Answer:
Explained below.
Explanation:
We are told that the surface material is ejected into space at a high speed. This means that it will have a likely high momentum as well.
Now, we can say that the total momentum is conserved because the entire asteroid system behaves like an isolated system.
Also, as the surface material is moving with the high momentum like we established earlier, it will cause the asteroid to move with a speed in an opposite direction which also means deflection in an opposite direction.
Answer:
Explained below.
Explanation:
The material ejected from the surface of the asteroid would have a significant momentum. Since the asteroid and all its material is an isolated system, the ejection would cause an oppositely directed change in momentum of the asteroid, according to the law of conservation of momentum.
The ejected material is analogous to gases expelled from a rocket, and the asteroid is analogous to a rocket.
Sometimes a person cannot clearly see objects close up or far away. To correct this type of vision, bifocals are often used. The top half of the lens is used to view distant objects and the bottom half of the lens is used to view objects close to the eye. Bifocal lenses are used to correct his vision. A diverging lens is used in the top part of the lens to allow the person to clearly see distant objects.
1. What power lens (in diopters) should be used in the top half of the lens to allow her to clearly see distant objects?
2. What power lens (in diopters) should be used in the bottom half of the lens to allow him to clearly see objects 25 cm away?
Answer:
1) P₁ = -2 D, 2) P₂ = 6 D
Explanation:
for this exercise in geometric optics let's use the equation of the constructor
[tex]\frac{1}{f} = \frac{1}{p} + \frac{1}{q}[/tex]
where f is the focal length, p and q are the distance to the object and the image, respectively
1) to see a distant object it must be at infinity (p = ∞)
[tex]\frac{1}{f_1} = \frac{1}{q}[/tex]
q = f₁
2) for an object located at p = 25 cm
[tex]\frac{1}{f_2} = \frac{1}{25} + \frac{1}{q}[/tex]
We can that in the two expressions we have the distance to the image, this is the distance where it can be seen clearly in general for a normal person is q = 50 cm
we substitute in the equations
1) f₁ = -50 cm
2)
[tex]\frac{1}{f_2} = \frac{1}{25} + \frac{1}{50}[/tex]
[tex]\frac{1}{f_2}[/tex] = 0.06
f₂ = 16.67 cm
the expression for the power of the lenses is
P = [tex]\frac{1}{f}[/tex]
where the focal length is in meters
1) P₁ = 1/0.50
P₁ = -2 D
2) P₂ = 1 /0.16667
P₂ = 6 D
A go-cart is traveling at a rate of 25 m/sec for 20 seconds. How far will the go cart travel?
Answer:
Distance travel by go-cart = 500 meter
Explanation:
Given:
Speed of go cart = 25 m/s
Time travel = 20 seconds
Find:
Distance travel by go-cart
Computation:
Distance = Speed x time
Distance travel by go-cart = Speed of go cart x Time travel
Distance travel by go-cart = 25 x 20
Distance travel by go-cart = 500 meter
Instead of changing the focal length of the lens, the eyes of amphibians work in a different manner: a set of muscles changes the shape of the eye which increases the distance between the front of the eye and the retina. The world's largest frog, the Goliath frog of west Africa, has an eye with a maximum size similar to a human's: 2.5cm. However, unlike a human, where the focal length is also 2.5cm, the focal length of the Goliath frog's eye is 2.146 cm. What is the maximum distance this frog can see
Answer:
The answer is "15.56 cm".
Explanation:
[tex]v= 2.5 \ cm\\\\f= 2.154 \ cm[/tex]
Calculating object of length is x so:
[tex]u= -x[/tex]
Using formula:
[tex]\to \frac{1}{v}-\frac{1}{u}=\frac{1}{f}\\\\\to \frac{1}{2.5}-\frac{1}{-x}=\frac{1}{2.154}\\\\\to \frac{1}{x}=\frac{1}{2.154}-\frac{1}{2.5}\\\\\to \frac{1}{2.5}-\frac{1}{-x}=\frac{1}{2.154}\\\\\to x= 15.56 \ cm[/tex]
16. Two electric bulbs marked 100W 220V and 200W 200V have tungsten
filament of same length. Which of the two bulbs will have thicker
filament?
Answer:
The second bulb will have thicker filament
Explanation:
Given;
First electric bulb: Power, P₁ = 100 W and Voltage, V₁ = 220 V
Second electric bulb: Power, P₂ = 200 W and Voltage, V₂ = 200 V
Resistivity of tungsten, ρ = 4.9 x 10⁻⁸ ohm. m
Resistance of the first bulb:
[tex]P = IV = \frac{V}{R} .V = \frac{V^2}{R} \\\\R = \frac{V^2}{P} \\\\R_1 = \frac{V_1^2}{P_1} = \frac{(220)^2}{100} = 484 \ ohms[/tex]
Resistance of the second bulb:
[tex]R_2 = \frac{V_2^2}{P_2} = \frac{(200)^2}{200} = 200 \ ohms[/tex]
Resistivity of the tungsten filament is given by the following equation;
[tex]\rho = \frac{RA}{L}[/tex]
where;
L is the length of the filament
R is resistance of each filament
A is area of each filament
[tex]A = \pi r^2[/tex]
where;
r is the thickness of each filament
[tex]\rho = \frac{R (\pi r^2)}{L} \\\\\frac{\rho L}{\pi} = Rr^2 \\\\Recall ,\ \frac{\rho L}{\pi} \ is \ constant \ for \ both \ filaments\\\\R_1r_1^2 = R_2r_2^2\\\\(\frac{r_1}{r_2} )^2 = \frac{R_2}{R_1} \\\\\frac{r_1}{r_2} = \sqrt{\frac{R_2}{R_1} } \\\\\frac{r_1}{r_2} = \sqrt{\frac{200}{484} } \\\\\frac{r_1}{r_2} = 0.64\\\\r_1 = 0.64 \ r_2\\\\r_2 = 1.56 \ r_1[/tex]
Therefore, the second bulb will have thicker filament
In many places on Earth, humans are responsible for the removal of grasses, shrubs, trees, and other plants with roots that hold soil in place. This activity is best described by which of the following? *
A) deforestation
B) urbanization
C) air pollution
D) rise in sea level
Which two chemical equations show double-replacement reactions?
A. C+02 - CO2
B. 2Li + CaCl2 - 2LiCl + Ca
I C. Ca(OH)2 + H2S04 - CaSO4 + 2H20
D. Na2CO3 + H2S - H2CO3 + Na2S
The two chemical equations show double-replacement reactions are Ca(OH)2 + H2S04 - CaSO4 + 2H20 and Na2CO3 + H2S - H2CO3 + Na2S.
What is double replacement reaction?A double replacement reaction have two ionic compounds that are exchanging anions or cations.
From the given options, we can choose the following based on their exchange of anions or cations.
Ca(OH)2 + H2S04 - CaSO4 + 2H20Na2CO3 + H2S - H2CO3 + Na2SThus, the two chemical equations show double-replacement reactions are Ca(OH)2 + H2S04 - CaSO4 + 2H20 and Na2CO3 + H2S - H2CO3 + Na2S.
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#SPJ2
Batteries are not perfect. They can't deliver infinite current. As the current load on a battery gets larger, the voltage output gets smaller.
a. True
b. False
NO LINKS PLEASE
At what speed do a bicycle and its rider, with a combined mass of 90 kg
k
g
, have the same momentum as a 1500 kg
k
g
car traveling at 4.0 m/s
m
/
s
?
Answer:
2
Explanation:
why the speed of light decreases as it passes from air into another substance?
Answer:
If light enters any substance with a higher refractive index (such as from air into glass) it slows down. The light bends towards the normal line. If light travels enters into a substance with a lower refractive index (such as from water into air) it speeds up. The light bends away from the normal line.
In a nuclear fusion reaction, atoms:
split apart.
combine.
explode.
cool down.
A spring has a spring constant of 450 N/m. How much must this spring be stretched to store 49 J of potential energy?
Answer:
W = 1/2 K x^2
x^2 = 2 * W / K = 2 * 49 J / (N/m) = .218 / m^2
x = .467 m
In a movie production, a stunt person must leap from a balcony of one building to a balcony 3.0 m lower on another building. If the buildings are 2.0 m apart, what is the minimum horizontal velocity the stunt person must have to accomplish the jump? Assume no air resistance and that ay = −g = −9.81 m/s2 . (Ans. 2.6m/s) PLS SHOW WORK
This question involves the concept of semi-projectile motion. It can be solved using the equations of motion in the horizontal and the vertical motion.
The minimum horizontal velocity required is "2.6 m/s".
First, we will analyze the vertical motion of the stunt person. We will use the second equation of motion in the vertical direction to find the time interval for the motion.
[tex]h=v_it+\frac{1}{2}gt^2[/tex]
where,
h = height = 3 m
vi = initial vertical speed = 0 m/s
t = time interval = ?
g = acceleration due to gravity = 9.81 m/s²
therefore,
[tex]3\ m = (0\ m/s)(t) + \frac{1}{2}(9.81\ m/s^2)t^2\\\\t^2 = \frac{(3\ m)(2)}{9.81\ m/s^2}\\\\t = \sqrt{0.611\ s^2}[/tex]
t = 0.78 s
Now, we will analyze the horizontal motion. We assume no air resistance, so the horizontal motion will be uniform. Hence, using the equation of uniform motion here:
[tex]s = vt\\\\v = \frac{s}{t}[/tex]
where,
s = horizontal distance = 2 m
t =0.78 s
v = minimum horizontal velocity = ?
Therefore,
[tex]v = \frac{2\ m}{0.78\ s}[/tex]
v = 2.6 m/s
Learn more about equations of motion here:
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The attached picture shows the equations of motion in the horizontal and vertical directions.
Protons, neutrons, electrons, and a nucleus are
A box having a weight of 8 lb is moving around in a circle of radius rA = 2 ft with a speed of (vA)1 = 5 ft/s while connected to the end of a rope. If the rope is pulled inward with a constant speed of vr = 4 ft/s, determine the speed of the box at the instant rB = 1 ft. How much work is done after pulling in the rope from A to B? Neglect friction and the size of the box
Answer:
W = 1.875 J
Explanation:
For this exercise let's use the relationship between work and kinetic energy
W = ΔK
The kinetic energy of rotational motion is
K₀ = ½ I w²
we can assume that the box is small, so it can be treated as a point object, with moment of inertia
I = m rₐ²
angular and linear velocity are related
v = w r
w = v / r
we substitute in the equation, for point A
K₀ = ½ (m rₐ²) (v / rₐ)²
K₀ = ½ m v²
For the final point B, as the system is isolated the angular momentum is conserved
initial L₀ = Io wo
final L_f = I_f w_f
L₀ = L_f
I₀ w₀ = I_f w_f
(m rₐ²) w₀ = (m [tex]r_{b} ^2[/tex]) w_f
w_f = (rₐ/r_b)² w₀
with this value we find the final kinetic energy
K_f = ½ I_f w_f²
K_f = ½ (m [tex]r_{b}^2[/tex]) ( (rₐ / r_b)² w₀) ²
K_f = ½ m [tex]\frac{r_a^4}{r_b^2} \ w_o^2[/tex]
we substitute in the realcion of work
W = K_f - K₀
W = ½ m [tex]( \( \frac {r_a^2 }{r_b} )^2[/tex] w₀² - ½ m v²
W = ½ m [tex]\frac{r_a^4}{r_b^2} ( \frac{v}{r_a} ) ^2[/tex] - ½ m v²
W = ½ m [tex]\frac{r_a^2}{r_b^2} \ v^2[/tex] - ½ m v2
W = ½ m v² (([tex]( \ (\frac{r_a}{r_b})^2 -1)[/tex]
let's calculate
W = ½ ( [tex]\frac{8}{32}[/tex] ) 5 ((2/1)² -1)
W = 0.625 (3)
W = 1.875 J
The moon does not stay at the same distance from the earth.why?
Answer:
The moon does not stay at the same distance of the earth because the ortbit of the moon is slightly elliptical. If earth is not tilted at an angle of 66.5°, there will be no change in the season and the earth will have equal length of days and night.
Explanation:
mark me brainlest
A 20 ft ladder leans against a wall. The bottom of the ladder is 3 ft from the wall at time t=0 and slides away from the wall at a rate of 2ft/sec. Find the velocity of the top of the ladder at time t=1.
Answer: 0.516 ft/s
Explanation:
Given
Length of ladder L=20 ft
The speed at which the ladder moving away is v=2 ft/s
after 1 sec, the ladder is 5 ft away from the wall
So, the other end of the ladder is at
[tex]\Rightarrow y=\sqrt{20^2-5^2}=19.36\ ft[/tex]
Also, at any instant t
[tex]\Rightarrow l^2=x^2+y^2[/tex]
differentiate w.r.t.
[tex]\Rightarrow 0=2xv+2yv_y\\\\\Rightarrow v_y=-\dfrac{x}{y}\times v\\\\\Rightarrow v_y=-\dfrac{5}{19.36}\times 2=0.516\ ft/s[/tex]
Suppose a diode consists of a cylindrical cathode with a radius of 6.200×10−2 cm , mounted coaxially within a cylindrical anode with a radius of 0.5580 cm . The potential difference between the anode and cathode is 400 V . An electron leaves the surface of the cathode with zero initial speed (vinitial=0). Find its speed vfinal when it strikes the anode.
Answer:
The final speed will be "[tex]1.185\times 10^7 \ m/sec[/tex]".
Explanation:
The given values are:
Potential difference,
Δv = 400 v
Radius,
r = 0.5580 cm
As we know,
⇒ [tex]W=e \Delta v[/tex]
and,
⇒ [tex]\frac{1}{2}mv^2=e \Delta v[/tex]
then,
⇒ [tex]v=\sqrt{\frac{2e \Delta v}{m} }[/tex]
On substituting the values, we get
⇒ [tex]=\sqrt{\frac{2\times 1.6\times 10^{-19}\times 400}{9.11\times 10^{-31}} }[/tex]
⇒ [tex]=\sqrt{\frac{1.6\times 10^{-19}\times 800}{9.11\times 10^{-31}}}[/tex]
⇒ [tex]=1.185\times 10^7 \ m/sec[/tex]
A copper wire of resistivity 2.6 × 10-8 Ω m, has a cross sectional area of 35 × 10-4 cm2
. Calculate
the length of this wire required to make a 10 Ω coil.
Answer:
the length of the wire is 134.62 m.
Explanation:
Given;
resistivity of the copper wire, ρ = 2.6 x 10⁻⁸ Ωm
cross-sectional area of the wire, A = 35 x 10⁻⁴ cm² = ( 35 x 10⁻⁴) x 10⁻⁴ m²
resistance of the wire, R = 10Ω
The length of the wire is calculated as follows;
[tex]R = \frac{\rho L}{A} \\\\L = \frac{RA}{\rho} \\\\L= \frac{10 \times (35\times 10^{-4}) \times 10^{-4}}{2.6 \times 10^{-8}} \\\\L = 134.62 \ m[/tex]
Therefore, the length of the wire is 134.62 m.
Help me please with both questions?
Answer:
question #1 is A
Question #2 is C
Explanation:
. Estimate the buoyant force that air exerts on you. (To do this, you can estimate your volume by knowing your weight and by assuming that your weight density is a bit less than that of water.)
Answer:
[tex]0.886[/tex] N buoyant force is exerted by air
Explanation:
My weight is [tex]75[/tex] Kg
Weight = mass * gravity
As we know
Buoyant Force is equal to the product of density * acceleration due to gravity and volume of the body
Assuming weight density is a bit less than that of water or equal to water i.e [tex]997.77[/tex] kg/m3
Volume is equal to mass / density
[tex]= 75[/tex] Kg * g/[tex]997.777[/tex]
[tex]= 0.0751[/tex] * g
Buoyant Force
= Volume * g * density
[tex]= 0.0751 * 9.8 * 1.2041[/tex]kg/m3
[tex]= 0.886[/tex] N
When a 20 kg explosive detonates and sends a 5 kilogram piece traveling to the right at 105 m/s
what is the speed and direction of the other 15 kilogram piece of the explosive!
Answer:
speed: 35m/s
direction: left
Explanation:
Assuming the right side is the positive direction:
before explosion:
P = mv = 0
after explosion:
P' = 15P + 5P
(Set the velocity of the 15kg piece after explosion as v1' and the velocity of the 5kg piece after explosion as v2')
P' = 0.75mv1' + 0.25mv2'
P' = (15kg)v' + (5kg)(105m/s)
P' = 525kg/m/s + (15kg)v1'
P = P'
525kg/m/s + (15kg)v1' = 0
(15kg)v1' = -525kg/m/s
v1' = -35m/s
speed = |-35| = 35m/s
direction is to the left since the right side is the positive direction.
Which one the answer to this question
A wave has a frequency of 67 Hz and a wavelength of 7.1 meters. What is the speed of this
wave?
Answer:
475.7 m/s
Explanation:
Given,
Frequency ( f ) = 67 Hz
Wavelength ( λ ) = 7.1 m
To find : Speed ( v ) = ?
Formula : -
v = f λ
v
= 67 x 7.1
= 475.7 m/s
Therefore,
the speed of the wave is 475.7 m/s.
Tameika makes a table about sensory organs
Eye
skin
brain
tongue
Which organ should be removed from the table?
A. eye
B. skin
C. brain
D. tongue
Answer:
I think its d
Explanation:
I'm not sure I'm sorry if I'm wrong
A 10 kg box initially at rest is pulled with a 50 N horizontal force for 4 m across a level surface. The force of friction
acting on the box is a constant 20 N. How much work is done by the gravitational force?
A. 03
OB. 10 J
C. 100
D. 50 J
Answer:
B i think
Explanation:
...
Tony ran 600 meters in 60 seconds. What was Tony's speed during the
race?
If a virtual image is formed 10.0 cm along the principle axis from a convex mirror of focal length-15.0 cm, how far is the object from the mirror
Answer:
U=30cm
Explanation:
All you have to do is to put
Mirror formula , 1/f=1/u + 1/v
You should be careful in sign convention .
Virtual image is negative
we take focal length of convex lens negative even if its not given and so on...
The ear drum vibrates when struck by sound waves and directly sends a message to the brain that is then recognized as sound
True or False
Answer:
true
Explanation:
A carnival ride starts at rest and is accelerated from an initial angle of zero to a final angle of 6.3 rad by a rad counterclockwise angular acceleration of 2.0 s2 What is the angular velocity at 6.3 rad?
The final angular velocity of the carnival ride at a displacement of 6.3 rad is 25.2 rad/s.
Final angular velocity of the carnival ride
The final angular velocity of the carnival ride is determined by applying third kinematic equation as shown below;
ωf = ωi + 2αθ
where;
ωf is the final angular velocity of the carnival ride = ?ωi is the initial angular velocity of the carnival ride = 0α is the angular acceleration = 2.0 rad/s²θ is the angular displacement of the carnival ride = 6.3 radωf = 0 + 2(2.0) x 6.3
ωf = 25.2 rad/s
Thus, the final angular velocity of the carnival ride at a displacement of 6.3 rad is 25.2 rad/s.
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Answer: 5.0 rad/s
Explanation: Because that’s what khan said so try it out.
A woman shouts at a boy who is underwater what happens to the speed of the sound wave as it moves from the air into the water
Answer:
B. it increases
Explanation:
As shown in the table provided, the speed of sound in water (1493 m/s) is greater than the speed of sound in air (346 m/s).
Answer:
B is the correct answer.
Explanation:
