Answer:
The final speed of the semi truck is 30.11 m/s.
Explanation:
Given;
mass of the motorcycle, m₁ = 200 kg
initial speed of the motorcycle, u₁ = 40 m/s
mass of the semi truck , m₂ = 18,000 kg
initial speed of the semi truck, u₂ = 30 m/s (same direction as u₁)
If the two vehicles stick together after collision, the final speed of the motorcycle will be equal to final speed of the semi truck = v
Apply the principle of conservation of linear momentum;
m₁u₁ + m₂u₂ = v(m₁ + m₂)
(200 x 40) + (18,000 x 30) = v(200 + 18,000)
8000 + 540,000 = v(18,200)
548,000 = v(18,200)
[tex]v = \frac{548,000}{18,200} \\\\v = 30.11 \ m/s[/tex]
Therefore, the final speed of the semi truck is 30.11 m/s.
Type your answers to these 4 questions in the submission area.
1. What do you think a plate boundary is?
2. What is an earthquake?
3. What happens during an earthquake?
4. What do you think about the plates that make up Earth’s outer layer?
Squamous cells are basically clear, dead cells, filled with an ingredient of keratin
called eleidin.
True
False
Squamous cells, which are tiny, flat cells that resemble fish scales, are present in the tissue that makes up the skin's surface, and the lining of the digestive and respiratory systems.
What are the characteristics of Squamous cells?Simple squamous epithelial cells serve as filtration and diffusion mediators. They enable facile transmembrane transit of tiny molecules (i.e., across the membrane and through the cell) because of their straightforward and thin construction.
The centre of squamous epithelial cells, which are big, flattened cells loaded with cytoplasm, is a small, spherical nucleus. They are spherical and unevenly flat.
Therefore, The lumen is a free area that is surrounded by simple squamous epithelium. Simple squamous cells join together to produce a layered structure.
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A space expedition discovers a planetary system consisting of a massive star and several spherical planets. The planets all have the same uniform mass density. The orbit of each planet is circular. In the observed planetary system, Planet A orbits the central star at the distance of 2R and takes T hours to complete one revolution around the star. Planet B orbits the central star at the distance of R. Which of the following expressions is correct for the number of hours it takes Planet B to complete one revolution around the star?
a. 1/√8T
b. 1/2T
c. 1/√4T
d. 2T
e. √8T
Answer:
[tex]T_B=(\frac{T}{\sqrt{8}})[/tex]
Explanation:
Distance of Planet A from Central star 2R
Time of Resolution T_A=T
Distance of Planet B from orbit star R
Generally the equation for Kepler's law of periods is given by
[tex]\frac{T_A^2}{T_B^2}=\frac{R_A^3}{R_B^3}[/tex]
[tex]T_B^2=T_A^2 \frac{R_A^3}{R_B^3}[/tex]
[tex]T_B^2=T_A^2 (\frac{R_A}{R_B})^3[/tex]
[tex]T_B^2=T^2 (\frac{R}{2R})^3[/tex]
[tex]T_B^2=T^2 (\frac{1}{R})^3[/tex]
[tex]T_B^2=(\frac{T^2}{8})[/tex]
Therefore the following expressions is correct for the number of hours it takes Planet B to complete one revolution around the star
[tex]T_B=(\frac{T}{\sqrt{8}})[/tex]
The number of hours that it takes for planet B to complete one revolution around the star is : ( A ) ( [tex]\frac{T}{\sqrt{8} }[/tex] )
Given data :
Distance of Planet A from massive star = 2R
Time taken by Planet A to orbit the massive star = Tₐ
Distance of Planet B from massive star = R
Time taken by Planet B to orbit the massive star = T[tex]_{b}[/tex]
To determine the proper expression exhibiting the correct number of hours it will take for Planet B to complete a revolution
we will apply Kepler's law of periods
[tex]\frac{T^{2} _{a} }{T_{b} ^{2} } = \frac{R_{a} ^{3} }{R_{b} ^{3} }[/tex]
∴ [tex]T_{b} ^{2} = T_{a} ^{2} ( \frac{R_{a} }{R_{b} } )^{3}[/tex]
[tex]T_{b} ^{2} = T^2 ( 1/R)^3[/tex]
Hence T[tex]_{b}[/tex] = ( [tex]\frac{T}{\sqrt{8} }[/tex] )
There we can conclude that The number of hours that it takes for planet B to complete one revolution around the star is : ( A ) ( [tex]\frac{T}{\sqrt{8} }[/tex] )
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In the first 2 seconds of a race, an athlete accelerates constantly from 0 m/s and reaches a
speed of 9 m/s.
Calculate the acceleration of the athlete. You do not need to state the unit here.
Given :
Initial speed, u = 0 m/s.
Final speed, v = 9 m/s.
Time taken to reach that speed, t = 2 sec.
To Find :
The acceleration of the athlete.
Solution :
We know, in uniform acceleration motion :
[tex]v= u+at\\\\a = \dfrac{v-u}{t}\\\\a = \dfrac{9-0}{2}\ m/s\\\\a = 4.5 \ m/s^2[/tex]
Therefore, acceleration of the athlete is 4.5 m/s².
On a sunny day, a family decided to take sail on a nearby lake. During sailing the wind applies a force of 25N north on the sails of the sailboat. The
water however exerts a force of 20 N east. If the total mass of both the sailboat and the crew (family on board) is 20kg, what is the magnutude and
direction of the acceleration produced?
Answer:
its n
Explanation:
The acceleration of the boat is obtained as 1.6 ms-2.
We have to obtain the resultant force acting on the boat in order to obtain the acceleration of the boat.
Resultant force on the boat = √(25)^2 + (20)^2 = √625 + 400 = 32 N
Now recall that from Newton's laws; F = ma
m= mass of the body
a = acceleration of the body
32 N = 20Kg × a
a = 32 N/20Kg
a = 1.6 ms-2
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If you Blow up a balloon, will the composition of Air inside the balloon changes? Why? Why not?
Answer:
it change by the pressure of the air and how its pulling the elastic even more
Explanation:
therefore it changes shape and has more differences than before sometimes the colour of it will go lighter if a lot of air and darker if not a lot
Which items in this image are electrically conductive?
Check all that apply
the power lines themselves
the wooden pole that supports the lines
the rubber soles on the worker's boots
the metal tools the worker uses
the wooden ladder leaning against the lines
all except the rubber boots.
The answers should be The power lines themselves and The metal tools the worker uses (the 1st and 4th choices).
(For anyone curious, the image I attached to this answer is the image given for this problem.)
Humans have three types of cone cells in their eyes, which are responsible for color vision. Each type absorbs a certain part of the visible spectrum. Suppose a particular cone cell absorbs light with a wavelength of . Calculate the frequency of this light.
Answer:
This question is incomplete
Explanation:
This question is incomplete because of the absence/omission of the wavelength but the completed question is attached below.
The formula for frequency is f = c/λ
where f is the frequency (unknown)
c is the speed of light (3.0 × 10⁸ m/s)
λ is the wavelength (599 nm)
f = (3 × 10⁸) ÷ 599
f = 500834.73 Hz or 5 × 10⁵ Hz
A light ray in glass (n=1.5) hits the air-glass interface at an angle of 10 degrees from the normal. What angle from the normal is the light ray in the air (n=1.0)? (You can use the small angle approximation.)
Answer:
The angle from the normal is 15.1°.
Explanation:
We can find the angle by using Snell's law:
[tex] n_{1}sin(\theta_{1}) = n_{2}sin(\theta_{2}) [/tex]
Where:
n₁: is the first medium (glass) = 1.5
n₂: is the second medium (air) = 1.0
θ₁: is the first angle (in the glass) = 10°
θ₂: is the second angle (in the air) =?
[tex] \theta_{2} = arcsin(\frac{n_{1}sin(\theta_{1})}{n_{2}}) = arcsin(\frac{1.5*sin(10)}{1.0}) = 15.1 ^{\circ} [/tex]
Therefore, the angle from the normal is 15.1°.
I hope it helps you!
Two rope tows operate on the same ski slope. Tow A with a heavier load pulls as fast as Tow B. Which does the most work? Which has the most power?
Answer:
tow a has the most power but tow b does the most work
Explanation:
Tow A with a heavier load pulls as fast as Tow B, Then A does The most work, and A also has more power.
What is power?It is defined as the Rate of work done or the ratio of work done and time.
If the amount of work(W) is down over a time interval of (t) then the power is given by,
P=W/t
Where W=is work done by a force is known as a product of force and distance.
t= time interval in s.
Force
F= force which is known as the rate of momentum.=mv/t
where m= mass of the body in kg.
v= velocity in m/s.
Here in the question, given that A pulls heavier weight than B that means A applied more force which results in more work.
When more work is done by A than by B then the power of an also greater than B because they move at the same velocity.
Hence, The Work done as well as the power of A is more.
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Amateur radio operators in the United States can transmit on several bands. One of those bands consists of radio waves with a wavelength near . Calculate the frequency of these radio waves.
This question is not complete, the complete question is;
Amateur radio operators in the United States can transmit on several bands. One of those bands consists of radio waves with a wavelength near 20 m . Calculate the frequency of these radio waves.
Answer: frequency of the radio waves is 15 MHz
Explanation:
Given that;
wavelength λ = 20 m
we know that; Speed of light C = 3 × 10⁸ m/s
frequency f = ?
we know that;
frequency f = Speed of light C / wavelength λ
so we substitute
frequency f = (3 × 10⁸m/s) / (20m)
frequency f = 15 × 10⁶ Hz
we also know that 1 Hertz = 0.000001 Megahertz
so
frequency f = 15 × 10⁶ Hz × 0.000001
frequency f = 15 MHz
Therefore frequency of the radio waves is 15 MHz
why is there a difference between potential and kinetic energy?
Potential energy is stored energy because it has the potential to do something which laters turns into kinetic energy which is the moving energy.
Certain neutron stars (extremely dense stars) are believed to be rotating at about 1 rev/s. If such a star has a radius of 15 km, what must be its minimum mass so that material on its surface remains in place during the rapid rotation
Answer:
1.9981 * 10^24 Kg
Explanation:
Given that
R = 15 * 10^3
ω = 1 rev/s = 2π rad/s
From;
g =GM/R^2 - Rω^2
If g = 0
M = R ^3ω^2/G
M = (15 * 10^3)^3 * (2π)^2/6.67 * 10^-11
M = 1.9981 * 10^24 Kg
Ignoring mass and weight contributed by fuel, what happens when the space shuttle takes off and moves away from earth?
Answer:
Its mass increases and weight decreases. Its mass remains constant and weight decreases.
Please could you help me. A race car travels along a straight length of track at 65 m/s. The race car and driver have a
combined mass of 580 kg. Calculate the total energy in the kinetic energy store of the car and
driver. Make sure you give the correct symbol for the unit. *
Answer:
1.23×10⁶ J
Explanation:
From the question given above, the following data were obtained:
Velocity (v) = 65 m/s
Mass (m) = 580 Kg
Kinetic energy (KE) =?
The kinetic energy can be obtained by using the following formula :
KE = ½mv²
Where:
KE => is the kinetic energy.
m => is the mass
v => is the velocity
With the above formula, we can obtain the kinetic energy as follow:
Velocity (v) = 65 m/s
Mass (m) = 580 Kg
Kinetic energy (KE) =?
KE = ½mv²
KE = ½ × 580 × 65²
KE= 290 × 4225
KE = 1.23×10⁶ J
Thus, the kinetic energy is 1.23×10⁶ J
An object with velocity 141 ft/s has a kinetic energy of 1558.71 ft∙lbf, on a planet whose gravity is 31.5 ft/s2. What is its mass in pounds
Answer:
The mass of the object is 5.045 lbm.
Explanation:
Given;
kinetic energy of the object, K.E = 1558.71 ft.lbf
velocity of the object, V = 141 ft/s
The kinetic energy of the object is calculated as;
[tex]K.E = \frac{1}{2} mV^2\\\\mV^2 = 2K.E\\\\m = \frac{2K.E}{V^2} \\\\1 \ lbf = 32.174 \ lbm.ft/s^2\\\\m = \frac{2 \ \times \ 1558.71 \ ft.lbf \ \times \ 32.174 \ lbm.ft/s^2 }{(141 \ ft/s)2 \ \ \times \ \ \ \ 1 \ lbf\ }[/tex]
[tex]m = \frac{(2 \ \times \ 1558.71 \ \times \ 32.174) \ lbm.ft^2/s^2 }{(141 )^2\ ft^2/s^2 }\\\\m = \frac{(2 \ \times \ 1558.71 \ \times \ 32.174) \ lbm }{(141 )^2 }\\\\m = 5.045 \ lbm[/tex]
Therefore, the mass of the object is 5.045 lbm.
A working fluid enters a steady flow system with a velocity of 30m/s and leaves at
140m/s. The mass flow rate is 9 kg/s. The properties of the fluid entry are 13.8 bar,
0.122 m3
/kg, and internal energy 42.2 kJ/kg, and exit properties are 1.035 bar, 0.805
m3
/kg and internal energy 208 kJ/kg. Determine the work transfer in kW from the
system
Answer:
Pretty sure it’s A
Explanation:
Would ike brainliest
Block A of mass M is at rest and attached to the top of a spring. The block compresses the spring a distance d from its uncompressed length at equilibrium. Block B of mass 2M is released from rest at a height h above block A . The two blocks collide and stick together. Express all answers in terms of d , h , M , and physical constants, as appropriate.
Required:
a. Derive an expression for the spring constant k of the spring.
b. Derive an expression for the speed of block B just before it collides with block A.
c. Derive an expression for the speed of the blocks immediately after the collision.
d. Derive an expression for the maximum compression of the spring after the collision.
e. If the collision between the blocks was elastic, would the maximum compression of the spring be greater than, less than, or equal to that found in part (d)? Justify your answer.
Answer:
a) k = Mg / d , b) v = √2gh , c) v_{f} = [tex]\frac{2}{3} \ \sqrt{2gh}[/tex], d) x² + 6d x - [tex]\frac{8}{3}[/tex] dh = 0
e)the spring must compress a greater distance.
Explanation:
a) when the block of mass M is placed on the spring, we have an equilibrium condition,
∑ F = 0
[tex]F_{e}[/tex]- W = 0
k d = Mg
k = Mg / d
b) let's use the concepts of energy to find the velocity of the block just before the collision
starting point. Position when released
Em₀ = U = m g h
lowest point. Right at the point of shock
[tex]Em_{f}[/tex] = K = ½ m v²2
as there is no friction, energy is conserved
Em₀ = Em_{f}
mg h = ½ m v²
v = √2gh
c) The velocity of the two blocks after the collision, we define a system formed by the two blocks, in such a way that the forces during the collision are internal and the moment is conserved
initial instant. Just before the crash
p₀ = 2M v + M 0
final instant. Just after the shock, before the spring compression begins
p_{f} = (2M + M) v_{f}
the moment is preserved
p₀ = p_{f}
2M v = 3M v_{f}
v_{f} = ⅔ v
v_{f} = [tex]\frac{2}{3} \ \sqrt{2gh}[/tex]
d) now we work with the joined system after the collision, let's use the concepts of energy
starting point. After shock, before beginning spring compression
Em₀ = K = ½ (3M) [tex]v_{f}^2[/tex]
Em₀ = 3/2 M (\frac{2}{3} \ \sqrt{2gh})²
Em₀ = 4/3 M gh
final point. With the spring fully compressed
Em_f = K_e + U = ½ k x² + (3M) g x
in this case we have taken the zero of gravitational potential energy at the point where the blocks collide, as there is no friction, the energy is conserved
Em₀ = Em_f
4/3 M g h = ½ k x² + 3M g x
½ k x² + 3Mg x - 4/3 Mgh = 0
we substitute the expression for k
[tex]\frac{1}{2}[/tex] ([tex]\frac{Mg}{d}[/tex]) x² + 3Mg x - [tex]\frac{4}{3}[/tex] Mgh = 0
[tex]\frac{x^{2} }{2d}[/tex] + 3 x - [tex]\frac{4}{3}[/tex]h = 0
to find the value of the spring compression, the second degree equation must be solved
x² + 6d x - [tex]\frac{8}{3}[/tex] dh = 0
x = [-6d ±[tex]\sqrt{(36 d^{2} - 4 \frac{8}{3} dh) }[/tex] ] / 2
x = [-6d ± 6d [tex]\sqrt{ 1 - \frac{32}{3 \ 36} \ \frac{h}{d} }[/tex] ]/2
x = 3d ( -1± [tex]\sqrt{ 1 - 0.296 \frac{h}{d} }[/tex] )
e) If the collision elastic force would not lose any part of the kinetic energy during the collision, therefore the speed of the block of mass M would be much higher and therefore the spring must compress a greater distance.
What is the most common frame of reference used by humans
Answer:
All measurements of motion will be compared to a frame of reference. Therefore, the most commonly used frame of reference is Earth itself, even though it moves.
Explanation:
Convert 75.170 seconds into hours.
Answer:0.02088056
Explanation:
Two carts connected by a 0.50 m spring hit a wall, compressing the spring to 0.25 m. The spring constant k is
N
200
m
What is the elastic potential energy stored from the spring's compression?
Khan aced my?
Answer:
6.25J
Explanation:
bc that’s the answer on khan
Two carts connected by a 0.50 m spring hit a wall, compressing the spring to 0.25 m. If the spring constant k is 200 N/m, then the elastic potential energy stored from the spring's compression would be 6.25 J.
What is the spring constant?The spring constant is used to define the stiffness of the spring, the greater the value of the spring constant stiffer the spring and it is more difficult to stretch the spring.
The mathematical relation for calculating the spring constant is as follows
F = - Kx
as given in the problem Two carts connected by a 0.50 m spring hit a wall, compressing the spring to 0.25 m. If the spring constant k is 200 N/m,
The elastic potential energy stored is given by
E = 1/2 kx²
=0.5×200×0.25²
=6.25 J
Thus, the elastic potential energy stored from the spring's compression would be 6.25 J.
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Gravity does 207 N of work when a 45 N object falls to the
ground. From what height does the object fall?
Answer: 4.6
Explanation:
Answer:
4.6
Explanation:
When the 5.0 kg cylinder fell 100 m, the final temperature of the water was °C and the change in temperature was °C.
When the 5.0 kg cylinder fell 500 m, the final temperature of the water was °C and the change in temperature was °C.
Answer:
26.17
1.17
30.86
5.86
Explanation:
said they were correct on my assignment
PLEASE NEED HELP ASAP
Answer:
My youngins heartless, so they ain't playin' no games
We really want 'em dead, he got hit up close range
He fuc__ked up in the head, he wanna see some more brains
On that corner, I couldn't stay up out that do_0pe game
My cousin got indicted dealin' coc__aine
She an Insta>:gram addict, she want more fame
I used to starve, now I'm blowing up like pro__pane
Told my inner self, "I promise you I won't change"
Explanation:
A 700N mountain climber scales a 100m cliff. How much work is done by the mountain climber?
Answer:
70000N/m
Explanation:
Work is Force multiplied by Displacement.
Multiply 700 by 100 and you get 70000N/m
*Note: N is newtons and m is meters.
Answer: 7000N/m
Explanation: 700 x 100 = 7000
Describe what determines magnetism and how an electromagnet works.
Answer:
Magnetism is the force exerted by magnets when they attract or repel each other
hope it helps you:)
jack goes an average of 40 m/min. How far does he jog in 1.2 hours?
Answer:
2880 (m)
Explanation:
1. the formula of distance is: L=v*t, where L - the required distance; v - average velocity; t - elapsed time.
2. if t=1.2 hours, it means 72 minutes, then
3. L=40*72=2880 m.
Help please..................
What body part is the metric system based upon?
legs
arms
feet
fingers
The metric system is an internationally used system of checking various quantiles if weight tat includes the length, temperature, mass, etc., and is used in everyday life.
As nearly all the units are based on the human body they include the thumb joint was taken as yardsticks and was once used to define distance.Hence the option D is correct.
Learn more about the part is the metric system based upon.
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Let's say you got tired and you are now pushing with a force of 36 N. Find the acceleration of the box.
This question is incomplete, the complete question is;
A 13 kg box rests on the flat surface. The coefficient of static friction and coefficient of kinetic friction is μ_s = 0.450 and μ_k = 0.380 respectively.
Let's say you got tired and you are now pushing with a force of 36 N. Find the acceleration of the box.
Answer: the acceleration of the box is 0.9548 m/s² {backward}
Explanation:
Given that;
mass m = 13 kg
pushing the box with a force of 36 N
the acceleration of the box = ?
ones the box starts moving, kinetic frictional force starts acting.
so
f_k = μ_k × m × g
we substitute
f_k = 0.380 × 13 × 9.8
f_k = 48.412 N
force needed to keep it in motion is 48.412 N
so
Acceleration = Net Force / mass
Acceleration = (48.412 N - 36 N) / 13
Acceleration = 12.412 / 13
Acceleration = 0.9548 m/s² {backward}
Therefore, the acceleration of the box is 0.9548 m/s² {backward}