d[/tex] quadrant)
The worker is applying a horizontal force, which means that the force of friction (f) will also be horizontal. So,
f = μ[tex]_k[/tex] N = 0.20 × m g = 0.20 × 30.0 kg × 9.8 m/s² = 58.8 N
then the work done by the force of friction will be,
W = f d = 58.8 N × 15 m
W = 882 J
A particular X-ray photon has initial energy of 60 keV when it enters the body. This photon transmits through 30 cm of soft tissue and exits the body. Calculate its energy (in keV) after it has exited the body.
Answer:
E = 7.334 KeV
Explanation:
given,
initial energy = 60 keV
Δ x = 30 cm
[tex]E = E_0e^{-\mu \Delta x}[/tex]
μ(for soft tissue) = 0.7 cm⁻¹ (taken from table)
[tex]E = E_0e^{-0.7\times 20}[/tex]
E = 60 × 0.1224
E = 7.334 KeV
energy of x-ray photon after travelling through 30 cm soft tissue in body is E = 7.334 KeV
A car traveling 85 km/h slows down at a constant 0.47 m/s^2 just by "letting up on the gas." Calculate the distance it travels during the first second. Express your answer to two significant figures and include the appropriate units. Calculate the distance it travels during the fifth second.
Answer:
Explanation:
u = 85 km / h = 23.61 m /s
Acceleration a = - 0.47 m /s²
Distance travelled in n th sec
S_n = u + ( 2t -1 )a/2
S_n is distance travelled in n th seond
Distance traveled in 1 st second
= 23.61 - .5 x .47
= 23.37 m
=23 m
Distance travelled when t = 5 th
= 23.61 - (2x5 -1)/2 x .47
= 23.61 - 4.5 x .47
23.61 -2.115
=21.495
21 m is distance travelled in 5 th second .
You comb your hair and the comb becomes negatively charged. Strictly speaking, how will the mass of your hair change? A. it wil increase B. it will decrease C. it wil not change
Answer:
B. it will decrease
Explanation:
If the comb is negatively charged, it's because electrons went from the hair to the comb. These electrons have a certain mass, that is why the mass of the hair will decrease.
The mass of your hair will not change significantly when combing and the comb becomes negatively charged because the mass of electrons is extremely small. Small pieces of paper will be attracted to a negatively charged comb due to electrostatic forces.
Explanation:When you comb your hair and the comb becomes negatively charged, it does so by gaining excess electrons from your hair. However, the mass of electrons is extremely small, so when considering the mass change of your hair due to the transfer of electrons, it's accurate to say that the mass will not change significantly enough to be noticeable or measurable with standard equipment. Therefore the correct answer is C. the mass will not change.
In the contexts of the related information provided, when you bring a negatively charged comb close to small pieces of paper, they will be attracted to the comb because of electrostatic forces. This is because the paper, being neutral, becomes polarized in the presence of the charged comb, leading to an attraction. Thus, the observed effect would be b. Pieces of paper are attracted to the comb.
A blue car pulls away from a red stop-light just after it has turned green with a constant acceleration of 0.5 m/s 2 . A green car arrives at the position of the stop-light 5 s after the light had turned green. If t = 0 when the light turns green, at what time does the green car catch the blue car if the green car maintains the slowest constant speed necessary to catch up to t
Answer:
The green car will catch the blue one after 10 s of the stop-light turning green.
Explanation:
The equation for the position of objects moving in a straight line is as follows:
x = x0 + v0 · t + 1/2 · a · t²
Where:
x = position of the object at time t
x0 = initial position
t = time
a = acceleration
v0 = initial velocity
For the blue car, this will be its position at time t
x = 0 m + 0 m/s · t + 1/2 · 0.5 m/s² · t²
x = 0.25 m/s² · t²
For the green car, that moves at constant speed, its position will be:
x = 0 m + v · t (where v = velocity)
x = v · t
Now let´s find how much distance the blue car has traveled until the green car arrives at the stop-light:
x = 0.25 m/s² · (5s)² = 6.25 m
When the persecution begins (5 s after the stop-light turns green), the blue car is 6.25 ahead of the green car.
When the green car catches the blue one, its position is the same as the position of the blue car.
Then:
Position of the blue car = position of the green car
6.25 m + 0.25 m/s² · t² = v · t
0.25 m/s² · t² - v · t + 6.25 m = 0
Let´s use the formula for solving quadratic equations:
a = 0.25
b = -v
c = 6.25
(-b ± √(b²-4·a·c))/2·a
Let´s replace with the data:
(v ± √(v² - 4 · 0.25 m/s² · 6.25 m))/2· 6.25 m
(v ± √(v² - 6.25 m²/s²))/12.5 m
The minimum value of v that solves this equation is the value that makes the content inside the root to be null. A lower value than that and the content inside the root will be negative and have no solution (real solution).
Then:
v² - 6.25 m²/s² = 0
v² = 6.25 m²/s²
v = 2.5 m/s
Now the quadratic equation will be as follows:
0.25 m/s² · t² - 2.5 m/s · t + 6.25 m = 0
Solving this equation:
t = 5 s
Remember that this is the time that takes the green car to catch the blue one after the green car arrived at the stop-light (5 s after it turned green). If we count from when the light turns green, the elapsed time will be 10 s.
A car is traveling 20 m/s and slows down at a uniform rate. It stops in 6 seconds. How far has it traveled in this interval?
Answer:
Car travel a distance of 60.06 m in 6 sec
Explanation:
We have given initial velocity v = 20 m/sec
Time = 6 sec
As the car stops finally so final velocity v = 0
From the first equation of motion
v = u+at (as the car velocity is slows down means it is a case of deceleration)
So v = u-at
[tex]0=20-a\times 6[/tex]
[tex]a=3.33m/sec^2[/tex]
Now from second equation of motion [tex]s=ut-\frac{1}{2}at^2=20\times 6-\frac{1}{2}\times 3.33\times 6^2=60.06m[/tex]
A lead ball is dropped into a lake from a diving board 16.0 ft above the water. It hits the water with a certain velocity and then sinks to the bottom with this same constant velocity. If it reaches the bottom 5.30 s after it is dropped, how deep is the lake (in feet)?
Answer:
138.46 ft
Explanation:
When the ball is dropped until the moment it hits the water, the ball moves in a uniform acceleration motion. Therefore, the equation that describes the movement of the ball is:
[tex]X = \frac{1}{2}*g*t^{2} + V_{0} *t + x_0[/tex]
Where X is the distance that the ball has fallen at a time t. [tex]V_0[/tex] is the initial velocity, which is 0 ft/s as the ball was simply dropped. [tex]x_0[/tex] is the initial position, we will say that this value is 0 in the position where the ball was dropped for simplicity, and it increases as the ball is falling. Now, we replace x with 16 feets and solves for t:
[tex]16 ft = \frac{1}{2} * 32.2 \frac{ft}{s^{2}} *t^{2} +0 \frac{ft}{s} * t + 0 ft[/tex]
[tex]t = \sqrt{2* \frac{16 ft}{32.2 \frac{ft}{s^2}}} = 1 s[/tex]
The velocity that the ball will have at the moment the ball that the ball hits the water will be:
[tex]V = V_o+g*t=0\frac{ft}{s}+32.2\frac{ft}{s^2}*1s =32.2\frac{ft}{s}[/tex]
The time that will take the ball to reach the bottom from the top of the lake will be t = 5.3s - 1s = 4.3s. And as the ball will travel with constant velocity equal to 32.2 ft/s^2, the depth of the lake will be:
[tex]d = v*t = 32.2\frac{ft}{s} * 4.3s = 138.46 ft[/tex]
The depth of the lake is approximately 153.6 feet.
Step 1: Understanding the Problem
We have a lead ball that is dropped from a height of 16.0 ft. After being dropped, it hits the water and sinks to the bottom with a constant velocity. We are told that it reaches the bottom of the lake 5.30 seconds after it is dropped. Our goal is to calculate the depth of the lake.
Step 2: Analyzing the Time
The total time from the moment the ball is dropped until it hits the bottom of the lake is 5.30 seconds. This time includes:
The time it takes to fall to the water's surface.The time it takes to sink to the bottom of the lake once it hits the water.Step 3: Calculate the Fall Time
We can first determine the time it takes for the ball to drop 16.0 ft. The formula for the distance fallen under gravity is:
[tex]d = \frac{1}{2} g t^2[/tex]
where:
[tex]d[/tex] is the distance (16.0 ft),[tex]g[/tex] is the acceleration due to gravity (approximately 32 ft/s²), and [tex]t[/tex] is the time in seconds.Rearranging the formula for time:
[tex]t = \sqrt{\frac{2d}{g}} = \sqrt{\frac{2 \times 16.0\text{ ft}}{32 \text{ ft/s}^2}} = \sqrt{1} = 1 \text{ second}[/tex]
Step 4: Calculate Time to Sink
If the total time from the drop to reaching the bottom is 5.30 seconds, the time taken to sink in the water is:
[tex]\text{Time to sink} = 5.30 \text{ s} - 1 \text{ s} = 4.30 \text{ s}[/tex]
Step 5: Calculate Depth of the Lake
Since the ball sinks at a constant velocity, we need to find the velocity of the ball once it is in the water. The velocity of the ball just before entering the water can be found using the formula:
[tex]v = g t = 32 \text{ ft/s}^2 \times 1 \text{ s} = 32 \text{ ft/s}[/tex]
This means the ball will sink at a velocity of 32 ft/s. Now, we can calculate the depth of the lake (d):
[tex]d = \text{velocity} \times \text{time} = 32 \text{ ft/s} \times 4.30 \text{ s} = 137.6 \text{ ft}[/tex]
Step 6: Add the Initial Drop
Finally, to find the total depth of the lake, we need to add the initial drop of 16.0 ft:
[tex]\text{Total depth} = 137.6 \text{ ft} + 16.0 \text{ ft} = 153.6 \text{ ft}[/tex]
A ball is thrown vertically upward with an initial speed of 30.0 m/s at a height of 1.5 m above the ground level. In the absence of air resistance, (a) how high does the ball go, (b) how much time does it take for the ball to reach its maximum height, and (c) what is the total time the ball is in the air before striking the ground?
Answer: Ok, this problems gives the next info:
Initial velocity = 30m/s
initial position = 15 m
So the only force in our problem is the gravitational, ence the acceleration will be:
a(t) = -9.8 [tex]\frac{m}{s^{2} }[/tex] constant.
for the velocity we must integrate the acceleration over time, and add the integration constant, in this case the initial velocity.
we get v(t) = -9.8 [tex]\frac{m}{s^{2} }[/tex]*t + 30m/s
for the position we integrate over time again, this time the integration constant will be the initial position.
x(t) = [tex]\frac{9.8}{2}[/tex] [tex]\frac{m}{s^{2} }[/tex]*[tex]t^{2}[/tex]+ 30m/s*t + 1.5m
and start doing some resolutions.
(a) how high does the ball go
in this problem you need to obtain the time where the ball stops goin up and starts going down, and put that time in the position equation.
For this, we see v(t0) = 0 so t0 = 30/9.8 = 3.06s
then x(3.06) = 50 meters.
(b) how much time does it take for the ball to reach its maximum height.
Well, we already obtained it it, is 3.06 seconds.
(c) what is the total time the ball is in the air before striking the ground?
here you must see when x(t1) = 0, because if the position is zero, then it means that the ball striking the ground.
As the position is a quadratic function of the time, we must use the bashkara equation so t = [tex]\frac{-30 +- \sqrt{30^{2} +4*4.9*1.5} }{-2*4.9}[/tex]
this gives us two times, we only took the positive one, because is the one that makes physical sense.
then t = 6.172 seconds.
A 1.0-μm-diameter oil droplet (density 900 kg/m^3) is negatively charged with the addition of 27 extra electrons. It is released from rest 2.0 mm from a very wide plane of positive charge, after which it accelerates toward the plane and collides with a speed of 3.5 m/s . What is the surface charge density of the plane?
Answer:
surface charge density = 5.91 µC/m²
Explanation:
given data
diameter = 1 μm
radius = 0.5 × [tex]10^{-6}[/tex] m
speed = 3.5 m/s
density 900 kg/m³
distance = 2 mm = 2 × [tex]10^{-3}[/tex] m
to find out
surface charge density of the plane
solution
volume is express as
volume = [tex]\frac{4}{3} \pi r^3[/tex]
volume = [tex]\frac{4}{3} \pi (0.5*10^{-6})^3[/tex]
volume = 5.23 × [tex]10^{-19}[/tex] m³
and
mass = density × volume
mass = 900 ×5.23× [tex]10^{-19}[/tex]
mass = 4.712 × [tex]10^{-16}[/tex] kg
and
from motion of equation
v² -u² = 2×a×s
here v is speed 3.5 and u initial is 0 and a is acceleration and s is distance
3.5² = 2×a×2 × [tex]10^{-3}[/tex]
acceleration = 3062.5 m/s²
so from newton second law
Force = mass × acceleration
force = 4.712 × [tex]10^{-16}[/tex] ×3062.5
force = 1.44305 × [tex]10^{-12}[/tex] N
so
surface charge density = [tex]\frac{2\epsilon F}{q}[/tex]
here q is charge with addition 27 extra electron and f is force and ∈ = 8.85× [tex]10^{-12}[/tex]
surface charge density = [tex]\frac{2\epsilon F}{q}[/tex]
surface charge density = [tex]\frac{2*8.85*10^{-12}*1.443*10^{-12}}{27*1.60*10^{-19}}[/tex]
surface charge density = 5.91 × [tex]10^{-6}[/tex] C/m²
so surface charge density = 5.91 µC/m²