Answer:
the speed of the bullet before striking the block is 302.3 m/s.
Explanation:
Given;
mass of the bullet, m₁ = 28.3 g = 0.0283 kg
mass of the wooden block, m₂ = 5004 g = 5.004 kg
initial velocity of the block, u₂ = 0
final velocity of the bullet-wood system, v = 1.7 m/s
let the initial velocity of the bullet before striking the block = u₁
Apply the principle of conservation of linear momentum to determine the initial velocity of the bullet.
m₁u₁ + m₂u₂ = v(m₁ + m₂)
0.0283u₁ + 5.004 x 0 = 1.7(0.0283 + 5.004)
0.0283u₁ = 8.5549
u₁ = 8.5549 / 0.0283
u₁ = 302.3 m/s
Therefore, the speed of the bullet before striking the block is 302.3 m/s.
A digital signal differs from an analog signal because it a.consists of a current that changes smoothly. b. consists of a current that changes in pulses. c.carries information. d. is used in electronic devices.
Answer:
d.it is used in electronic devices
The lever of a car lift has an area of 0.2 meters squared, and the area of the lift under the car is 8
meters squared. If you push with a force of 3 newtons, how much force will be applied to the
car?
Answer:
THE ANSWER IS SOMETHING LIKE 55
A string that is under 50.0N of tension has linear density 5.0g/m. A sinusoidal wave with amplitude 3.0cm and wavelength 2.0m travels along the string. What is the maximum speed of a particle on the string
Answer:
9.42 m/s
Explanation:
Applying,
V' = Aω.............. Equation 1
Where V' = maximum speed of the string, A = Amplitude of the wave, ω = angular velocity.
But,
ω = 2πf................. Equation 2
Where f = frequency, π = pie
And,
f = v/λ................ Equation 3
Where, λ = wave length, v = velocity
Also,
v = √(T/μ)................. Equation 4
Where T = Tension, μ = linear density.
From the question,
Given: T = 50.0 N, μ = 5.0 g/m = 0.005 kg/m
Substitute into equation 4
v = √(50/0.005)
v = √(10000)
v = 100 m/s
Also Given: λ = 2.0 m
Substitute into equation 3
f = 100/2
f = 50 Hz.
Substitute the value of f into equation 2
Where π = constant = 3.14
ω = 2(3.14)(50)
ω = 314 rad/s
Finally,
Given: A = 3.0 cm = 0.03 m
Substitute into equation 1
V' = 0.03(314)
V' = 9.42 m/s
Two identical satellites orbit the earth in stable orbits. Onesatellite orbits with a speed vat a distance rfrom the center of the earth. The second satellite travels at aspeed that is less than v.At what distance from the center of the earth does the secondsatellite orbit?At a distance that is less than r.At a distance equal to r.At a distance greater than r.Now assume that a satellite of mass m is orbiting the earth at a distance r from the center of the earth with speed v_e. An identical satellite is orbiting the moon at thesame distance with a speed v_m. How does the time T_m it takes the satellite circling the moon to make onerevolution compare to the time T_e it takes the satellite orbiting the earth to make onerevolution?T_m is less than T_e.T_m is equal to T_e.T_m is greater than T_e.
Answer:
a. At a distance greater than r
b. T_m is greater than T_e.
Explanation:
a. Two identical satellites orbit the earth in stable orbits. One satellite orbits with a speed vat a distance r from the center of the earth. The second satellite travels at a speed that is less than v. At what distance from the center of the earth does the second satellite orbit?
Since the centripetal force on any satellite, F equals the gravitational force F' at r,
and F = mv²/r and F' = GMm/r² where m = mass of satellite, v = speed of satellite, G = universal gravitational constant, M = mass of earth and r = distance of satellite from center of earth.
Now, F = F'
mv²/r = GMm/r²
v² = GM/r
v = √GM/r
Since G and M are constant,
v ∝ 1/√r
So, if the speed decreases, the radius of the orbit increases.
Since the second satellite travels at a speed less than v, its radius, r increases since v ∝ 1/√r.
So, the distance the second satellite orbits is at a distance greater than r
b. An identical satellite is orbiting the moon at the same distance with a speed v_m. How does the time T_m it takes the satellite circling the moon to make one revolution compare to the time T_e it takes the satellite orbiting the earth to make one revolution?
Since the speed of the satellite, v = √GM/r where M = mass of planet
Since the satellite is orbiting at the same distance, r is constant
So, v ∝ √M
Since mass of earth M' is greater than mass of moon, M", the speed of satellite circling moon, v_m is less than v the speed of satellite circling earth at the same distance, r
Now, period T = 2πr/v where r = radius of orbit and v = speed of satellite
Since r is constant for both orbits, T ∝ 1/v
Now, since the speed of the speed of the satellite on earth orbit v is greater than the speed of the satellite orbiting the moon, v_m, and T ∝ 1/v, it implies that the period of the satellite orbiting the earth, T_e is less than the period of the satellite orbiting the moon, T_m since there is an inverse relationship between T and v. T_e is less T_m implies T_m is greater than T_e
So, T_m is greater than T_e.
If the pressure of a gas is really due to the random collisions of molecules with the walls of the container, why do pressure gauges – even very sensitive ones – give perfectly steady readings? Shouldn't the gauge be continually jiggling and fluctuating? Explain.
Answer:
there is no fluctuation in the measurement because the quantity of molecule is too large and a quantity of some molecules is imperceptible.
Explanation:
The pressure measurement is carried out by calibrating the force exerted by the air on a surface of known area, suppose a small area 1 mm² = 0.01 cm²
To find out if the random movement of air molecules affects the pressure reading, let's calculate the number of molecules that reaches the pressure gauge.
In a system at atmospheric pressure and in a volume of 1 m³ (walls of 1 m each) there is one mole of air molecules, this mole is evenly distributed, so how many molecules fall on our surface
# _molecule = 6.02 10²³ 0.01 10⁻⁴ / 1
#_molecular = 6.02 10¹⁷ molecules per second
therefore the variation of the number of molecules is not very important
Consequently there is no fluctuation in the measurement because the quantity of molecule is too large and a quantity of some molecules is imperceptible.
When landing after a spectacular somersault, a 40.0-kg gymnast decelerates by pushing straight down on the mat. Calculate the force she must exert if her deceleration is 7.00 times the acceleration due to gravity. Explicitly show how you follow the steps in the Problem-Solving Strategy for Newton’s laws of motion.
Answer:
[tex]F=3139.2N[/tex]
Explanation:
Mass [tex]m=40.0kg[/tex]
Acceleration [tex]a=7g=68.67m/s^2[/tex]
With g as 9.81
Generally the equation for Force is mathematically given by
[tex]F_net=F-w\\\\F-w=ma\\\\F=ma-w\\\\[/tex]
[tex]F=ma-mg\\\\F=m(a+g)[/tex]
Therefore
[tex]F=m*8g[/tex]
[tex]F=40*8*9.81[/tex]
[tex]F=3139.2N[/tex]
What is 3*10^-6 divided by 2.5*10^6 expressed in standard notation?
Answer:
1.2 x 10^-12
Explanation:
3/2.5 x 10^-6/10^6
1.2 x 10^-6 x 10^-6
1.2 x 10^-12
instrument used in measurement Amount of substance
Answer:
For liquids: A measuring cylinder is used.
For solid: Over flow can is used
Answer:
i think a measuring cylinder
A car moving in a straight line uniformly accelerated speed increased from 3 m / s to 9 m / s in 6 seconds. With what acceleration did the car move?
a.
2 m/s2
b.
1 m/s2
c.
0 m/s2
d.
3 m/s2
Answer:
b) 1 m/s
I am sure...........
A car moves at a constant speed of 90km/h from a starting point. Another car moves at 70km/h after 2hours from the same starting point. if both cars moves in the same direction, after how many hours will the distance between the first car and the second car to be 40 km.
Answer:
400
Explanation:
A boy throws a ball straight up with a speed of 21.5 m/s. The ball has a mass of 0.19 kg. How much gravitational potential energy will the ball have at the top of its flight? (Assume there is no air resistance.) A. 43.9 J B. 37.5 J C. 48.5 J D. 41.2 J
Answer:
Explanation:
The equation fo potential energy is PE = mgh, where m is the mass of the ball, g is the pull of gravity (constant at 9.8), and h is the max height of the ball. What we do not have here is that height. We need to first solve for it using one-dimensional equations. What we have to know above all else, is that the final velocity of an object at its max height is always 0. That allows us to use the equation
[tex]v_f=v_0+at[/tex] where vf is the final velocity and v0 is the initial velocity. We will find out how long it takes for the object to reach that max height first and then use that time to find out what that max height is. Baby steps here...
0 = 21.5 + (-9.8)t and
-21.5 = -9.8t so
t = 2.19 seconds (Keep in mind that if I used the rules correctly for sig fig's, the answer you SHOULD get is not one shown, so I had to adjust the sig fig's and break the rules. But you know what they say about rules...)
Now we will use that time to find out the max height of the object in the equation
Δx = [tex]v_0t+\frac{1}{2}at^2[/tex] and filling in:
Δx = [tex]21.5(2.19)+\frac{1}{2}(-9.8)(2.19)^2[/tex] which simplifies down a bit to
Δx = 47.1 - 23.5 so
Δx = 23.6 meters.
Now we can plug that in to the PE equation to find the PE of the object:
PE = (.19)(9.8)(23.6) so
PE = 43.9 J
Which option is the best blackbody radiator?
A.
The Sun
B.
A red laser pointer
C.
A tennis ball
D.
Boiling water
Answer:
A. The Sun
Explanation:
The Sun is to be considered a perfect black body.
A block of mass m is moved over a distance d. An applied force F is directed perpendicularly to the block’s displacement. How much work is done on the block by the force F?
zero
Explanation:
Work W is defined as
W = F•d = Fdcos(theta)
and it is a dot product of the force and displacement and theta is angle between F and d Since the force is perpendicular to d, angle is 90° thus cos90 = 0. Hence work is zero.
As shown in Fig. 4, an ideal gas of monatomic molecules expands from its initial state A to a state B through an isobaric process and then further expands to a volume C. Find the work done by the gas, increase in internal energy, and the energy transferred by heat to the gas over the whole process
Calculate the period of a spring if it has a mass of 5 kg and a spring constant of 6 N/m
Answer: The period of a spring if it has a mass of 5 kg and a spring constant of 6 N/m is 5.73 sec.
Explanation:
Given: Mass = 5 kg
Spring constant = 6 N/m
Formula used to calculate period is as follows.
[tex]T = 2 \pi \sqrt\frac{m}{k}[/tex]
where,
T = period
m = mass
k = spring constant
Substitute the values into above formula as follows.
[tex]T = 2 \pi \sqrt\frac{m}{k}\\= 2 \times 3.14 \times \sqrt\frac{5}{6}\\= 5.73 s[/tex]
Thus, we can conclude that the period of a spring if it has a mass of 5 kg and a spring constant of 6 N/m is 5.73 sec.
In young Goodman’s Brown hawthornes reveals his feelings about his Puritan ancestors when
Answer:
In "Young Goodman Brown," Hawthorne reveals his feelings about his Puritan ancestors when the dark man reveals that he helped Brown's forebears persecute others.
Explanation:
hey mate i know it is right so don't worry this is the correct
Answer:
In "Young Goodman Brown," Hawthorne reveals his feelings about his Puritan ancestors when the dark man reveals that he helped Brown's forebears persecute others. so that means the answer is D.
Explanation:
A. brown strives to resist his dark mission
B. faith expresses her anxieties about young brown's departure
C. brown discovers his catechism teacher is on speaking terms with the devil
D. the dark man reveals that he helped brown's forebears persecute others(correct answer)
The image shows the right-hand rule being used for a current-carrying wire.
An illustration with a right hand with fingers curled and thumb pointed up.
Which statement describes what the hand shows?
When the current flows down the wire, the magnetic field flows out on the left side of the wire and in on the right side of the wire.
When the current flows up the wire, the magnetic field flows out on the left side of the wire and in on the right side of the wire.
When the current flows down the wire, the magnetic field flows in on the left side of the wire and out on the right side of the wire.
When the current flows up the wire, the magnetic field flows in on the left side of the wire and out on the right side of the wire.
Answer:
The answer is (D): When the current flows up the wire, the magnetic field flows in on the left side of the wire and out on the right side of the wire.
Explanation:
The efficiency of a machine can be increased by
Explanation:
the efficiency of a machine can be increased by reducing the friction
please mark the brainliest
According to the model, when was the universe at its most dense?
A) During the Dark Ages where matter increased in mass.
B) Just before the Big Bang where all matter existed in a singularity.
C) During the nuclear fusion events, as the atoms become more massive.
D) Current day, as the number of galaxies, solar systems, and planets have increased.
Answer:
The Answer is D
Explanation:
Hope this helps!!!!
A scenario where reaction time is important is when driving on the highway. During the delay between seeing an obstacle and reacting to avoid it (or to slam on the brakes!) you are still moving at full highway speed. Calculate how much distance you cover in meters before you start to put your foot on the brakes if you are travelling 65 miles per hour.
Answer:
66.83 meters
Explanation:
After a quick online search, it seems that scientists calculate the average reaction time of individuals as 2.3 seconds between seeing an obstacle and putting their foot on the brakes. Now that we have this reaction time we need to turn the miles/hour into meters/second.
1 mile = 1609.34 meters (multiply these meters by 65)
65 miles = 104,607 meters
1 hour = 3600 seconds
Therefore the car was going 104,607 meters every 3600 seconds. Let's divide these to find the meters per second.
[tex]\frac{104,607}{3600} = \frac{29.0575 meters}{1 second}[/tex]
Now we simply multiply these meters by 2.3 seconds to find out the distance covered before the driver puts his/her foot on the brakes...
29.0575m * 2.3s = 66.83 meters
The latent heat of vaporization of water is roughly 10 times the latent heat of fusion of water. The amount of heat required to boil away 1 kg of water is __________ the amount of heat required to melt 1 kg of ice.
Answer:
The amount of heat required to boil away 1 kg of water is 10 times the amount of heat required to melt 1 kg of ice
Explanation:
let the latent heat of fusion of ice = L
then, the latent heat of vaporization of water = 10L
The heat of fusion of 1 kg of ice = 1 x L = L
The heat of vaporization 1 kg of water = 1 x 10L = 10L
Therefore, the amount of heat required to boil away 1 kg of water is 10 times the amount of heat required to melt 1 kg of ice
why are cows is important?
Answer:
cause they give u milk
Explanation:
Answer:
Cows are important as they provide humans many things for survival. They provide milk, meat, and leather, all of these are important resources.
a. A horse pulls a cart along a flat road. Consider the following four forces that arise in this situation.
1. the force of the horse pulling on the cart
2. the force of the cart pulling on the horse
3. the force of the horse pushing on the road
4. the force of the road pushing on the horse
b. Suppose that the horse and cart have started from rest; and as time goes on, their speed increases in the same direction. Which one of the following conclusions is correct concerning the magnitudes of the forces mentioned above?
1. Force 1 exceeds Force 2.
2. Force 2 is less than Force 3.
3. Force 2 exceeds Force 4.
4. Force 3 exceeds Force 4.
5. Forces 1 and 2 cannot have equal magnitudes.
Answer:
a) F₁ = F₂, F₃ = F₄, b) the correct answer is 3
Explanation:
a) In this exercise we have several action and reaction forces, which are characterized by having the same magnitude, but different direction and being applied to different bodies
Forces 1 and 2 are action and reaction forces F₁ = F₂
Forces 3 and 4 are action and reaction forces F₃ = F₄
as it indicates that the
b) how the car increases if speed implies that force 1> force3
F₁ > F₃
therefore the correct answer is 3
If 1.02 ✕ 1020 electrons move through a pocket calculator during a full day's operation, how many coulombs of charge moved through it?
Answer:
Explanation:
one electron has [tex]1.60217662*10^{-19}~coulombs~then\\\\1.02*10^{20}~electrons------->1.02*10^{20}*1.60217662*10^{-19}~coulombs= 16.3422~coulombs[/tex]
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Answer:
ok
Explanation:
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N 4. Which of the following can cause a short circuit?
Answer:
pretty sure its A
Explanation:
please give brainliest if i'm correct
Answer:
A.
Explanation:
I have done this and that was correct
hope this helps
15. A car travelling towards the right has a mass of 1332 kg and has a speed of 25 m/s. A truck is
travelling towards the left with a mass of 3000 kg and a speed of 15 m/s. They collide head
on with each other. What is the total momentum after the crash? In which direction will the
vehicles travel after the collision?
Explanation:
Given that,
The mass of a car, m₁ = 1332 kg
The speed of the car, u₁ = 25 m/s (right)
The mass of a truck, m₂ = 3000 kg
The speed of the truck, u₂ = -15 m/s
The total momentum after the crash is given by :
p=m₁u₁ + m₂u₂
Put all the values,
P = 1332(25) + 3000(-15)
= −11700 kg-m/s
So, the total momentum after the crash is equal to 11700 kg-m/s and it is in the left direction.
Calculate the equivalent resistance
why the walls of tyres becomes warm as the car moves
Answer:
the particles vibrate inside the tyre
Explanation:
as the car moves kinetic energy is transfered in the tyres which causes the particles to vibrate inside the tyre so the kinetic store is. transferred into thermal
A ship is flying away from Earth at 0.9c (where c is the speed of light). A missile is fired that moves toward the Earth at a speed of 0.5c relative to the ship. How fast does the missile move relative to the Earth
Answer:
the required speed with which the missile move relative to the Earth is -0.727c
Explanation:
Given the data in the question;
relative velocity relation;
u' = u-v / 1 - [tex]\frac{uv}{c^2}[/tex]
so let V[tex]_B[/tex] represent the velocity as seen by an external reference frame; u=V[tex]_B[/tex]
and let V[tex]_A[/tex] represent the speed of the secondary reference frame; v=V[tex]_A[/tex]
hence, u' is the speed of B as seen by A
so
u' = V[tex]_B[/tex]-V[tex]_A[/tex] / 1 - [tex]\frac{V_BV_A}{c^2}[/tex]
now, given that; V[tex]_A[/tex] = 0.9c and V[tex]_B[/tex] = 0.5c
we substitute
u' = ( 0.5c - 0.9c ) / 1 - [tex]\frac{(0.5c)(0.9c)}{c^2}[/tex]
u' = ( 0.5c - 0.9c ) / 1 - [tex]\frac{c^2(0.5)(0.9)}{c^2}[/tex]
u' = ( 0.5c - 0.9c ) / 1 - (0.5 × 0.9)
u' = ( -0.4c ) / 1 - 0.45
u' = -0.4c / 0.55
u' = -0.727c
Therefore, the required speed with which the missile move relative to the Earth is -0.727c
