Answer:
F = 580.75 N
Explanation:
The force can be calculated through the following formula:
[tex]F = PA[/tex]
where,
F = Force = ?
P = Pressure = (72 mm Hg)(133.322 Pa/1 mm Hg) = 9599.18 Pa
A = Area = (605 cm²)(1 m²/10⁴ cm²) = 0.0605 m²
Therefore,
[tex]F = (9599.18\ Pa)(0.0605\ m^2)[/tex]
F = 580.75 N
Using a 2.5 meter long lever, we want to lift a body weighing 15 kilograms. If the lever support so that the pitch of the load is 55 cm, determine the force we must apply to be lift that burden? HURRYYYYY PLEASE ITS A TEST
Answer:
F = 668.9N
Explanation:
Using the formula;
m1d1 = m2d2
m1 and m2 are the masses
d1 and d2 are the distances
Substitute;
15 * 2.5 = x * (55/100) [55cm was converted to metres]
37.5 = 0.55x
Swap
0.55x = 37.5
x = 37.5/0.55
x = 68.18kg
Since F = mg
F = 68.18 * 9.81
F = 668.9N
This gives the required force
paano matutugunan o matutulungan ng pamahalaan at ng mga guro yubg mga estudyanteng nakararanas nag stress at anxiety.
Answer: how the government and teachers can address or help students experiencing stress and anxiety.
Explanation:
An old fashioned string of 80 Christmas lights is wired in series. Each bulb has a resistance of 2 Ohms and the entire string is plugged into a 120V outlet. What is the current passing through each of the bulbs?
The sum of the resistance = 2 ohms x 80 lights = 160 ohms.
Current = Total voltage / total resistance:
Current = 120V / 160 ohms
Current = 0.75 Amps
You have two flashlights that have 0.500-A currents flowing through them. One flashlight has a single 1.50-V battery in its circuit, and the second flashlight has three 1.50-V batteries connected in a chain that provides 4.50 V. How much power is the battery in the first flashlight providing
Answer: 0.75 W
Explanation:
Given
Current flowing [tex]I=0.5\ A[/tex]
The first flashlight has a single 1.5 V battery
The second flashlight has three 1.5 V batteries connected to provide 4.5 V
Power is the product of voltage and current.
For the first flashlight
[tex]\Rightarrow P=1.5\times 0.5\\\Rightarrow P=0.75\ W[/tex]
Thus, the first flashlight providing 0.75 W of power.
An electron that has an instantaneous velocity of is moving through the uniform magnetic field (a) Find the force on the electron due to the magnetic field. (b) Repeat your calculation for a proton having the same velocity
Answer:
Explanation:
From the given information:
The missing value are as follows:
The velocity [tex]v ^{\to} = (2.0 \times 10^6 \ m/s) \hat i + (3.0 \times 10^6 \ m.s) \hat j[/tex]
The uniform magnetic field [tex]B^{\to} = (0.030 T)\hat i -(0.15 T) \hat j[/tex]
The force on electron as a result of the Magnetic field is:
[tex]F^\to = q(v^\to \times B^\to)[/tex]
here;
Change of electron [tex]q = -1.6 \times 10^{-19} C[/tex]
Then,
[tex]v^{\to} \times B^{\to} = \left|\begin{array}{ccc}2\times10^6&3\times 10^6&0\\0.03&-0.15&0\end{array}\right|[/tex]
[tex]= \hat i(0-0) -\hat j ( 0-0) +\hat k (-3\times 10^5 -0.9\times 10^5) \\ \\ =-3.9 \times 10^5 \ k[/tex]
∴
[tex]F^\to = q(v^\to \times B^\to)[/tex]
[tex]F^{\to} = -1.6 \times 10^{-19} \times -3.9 \times 10^5 \\ \\ F^{\to} = 6.24 \times 10^{-14 } \ N[/tex]
For the proton with the same velocity:
[tex]q = 1.6 \times 10^{-19 } \ C \\ \\ F = q (v^\to * B^\to) \ \\ \\ F^\to = q(-3.9 \times 10^5) \\ \\ F = (1.6 \times 10^{-19}) (-3.9 \times 10^5) \\ \\ \mathbf{F ^{\to} = -6.24 \times 10^{-14} \ N}[/tex]
A sound wave with a frequency of 700 Hz and a wavelength of 5 m travels through a liquid. How fast does sound travel through the liquid?
A.
140 m/s
B.
0.007 m/s
C.
3500 mHz
D.
3500 m/s
source of sinusoidal electromagnetic waves radiates uniformly in all directions. At a distance of 10.0 m from this source, the amplitude of the electric field is measured to be 3.50 N>C. What is the electric-field amplitude 20.0 cm from the source
Answer:
[tex]175\ \text{N/C}[/tex]
Explanation:
[tex]E_1[/tex] = Initial electric field = 3.5 N/C
[tex]E_2[/tex] = Final electric field
[tex]r_1[/tex] = Initial distance = 10 m
[tex]r_2[/tex] = Final distance = 20 cm
Electric field is given by
[tex]E=\sqrt{\dfrac{2P}{\pi r^2c\varepsilon_0}}[/tex]
So,
[tex]E\propto \dfrac{1}{r}[/tex]
[tex]\dfrac{E_2}{E_1}=\dfrac{r_1}{r_2}\\\Rightarrow E_2=E_1\dfrac{r_1}{r_2}\\\Rightarrow E_2=3.5\dfrac{10}{0.2}\\\Rightarrow E_2=175\ \text{N/C}[/tex]
The electric field amplitude at the required point is [tex]175\ \text{N/C}[/tex].
There are three 20.0 Ω resistors connected in series across a 120 V generator
Answer:
That is equal to R1 + R2. If three or more unequal (or equal) resistors are connected in series then the equivalent resistance is: R1 + R2 + R3 +…, etc. One important point to remember about resistors in series networks to check that your maths is correct.
In the experiment, a meter is hooked up to a speaker to monitor the amplitude of the received sound. Suppose the background signal level is 13 mV, the signal is 91 mV with no attenuator and is 25 mV with an attenuator in place. Calculate pt/pi including the background correction.
Answer:
The answer is "[tex]15.38\%[/tex]"
Explanation:
Background[tex]= 13 \ mv\\\\[/tex]
corrected signal[tex]= 91 \ mv-13\ mv= 78\ mv\\\\[/tex]
with attenuator[tex]=25\ mv-13\ mv= 12\ mv\\\\[/tex]
[tex]\to \frac{p_t}{p_i}=\frac{12}{78}\times 100= 15.38\%[/tex]
A 0.413 kg block requires 1.09 N
of force to overcome static
friction. What is the coefficient
of static friction?
(No unit)
PLEASE HELP!
Answer:
static friction=0.126
g Monochromatic light with wavelength 633 nn passes through a narrow slit and a patternappears on a screen 6.0 m away. The distance on the screen between the centers of thefirst minima on either side of the screen is 32 mm. How wide (in mm) is the slit
Answer:
d = 1.19 x 10⁻⁴ m = 0.119 mm
Explanation:
This problem can be solved by using Young's double-slit experiment formula:
[tex]Y = \frac{\lambda L}{d}[/tex]
where,
Y = fringe spacing = 32 mm = 0.032 m
L = slit to screen distance = 6 m
λ = wavelength of light = 633 nm = 6.33 x 10⁻⁷ m
d = slit width = ?
Therefore,
[tex]0.032\ m = \frac{(6.33\ x\ 10^{-7}\ m)(6\ m)}{d}\\\\d = \frac{(6.33\ x\ 10^{-7}\ m)(6\ m)}{0.032\ m}[/tex]
d = 1.19 x 10⁻⁴ m = 0.119 mm
24. A anvil with a mass of 60 kg falls from a height of 9.5 m. How fast is it going right
before it hits the ground?
V= I*R
V = voltage (measured in volts) V
I = current (measured in amperes) A
R = resistance (measured in Ohms) Ω
So they give us this
V=IR
V= 1.8
I=0.4
R=?
So we insert the thing that we know.
1.8=0.4*R
We need to leave our unknown value alone. So if our value of 0.4 is multiplying the unknown value it passes to the other side dividing.
So we have this.
Lastly we solve.
R=4.5ohms
The formula to find R is V=IR
V/I=R
So the resistance will be the Voltage divided by the Current
Two blocks are connected by a massless rope that passes over a 1 kg pulley with a radius of 12 cm. The rope moves over the pulley without slipping. The mass of block A is 2.1 kg and the mass of block B is 4.1 kg. Block A is also connected to a horizontally-mounted spring with a spring constant of 358 J/m2. What is the angular frequency (in rad/s) of oscillations of this system
Answer:
[tex]F=1.159[/tex]
Explanation:
From the question we are told that:
Mass of pulley [tex]M=1kg[/tex]
Radius [tex]r=12cm[/tex]
Mass of block A [tex]M_a=2.1kg[/tex]
Mass of block B [tex]m_b=4.1kg[/tex]
Spring constant[tex]\mu= 358 J/m2[/tex]
Generally the equation for Torque is mathematically given by
Since [tex]\sumF=ma[/tex]
At mass A
[tex]T_2-f_3=2.1a[/tex]
At mass B
[tex]4.8-T_1=4.1a[/tex]
At Pulley
[tex]R(T_1-T_2)=\frac{1*1*R^2}{2}\frac{a}{R}[/tex]
[tex]R(T_1-T_2)=0.55a[/tex]
Therefore the equation for total force F
At mass A+At mass B+At Pulley
[tex](T_2-f_3+4.8-T_1+R(T_1-T_2)=2.1a+4.1a+0.55a[/tex]
[tex](T_2-f_3+4.8-T_1+R(T_1-T_2)=2.7a+4.8a+0.55a[/tex]
[tex]-f_3+4.1=6.75a[/tex]
[tex]-f_3=6.75a+4.8[/tex]
Since From above equation
[tex]M_{eff}=6.7kg[/tex]
Therefore
[tex]T=2\pi \sqrt{{\frac{M_{eff}}{k}}[/tex]
[tex]T=2\pi \sqrt{{\frac{6.75}{\mu}}[/tex]
[tex]T=0.862s[/tex]
Generally the equation for frequency is mathematically given by
[tex]F=\frac{1}{T} \\F=\frac{1}{0.862}[/tex]
[tex]F=1.159[/tex]
5. Stopping a fast-moving object is harder than stopping a slow-moving
one.
True
False
A closed loop conductor that forms a circle with a radius of 2.0 m is located in a uniform but changing magnetic field. If the maximum emf Induced in the loop is 5.0 V what is the maximum rate at which the magnetic field strength is changing if the magnetic field is oriented perpendicular to the plane in which the loop lies
Answer:
The maximum rate at which the magnetic field strength is changing if the magnetic field is oriented perpendicular to the plane in which the loop lies 0.398 T/s.
Explanation:
Given;
radius of the circular loop, r = 2.0 m
maximum induced emf, E = 5.0 V
The emf induced in a magnetic field is given as;
[tex]emf = \frac{d\phi}{dt} \\\\\phi = AB\\\\emf = A\frac{dB}{dt} \\\\\frac{dB}{dt} = \frac{emf}{A} \\\\where;\\A \ is \ the \ area \ circular \ l00p = \pi r^2 = \pi (2)^2 = 4\pi \ m^2\\\\\frac{dB}{dt} = \frac{5}{4\pi} \\\\\frac{dB}{dt} = 0.398 \ T/s[/tex]
Therefore, the maximum rate at which the magnetic field strength is changing if the magnetic field is oriented perpendicular to the plane in which the loop lies 0.398 T/s.
Two particles in a high-energy accelerator experiment are approaching each other head-on, each with a speed of 0.9520c as measured in the laboratory. What is the magnitudeof the velocity of one particle relative to the other
Answer:
the magnitude of the velocity of one particle relative to the other is 0.9988c
Explanation:
Given the data in the question;
Velocities of the two particles = 0.9520c
Using Lorentz transformation
Let relative velocity be W, so
v[tex]_r[/tex] = ( u + v ) / ( 1 + ( uv / c²) )
since each particle travels with the same speed,
u = v
so
v[tex]_r[/tex] = ( u + u ) / ( 1 + ( u×u / c²) )
v[tex]_r[/tex] = 2(0.9520c) / ( 1 + ( 0.9520c )² / c²) )
we substitute
v[tex]_r[/tex] = 1.904c / ( 1 + ( (0.906304 × c² ) / c²) )
v[tex]_r[/tex] = 1.904c / ( 1 + 0.906304 )
v[tex]_r[/tex] = 1.904c / 1.906304
v[tex]_r[/tex] = 0.9988c
Therefore, the magnitude of the velocity of one particle relative to the other is 0.9988c
4- What force must be applied to a surface area of 0.0025m , to create a pressure ol
200.000Pa?
A man on the Moon observes two spaceships coming toward him from opposite directions at speeds of 0.600c and 0.600c. What is the relative speed of the two ships as measured by a passenger on either one of the spaceships
Answer:
If we use the equation for the transformation of velocities for moving frames:
v' = (v - u) / (1 - u * v / c^2) where we measure the speed of v' approaching from the left where v is in a frame moving at -u towards v'
v' = (.6 c - (-.6 c)) / (1 - (-.6 c) * .6 c / c^2) = 1.2 c / (1 + .6 * .6)
or v' = 1.2 c / (1 + .36) = .88 c
v is approaching from the left at .6 c in the reference frame and the other frame approaches from the right at -.6 c with speed u (-.6 c) and we measure the speed of v as seen in the frame moving to the left
What is the speed of a wave if it has a wavelength of
42 m and a frequency of 7 hertz?
Answer:
♕ [tex]\large{ \red{ \tt{Step - By - Step \: Explanation}}}[/tex]
☃ [tex] \underline{ \underline{ \blue{ \large{ \tt{G \: I \: V \: E\: N}}}}} : [/tex]
Frequency ( f ) = 7 HertzWavelength ( λ ) = 42m♨ [tex] \underline {\underline{ \orange{ \large{ \tt{T \: O \: \: F \: I \: N\: D}}}} }: [/tex]
Wave velocity ( v )☄ [tex]\underline{ \underline{ \large{ \pink{ \tt{S\: O \: L \: U \: T\: I \: O \: N}}}}}: [/tex]
✧ [tex] \red{ \boxed{ \large{ \purple{ \sf{Wave \: velocity(v) = Frequency(f) \times Wavelength(λ)}}}}}[/tex]
~Plug the known values and then multiply!
↦ [tex] \large{ \tt{7 \times 42}}[/tex]
↦ [tex] \boxed{ \boxed{ \large{ \bold{ \tt{294 \: m {s}^ {- 1} }}}}}[/tex]
☥ [tex] \large{ \boxed{ \boxed{ \large{ \tt{Our \: Final \: Answer : \underline{ \large{ \tt{294 \: m {s}^{ - 1}}}}}}}}} [/tex]
---------------------------------------------------------------
❁ [tex] \underline{ \large{ \red{ \tt{D\: E\: T \: A \: I \: L\: E \: D \: \: I\: N \: F \: O}}}} : [/tex]
Frequency ( f ) : The number of complete waves , set up in a medium in one second is called frequency of the wave. The SI unit of frequency is Hertz ( Hz ). For example : if a sound wave completes 15 compressions and 15 rarefactions in one second , it's frequency is 15 Hz.Wavelength ( λ ) : The distance between two consecutive troughs or crests in a transverse wave or the distance between two consecutive compressions or rarefactions in a longitudinal wave us called wavelength. It is the distance travelled by a wave in a time equal to it's time period. It's SI unit is metre ( m ).Wave velocity ( v ) : The velocity with which a wave propagates in a medium is called wave velocity. It's SI unit is m/s.# KILL : Excuses
KISS : Opportunities
MARRY : Goals
♪ Hope I helped! ♡
☂ Have a wonderful day / night ! ツ
✎ [tex] \underbrace{ \overbrace{ \mathfrak{Carry \: On \: Learning}}}[/tex] ✔
▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁
A 35.0 g bullet strikes a 5.3 kg stationary wooden block and embeds itself in the block. The block and bullet fly off together at 7.1 m/s. What was the original speed of the bullet? (WILL GIVE BRAINLIEST)
Answer:
= 1200m/s or 1.2 x [tex]10^{3}[/tex] m/s
Explanation:
A 5.0-m radius playground merry-go-round with a moment of inertia of 1,630 kg m2 is rotating freely with an angular speed of 1.6 rad/s. Two people, each having a mass of 69.5 kg, are standing right outside the edge of the merry-go-round and step on it with negligible speed. What is the angular speed of the merry-go-round right after the two people have stepped on
Answer:
The right solution is "0.511".
Explanation:
Given:
Initial moment of inertia,
= 1630 kg.m²
Radius,
= 5 m
Angular speed,
= 1.6 rad/s
Now,
The moment of inertia after stepping on will be:
= [tex]1630+2\times (69.5\times (5)^2)[/tex]
= [tex]1630+2\times (69.5\times 25)[/tex]
= [tex]5105 \ Kg.m^2[/tex]
hence,
As per the question, the angular speed is conserved, then
⇒ [tex]1630\times 1.6=5105\times \omega'[/tex]
[tex]2608=5105\times \omega'[/tex]
[tex]\omega'=\frac{2608}{5105}[/tex]
[tex]=0.511[/tex]
A 17-mm-wide diffraction grating has rulings of 530 lines per millimeter. White light is incident normally on the grating. What is the longest wavelength that forms an intensity maximum in the fifth order
Answer:
377 nm
Explanation:
Number of lines per meter is, [tex]N &=530 \times 1000 \\ &=530000 \text { lines } / \mathrm{m} \end{aligned}[/tex]
Grating element is, [tex]d=\frac{1}{N}[/tex]
[tex]=1.8868 \times 10^{-6} \mathrm{~m}[tex]
Order is, n=5
Condition for maximum intensity is, [tex]d \sin \theta=n \lambda[/tex]
[tex]\lambda &=\frac{1.8868 \times 10^{-6}}{5(\sin 90)} \\ &=0.377 \times 10^{-6} \mathrm{~m} \\ &=377 \mathrm{~nm}[/tex]
Problem 3:A starship voyages to a distant planet 10 ly away. The astronauts in the starship reach theplanet and then they immediately return to the Earth at the same speed. The round trip from the Earthto the planet and back to the Earth takes 25 years.(a) What is the speed of the starship
Answer:
speed = 0.8c
Explanation:
Given :
Distance from earth to the distant planet = 10 ly
Time taken by the astronauts for the entire journey = 25 years
The time taken to reach the planet is [tex]$t_1=\frac{25}{2}$[/tex]
= 12.5 years
Therefore, speed of the starship can be calculated by :
[tex]$\text{Speed} = \frac{\text{distance}}{\text{time}}$[/tex]
[tex]$v=\frac{10 \times c \times 3.15 \times 10^7}{12.5 \times 3.15 \times 10^7}$[/tex]
[tex]$=0.8c$[/tex]
Therefore the speed of the starship is 0.8c
Are surface currents warm or cold?
A:war m
B:cold
Answer:
Cold
Explanation:
Im pretty sure im sorry if I am wrong
If an electromagnetic wave has a frequency of 6×10^5 hz, what is its wavelength? what is its wavelength? A. 2 x 10^12m, B. 5 x 10^14m, C. 5 x 10^2m, 2 x 10^-3m
Answer:
5*10^2
Explanation:
A p e x
At what speed does light travel through water. n=1.33
Answer:
"In water, the speed of light is v01.33=2.26 x 108 m/s."
Explanation:
ADHD is triggered ________ by genetic and neurological factors and ________ by social or environmental ones.
A. less; more
B. somewhat; somewhat
C. little; mostly
D. more; less
Answer:
D More: Less
Explanation:
ADHD is triggered more by genetic and neurological factors and less by social or environmental ones.
What is ADHD?ADHD is one of the most common neurodevelopmental disorders of childhood. It is usually first diagnosed in childhood and often lasts into adulthood. Children with ADHD may have trouble paying attention, controlling impulsive behaviors (may act without thinking about what the result will be), or be overly active.
Moreover, ADHD tends to run in families and, in most cases, it's thought the genes you inherit from your parents are a significant factor in developing the condition. Research shows that parents and siblings of someone with ADHD are more likely to have ADHD themselves.
Therefore, adults with attention deficit hyperactivity disorder (ADHD or ADD) lead anxious lives. The nature of ADHD often makes day-to-day life stressful, creating situations and environments fraught with uncertainty – anxiety's primary fuel.
Learn more about ADHD:
https://brainly.com/question/13719580
#SPJ2
Which of the following is form of energy:
a) Power
b) Light
C) pressure
d) None
Answer:
Explanation:
b) light
Help me please I don’t understand
Answer:
a. hydroelectric power plant
An irrigation canal has a rectangular cross section. At one point where the canal is 18.2 m wide and the water is 3.55 m deep, the water flows at 2.55 cm/s . At a second point downstream, but on the same level, the canal is 16.3 m wide, but the water flows at 11.6 cm/s . How deep is the water at this point
Answer:
Explanation:
Rate of volume flow at two points will be same at two points .
A₁ V₁ = A₂V₂
A₁ and A₂ are area of cross section at two points and V₁ and V₂ are velocities .
A₁ = 18.2 x 3.55 = 64.61 m²
V₁ = 2 .55 x 10⁻² m/s
A₂ = 16.3 x d = 16.3 d m²
d is depth at second point .
V₂ = 11.6 x 10⁻² m/s
64.61 m² x 2 .55 x 10⁻² m/s = 16.3 d m² x 11.6 x 10⁻² m/s
d = .87 m
so canal is .87 m deep.