Answer:
v = -17.94 cm
Explanation:
Given that,
The candle is placed at a distance of 50 cm, u = -50 cm
The focal length of the diverging lens, f = -28 cm (negative in case of a concave lens)
We need to find the image distance. We know that the lens formula is as follows:
[tex]\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}\\\\\dfrac{1}{v}=\dfrac{1}{f}+\dfrac{1}{u}\\\\\dfrac{1}{v}=\dfrac{1}{(-28)}+\dfrac{1}{(-50)}\\\\v=-17.94\ cm[/tex]
So, the image distance is equal to 17.94 cm.
what is friction and the types withe examples.
Explanation:
The answer is In the picture. Thanks.
What is the period of a wave with a speed of 20.0 m/s and a frequency of 10.0 Hz?
im confused hold on imma send you a link to the answerExplanation:
The fictional rocket ship Adventure is measured to be 65 m long by the ship's captain inside the rocket.When the rocket moves past a space dock at 0.5c. As rocket ship Adventure passes by the space dock, the ship's captain flashes a flashlight at 1.20-s intervals as measured by space-dock personnel.
Required:
How often does the flashlight flash relative to the captain?
Answer:
1.04 s
Explanation:
The computation is shown below:
As we know that
t = t' × 1 ÷ (√(1 - (v/c)^2)
here
v = 0.5c
t = 1.20 -s
So,
1.20 = t' × 1 ÷ (√(1 - (0.5/c)^2)
1.20 = t' × 1 ÷ (√(1 - (0.5)^2)
1.20 = t' ÷ √0.75
1.20 = t' ÷ 0.866
t' = 0.866 × 1.20
= 1.04 s
The above formula should be applied
g 1. Water flows through a 30.0 cm diameter water pipe at a speed of 3.00 m/s. All of the water in the pipe flows into a smaller pipe that is 10.0 cm in diameter. Determine: a) The speed of the water flowing through the 10.0 cm diameter pipe. b) The mass of water that flows through the larger pipe in 1.00 minute. c) The mass of water that flows through the smaller pipe in 1.00 minute.
Answer:
a) v₂ = 30 m/s
b) m₁ = 12600 kg
c) m₂ = 12600 kg
Explanation:
a)
Using the continuity equation:
[tex]A_1v_1 = A_2v_2[/tex]
where,
A₁ = Area of inlet = π(0.15 m)² = 0.07 m²
A₂ = Area of outlet = π(0.05 m)² = 0.007 m²
v₁ = speed at inlet = 3 m/s
v₂ = speed at outlet = ?
Therefore,
[tex](0.07\ m^2)(3\ m/s)=(0.007\ m^2)v_2\\\\v_2 = \frac{0.21\ m^3/s}{0.007\ m^2}[/tex]
v₂ = 30 m/s
b)
[tex]m_1 = \rho A_1v_1t[/tex]
where,
m₁ = mass of water flowing in = ?
ρ = density of water = 1000 kg/m³
t = time = 1 min = 60 s
Therefore,
[tex]m_1 = (1000\ kg/m^3)(0.07\ m^2)(3\ m/s)(60\ s)\\[/tex]
m₁ = 12600 kg
c)
[tex]m_1 = \rho A_1v_1t[/tex]
where,
m₂ = mass of water flowing out = ?
ρ = density of water = 1000 kg/m³
t = time = 1 min = 60 s
Therefore,
[tex]m_2 = (1000\ kg/m^3)(0.007\ m^2)(30\ m/s)(60\ s)\\[/tex]
m₂ = 12600 kg
Determine the poles of the magnet. Look at the three compass readings that are on top of the magnet. Label the
end the compass points away from as "S" (south), and the other end that the compass points toward as "N" (north).
Record these poles in Figure 1.
Continue
Intro
Answer:
the red pointer on the magnet ( grey region) : points towards north
red pointer outside the magnet ( white region) is pointing towards south
Explanation:
please see the attached image
The density of 1 kilogram of gold is
Answer:
0.02 kg/cm³
Explanation:
If you weigh 690 N on the earth, what would be your weight on the surface of a neutron star that has the same mass as our sun and a diameter of 15.0 km ? Take the mass of the sun to be ms = 1.99×1030 kg , the gravitational constant to be G = 6.67×10−11 N⋅m2/kg2 , and the free-fall acceleration at the earth's surface to be g = 9.8 m/s2 . Express your weight wstar in newtons.
Answer:
W' = 1.66 x 10¹⁴ N
Explanation:
First, we will calculate the mass:
[tex]W = mg[/tex]
where,
W = weight on earth = 690 N
m = mass = ?
g = acceleration due to gravity on earth = 9.8 m/s²
Therefore,
[tex]m = \frac{W}{g} = \frac{690\ N}{9.8\ m/s^2}\\\\m = 70.4\ kg[/tex]
Now, we will calculate the value of g on the neutron star:
[tex]g' = \frac{GM}{R^2}[/tex]
where,
g' = acceleration due to gravity on the surface of the neutron star = ?
G = Gravitational Constant = 6.67 x 10⁻¹¹ N.m²/kg²
M = Mass of the Neutron Star = 1.99 x 10³⁰ kg
R = Radius of the Neutron Star = 15 km/2 = 7.5 km = 7500 m
Therefore,
[tex]g' = \frac{(6.67\ x\ 10^{-11}\ N.m^2/kg^2)(1.99\ x\ 10^{30}\ kg)}{(7500\ m)^2}\\\\g' = 2.36\ x\ 10^{12}\ m/s^2[/tex]
Therefore, the weight on the surface of the neutron star will be:
[tex]W' = mg'\\W' = (70.4\ kg)(2.36\ x\ 10^{12}\ m/s^2)[/tex]
W' = 1.66 x 10¹⁴ N
Why would the kinetic energy at the bottom of the track be less than the potential energy at the top of the track?
Answer:
This is because of friction and heat lost.
Explanation:
Why does Marx’s workers’ paradise resolve the problems of capitalism?
A. Everything is free, and no one has to work.
B. Workers are divided into three classes, much like in Plato’s ideas.
C. There is no currency in the paradise so no economic problems.
D. People work for their own good instead of a factory owner’s.
9. Mr. Smith went skiing in Maine last weekend. He traveled 523 kilometers to Sugarloaf from
Leominster. His average speed was 109 km/hr. How long did it take Mr. Smith to hit the slopes?
Answer:
Time taken by Mr. smith = 4.80 hour (Approx.)
Explanation:
Given:
Distance travel by Mr. smith = 523 kilometer
Average speed of Mr. smith = 109 km/hr
Find;
Time taken by Mr. smith
Computation:
Time taken = Distance cover / Speed
Time taken by Mr. smith = Distance travel by Mr. smith / Average speed of Mr.
smith
Time taken by Mr. smith = 523 / 109
Time taken by Mr. smith = 4.798 hr
Time taken by Mr. smith = 4.80 hour (Approx.)
how can we know that atmosphere exert pressure explain with figure
The speed of surface waves in water decreases as the water becomes shallower. Suppose waves travel across the surface of a lake withn a speed of 2m/s and a wavelength of 1.5m. When these waves move into a shallower part of the lake, their speed decreases to 1.6m/s, though their frequency remains the same. Find the wavelength of the wave in the shallower water.
Answer:
The correct solution is "1.2 m".
Explanation:
The given values are:
Wavelength of waves,
λ = 1.5 m
Speed of waves on surface,
V = 2 m/sec
Speed of waves in water,
V₁ = 1.6 m/sec
As we know,
⇒ [tex]V=f\times \lambda[/tex]
or,
⇒ [tex]f=\frac{V}{\lambda}[/tex]
On substituting the values, we get
⇒ [tex]=\frac{2}{1.5}[/tex]
⇒ [tex]=1.33 \ Hz[/tex]
hence,
⇒ [tex]\lambda_1=\frac{V_1}{f}[/tex]
On substituting the values, we get
⇒ [tex]=\frac{1.6}{1.33}[/tex]
⇒ [tex]=1.2 \ m[/tex]
Two risks of exposure to High levels of UV radiation
Answer:
uv radiation cause cancer
uv radiation effect our eyes
Answer:
you can get
1:skin cancer
2:eye damage
3:skin damage
4:immune system suppression
choose which two u want
hope this helped
:)
Explanation:
While flying at an altitude of 5.75 km, you look out the window at various objects on the ground. If your ability to distinguish two objects is limited only by diffraction, find the smallest separation between two objects on the ground that are distinguishable. Assume your pupil has a diameter of 4.0 mm and take ???? = 460 nm.
Answer:
the smallest separation between two objects is 0.8067 m
Explanation:
Given the data in the question;
Altitude h = 5.75 km = 5750 m
Diameter D = 4.0 mm = 0.004 m
λ = 460 nm = 4.6 × 10⁻⁷ m
Now, Using Rayleigh criterion for Airy disks resolution.
we know that, Minimum angular separation for resolving two points is;
θ = 1.22λ / D
so we substitute
θ = (1.22 × 4.6 × 10⁻⁷) / 0.004
θ = 5.612 × 10⁻⁷ / 0.004
θ = 1.403 × 10⁻⁴ rad
so minimum separation [tex]d_{min[/tex] = θh
so we substitute
[tex]d_{min[/tex] = (1.403 × 10⁻⁴) × 5750 m
[tex]d_{min[/tex] = 0.8067 m
Therefore, the smallest separation between two objects is 0.8067 m
A right triangle has side lengths of 4 centimeters and 5 centimeters. What is the length of the hypotenuse?
O A. 3 cm
B. 4 cm
C. 5 cm
D. 41 cm
How much heat is needed to warm 365 mL of water in a baby bottle from 240C to 38C
Answer:98.6 degrees Fahrenheit
Explanation:
The first extra-solar planet around a main sequence star (i.e., not a neutron star or white dwarf) was found around the star 51 Pegasi in 1995. The large planet causes a measureable motion of the star around the center of mass of the system. Pegasi 51's orbital motion had a period of 4 days indicating a very large planet very close to the star. A large planet of mass M around a small star of mass 4M. The distance between planet and star is L. Both the planet and star will orbit around the center of mass of the system (marked by the red X).
Required:
Where is the center of mass of the star-planet system?
Answer:
[tex]r_{cm}[/tex]= 1/5 L
Explanation:
To find the center of mass of the system let's use
[tex]r_{cm} = \frac{1}{M}[/tex] ∑ r_i x_i
where m is the total mass of the system
let's apply this expression to our case
Let's set the reference frame on the star
[tex]r_{cm} = \frac{1}{M +4M} ( 4M 0 + M L)[/tex]
r_{cm} = [tex]\frac{1}{5}[/tex] L
[tex]r_{cm}[/tex]= 1/5 L
Hey, the breast center is 1/5 of the distance between the star and the planet.
If acceleration is zero what statement about velocity is
true *
A)Velocity is zero
B)Velocity is constant
C)Velocity cannot be determined
D) Velocity is changing
Answer: A
Velocity is zero because the acceleration isn't affected, and velocity is the rate of change, so it can't be any other options.
Answer:
B)Velocity is constantExplanation:
If an object moves with a velocity and there is no acceleration, then the velocity remains constant. His velocity after five second will be equal to his initial velocity.#keeplearning dude:)Imagine you are on a space mission and you are 6 AU's from the Sun and you use a light sensor to measure the brightness of the Sun. The amount of sunlight received per square centimeter would be different by what factor compared to the same measurement on Earth at AU
Answer:
36 times less.
Explanation:
The distance from you to the sun is 6AU's, and from the sun to the earth is 1 AU.
Therefore,
At Earth sunlight received per unit cm² is:
[tex]I_{earth} = \dfrac{I_o}{4 \pi \times (1)^2}[/tex]
[tex]I_{earth} = \dfrac{I_o}{4 \pi}[/tex]
[tex]I_{me} =\dfrac{I}{4 \pi (6)^2}[/tex]
[tex]I_{me} = \dfrac{I_o}{36(4 \pi)}[/tex]
Thus, [tex]I_{earth} = 36 \times I_{me}[/tex]
Thus, the right answer is 36 times less.
Earth's neighboring galaxy, the Andromeda Galaxy, is a distance of 2.54×107 light-years from Earth. If the lifetime of a human is taken to be 90.0 years, a spaceship would need to achieve some minimum speed vmin to deliver a living human being to this galaxy. How close to the speed of light would this minimum speed be? Express your answer as the difference between vmin and the speed of light c.
Answer:
[tex]0.0018833\ \text{m/s}[/tex]
Explanation:
[tex]d[/tex] = Distance of Andromeda Galaxy from Earth = [tex]2.54\times 10^7\ \text{ly}[/tex]
[tex]t[/tex] = Time taken = [tex]90\ \text{years}[/tex]
[tex]c[/tex] = Speed of light = [tex]3\times 10^8\ \text{m/s}[/tex]
We have the relation
[tex]t=t_o\sqrt{1-\dfrac{v^2}{c^2}}\\\Rightarrow 90=2.54\times 10^7\sqrt{1-\dfrac{v^2}{c^2}}\\\Rightarrow \dfrac{90^2}{(2.54\times 10^7)^2}=1-\dfrac{v^2}{c^2}\\\Rightarrow 1-\dfrac{90^2}{(2.54\times 10^7)^2}=\dfrac{v^2}{c^2}\\\Rightarrow v=c\sqrt{1-\dfrac{90^2}{(2.54\times 10^7)^2}}[/tex]
[tex]c-v=c(1-\sqrt{1-\dfrac{90^2}{(2.54\times 10^7)^2}})\\\Rightarrow c-v=3\times 10^8(1-\sqrt{1-\dfrac{90^2}{(2.54\times 10^7)^2}})\\\Rightarrow c-v=0.0018833\ \text{m/s}[/tex]
The required answer is [tex]0.0018833\ \text{m/s}[/tex].
what is force,momentum,and velocity.
Answer:
A force is a push or pull upon an object resulting from the object's interaction with another object.
Momentum is force or speed of movement.
Velocity defines the path of the motion of the frame or the object
A whole set of birdfeeders are designed using conservation of Angular Momentum to spin when a squirrel jumps on them. This can throw the squirrel off (though not all squirrels give up that easily - see this video for an example). A bird, landing, doesn't cause the same problem. A squirrel, with a mass of 3.00 kg launches itself at the bird feeder with a velocity of 3.40 m/s. The bird feeder has a radius of 6.30 cm and a Moment of Inertia of 2.00 kg m2. Initially the bird feeder is not rotating at all, but starts rotating when the squirrel lands on the outer edge (at the same radius as described above). You can assume that the squirrel is small compared to the size of the bird feeder radius (not true in the video, but it does make this a bit easier for out calculations). What is the angular velocity of the bird feeder - squirrel system after the squirrel lands on it
Answer:
w = 0.319 rad / s
Explanation:
This is an angular momentum problem, let's form a system composed of the feeder and the squirrel, therefore the forces during the collision are internal and the angular momentum is conserved.
initial instant. Before the squirrel jumps
L₀ = m v r
final instant. After the trough and the squirrel are together
L_f = (I_fetter + I_ardilla) w
angular momentum is conserved
L₀ = L_f
m v r = (I_fetter + I_ardilla) w
w = [tex]\frac{mvr}{I_{fetter} + I_{ardilla} }[/tex]
the moment inercial ofbody is
I_thed = 2.00 kg m²
We approach the squirrel to a specific mass
I_ardilla = m r²
we substitute
w = m v r / ( I_[feefer + m r²)
let's calculate
w = 3 3.40 6.30 10⁻² / (2.00 + 3.00 (6.30 10-2)² )
w = 0.6426 / 2.0119
w = 0.319 rad / s
If a runner is running at 100 meters per minute at the exact moment they cross the finish line of a race, it is the
instantaneous speed.
O True
O False
Answer:
Hello! Your answer would be, O False
Explanation:
Hope I helped! Brainiest plz!♥ Have a nice morning! Hope you make a 100%! -Abby
Answer:
False is the correct answer.
Explanation:
plz mark me as brainliest.
A heating coil operates on 220 V if it draws 15.0 A. Find it's resistance
Answer:
R ≈ 15 ohms
Explanation:
Using ohm's law equation,
I = V/R, to solve for the resistance of the heating coil.
R = V/I
Known:
V = 220 v = 220 kgm^2s^-3A^-1
I = 15 A
Unknown:
R =?
Solution:
R = (220 kgm^2s^-3A^-1)/ 15.0 A
R = 14.6 kgm^2s^-3A^-2
R ≈ 15 kgm^2s^-3A^-2
R ≈ 15 ohms
A sound wave travels with a velocity of 1.5 m/s and has a frequency of 500 Hz. What is its wavelength?
Pause
He
When an unbalanced force of 10 N is applied to an object whose mass is 4.0 kg, the acceleration of the object will be:
OA. 40 m/s
OB. 9.8 m/s2
OC 2.5 m/s2
OD. 0.40 m/s2
Answer:
C
Explanation:
a=f/m
10Kgm/s2/4kg
2.5m/s2
When an unbalanced force of 10 N is applied to an object whose mass is 4.0 kg, the acceleration of the object will be 2.5 m/s².
What is acceleration?The rate at which an item changes its velocity is known as acceleration, a vector quantity. If an object's velocity is changing, it is acceleration. The net acceleration that objects get as a result of the combined action of gravity and centrifugal force is known as the Earth's gravity, or g. It is a vector quantity whose strength or magnitude is determined by the norm and whose direction correlates with a plumb bob.
Given in the question, force 10 N and mass 4.0 Kg the acceleration is,
a = 10/4 = 2.5 m/sec²
When an unbalanced force of 10 N is applied to an object whose mass is 4.0 kg, the acceleration of the object will be 2.5 m/s².
To learn more about acceleration refer to the link:
brainly.com/question/12550364
#SPJ2
two bodies of masses 10 kg and 4 kg are moving in opposite direction with the same velocity of 5 m per second if they Collide and stick together after the impact calculate their velocities
Answer:
dasd
Explanation:
dsdas
pls help
What is the frequency range of UV light? Of infrared light?
Answer:
violet region (UV) is wavelength less than 400 nm, the most common is between 200 and 400 nm,
The infrared (IR) range wavelengths greater than 700 nm
Explanation:
The spectral region near the visible is what humans can see with our eyes.
The range of the ultra violet region (UV) is wavelength less than 400 nm, the most common is between 200 and 400 nm, this radiation is responsible for tanning and skin burns.
The infrared (IR) range is a very wide range that begins at wavelengths greater than 700 nm and continues up to approximately 25,000 nm (1012 Hz).
Can you please help me ?
Answer:
Distance is 300 and displacement is 100
Explanation:
Distance= 100+100+100=300
Displacement=100
plz mark me as brainliest.
A 0.5kg ball of clay originally moving at 6 m/s strikes a wall and comes to rest in 0.25s, what is the magnitude of the impulse given to the ball of clay?
A) 0.75 kg m/s
B) 1.5 kg m/s
C) 3.0 kg m/s
D) 12 kg m/s
Answer:
C I did USA testprep
Explanation:
