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
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
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.
5. Stopping a fast-moving object is harder than stopping a slow-moving
one.
True
False
Can someone help me
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]
Why is it important for equipment for sport to be strong? To protect us
Answer:
To protect us.
Explanation:
For ex. your dunking on a basketball hoop if that wasn't strong you would fall on your back and get injured.
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].
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
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
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
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:
The atmospheric features of Neptune are easier to see than those of Uranus because A. Neptune has greater warmth and less haze. B. Neptune has more methane. C. The atmosphere of Uranus rotates differentially. D. Uranus has no significant atmosphere.
Answer:
Option B is the correct answer (Neptune has more methane)Explanation:
From the options given,
The atmospheric features of Neptune are easier to see than those of Uranus because Neptune has more methane
Neptune has small amount of methane and water which gives it blue colour and white patches which distinguish it from uranus
For more information, visit
http://abyss.uoregon.edu/~js/ast121/lectures/lec20.html
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:
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.
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 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
1. Pam has a mass of 48.3 kg and she is at rest on
smooth, level, frictionless ice. Pam straps on
a rocket pack. The rocket supplies a constant
force for 27.3 m and Pam acquires a speed of
62 m/s.
What is the magnitude of the force?
Answer in units of N.
2. What is Pam’s final kinetic energy?
Answer in units of J.
3. A child and sled with a combined mass of 55.7
kg slide down a frictionless hill that is 11.3 m
high at an angle of 29 ◦
from horizontal.
The acceleration of gravity is 9.81 m/s
3. If the sled starts from rest, what is its speed
at the bottom of the hill?
Answer in units of m/s
Answer:
1. F = 3400 N = 3.4 KN
2. [tex]K.E_f=92832.6\ J = 92.83\ KJ[/tex]
3. v = 14.9 m/s
Explanation:
1.
First, we will calculate the acceleration of Pam by using the third equation of motion:
[tex]2as = v_f^2-v_i^2[/tex]
where,
a = acceleration = ?
s = distance = 27.3 m
vf = final speed = 62 m/s
vi = initial speed = 0 m/s
Therefore,
[tex]2a(27.3\ m) = (62\ m/s)^2-(0\ m/s)^2\\\\a = 70.4\ m/s^2[/tex]
Now, we will calculate the force by using Newton's Second Law of Motion:
F = ma
F = (48.3 kg)(70.4 m/s²)
F = 3400 N = 3.4 KN
2.
Final kinetic energy is given as:
[tex]K.E_f = \frac{1}{2}mv_f^2\\\\K.E_f = \frac{1}{2} (48.3\ kg)(62\ m/s)^2[/tex]
[tex]K.E_f=92832.6\ J = 92.83\ KJ[/tex]
3.
According to the law of conservation of energy:
[tex]Potential\ Energy\ at\ top = Kinetic\ Energy\ at\ bottom\\mgh = \frac{1}{2}mv_2 \\\\v = \sqrt{2gh}[/tex]
where,
v = speed at bottom = ?
g = acceleration due to gravity = 9.81 m/s²
h = height at top = 11.3 m
Therefore,
[tex]v = \sqrt{(2)(9.81\ m/s^2)(11.3\ m)}[/tex]
v = 14.9 m/s
. A small car of mass m and a large car of mass 2m drive around a highway curve of radius R. Both cars travel at the same speed (v). The
centripetal acceleration (Grad) of the large car is the centripetal acceleration of the small car.
four times
twice
half
equal to
Complete question is;
A small car of mass m and a large car of mass 2m drive around a highway curve of radius R. Both cars travel at the same speed (v). The
centripetal acceleration (Grad) of the large car is the centripetal acceleration of the small car. How does the Force of the small car FS compare to the force of the large car FL as they round the curve.
four times
twice
half
equal to
Answer:
Half
Explanation:
Formula for centripetal force is given as;
F = mv²/R
Where;
v is velocity
R is radius
Now, centripetal acceleration is given by;
a = v²/R
Since they both travel with the same velocity V and radius remains the same, we can say that;
F = ma
For the small car;
FS = ma
For the big car;
FL = 2ma
This means the force of the small car is half of that of the Large car
Thus;
FS = ½FL
Are surface currents warm or cold?
A:war m
B:cold
Answer:
Cold
Explanation:
Im pretty sure im sorry if I am wrong
Moving current has electrical energy.
Which of the following is form of energy:
a) Power
b) Light
C) pressure
d) None
Answer:
Explanation:
b) light
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.
Help me please I don’t understand
Answer:
a. hydroelectric power plant
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
A 1460 kg car moving north at 27.0 m/s collides with a 2165 kg car moving east at 18.0 m/s. They stick together. In what direction and with what speed do they move after the collision?
Answer in degrees north of east
AND
the speed after the collision in m/s
Answer:
Solution given:
North car
mass[m1]=1460kg
velocity[u1]=27 m/s
mass[m2]=2165kg
velocity [u2]=18m/s
let v be velocity after collision
we have
From the principle of conservation of linear momentum
m1u1+m2u2=(m1+m2)v
1460*27+2165*18=(1460+2165)v
v=[tex] \frac{78390}{3625} [/tex]
v=21.6m/s
the speed after the collision in 21.6 m/s.
For angle.
Tan angle =[tex] \frac{m1u1}{m2u2} [/tex]
Tan angle =[tex] \frac{1460*27}{2165*18} [/tex]
Tan angle=327.74
angle=Tan-¹(327.74)=89.82=90°
in degrees north of east is 90°
The door is 2 m tall. How tall is it in inches? Note: There are 2.54 cm in 1 inch.
A. 78.7 in
B. 500 in
C. 787.4 in
D. 201.4 in
Answer:
Height of the door = 2m = 2000 cm
1 in = 2.54 cm
So 1 cm = 1/2.54 in
2000 cm = 200000/ 254
=
787.401574803
So no.c is correct
The door is 78.7 inch tall. Hence, option (A) is correct.
What is unit of length?Any arbitrarily selected and widely used reference standard for length measurement is referred to as a unit of length. The metric system, which is adopted by every nation on earth, is the most widely utilized in modern times.
The American customary units are also in use in the United States. In the UK and several other nations, British Imperial units are still used sometimes. There are SI units and non-SI units in the metric system.
Given that: the height of the door is = 2 meter
= 2*100 centimeter
= 200 centimeter.
There are 2.54 centimeter in 1 inch.
Hence, the height of the door is = 2 meter = 200 centimeter
= (200/2.54) inch
= 78.7 inch.
The door is 78.7 inch tall.
Learn more about length here:
https://brainly.com/question/17139363
#SPJ2
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]
4- What force must be applied to a surface area of 0.0025m , to create a pressure ol
200.000Pa?
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
An ideal massless spring with a spring constant of 2.00 N/m is attached to an object of 75.0 g. The system has a small amount of damping. If the amplitude of the oscillations decreases from 10.0 mm to 5.00 mm in 15.0 s, what is the magnitude of the damping constant b
Answer: 0.00693
Explanation:
Given
Spring constant [tex]k=2\ N/m[/tex]
Mass of object [tex]m=75\ g[/tex]
The amplitude of the oscillation decreases from 10 mm to 5 mm in 15 s
Equation of amplitude for the ideal spring-mass system is
[tex]\Rightarrow A=A_oe^{-\frac{bt}{2m}}\quad \quad [\text{b=damping constant}]\\\text{Insert the values}\\\\\Rightarrow 5=10e^{\frac{b\times 15}{2\times 0.075}}\\\\\Rightarrow e^{-\frac{b\times 15}{2\times 0.075}}=0.5\\\\\text{Taking natural log both sides}\\\\\Rightarrow \ln \left(e^{-\frac{b\times 15}{2\times 0.075}}\right)=\ln 0.5\\\\\Rightarrow -\dfrac{15b}{0.15}=-0.693\\\\\Rightarrow b=0.00693[/tex]