ANSWER:
51.96 N
STEP-BY-STEP EXPLANATION:
The horizontal component of the pulling force of Force is FAx since the axis is horizontal, therefore:
[tex]\begin{gathered} F_x=F\cdot\cos _{}\theta \\ \text{ replacing} \\ F_x=60\cdot\cos 30 \\ F_x=51.96\text{ N} \end{gathered}[/tex]The magnitude of the component of the 60 newton is 51.96 N
If an object at rest accelerates by 5 meters/sec/sec, what will be its speed after 15 seconds?
Given:
The initial velocity of the object is
[tex]v_i=\text{ 0 m/s}[/tex]The acceleration is a = 5m/s^2
The time duration is t = 15 s
To find the final velocity of the object.
Explanation:
The final velocity of the object will be
[tex]\begin{gathered} v_f=v_i+at \\ =0+5\times15 \\ =75\text{ m/s} \end{gathered}[/tex]Hello,I was looking for some help on this physics question.
Given:
Height = 1.5 m
Given that a book is dropped from a height of 1.5 meters above the floor. How fast is the book moving when it hits the floor?
To find the speed of the book when it hits the floor, apply the kinematics equation:
[tex]vf^2=v_i^2+2ad[/tex]Now, to find the final velocity, vf, we have:
[tex]vf^2=vi^2+2ad[/tex]Where:
vf is the final velocity
vi is the initial velocity = 0 m/s
a is acceleration due to gravity = 9.8 m/s
d is the height = 1.5 m
Thus, we have:
[tex]\begin{gathered} vf^2=0^2+2(9.8)*1.5 \\ \\ v_f^2=29.4 \\ \\ v_f=\sqrt{29.4} \\ \\ v_f=5.42\text{ m/s} \end{gathered}[/tex]Therefore, the velocity when the book hits the floor is 5.42 m/s
ANSWER:
5.42 m/s
Two charges separated a distance of 1.0 meter exert a 2.0-N force on each other. If the charges are pushed to a separation of 1/3 meter, the force on each charge will be?
ANSWER:
18 N
STEP-BY-STEP EXPLANATION:
Force between two charge particles is given by:
[tex]F=k\frac{q_1\cdot q_2}{d^2}[/tex]We can make the following relationship:
[tex]\begin{gathered} F_1=k\frac{q_1\cdot q_2}{d_1^2} \\ \\ F_2=k\frac{q_1\cdot q_2}{d_2^2} \\ \\ \frac{F_2}{F_1}=\frac{k\frac{q_1\cdot q_2}{d_2^2}}{k\frac{q_1\cdot q_2}{d_1^2}} \\ \\ \frac{F_2}{F_1}=\frac{d_2^2}{d_1^2} \\ \\ \frac{F_{2}}{F_{1}}=\left(\frac{d_1}{d_2}\right)^2 \end{gathered}[/tex]Therefore, we can establish the following:
[tex]\begin{gathered} \frac{F_2}{F_1}=\left(\frac{d_1}{d_2}\right)^2 \\ \\ \text{ We replacing:} \\ \\ \frac{F_2}{2}=\left(\frac{1}{\frac{1}{3}}\right)^2 \\ \\ F_2=2\cdot9 \\ \\ F_2=18\text{ N} \end{gathered}[/tex]Therefore, the force will now be 18 N at each charge.
A student is trying to determine if she can use Newton’s equations to calculate force in each of the situations shown in the table. Use what you know about predicting an object’s motion to choose the correct entry from each drop-down menu to complete the table.
In the first option, we have two masses, the gravitational constant and the distance between the masses, therefore we can calculate the force using Newton's law of universal gravitation.
In the second case, we only have the gravitational constant and no masses, therefore the force cannot be determined.
In the third case, we also don't have the masses, therefore the force cannot be determined.
MS . madero sells tables . She charges $26 per square foot of the table stop and charges a fixed fee of $75 . Which equation could be used to find the area x , in square feet , of the tablestop of a table Ms. madero sells for $244 .
Given:
Charge per square foot of table stop = $26
Fixed charge = $75
Cost of table = $244
Let's write an equation that could be used to find the area x, in square feet of a tablestop.
Use the slope-intercept form:
y = mx + b
Let y represent cost of the table = $244
Let b represent the fixed fee = $75
Let m represent the charge per square foot = $26
Therefore, we have the equation:
244 = 26x + 75
The equation that could be used to find the area x, in square feet of thr tablestop of a table is:
26x + 75 = 244
ANSWER:
26x + 75 = 244
what is the kinetic energy of a 1,719.44kg car that travels at 9.22 m/s?
Given data:
* The mass of the car is 1719.44 kg.
* The velocity of the car is 9.22 m/s.
Solution:
The kinetic energy of the car in terms of mass and velocity is,
[tex]K=\frac{1}{2}mv^2[/tex]where m is the mass of the car, v is the velocity of the car, and K is the kinetic energy,
Substituting the known values,
[tex]\begin{gathered} K=\frac{1}{2}\times1719.44\times(9.22)^2 \\ K=73083.42\text{ J} \\ K=73.08\times10^3\text{ J} \\ K=73.08\text{ kJ} \end{gathered}[/tex]Thus, the kinetic energy of the car is 73083.42 J or 73.08 kJ.
Can I please get this problem asap? I am unsure how to do it.
Given:
Moment of intertia of first disk = 1.0 kg.m²
Spinning counterclockwise at = 20.0 rad/s
Rotational intertia of second disk = 4.0 kg.m²
Spinning clockwise at 15.0 rad/s
Let's find the angular velocity of the two disk system.
To find the angular velocity, apply the formula for angular momentum conserve:
[tex]I_1w_1-I_2w_2=(I_1+I_2)w[/tex]Rewrite the formula for angular velocity, w:
[tex]w=\frac{I_1w_1-I_2w_2}{I_1+I_2}[/tex]Where:
I₁ = 1.0 kg.m²
w₁ = 20.0 rad/s
I₂ = 4.0 kg.m²
w₂ = 15.0 rad/s
Input values into the equation and evaluate:
[tex]\begin{gathered} w=\frac{(1.0\times20.0)-(4.0\times15.0)}{1.0+4.0} \\ \\ w=\frac{20.0-60.0}{5.0} \\ \\ w=\frac{-40.0}{5.0} \\ \\ w=-8.0\text{ rad/s} \end{gathered}[/tex]Therefore, the angular velocity for the two disk system is 8.0 rad/s clockwise.
ANSWER:
8.0 rad/s clockwise.
Finding where a function is increasing, decreasing, or constant given the graph: Interval notation
We are asked to determine the intervals where the function is decreasing. To do that we need to remember that a funtion decreases when the output variable decreases as the input variables increases. Graphically, this can be seen where the graph has a negative slope. In this case, the function decreses in the interval:
[tex]\lbrack-2,1\rbrack[/tex]We use brackets to indicate that the extreme points of the interval are included in the interval.
Question 11Not yetWavelength is determined by dividing a wave's speed by its frequency. If a wave has a frequency of 1 hertz and aspeed of 20 m/s, what is its wavelength?gi**76°F5)9:058/31
We have the next information
v=speed=20m/s
f=frequency=1 hz
we have the next formula
[tex]\lambda=\frac{v}{f}[/tex]we substitute the values
[tex]\lambda=\frac{20}{1}[/tex][tex]\lambda=20m[/tex]The wave length is 20m
A BMX biker leaves the end of a horizontal ledge with a velocity of 27 m/s and lands 7.2 m from the base of the ledge. How high is the ledge?
Given data
*The given velocity of the BMX biker is u = 27 m/s
*The horizontal range from the base of the ledge is R = 7.2 m
The formula for the time taken by the BMX bike is to reach 7.2 m from the base of the ledge is given by the relation as
[tex]\begin{gathered} R=ut \\ t=\frac{R}{u} \end{gathered}[/tex]Substitute the known values in the above expression as
[tex]\begin{gathered} t=\frac{7.2}{27} \\ =0.26\text{ s} \end{gathered}[/tex]The formula for the height of the ledge is given by the equation of motion as
[tex]h=\frac{1}{2}gt^2[/tex]Substitute the known values in the above expression as
[tex]\begin{gathered} h=\frac{1}{2}\times9.8\times(0.26)^2 \\ =0.33\text{ m} \end{gathered}[/tex]Hence, the height of the ledge is h = 0.33 m
A converging lens has a focal length of 0.36 m. If an object is placed at 0.21 m from the lens, where will its image be located?
ANSWER
[tex]0.504\text{ }m\text{ in front of the lens}[/tex]EXPLANATION
We want to determine the position where the image will be formed.
To do this, we apply the formula:
[tex]\frac{1}{f}=\frac{1}{u}+\frac{1}{v}[/tex]where f = focal length
u = distance of the object from the lens
v = distance of the image from the lens
Solving for v, we have the position that the image is formed:
[tex]\begin{gathered} \frac{1}{0.36}=\frac{1}{0.21}+\frac{1}{v} \\ \frac{1}{0.36}=\frac{v+0.21}{0.21v} \\ 0.21v=0.36(v+0.21) \\ 0.21v=0.36v+0.0756 \\ 0.21v-0.36v=0.0756 \\ -0.15v=0.0756 \\ v=\frac{0.0756}{-0.15} \\ v=-0.504\text{ }m \end{gathered}[/tex]Since this distance is negative, we say that the image formed is a virtual image and it is formed 0.504 m in front of the lens.
Which of the following diagrams illustrates diffraction of a sound wave around a wall?A.B.C.D.
ANSWER:
STEP-BY-STEP EXPLANATION:
Sound is a wave that can't go through walls, but it can engulf them, because when the sound comes closer and hits the wall, we would get a doppler effect equal to the sound you hear when the sound source is moving towards you. This means that the sound waves reaching you have a shorter wavelength and a higher frequency.
Therefore, the correct answer is diagram D.
A. O 33.836 kg m/sB. O31.903 kg m/sC. O 12.883 kg m/sD. O 19.2 kg m/sE. O 23.979 kg m/s6. An object of mass 9.5 kg was acted by a force of 14.2 N for3.6 seconds. If its initial speed was 4.0 m/s,calculate its final speed. (1 point)A) 14.189B) 16.584C) 2.015D)3.015E) 9.381
Given:
The mass of the object is m = 9.5 kg
The force acting on it is F = 14.2 N
The time duration is t = 3.6 s
The initial speed is
[tex]v_i=\text{ 4 m/s}[/tex]To calculate the final speed.
Explanation:
An impulse of an object is calculated as
[tex]Impulse\text{ = F}\times t[/tex]It can also be written as
[tex]\begin{gathered} Impulse\text{ = change in momentum} \\ =mv_f-mv_i \end{gathered}[/tex]So, the final velocity can be calculated by equating both the formula
[tex]\begin{gathered} F\times t=mv_f-mv_i \\ mv_f=F\times t+mv_i \\ v_f=\frac{(F\times t)+mv_i}{m} \end{gathered}[/tex]On substituting the values, the final speed will be
[tex]\begin{gathered} v_f=\frac{(14.2\times3.6)+(9.5\times4)}{9.5} \\ =9.381\text{ m/s} \end{gathered}[/tex]Thus, the correct choice is E
The vector A= 90m/s towards north and vector B = 125 m/s towards west. Find the magnitude of the resultant vector using pythagorean theorem
154.03 m/s
Explanation
The Pythagorean theorem states that the sum of the squares on the legs of a right triangle is equal to the square on the hypotenuse (the side opposite the right angle)
[tex]a^2+b^2=c^2[/tex]so, to add the vectors we solve use tht P.T and solve for c
s
Step 1
a) let
[tex]\begin{gathered} a=\text{ Vector A=90 }\frac{m}{s} \\ b=\text{ Vector B=125 }\frac{m}{s} \\ c=\text{ unknown= maginutude of the resultant} \end{gathered}[/tex]b) now, replace in the formula:
[tex]\begin{gathered} a^2+b^2=c^2 \\ 90^2+125^2=c^2 \\ 8100+15625=c^2 \\ 23725=c^2 \\ square\text{ root in both sides} \\ \sqrt{23,725}=\sqrt{c^2} \\ 154.03=c \end{gathered}[/tex]therefore, the answer is
154.03 m/s
I hope this helps you
which of the following is most suitable when handling nuclear fuel?
The best suitable equipment for handling nuclear fuel is lab coat,
If we wanted to convert 594300 into scientific notation how would we shift the decimal point
Answer: 5.943 x 10^5
Explanation:
The given expression is
594300
To convert it to scientific notation, we would place a decimal point between the first and second digits. We would count the number of digits to the right of the decimal point. This number would represent the exponent to which it would be raised.
Looking at the given value, the decimal point is between 5 and 9. Number of digits to the right is 5. Thus, in scientific notation, it becomes
5.943 x 10^5
What is deuterium?A.a hydrogen atom with 1 proton and 2 neuronsB.a hydrogen atom with 1 proton and 1 neutronC.a helium atom with 2 proton 1 neutronD. The stable and product of fusion
Deuterium is an isotop of hydrogen that contains one proton and one neutron. The most common isotops of hydrogen contain one proton and one electron and no neutrons.
I have to find (%deviation from expected value (show calculations) ) things for the relation between Centripetal force and frequency.
We are asked to determine the frequency of the experimental setting. To do that we will use the fact that the rubber stopper is moving in a circular motion, therefore, there is a centripetal force acting on the rubber stopper. The centripetal force is given by:
[tex]F_c=\frac{m_sv^2}{r}[/tex]Where:
[tex]\begin{gathered} m_s=\text{ mass of the stopper} \\ v=\text{ linear velocity } \\ r=\text{ radius} \end{gathered}[/tex]The linear velocity is given by:
[tex]v=\frac{2^\pi r}{T}[/tex]Where:
[tex]T=\text{ period}[/tex]Now, we substitute in the formula for the centripetal force:
[tex]F_c=\frac{m_s}{r}(\frac{2\pi r}{T})^2[/tex]Solving the square:
[tex]F_C=\frac{m_s}{r}\frac{4\pi^2r^2}{T^2}[/tex]Simplifying:
[tex]F_C=\frac{4\pi^2m_sr}{T^2}[/tex]We have that the period and the frequency are related by the following formula:
[tex]f=\frac{1}{T}[/tex]Therefore, we have:
[tex]F_C=4\pi^2m_srf^2[/tex]We can solve for the frequency:
[tex]\frac{F_c}{4\pi^2m_sr}=f^2[/tex]Now, we take the square root to both sides:
[tex]\sqrt{\frac{F_c}{4\pi^2m_sr}}=f^[/tex]Simplifying:
[tex]\frac{1}{2\pi}\sqrt{\frac{F_c}{m_sr}}=f[/tex]Now, the centripetal force is related to the hanging mass by the fact that the tension on both ends must be equal. This means that the centripetal force is equivalent to the weight of the hanging mass:
[tex]F_c=m_hg[/tex]Substituting we get:
[tex]\frac{1}{2\pi}\sqrt{\frac{m_hg}{m_sr}}=f[/tex]This formula will allow us to calculate the frequency for each trial, given that the hanging mass is constant.
For example, let's take the first trial for the 20g mass. Substituting the values:
[tex]\frac{1}{2\pi}\sqrt{\frac{(20g)(9.8\frac{m}{s^2})}{(15.3g)(0.265m)}}=f[/tex]Solving the operation:
[tex]1.11\text{ rev/s}=f[/tex]Since the frequency is the number of revolutions per unit of time if we multiply by the total time we should get the expected number of spins:
[tex]1.11\text{ rev/s}\times10s=11rev[/tex]Now, the percentage of deviation from this value with respect to the experimental value can be determined using the following formula:
[tex]D=\frac{T.Value-E.value}{E.value}\times100[/tex]Where:-
[tex]\begin{gathered} E.value=\text{ experimental value} \\ T.value=\text{ theoretical value} \end{gathered}[/tex]Now, we substitute:
[tex]D=\frac{11-26}{26}\times100[/tex]Solving we get:
[tex]D=-57.69\%[/tex]The same procedure is done for the other trials.
a bicycle wheel with a diameter of 26 inches is rotating 3 revolutions per second. what is the linear speed in feet/sec? use 3.14 for pi and round to the nearest tenth.
Given:
Diameter of the wheel = 26 inches
It rotates at 3 revolutions per second.
Let's find the linear speed of the wheel in feet/sec.
To find the linear speed, let's first find the circumference of the wheel.
Apply the formula:
[tex]c=2\pi r[/tex]Where:
π = 3.14
r is the radius = diameter/2 = 26/2 = 13 inches
Hence, we have:
[tex]\begin{gathered} C=2\times3.14\times13 \\ \\ C=81.64\text{ inches} \end{gathered}[/tex]Now, to find the linear speed, we have:
[tex]Linearspeed=3\text{ }\times81.64\text{ = }244.9\text{ inches/sec}[/tex]Now, convert from inches/sec to feet/sec.
Where:
1 inch/sec = 0.0833333 ft/sec
Thus, we have:
244.9 in/sec = 244.9 x 0.0833333 = 20.4 feet/sec
Therefore, the linear speed in feet/sec is 20.4 feet/sec.
ANSWER:
20.4 feet/sec
A 5.0 n force is applied to a 3.0 kg ball to change its velocity from +9.0m/s to +3.0m/s. The impulse on the ball is ____ n s
Take into account that the impulse is given by:
[tex]I=m\Delta v=m(v-v_o)[/tex]where,
m: mas of the ball = 3.0kg
v: final speed = 3.0m/s
vo: initial speed = 9.0m/s
replace the previous values into the formula for I and simplify:
Hence, the impulse is -18 kg*m/s (it's negative due to the force is opposite to the motion of the ball)
D. O8356.675 JE. 5056.506 J2. A bullet of mass 0.0555 kg is fired from a12 m tall building horizontally with a speedof 600 m/s. Calculate its mechanical energyjust after it is fired. (1 point)A. O 4399.788 JB. O 7576.232 JC. 2900.974 JD. O9996.527 JE. O 11970.179 J4
Given:
The mass of the bullet is m = 0.0555 kg
The height of the building is h = 12 m
The speed of the bullet is v = 600 m/s
To find the mechanical energy just after it fired.
Explanation:
Mechanical energy is the sum of potential energy and kinetic energy.
The potential energy of the bullet can be calculated by the formula
[tex]P.E.\text{ = mgh}[/tex]Here, g = 9.8 m/s^2 is the acceleration due to gravity.
On substituting the values, the potential energy will be
[tex]\begin{gathered} P.E.\text{ = 0.0555}\times9.8\times12 \\ =6.5268\text{ J} \end{gathered}[/tex]The kinetic energy can be calculated as
[tex]\begin{gathered} K.E.\text{ = }\frac{1}{2}mv^2 \\ =\frac{1}{2}\times0.0555\times(600)^2 \\ =\text{ 9990 J} \end{gathered}[/tex]Thus, the mechanical energy will be
[tex]\begin{gathered} E=P.E.+K.E. \\ =6.5268+9990 \\ =9996.5268\text{ J} \\ =9996.527\text{ J} \end{gathered}[/tex]Hence, the correct option is D
A tennis ball with a velocity of 11.3 m/s to the right is thrown perpendicularly at a wall. After striking the wall, the ball rebounds in the opposite direction with a velocity of −9.71 m/s to the left. If the ball is in contact with the wall for 0.011 s, what is the average acceleration of the ball while it is in contact with the wall? Answer in units of m/s^2
The average acceleration of the ball while it is in contact with the wall for 0.011 s is 1910 m / s²
a = ( v - u ) / t
a = Acceleration
v = Final velocity
u = Initial velocity
t = Time
u = 11.3 m / s
v = - 9.71 m / s
t = 0.011 s
a = ( - 9.71 - 11.3 ) / 0.011
a = 21.01 / 0.011
a = 1910 m / s²
Average acceleration can be called as the rate of change in velocity with respect to the change in time. Here change in time is not mentioned because the time starts from 0.
Therefore, the average acceleration of the ball while it is in contact with the wall for 0.011 s is 1910 m / s²
To know more about Average acceleration
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On a hot summer night (32°C), you are listening to a rock group in a stadium 350 m away. A friend of yours is sitting in an air-conditioned house across the country, listening to the broadcast on the radio. If the signal travels 30,000 km up to a satellite that retransmits it, who hears the concert first (speed of signal = 3.00 x 108 m/s)?
In ou case we have
[tex]\text{Temp}=331.4+(0.606\cdot32)=350.8\text{ m/s}[/tex]Then for the case of the friend
[tex]\text{Friend}=300,000,000\text{ m/s}[/tex]Therefore the person who hears the concert first is your friend
ANSWER
Your friend
Europa, a satellite of Jupiter, has an orbital diameter of 1.34 × 109m and a period of 3.55 days. What is the mass of Jupiter?
Given,
The orbital diameter is
[tex]d=1.34\times10^9m[/tex]The radius is
[tex]r=\frac{d}{2}=\frac{1.34\times10^9}{2}=0.67\times10^9m[/tex]The time period is T=3.55 days=3.55x86400 sec.
We know,
[tex]\begin{gathered} T=\frac{2\pi r^{\frac{3}{2}}}{\sqrt[\square]{GM}}^{}^{} \\ \Rightarrow3.55\times86400=\frac{2\pi\text{ (0.67}\times10^9)^{\frac{3}{2}}}{\sqrt[]{6.67\times10^{-11}\times M}} \\ \Rightarrow9.4\times10^{10}=\frac{1.18\times10^{28}}{6.67\times10^{-11}M} \\ \Rightarrow M=1.89\times10^{27}kg \end{gathered}[/tex]The answer is
[tex]1.89\times10^{27}kg[/tex]A 250-g block of ice is removed from the refrigerator at -8.0°C. How much thermal energy does the ice absorb as it warms to room temperature (22°C)? The heat of fusion of water is 3.34x10^5 J/kg. )(C ice=2060 J/(kg • c))( C water 4180 J/(kg • c)
Give;:
Mass of block = 250 g
Tempearture of refrigerator = -8.0°C
heat of fusion of water is 3.34x10^5 J/kg. c
specific heat of water = 4180 J/kg
Let's find the thermal energy the ice absorbs as it warns th room temperature of 22°C).
Apply the formula:
[tex]Q=mc(T_2-T_1)+h_fm_{}[/tex]Where:
c = 4180 J/kg
m = 250 g = 0.2450kg
T2 = 22°C
T1 = -8.0°C
hf = 3.34x10⁵ J/kg. c
Thus, we have:
[tex]\begin{gathered} Q=0.250\ast4180(22-(-8.0))+(3.34\ast10^5\ast0.250) \\ \\ Q=0.250\ast4180(22+8.0))+(3.34\ast10^5\ast0.250) \\ \\ Q=1045(30.0)+(83500) \\ \\ Q=114850\text{ J} \end{gathered}[/tex]Therefore, the thermal energy the ice absorbs is 114850 Joules.
ANSWER:
114850 J
Calculate the rotational kinetic energy of a 9-kg motorcycle wheel if its angular velocity is 137 rad/s and its inner radius is 0.280 m and outer radius 0.330 m.
The rotational kinetic energy = 7909.73 Joules
Explanation:The mass of the motorcycle, m = 9kg
The angular velocity, w = 137 rad/s
Inner radius, r = 0.280 m
Outer radius, R = 0.330 m
The rotational kinetic energy is calculated as:
[tex]KE_{\text{rot}}=\frac{m}{4}(r^2+R^2)w^2[/tex]Substitute m = 9, r = 0.280, R = 0.330, and w = 137 rad/s into the formula above
[tex]\begin{gathered} KE_{\text{rot}}=\frac{9}{4}(0.280^2+0.33^2)137^2 \\ KE_{\text{rot}}=\frac{9}{4}(0.1873)(18769) \\ KE_{\text{rot}}=7909.73\text{ }Joules\text{ } \end{gathered}[/tex]The ACE towing company tows a disabled 1050-kg automobile off the road ata constant speed. If the tow line makes an angle of 10.0° with the vertical asshown, what is the tension in the line supporting the car?
Answer:
10,448.74 N
Explanation:
First, we know that the mass of the car is 1050 kg and the tow line makes an angle of 10.0° with the vertical and we want to know the tension in the line, so
The given
m = 1050 kg
θ = 10.0°
g = 9.8 m/s²
The unknown
T = ?
To write the formula, we need to draw a free body diagram as follows
They are moving at a constant speed, so there is no acceleration and the sum of the forces is equal to 0.
In the vertical direction, we have the following equation
[tex]\begin{gathered} T\cos\theta-mg=0 \\ T\cos\theta=mg \\ T=\frac{mg}{cos\theta} \end{gathered}[/tex]Therefore, the formula is
[tex]T=\frac{mg}{cos\theta}[/tex]Replacing the values, we get
[tex]T=\frac{(1050kg)(9.8\text{ m/s}^2)}{cos(10)}=10448.74\text{ N}[/tex]So, the answer is 10448.74 N
X-rays typically have a wavelength of around 1x10^(-9)m. What is the frequency of the X-ray
The speed v of a wave is related to its wavelength λ and its frequency f as follows:
[tex]v=\lambda f[/tex]X-rays are electromagnetic radiation, whose speed is the same as the speed of light c:
[tex]c\approx300,000,000\frac{m}{s}[/tex]Isolate f from the equation and substitute the values for the speed and the wavelength to find the frequency of x-rays:
[tex]\begin{gathered} f=\frac{v}{\lambda} \\ =\frac{300,000,000\frac{m}{s}}{1\times10^{-9}m^{}} \\ =3\times10^{17}\frac{1}{s} \\ =3\times10^{17}Hz \end{gathered}[/tex]Therefore, the frequency of the X-ray, is:
[tex]3\times10^{17^{}}Hz[/tex]The ohm is another name for which of the following?a.)ampere per volt b.)volt per ampere
Ohm is the unit of resistance.
Resistance = voltage/current
Unit of voltage is volt
unit of current is ampere
Thus,
ohm = volt/ampere
The correct option is
b.)volt per ampere
So I got stuck on this question
The initial speed is 0, then while he falls down the speed increase