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.
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WILL GIVE BRAINLIEST
Imagine you are holding a box in your hand, the force required for you to hold the box with mass 4.54kg and you want to accelerate the box 2m/s2. Move the box in your hand a) upward vertically, b) downward vertically, and c) horizontally. Calculate the magnitudes of forces in the x and y direction.
Answer:
(a) Fₓ = 0 N
[tex]F_y[/tex] = 9.08 N
(b)
(a) Fₓ = 0 N
[tex]F_y[/tex] = 9.08 N
(c) [tex]F_y[/tex] = 0 N
Fₓ = 9.08 N
Explanation:
The magnitude of the force will remain the same in each case, which is given as follows:
F = ma (Newton's Second Law)
where,
F = force = ?
m = mass = 4.54 kg
a = acceleration = 2 m/s²
Therefore,
F = (4.54 kg)(2 m/s²)
F = 9.08 N
Now, we come to each scenario:
(a)
Since the motion is in the vertical direction. Therefore the magnitude of the force in x-direction will be zero:
Fₓ = 0 N
For upward direction the force will be positive:
[tex]F_y[/tex] = 9.08 N
(b)
Since the motion is in the vertical direction. Therefore the magnitude of the force in x-direction will be zero:
Fₓ = 0 N
For upward direction the force will be negative:
[tex]F_y[/tex] = - 9.08 N
(c)
Since the motion is in the horizontal direction. Therefore the magnitude of the force in y-direction will be zero:
[tex]F_y[/tex] = 0 N
Fₓ = 9.08 N
the caste system is an example of
Answer:
It is example of Closed system
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].
What can be the maximum value of the original kinetic energy of disk AA so as not to exceed the maximum allowed value of the thermal energy
The question is incomplete. The complete question is :
In your job as a mechanical engineer you are designing a flywheel and clutch-plate system. Disk A is made of a lighter material than disk B, and the moment of inertia of disk A about the shaft is one-third that of disk B. The moment of inertia of the shaft is negligible. With the clutch disconnected, A is brought up to an angular speed ?0; B is initially at rest. The accelerating torque is then removed from A, and A is coupled to B. (Ignore bearing friction.) The design specifications allow for a maximum of 2300 J of thermal energy to be developed when the connection is made. What can be the maximum value of the original kinetic energy of disk A so as not to exceed the maximum allowed value of the thermal energy?
Solution :
Let M.I. of disk A = [tex]$I_0$[/tex]
So, M.I. of disk B = [tex]$3I_0$[/tex]
Angular velocity of A = [tex]$\omega_0$[/tex]
So the kinetic energy of the disk A = [tex]$\frac{1}{2}I_0\omega^2$[/tex]
After coupling, the angular velocity of both the disks will be equal to ω.
Angular momentum will be conserved.
So,
[tex]$I_0\omega_0 = I_0 \omega + 3I_0 \omega$[/tex]
[tex]$I_0\omega_0 = 4I_0 \omega$[/tex]
[tex]$\omega = \frac{\omega_0}{4}$[/tex]
Now,
[tex]$K.E. = \frac{1}{2}I_0\omega^2+ \frac{1}{2}3I_0\omega^2$[/tex]
[tex]$K.E. = \frac{1}{2}I_0\frac{\omega_0^2}{16}+ \frac{1}{2}3I_0\frac{\omega_0^2}{16}$[/tex]
[tex]$K.E. = \frac{1}{2}I_0\omega_0^2 \left(\frac{1}{16}+\frac{3}{16}\right)$[/tex]
[tex]$K.E. = \frac{1}{2}I_0\omega_0^2\times \frac{1}{4}$[/tex]
[tex]$\Delta K = \frac{1}{2}I_0\omega_0^2 - \frac{1}{2}I_0\omega_0^2 \times \frac{1}{4} $[/tex]
[tex]$2300=\frac{3}{4}\left(\frac{1}{2}I_0\omega_0^2\right)$[/tex]
[tex]$\frac{1}{2}I_0\omega_0^2=2300 \times \frac{4}{3 } \ J $[/tex]
Therefore, the maximum initial K.E. = 3066.67 J
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
What would be the net force and in what direction for the motion to be? The first one* and explain
Answer:
11. 5 N to the left.
12. 10 N to the right.
Explanation:
11. Determination of the net force.
Force in the left direction (Fₗ) = 10 N
Force in the right direction (Fᵣ) = 5 N
Net force (Fₙ) =?
The net force can be obtained as follow:
Fₙ = Fₗ – Fᵣ (Since the forces act in opposite direction)
Fₙ = 10 – 5
Fₙ = 5 N to left
Thus, the net force is 5 N to the left.
12. Determination of the net force.
Force in the right direction 1 (Fᵣ₁) = 5 N
Force in the right direction 2 (Fᵣ₂) = 5 N
Net force (Fₙ) =?
The net force can be obtained as follow:
Fₙ = Fᵣ₁ + Fᵣ₂ (Since the forces act in the same direction)
Fₙ = 5 + 5
Fₙ = 10 N to the right
Thus, the net force is 10 N to the right.
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
If a 4 Ohm resistor and a 12 Ohm resistor are connected in parallel, what is the total
resistance?
Rt = 3 ohms
Explanation:
Let R1 = 4-ohm resistor
R2 = 12-ohm resistor
For 2 resistors connected in parallel, the total resistance Rt is given by
1/Rt = 1/R1 + 1/R2
or
Rt = R1R2/(R1 + R2)
= (4 ohms)(12 ohms)/(4 ohms + 12 ohms)
= 48 ohms^2/16 ohms
= 3 ohms
please teach me reflection of light
I’ll give Brainliest, Identify the force acting as the
centripetal force when a car circulates
around a flat circular track
Friction
Gravitation
Normal
tension
Answer:
Friction hope this helps <3
Moving current has electrical energy.
A beam of electrons is accelerated across a potential of 14.60 kV before passing through two slits. The electrons form an interference pattern on a screen 4.40 m in front of the slits. The first-order maximum is 8.80 mm from the central maximum. What is the distance between the slits
Answer:
[tex]5.08\times 10^{-9}\ \text{m}[/tex]
Explanation:
h = Planck constant = [tex]6.626\times 10^{-34}\ \text{Js}[/tex]
[tex]\lambda[/tex] = Wavelength
m = Mass of electron = [tex]9.11\times 10^{-31}\ \text{kg}[/tex]
[tex]\Delta V[/tex] = Potential difference = 14.6 kV
e = Charge of electron = [tex]1.6\times 10^{-19}\ \text{C}[/tex]
d = Distance between slits
[tex]\sin\theta=\dfrac{8.8\times 10^{-3}}{4.4}[/tex]
We have the relation
[tex]\dfrac{h}{\lambda}=\sqrt{2me\Delta V}\\\Rightarrow \lambda=\dfrac{h}{\sqrt{2me\Delta V}}\\\Rightarrow \lambda=\dfrac{6.626\times 10^{-34}}{\sqrt{2\times 9.1\times 10^{-31}\times 1.6\times 10^{-19}\times 14.6\times 10^3}}\\\Rightarrow \lambda=1.016\times 10^{-11}\ \text{m}[/tex]
Wavelength is given by
[tex]d\sin\theta=\lambda\\\Rightarrow d=\dfrac{\lambda}{\sin\theta}\\\Rightarrow d=\dfrac{1.016\times 10^{-11}}{\dfrac{8.8\times 10^{-3}}{4.4}}\\\Rightarrow d=5.08\times 10^{-9}\ \text{m}[/tex]
The distance between the slits is [tex]5.08\times 10^{-9}\ \text{m}[/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
I NEED HELP THIS QUESTION IS SO HARDDD!!
Answer:
c
Explanation:
search it up
Answer:
Option A.
Explanation:
Because this is a lunar eclipse it normally happens 2 times a year. This is a decently rare phenomenon due to the positions they have to be in to make a lunar eclipse. Therefore, it is option A.
Light of wavelength 580 nm is incident on a slit of width 0.30 mm. An observing screen is placed 2.0 m past the slit. Find the distance on the screen of the first order dark fringe from the center of the pattern.
Answer:
Y = 3.87 x 10⁻³ m = 3.87 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 = ?
L = slit to screen distance = 2 m
λ = wavelength of light = 580 nm = 5.8 x 10⁻⁷ m
d = slit width = 0.3 mm = 3 x 10⁻⁴ m
Therefore,
[tex]Y = \frac{(5.8\ x\ 10^{-7}\ m)(2\ m)}{3\ x\ 10^{-4}\ m}[/tex]
Y = 3.87 x 10⁻³ m = 3.87 mm
Darth Vader has stolen a Bugatti Chiron (high speed sports car) and is now driving at aconstant velocity of 120 mph down the street for 45 seconds. Suddenly, he sees a family ofgeese crossing the street 200mahead and presses the brakes immediately (assume constant acceleration as he comes to a stop). He just barely avoids running over the geese. Putanswers in terms of SI units.
Required:
a. What is his acceleration as he comes to a stop?
b. How long does it take Vader to stop?
c. What is the total distance traveled?
The acceleration is 1.2 m/s^2. The distance travelled is 1197 m. The time taken is 44.7 seconds.
What is acceleration?The rate at which the velocity changes is called the acceleration of the body.
Given that;
v = u - at
v = 0 m/s because the car comes to stop
u = at
a = u/t
a = 53.6 m/s/ 45 seconds
a = 1.2 m/s^2
The distance travelled is;
v^2 = u^2 - 2as
v = 0 m/s
u^2 = 2as
s = u^2/2a
a = ( 53.6 )^2/2 * 1.2
s = 1197 m
Time taken to stop is obtained from;
v = u - at
when v = 0 m/s
u = at
t = u/a
t = 53.6/1.2
t = 44.7 seconds
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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:
Help I only have 7 minutes help now
. 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
Two long anti-parallel current-carrying wires carry currents of 20 A and 5.0 A. If the wires are separated by 20 cm. Find the magnitude of the magnetic field a point equally separated between both wires.
Answer:
5 x 10⁻⁵ T
Explanation:
Magnetic field at a point d distance away from a current carrying wire having current of I
B = 10⁻⁷ x 2I / d
Here magnetic field due to current of 20 A
B₁ = 10⁻⁷ x 2 x 20 / 10 x 10⁻²
= 4 x 10⁻⁵ T .
Here magnetic field due to current of 5 A
B₂ = 10⁻⁷ x 2 x 5 / 10 x 10⁻²
= 10⁻⁵ T .
These magnetic fields are acting in the same direction
Total magnetic field
B = B₁ + B₂
= 4 x 10⁻⁵ + 10⁻⁵ = 5 x 10⁻⁵ T .
If the output work of a simple machine equals the input work, the machine is said to have ______ efficiency.
If the output work of a simple machine equals the input work, the machine is said to have 100% efficiency.
[tex]\circ \: \: { \underline{ \boxed{ \sf{ \color{green}{Happy\:learning.}}}}}∘[/tex]
Mass, volume and density are all properties of
Answer:
Properties of matter
Explanation:
All properties of matter are either extensive or intensive and either physical or chemical. Extensive properties, such as mass and volume, depend on the amount of matter that is being measured. Intensive properties, such as density and color, do not depend on the amount of matter.
Answer:
The pythagoream theorem
Explanation:
Water has a heat capacity of 4.184 J/g °C. If 50 g of water has a temperature of 30°C and a piece of hot copper is added to the water causing the temperature to increase to 70°C. What is the amount of heat absorbed by the water?
Answer:
Q = 8368 Joules.
Explanation:
Given the following data;
Mass = 50 g
Initial temperature = 30°C
Final temperature = 70°C
Specific heat capacity = 4.184 J/g °C
To find the amount of heat absorbed by the water;
Heat capacity is given by the formula;
[tex] Q = mcdt[/tex]
Where;
Q represents the heat capacity or quantity of heat.
m represents the mass of an object.
c represents the specific heat capacity of water.
dt represents the change in temperature.
dt = T2 - T1
dt = 70 - 30
dt = 40°C
Substituting the values into the equation, we have;
[tex] Q = 50*4.184*40[/tex]
Q = 8368 Joules.
Will give brainliest
Answer:
I believe the answer is 2.5 N.......
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
a fly wheel is rotating with an angular velocity of 2 rad/sec and acted by an acceleration of 1/π rad/s2 . how long will it take to complete three revolutions?
Explanation:
this is the answer for your question if you have any doubt you can join
Copper wires are used as connecting wires because
(1 Point)
Copper has very high melting point
Copper wires are very thick wires
Copper wire offers a lower resistance
Copper wire has an attractive colour
[tex]\huge{ \mathfrak{ \underline{ Answer }\: \: ✓ }}[/tex]
Copper wires are used as connecting wires in circuits because they offer very less resistance.
__________________________________________________________
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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