Answer:
I cant answer too complicated sorry
Explanation:
im sorry for not answering
Answer:
I
will
only
explain
Explanation:
The cannon is fired when an explosive charge is detonated, causing a sudden and immense increase in pressure. Is it not this pressure that causes the rapid acceleration of the cannonball and the recoil of the cannon, not an action–reaction between the cannon and the cannonball?
For purposes of this model, we can consider the expanding gas from the explosion to be part of the cannon, or as an intervening object between the cannon and the ball. So the gasses exert a force on the ball. The ball exerts a force back on the gasses. This is transferred to the cannon.
You could also imagine or build a (toy) "cannon" with a spring mechanism to propel the ball, rather than an explosion. You'd see very similar results.
In any case, the deeper point, which you will soon learn, is that momentum is a conserved quantity. Regardless of what mechanism applies the force on the ball and the cannon, after the ball is flying free the cannon must end up with as much backwards momentum as the ball has forward momentum.
If no cannonball is present when the charge is detonated, then the pressure dissipates much more quickly and the recoil is smaller but still present
Because air and exhaust gasses from the explosion are expelled from the cannon. These gasses have mass and carry momentum, therefore they exert a reaction force on the cannon just as a ball does.
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it's hard but not for memark me in brainliest answers please please please5/6 When switched on, the grinding machine accelerates from rest to its operating speed of 3450 rev/min in 6 seconds. When switched off, it coasts to rest in 32 seconds. Determine the number of revolutions turned during both the startup and shutdown periods. Also determine the number of revolutions turned during the first half of each period. Assume uniform angular acceleration in both cases.
Answer:
Δθ₁ = 172.5 rev
Δθ₁h = 43.1 rev
Δθ₂ = 920 rev
Δθ₂h = 690 rev
Explanation:
Assuming uniform angular acceleration, we can use the following kinematic equation in order to find the total angle rotated during the acceleration process, from rest to its operating speed:[tex]\Delta \theta = \frac{1}{2} *\alpha *(\Delta t)^{2} (1)[/tex]
Now, we need first to find the value of the angular acceleration, that we can get from the following expression:[tex]\omega_{f1} = \omega_{o} + \alpha * \Delta t (2)[/tex]
Since the machine starts from rest, ω₀ = 0.We know the value of ωf₁ (the operating speed) in rev/min.Due to the time is expressed in seconds, it is suitable to convert rev/min to rev/sec, as follows:[tex]3450 \frac{rev}{min} * \frac{1 min}{60s} = 57.5 rev/sec (3)[/tex]
Replacing by the givens in (2):[tex]57.5 rev/sec = 0 + \alpha * 6 s (4)[/tex]
Solving for α:[tex]\alpha = \frac{\omega_{f1}}{\Delta t} = \frac{57.5 rev/sec}{6 sec} = 9.6 rev/sec2 (5)[/tex]
Replacing (5) and Δt in (1), we get:[tex]\Delta \theta_{1} = \frac{1}{2} *\alpha *(\Delta t)^{2} = \frac{1}{2} * 6.9 rev/sec2* 36 sec2 = 172.5 rev (6)[/tex]
in order to get the number of revolutions during the first half of this period, we need just to replace Δt in (6) by Δt/2, as follows:[tex]\Delta \theta_{1h} = \frac{1}{2} *\alpha *(\Delta t/2)^{2} = \frac{1}{2} * 6.9 rev/sec2* 9 sec2 = 43.2 rev (7)[/tex]
In order to get the number of revolutions rotated during the deceleration period, assuming constant deceleration, we can use the following kinematic equation:[tex]\Delta \theta = \omega_{o} * \Delta t + \frac{1}{2} *\alpha *(\Delta t)^{2} (8)[/tex]
First of all, we need to find the value of the angular acceleration during the second period.We can use again (2) replacing by the givens:ωf =0 (the machine finally comes to an stop) ω₀ = ωf₁ = 57.5 rev/secΔt = 32 s[tex]0 = 57.5 rev/sec + \alpha * 32 s (9)[/tex]
Solving for α in (9), we get:[tex]\alpha_{2} =- \frac{\omega_{f1}}{\Delta t} = \frac{-57.5 rev/sec}{32 sec} = -1.8 rev/sec2 (10)[/tex]
Now, we can replace the values of ω₀, Δt and α₂ in (8), as follows:[tex]\Delta \theta_{2} = (57.5 rev/sec*32) s -\frac{1}{2} * 1.8 rev/sec2\alpha *(32s)^{2} = 920 rev (11)[/tex]
In order to get finally the number of revolutions rotated during the first half of the second period, we need just to replace 32 s by 16 s, as follows:[tex]\Delta \theta_{2h} = (57.5 rev/sec*16 s) -\frac{1}{2} * 1.8 rev/sec2\alpha *(16s)^{2} = 690 rev (12)[/tex]HELP a lot of points and brainliest Step 3: Measure the Speed of the Toy Car on the Lower Track
Calculate the change in time for each quarter of the track. Record the change in time in Table C of your Student Guide.
The change in time for the first quarter is
seconds.
The change in time for the second quarter is
seconds.
The change in time for the third quarter is
seconds.
The change in time for the fourth quarter is
seconds.
Answer:
is that 4 question each one of them we have to solve
An optical disk drive in your computer can spin a disk up to 10,000 rpm (about 1045 rad/s1045 rad/s ). If a particular disk is spun at 734.1 rad/s 734.1 rad/s while it is being read, and then is allowed to come to rest over 0.569 seconds0.569 seconds , what is the magnitude of the average angular acceleration of the disk
Answer:
[tex]1290.16\ \text{rad/s}^2[/tex]
Explanation:
[tex]\omega_i[/tex] = Initial angular velocity = 734.1 rad/s
[tex]\omega_f[/tex] = Final angular velocity = 0
t = Time = 0.569 seconds
[tex]\alpha[/tex] = Angular acceleration
From the kinematic equations of rotational motion we have
[tex]\omega_f=\omega_i+\alpha t\\\Rightarrow \alpha=\dfrac{\omega_f-\omega_i}{t}\\\Rightarrow \alpha=\dfrac{0-734.1}{0.569}\\\Rightarrow \alpha=-1290.16\ \text{rad/s}^2[/tex]
The magnitude of the average angular acceleration of the disk is [tex]1290.16\ \text{rad/s}^2[/tex].
g Suppose you have this brilliant idea: A Ferris wheel has radial metallic spokes between the hub and the circular rim (of radius roughly 16 m). These spokes move in the magnetic field of the Earth (5e-05 T), so each spoke acts like a rotating bar in a magnetic field. The magnetic field points perpendicular to the plane of the Ferris wheel. You plan to use the emf generated by the rotation of the Ferris wheel to power the light-bulbs on the wheel. Suppose the period of rotation for the Ferris wheel is 90 seconds. What is the magnitude of the induced emf between the hub and the rim
Answer:
[tex]4.46\times 10^{-4}\ \text{V}[/tex]
Explanation:
B = Magnetic field = [tex]5\times 10^{-5}\ \text{T}[/tex]
r = Radius of rim = 16 m
t = Time = 90 seconds
A = Area of rim = [tex]\pi r^2[/tex]
EMF is given by
[tex]\varepsilon=\dfrac{BA}{t}\\\Rightarrow \varepsilon=\dfrac{5\times 10^{-5}\times \pi\times 16^2}{90}\\\Rightarrow \varepsilon=0.000446=4.46\times 10^{-4}\ \text{V}[/tex]
The magnitude of the induced emf between the hub and the rim is [tex]4.46\times 10^{-4}\ \text{V}[/tex].
A pendulum is made up of a small sphere of mass 0.500 kg attached to a string of length 0.950 m. The sphere is swinging back and forth between point A, where the string is at the maximum angle of 35.0∘ to the left of vertical, and point C, where the string is at the maximum angle of 35.0∘ to the right of vertical. The string is vertical when the sphere is at point B. Calculate how much work the force of gravity does on the sphere from A to B.
Answer:
W = 0.842 J
Explanation:
To solve this exercise we can use the relationship between work and kinetic energy
W = ΔK
In this case the kinetic energy at point A is zero since the system is stopped
W = K_f (1)
now let's use conservation of energy
starting point. Highest point A
Em₀ = U = m g h
Final point. Lowest point B
Em_f = K = ½ m v²
energy is conserved
Em₀ = Em_f
mg h = K
to find the height let's use trigonometry
at point A
cos 35 = x / L
x = L cos 35
so at the height is
h = L - L cos 35
h = L (1-cos 35)
we substitute
K = m g L (1 -cos 35)
we substitute in equation 1
W = m g L (1 -cos 35)
let's calculate
W = 0.500 9.8 0.950 (1 - cos 35)
W = 0.842 J
Pesticides sprayed on the plants in fields become When they settle and drip onto the soil .
A. Nutrients
B.Pollutants
C.Acid rain
D.Ammonia
Answer:nutrients
Explanation:
Answer:
pollutants is the right answer.
A cyclist traveling at 5m/s uniformly accelerates up to 10 m/s in 2 seconds. Each tire of the bike has a 35 cm radius, and a small pebble is caught in the tread of one of them. (A) What is the angular acceleration of the pebble during those two seconds
Answer:
[tex]a=2.5\ m/s^2[/tex]
Explanation:
Given that,
Initial speed, u = 5 m/s
Final speed, v = 10 m/s
Time, t = 2 s
The radius of the tire of the bike, r = 35 cm
We need to find the angular acceleration of the pebble during those two seconds. It can be calculated as follows.
[tex]a=\dfrac{v-u}t{}\\\\a=\dfrac{10-5}{2}\\\\a=2.5\ m/s^2[/tex]
So, the required angular acceleration of the pebble is equal to [tex]2.5\ m/s^2[/tex].
Exact solutions for gravitational problems involving more than two bodies are notoriously difficult. One solvable problem involves a configuration of three equal-mass objects spaced in an equilateral triangle. Forces due to their mutual gravitation cause the configuration to rotate. Suppose three identical stars, each of mass M, form a triangle of side L. Find an expression for the period of their orbital motion.
Answer:
T = [tex]4\pi ^2 \ \sqrt{ \frac{L^3}{GM} }[/tex]
Explanation:
Let's analyze the situation, as the stars have the same mass and all are at the same distance, the magnitude of the force is constant, even when its direction changes, Let's use Newton's second law
F = m a
in this case the force is the universal attraction
F = G M₁ M₂ / r²
they relate it is centripetal
a = v² / r = w² r
we substitute
G M² / r² = M w² r
w² = [tex]\frac{G M}{r^3}[/tex]
angular velocity is related to frequency and period
w = 2π f = 2π / T
we substitute
[tex](\frac{2\pi }{T })^2 = \frac{G M}{r^3}[/tex]
T = [tex]4\pi^2 \ \sqrt{ \frac{r^3}{G M}}[/tex]
in this case r= L
T = [tex]4\pi ^2 \ \sqrt{ \frac{L^3}{GM} }[/tex]
a forward horizontal force of 50 N is applied to crate a second horizontal force of 180 N is applied to crate in the opposite direction determine the magnitude and direction of the resultant force acting on the crate
Answer:
130n on the 2nd horizontal
Explanation:
A 60 kg swimmer at a water park enters a pool using a 2 m high slide. Find the velocity of the swimmer
at the bottom of the slide.
Answer:
the velocity of the swimmer at the bottom of the slide is 6.26 m/s
Explanation:
The computation of the velocity of the swimmer at the bottom of the slide is given below:
v = √2gh
= √2 × 9.8 × 2
= 6.26 m/s
Hence, the velocity of the swimmer at the bottom of the slide is 6.26 m/s
The velocity of the swimmer at the bottom of the slide will be 6.26 m/s.The pace of displacement change with reference to time is referred to as the velocity
What is velocity?The change of displacement with respect to time is defined as the velocity. Velocity is a vector quantity. it is a time-based component. Velocity at any angle is resolved to get its component of x and y-direction.
The velocity is found as;
[tex]\rm v= \sqrt{2gh} \\\\ \rm v= \sqrt{2\times 9.81 \times 2}\\\\ \rm v= 6.26 \ m/sec[/tex]
Hence velocity of the swimmer at the bottom of the slide will be 6.26 m/s.
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7. Copper can be coated on the surface of iron but not on silver? why
Answer:
Silver has greater electronegativity (pull on electrons) than copper, so it will be reduced rather than oxidized. Silver ions will plate out on copper metal. Sir the coating of iron with copper surface can be referred as IRON-COPPER PLATING.
Explanation:
Iron is used for the electroplating of so copper because iron falls above copper in the electrochemical series whereas silver will fall off below the copper in electrochemical series that makes it not reliable for the electroplating purpose
prove that g is inversely proportional to the radius
Answer:The acceleration due to gravity g is inversely proportional to the square of the radius in the formula g = GM / R^2 where G is the gravitational constant = 6.67 x 10^-11 Nm^2/kg^2, M is the mass of the Earth and R is the radius of the Earth
Explanation:
How do you find the period of a sound wave?
Answer:
Period refers to the time for something to happen and is measured in seconds/cycle. In this case, there are 11 seconds per 33 vibrational cycles. Thus the period is (11 s) / (33 cycles) = 0.33 seconds. We now know that the period is 3.2 seconds and that the frequency is 0.31 Hz.
A consumer uses 3098 kWh in 29 days. The utility company charges AED 0.077592 per kWh for the electricity plus AED 0.029998 per kWh for the distribution of the electricity . What is the consumer's electric bill for the 29 days?
The consumer electric bill for 29 days is AED 331.31
Given that, The utility company charges AED 0.077592 per kWh for the electricity plus AED 0.029998 per kWh for the distribution of the electricity
Since, A consumer uses 3098 kWh in 29 days.
The utility company charges for electricity is,
[tex]=0.077592*3098=240.38[/tex]
The utility company charges for distribution of the electricity is,
[tex]=0.029998*3098=92.93[/tex]
So that, The consumer electric bill is,
[tex]=240.38+92.93=333.31[/tex]
Hence, the consumer electric bill for 29 days is AED 331.31
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Define threshing?
Brainliest For the right answer
Answer:
Process used for separating grains from the stalks is known as threshing, In this process, stalks are beaten to free the grain seeds.
Answer:
Threshing is extraction of wheat germ from the stalk. In today's usage the combine tractor cuts and threshes the wheat at the same time. Imagine a big lawn mower with a rotating drum inside.
The drum turns and shakes the germ out of the wheat, the seeds falling through small holes onto a conveyor belt one way, the leftover grass dumping out the other way. The grain is poured into a truck driving beside the combine.
In old times, grain had to be beaten out of the grass on a Threshing Floor.
In which direction do electrical and magnetic waves travel in relation to THE
DIRECTION OF WAVE PROPAGATION? *
Parallel
Perpendicular
Backwards
Longitudinal
Answer: perpendicular
Explanation:
Fill in the graph for 50 points
Answer:
Speed: 3, 4, 5, 6. Distance: 1, 2, 3, 4, 5
Answer:
Speed: 3, 4, 5, 6. Distance: 1, 2, 3, 4, 5
Explanation:
What is the de Broglie - equation
Answer:
is this it?
Explanation:
λ = h/mv, where λ is wavelength, h is Planck's constant, m is the mass of a particle, moving at a velocity v. de Broglie suggested that particles can exhibit properties of waves.
The phrase that best describes a field is:
a tower of power
a source of momentum
a sphere of influence
a lever arm
Answer:
a sphere of influence mmmmmmmmmmm not actually right
state three factors that determine the strength of electromagnetic
Answer:
Factors that affect the strength of electromagnets are the nature of the core material, strength of the current passing through the core, the number of turns of wire on the core and the shape and size of the core.Nov 28, 2020
Explanation:
Rewire each of the following using the correct prefix using 2 decimal places where applicable.
a.0.00000123N
b. 417 000 000 kg
c. 246800
d. 0,00088 mm
Answer:
a. 1.2×10^-6
b. 0.42×10^9
c. 246.8×10^3
d. 88
Explain how a transistor behaves as a switch
Answer:
With a zero signal applied to the Base of the transistor it turns “OFF” acting like an open switch and zero collector current flows. With a positive signal applied to the Base of the transistor it turns “ON” acting like a closed switch and maximum circuit current flows through the device.
Explanation:
please give me brainlist and follow
Which term does this explain?
This is a non-mathematical explanation of how nature works. It must be
supported by a large body of evidence.
Fact
Law
Theory
Hypothesis
A straight wire, labeled as Wire A, lies horizontally on a tabletop and is oriented to run north-south. A conventional current of 1.0 amperes runs in the wire directed towards the north. A second wire, labeled as Wire B, is also laid on the tabletop oriented north- south. Which of the following statements is true?
a. If Wire B carries no current in it and lies to the left (west) of wire A, then it will experience an attractive force to the right towards wire A).
b. If Wire B carries a northward conventional current and lies to the left (west) of wire A, then it will experience an attractive force to the right (towards Wire A).
c. If Wire B carries a northward conventional current and lies to the right (east) of wire A, then it will experience a repulsive force to the right (away from Wire A).
d. If Wire B carries a southward conventional current and lies to the left (west) of wire A, then it will experience a repulsive force to the left (away from Wire A).
Answer:
b. If Wire B carries a northward conventional current and lies to the left (west) of wire A, then it will experience an attractive force to the right (towards Wire A).
d. If Wire B carries a southward conventional current and lies to the left (west) of wire A, then it will experience a repulsive force to the left (away from Wire A).
Explanation:
Two parallel conductors experience attractive force when the current flowing in the conductors are in the same direction.
Also two parallel conductors experience repulsive force when the current flowing in the conductors are in opposite direction.
Therefore, b and d are the correct options.
b. If Wire B carries a northward conventional current and lies to the left (west) of wire A, then it will experience an attractive force to the right (towards Wire A).
d. If Wire B carries a southward conventional current and lies to the left (west) of wire A, then it will experience a repulsive force to the left (away from Wire A).
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I believe you are incorrect. A weathered mountain would appear more jagged.
I do believe with a lot of exposure to weather will make the mountain appear somewhat more jagged compared to a mountain that is less weathered.
If this is incorrect, please, don't refrain to tell me.
Which disciplines were developed based on the Greek questioning of the elements?
A. art and music
B. chemistry and biology
C. tragedy and comedy
Betty hits a baseball.
At which point along the baseball's path does it have the most gravitational potential energy?
Answer:
point 2
Explanation:
2. An object is dropped from rest. Calculate its velocity after 2.5s if it is dropped:
a.On Earth, where the acceleration due to gravity is 9.8m/s?
b. On Mars, where the acceleration due to gravity is 3.8m/s?
Answer:
a=24.5 b=9.5
Explanation:
Which of the following is necessary in order to classify the different types of
electromagnetic radiation?
Answer:
The EM spectrum is generally divided into seven regions, in order of decreasing wavelength and increasing energy and frequency. The common designations are: radio waves, microwaves, infrared (IR), visible light, ultraviolet (UV), X-rays and gamma rays.
Explanation:
HOPE THIS HELPS! ;)
In order to classify the different types of electromagnetic radiation, frequency is necessary.
What is electromagnetic radiation?The EM spectrum have seven different regions in terms of decreasing wavelength. The increase in energy and frequency denotes decreasing wavelength. The parts of EM spectrum are radio waves, microwaves, infrared (IR), visible light, ultraviolet (UV), X-rays and gamma rays.
Thus, in order to classify the different types of electromagnetic radiation, frequency is necessary.
Learn more about electromagnetic spectrum.
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What conversion takes place in a motor?
A. An electric current into a magnetic field
B. Mechanical energy into electric energy
C. Electric energy into mechanical energy
D. A lower voltage into a higher voltage
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