Answer:
Car first should be 30.13 m behind the second car.
Explanation:
Given that,
Constant speed = 108.5 Km/hr
Time = 1 sec
Let the distance covered by second car S and by first car S'
We need to calculate the distance covered by second car
Using equation of motion
[tex]S=ut-\dfrac{1}{2}at^2[/tex].....(I)
The distance covered by first car to avoid collision
[tex]S''=S'+S[/tex]
Put the value into the formula
[tex]S'+S=ut+ut-\dfrac{1}{2}at^2[/tex]...(II)
We need to calculate the distance covered by first car
Using equation (I) and (II)
[tex]S'=ut[/tex]
Put the value into the formula
[tex]S'=108.5\times\dfrac{5}{18}\times1[/tex]
[tex]S'=30.13\ m[/tex]
Hence, Car first should be 30.13 m behind the second car.
If A= +4j - 2i - 3k and C= -4i -2j - 3k , which of the following numbers is closest to the magnitude of A-C ?
The magnitude of Dot product of two vectors is 3.
Given vectors are,[tex]A= 4j - 2i - 3k[/tex] and [tex]C= -4i -2j - 3k[/tex]
The dot product two vectors shown below,[tex]A\cdot C=(-4*4)+(-2*-2)+(-3*-3)\\\\A\cdot C=-16+4+9\\\\A\cdot C=-16+13=-3[/tex]
Thus, The magnitude of Dot product of two vectors is 3.
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A communications satellite orbiting the earth has solar panels that completely absorb all sunlight incident upon them. The total area A of the panels is 10m2.
1) The intensity of the sun's radiation incident upon the earth is about I = 1.4kW/m2. Suppose this is the value for the intensity of sunlight incident upon the satellite's solar panels. What is the total solar power P absorbed by the panels?
2) What is the total force F on the panels exerted by radiation pressure from the sunlight?
Answer:
0.00004666N
Explanation:
We know that
intensity (I) = P/ A
Where
P= power
A= Area
So lets say that the power absorbed
Will be = Intensity x Area
Which Is = 1.4 x 10^3 x(10)
So
14000 Watt = 14 kWatt
However we know that radiation pressure is equal to
time-averaged intensity all over the speed of light in free space
So
P = (1.4 x 1000)/c
But
F= P x A
So
((1.4 x 1000)/(3 x1 0^8)) x 10
Which is
=0.000046666N
Explanation:
explain how ozone in the atmosphere affects visible light on earth
The pilot of a helicopter drops a lead brick from a height of 500 m. (assume no air resistance and g~10 m/s2) roles Evalu How long does it take to reach the ground?
Answer:
The value is [tex]t = 10 \ s[/tex]
Explanation:
From the question we are told that
The height is [tex]h = 500 \ m[/tex]
The acceleration is [tex]g = 10 \ m/s[/tex]
Generally from the second equation of motion we have that
[tex]h = ut + \frac{1}{2} a t^2[/tex]
Here the initial velocity u = 0 m/s given that the stone started from rest
So
[tex]500 = 0 (t) +0.5 * (10) t^2[/tex]
=> [tex]t = 10 \ s[/tex]
The second law of thermodynamics leads us to conclude what?
a. the total energy of the universe is constant
b. disorder in the universe is increasing with the passage of time
c. it is theoretically impossible to convert work into heat with 100% efficiency
d. the total energy in the universe is increasing with time
e. the total energy in the universe is decreasing with time
Answer:
b. disorder in the universe is increasing with the passage of time
Explanation:
The second law of thermodynamics states that the total entropy of an isolated system increases over time . The entropy of a system measures the randomness or disorder in the system . So the we can state, in terms of disorder , the second law of thermodynamics as follows .
The disorder in the universe is increasing with the passage of time .
The second law of thermodynamics leads us to conclude that disorder in the universe is increasing with the passage of time. So, here, option b is the correct option.
Entropy is a measure of the disorder or randomness in a system. The second law of thermodynamics states that in any spontaneous process, the total entropy of an isolated system (including both the system and its surroundings) tends to increase or, at best, remain constant. This means that over time, natural processes tend to lead to an increase in the overall disorder or randomness of the system. It's important to note that while the increase in entropy is a statistical tendency on a macroscopic scale, there can still be localized decreases in entropy within a system.
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Enter the following expression in the answer box below: 2gλ3m−−−−√, where λ is the lowercase Greek letter lambda.
Answer:
[tex]\sqrt{\dfrac{2g\lambda^3}{m}}[/tex]
Explanation:
We can write the expression here, but the point of the problem seems to be to see if you can manipulate the controls on the answer box to reproduce that expression.
[tex]\boxed{\sqrt{\dfrac{2g\lambda^3}{m}}}[/tex]
A boy throws a ball on a spring scales which oscillates about the equilibrium position with a period of T = 0.5 sec. The amplitude of the vibration is A = 2 cm. Assume the ball does not bounce from the scales’s surface afterwards. Assume the vibration of the scale is expressed mathematically as x(t) = Acos(t + ). Find:
Complete Question
A boy throws a ball on a spring scales which oscillates about the equilibrium position with a period of T = 0.5 sec. The amplitude of the vibration is A = 2 cm. Assume the ball does not bounce from the scales’s surface afterwards. Assume the vibration of the scale is expressed mathematically as x(t) = Acos(t + ). Find:
a) frequency
b) the maximum acceleration
c) the maximum velocity
Answer:
a
[tex]f = 2 \ Hz[/tex]
b
[tex]a_{max} = 3.2 \ m/s^2[/tex]
c
[tex]v_{max} = 0.25 \ m/s[/tex]
Explanation:
From the question we are told that
The period is [tex]T = 0.5 \ sec[/tex]
The amplitude is [tex]A = 2 \ cm = 0.02 \ m[/tex]
The vibration of the scale is [tex]Acos(wt + \phi )[/tex]
Generally the frequency is mathematically represented as
[tex]f = \frac{1}{T}[/tex]
=> [tex]f = \frac{1}{0.5}[/tex]
=> [tex]f = 2 \ Hz[/tex]
The maximum acceleration is mathematically represented as
[tex]a_{max} = A *(2 \pi f)^2[/tex]
=> [tex]0.02 * (2 * 3.142 * 2)^2[/tex]
=> [tex]a_{max} = 3.2 \ m/s^2[/tex]
The maximum velocity is mathematically represented as
[tex]v_{max} = A * (2 \pi f)[/tex]
=> [tex]v_{max} = 0.02 * (2 * 3.142 * 2)[/tex]
=> [tex]v_{max} = 0.25 \ m/s[/tex]
You travel down the highway at a steady rate of 75 mph (33.53 m/s)  for a total of 25 minutes, calculate how far you traveled in meters
Answer:
50280 meters
Explanation:
33.52 meters/seconds is 2011.2 meters/minutes (multiply by 60)
2011.2 meters/m * 25 minutes = answer
Calculate the change in momentum of a 0.5kg ball that strikes the floor at 15 m/s and bounces back up at 12 m/s
Answer:
The change in momentum is: [tex]13.5\,\frac{kg\,m}{s}[/tex]
Explanation:
Let's define that vectors pointing up are positive, and vectors pointing down negative.
Then we express the initial momentum of the ball as the product of its mass times the velocity, and include the negative sign since this momentum is pointing down (velocity vector is pointing down):
[tex]P_i=-0.5\,(15) \frac{kg\,m}{s} =-7.5\, \frac{kg\,m}{s}[/tex]
the final momentum is positive (pointing up) and given by the product:
[tex]P_f=0.5\,(12) \frac{kg\,m}{s} =6\, \frac{kg\,m}{s}[/tex]
Therefore, since the change in momentum is defined as the difference between the final momentum minus the initial one, we get:
[tex]P_f-P_i=6-(-7.5)\,\frac{kg\,m}{s} =13.5\,\frac{kg\,m}{s}[/tex]
What must the charge (sign and magnitude) of a particle of mass 1.50 gg be for it to remain stationary when placed in a downward-directed electric field of magnitude 670 N/CN/C ?
Answer:
q = -2.19 x 10⁻⁵ C
Explanation:
Given;
mass of the particle, m = 1.5 g = 0.0015 kg
magnitude of electric field, E = 670 N/C
Electric field is given by;
[tex]E = \frac{F}{q}[/tex]
where;
q is the magnitude of the
f is the force of the charge
f = mg
[tex]E = \frac{F}{q}\\\\E = \frac{mg}{q}\\\\q = \frac{mg}{E}\\\\q = \frac{0.0015*9.8}{670} \\\\q = 2.19*10^{-5} \ C[/tex]
Since the electric field is acting downward, the force on the charge must be acting upward. Therefore, the charge must be negative
q = -2.19 x 10⁻⁵ C
The standard deviation of Eric’s data is 0.8°C. Martha conducted the same experiment. Her average temperature was 35.1 with a standard deviation of 1.2°C. Her data are more precise than Eric's
Answer:
less precise than
Explanation:
Answer:
less precise than
Explanation:
its right on edge2020
If three resistors 2ohm, 3ohm and 4ohm are connected in a circuit. Calculate the equivalent resistance of the combination
You haven't described whether they're connected in series or in parallel. Actually, there are eight (8) different ways they can be arranged, and each way has a different equivalent resistance.
==> All 3 resistors in series. Equivalent resistance = (2+3+4) = 9.000 ohms
==> All 3 resistors in parallel.
Equivalent resistance = 1 / (1/2 + 1/3 + 1/4) = (12/13) ohm or 0.923 ohm
==> The 2 ohm resistor in series with (the 3 and the 4 in parallel)
Equivalent resistance = 2 + 1/(1/3 + 1/4) = 3.714 ohms
==> The 3 ohm resistor in series with (the 2 and the 4 in parallel)
Equivalent resistance = 3 + 1/(1/2 + 1/4) = 4.333 ohms
==> The 4 ohm resistor in series with (the 2 and the 3 in parallel)
Equivalent resistance = 4 + 1/(1/2 + 1/3) = 5.200 ohms
==> The 2 ohm resistor in parallel with (the 3 and the 4 in series)
Equivalent resistance = 1.556 ohms
==> The 3 ohm resistor in parallel with (the 2 and the 4 in series)
Equivalent resistance = 2.000 ohms
==> The 4 ohm resistor in parallel with (the 2 and the 3 in series)
Equivalent resistance = 2.222 ohms
True or False: If a point charge has electric field lines that point into it, the charge must be positive.
Answer:
False
Explanation:
If a point charge has electric field lines point into it,then charge must be negative because electric lines point into negative charges and point out of positive charges
What does the statement "all motion is relative"
mean?
Which of the following describes how the mechanical advantage of a simple machine affects the amount of force needed for work? A. By decreasing the distance over which the input force is applied, less force is needed to do the same amount of work. B. By increasing the distance over which the input force is applied, more force is needed to do the same amount of work. C. By decreasing the distance over which the input force is applied, more force is needed to do the same amount of work. D. By increasing the distance over which the input force is applied, less force is needed to do the same amount of work.
Answer:
The answer is a
Explanation:
Caroline, a piano tuner, suspects that a piano's B4 key is out of tune. Normally, she would play the key along with her B4 tuning fork and tune the piano to match, but her B4 tuning fork is missing! Instead, she plays the errant key along with her A4 tuning fork (which has a frequency of 440.0 Hz), displays the resulting waveform on a handheld oscilloscope, and measures a beat frequency of 15.9 Hz. Then, she plays the errant key along with her C5 tuning fork (which has a frequency of 523.3 Hz) and measures a beat frequency of 67.4 Hz. What frequency is being played by the out-of-tune key? If the B4 key is supposed to produce a frequency of 493.9, is the frequency of the key lower than it should be ("flat") or higher than it should be ("sharp")?
Answer:
455.9 Hz, Flat
Explanation:
The beat frequency is basically the difference between the frequency of the B4 note and the frequency of the A4 tuning fork. This means the B4 note is 15.9 Hz off of 440Hz, and 67.4 Hz off of 525.3 Hz. As B4 is between the two notes, it would make sense to find its frequency by adding 15.9 to 440 and subtracting 67.4 from 523.3, both if which give us a frequency 455.9 Hz for the B4 key. This is because the note doesn't change for the different turning forks so both differences should result in the same frequency. Because the note should be 493.9 frequency but instead has a frequency of 455.9 Hz, it is flat because the frequency is lower than it is supposed to be.
Hope this helped!
This question deals with the phenomenon of "Beat Frequency".
The frequency being played by the B4 key is "455.9 Hz", which is "lower (flat) than it should be".
When two waves with a slightly different frequency interfere with each other, they produce a pattern known as beat frequency. The beat frequency is the difference between the frequencies of the interfering wave. Since the frequency of B4 key produces beat frequency with both the C5 tuning fork (which has a frequency of 523.3 Hz) and the A4 (which has a frequency of 440 Hz). Hence, the frequency of the B4 key must be between the range of these frequencies, that is, 440 Hz to 523.3 Hz.
ANALYZING THE BEAT FREQUENCY WITH A4 TUNING FORK:
Beat Frequency with A4 tuning fork = Frequency of B4 - Frequency of A4
Frequency of B4 = Frequency of A4 + Beat Frequency with A4 tuning fork
Frequency of B4 = 440 Hz + 15.9 Hz
Frequency of B4 = 455.9 Hz
ANALYZING THE BEAT FREQUENCY WITH C5 TUNING FORK:
Beat Frequency with C5 tuning fork = Frequency of C5 - Frequency of C5
Frequency of B4 = Frequency of C5 - Beat Frequency with C5 tuning fork
Frequency of B4 = 440 Hz - 67.4 Hz
Frequency of B4 = 455.9 Hz
Both the answers validate each other.
Since the expected frequency of the B4 key is 493.9 Hz, which is greater than the actual frequency produced by the B4 key. Hence, we can say that "the frequency of the key lower than it should be ("flat")"
The attached picture shows the general formula of the beat frequency.
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A speeder passes a parked police car with a speed of 65 km/h in a 50km/h zone. One second after the speeder has passed the police car, the police begin his pursuit. The police car accelerates with constant acceleration of 2m/s^2.Required:a. How long does it take for the police to catch the speeding car? b. How far did the police car travel before police caught up with the speeder? c. What is the speed of the police car when catches up with the speeder?
Answer:
a) The police will take 18.056 seconds to catch the speedy car, b) The police will travel 326.019 meters before catching the speedy car, c) The speed of the police car when catches up with the speeder is 36.112 meters per second.
Explanation:
Let suppose that speeder moves in a uniform motion, whereas police car has an uniformly accelerated motion.
a) How long does it take for the police to catch the speeding car:
Kinematic equation of each vehicle's position are described:
Speeder
[tex]s_{A} = s_{A,o}+v_{A}\cdot t[/tex]
Police Car
[tex]s_{B} = s_{B,o}+v_{B,o}\cdot t + \frac{1}{2}\cdot a_{B}\cdot t^{2}[/tex]
If [tex]s_{A} = s_{B}[/tex], [tex]s_{A,o} = s_{B,o}[/tex], [tex]v_{A} = 18.056\,\frac{m}{s}[/tex], [tex]v_{B,o} = 0\,\frac{m}{s}[/tex] and [tex]a_{B} = 2\,\frac{m}{s^{2}}[/tex], the resulting expression is done:
[tex]v_{A} \cdot t = v_{B,o}\cdot t +\frac{1}{2}\cdot a_{B}\cdot t^{2}[/tex]
[tex]\frac{1}{2}\cdot a_{B}\cdot t^{2}+(v_{B,o}-v_{A})\cdot t = 0[/tex]
[tex]t \cdot \left(\frac{1}{2}\cdot a_{B}\cdot t +v_{B,o}-v_{A} \right)= 0[/tex]
[tex]t = 0\,s\,\wedge\, t = \frac{2\cdot (v_{A}-v_{B,o})}{a_{B}}[/tex]
[tex]t = \frac{2\cdot \left(18.056\,\frac{m}{s}-0\,\frac{m}{s} \right)}{2\,\frac{m}{s^{2}} }[/tex]
[tex]t = 18.056\,s[/tex]
The police will take 18.056 seconds to catch the speedy car.
b) How far did the police car travel before police caught up with the speeder?
The distance travelled by the police is: ([tex]s_{B,o} = 0\,m[/tex], [tex]a_{B} = 2\,\frac{m}{s^{2}}[/tex], [tex]v_{B,o} = 0\,\frac{m}{s}[/tex] and [tex]t = 18.056\,s[/tex])
[tex]s_{B} = 0\,m + \left(0\,\frac{m}{s} \right)\cdot (18.056\,s)+\frac{1}{2}\cdot \left(2\,\frac{m}{s^{2}} \right) \cdot (18.056\,s)^{2}[/tex]
[tex]s_{B} = 326.019\,m[/tex]
The police will travel 326.019 meters before catching the speedy car.
c) What is the speed of the police car when catches up with the speeder?
The speed of the police car is represented by the following formula:
[tex]v_{B} = v_{B,o} + a_{B}\cdot t[/tex]
Where [tex]v_{B}[/tex] is the speed of the police car, measured in meters per second.
Given that [tex]v_{B,o} = 0\,\frac{m}{s}[/tex], [tex]t = 18.056\,s[/tex] and [tex]a_{B} = 2\,\frac{m}{s^{2}}[/tex], the final speed of the police car when catches up with the speeder is:
[tex]v_{B} = 0\,\frac{m}{s}+\left(2\,\frac{m}{s^{2}} \right)\cdot (18.056\,s)[/tex]
[tex]v_{B} = 36.112\,\frac{m}{s}[/tex]
The speed of the police car when catches up with the speeder is 36.112 meters per second.
A wire carries a steady current of 2A.the Charge that passes a cross Section in 2s is
◈ Electric current :
Electric current is defined as the rate of charge flow. (per unit time)
I = Q/tI denotes current
Q denotes charge
t denotes time
Given : I = 2A and t = 2s
To Find : Charge flow in wire (Q)
[tex]:\implies\sf\:I=\dfrac{Q}{t}[/tex]
[tex]:\implies\sf\:2=\dfrac{Q}{2}[/tex]
[tex]:\implies\:\boxed{\bf{\red{Q=4\:C}}}[/tex]
A flow of electrical charge carriers, generally electrons or electron-deficient atoms, is referred to as current. The charge that will be in the wire is 4 C.
What is current?A flow of electrical charge carriers, generally electrons or electron-deficient atoms, is referred to as current. The capital letter I is a typical sign for current. The ampere, abbreviated as A, is the standard unit.
As it is given to us that the current in the wire is 2A, while the time for which the current flows through the wire is 2 seconds.
[tex]\rm I=\dfrac Qt[/tex]
Substitute the value we will get,
[tex]\rm 2\ A = \dfrac{Q}{2\ sec}\\\\Q = 4\ C[/tex]
Thus, the charge that will be in the wire is 4 C.
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A baseball is thrown from the outfield toward the catcher. When the ball reaches its highest point, which statement is true? (A) Its velocity is not zero, but its acceleration is zero. (B) Its velocity and its acceleration are both zero. (C) Its velocity is perpendicular to its acceleration. (D) Its acceleration depends on the angle at which the ball was thrown. (E) None of the above statements are true.
Answer:
C. Its velocity is perpendicular to its acceleration
Explanation:
Because acceleration is always perpendicular to the velocity when the velocity will change direction without change it's magnitude
Which is the least dense liquid?
Answer:
lamp oil.
Explanation: the ping pong ball doesnt sink through the lamp oil.
Public television station KQED in San Francisco broadcasts a sinusoidal radio signal at a power of 777 kW. Assume that the wave spreads out uniformly into a hemisphere above the ground.
A) At a home 5.50km away from the antenna, what average pressure does this wave exert on a totally reflecting surface? answer in PA units
B) At a home 5.50km away from the antenna, what are the amplitudes of the electric and magnetic fields of the wave? answer in N/C, T
C) At a home 5.50km away from the antenna, what is the average density of the energy this wave carries? answer in J/m^3
D) For the energy density in part (c), what percentage is due to the electric field?
E) For the energy density in part (c), what percentage is due to the magnetic field?
Answer:
A. Using
I= P/ A
So = P/ 2π R²
So=777E3 watt/ 2*πx (²5.5E3m)
4.3*10^ -3W/m²
So pressure= 2I/c
=2* 4.3*10^ -3W/m²/ 3*10^ 8
2.7*10^-11p.a
B.
Using Emax= √2I/Eoc
So
√ 2*4.3*10^-3/8.8E-12*3*10^8
= 5.48V/m
So
Bmax= Emax/c
5.58/3E8
= 1.83*10-9T
D.average density is given by
= Eo(Emax/√2)²
= 8.85E-12(5.48/√2)²
1.3*10^10W/m²
Challenge Problem
5.
A man is standing on the top of a hill and sees a flagpole he knows is 45 feet high. The angle of
depression to the bottom of the pole is 12 degrees, and the angle of elevation to the top of the
pole is 16 degrees. Find his distance from the pole.
Answer: his distance from the pole is 90.16 ft
Explanation:
based on the diagram of the question i will upload along this answer;
45 - d is equal to part of the pole below the horizontal line
d/x = tan(16) ; x = d/tan(16)............equ1
(45-d) / x = tan(12); x = (45-d) / tan(12)-----------equ2
∴ d/tan(16) = (45-d) / tan(12)
d.tan(12) = (45-d).tan(16)
d.tan12 = (45×tan16) - d.tan16
0.2125d = 12.9034 - 0.2867d
0.2125d + 0.2867d = 12.9034
0.4992d = 12.9034
d = 12.9034/0.4992
d = 25.85 ft
now substitute value of d into any of our previous equation, lets take equation 1
x = 25.85 / tan(16)
x = 25.85 / 0.2867
x = 90.16 ft
Therefore his distance from the pole is 90.16 ft
A particle is moving along the x-axis. Its position as a function of time is given as x=bt-ct^2a) What must be the units of the constants b and c, if x is in meters and t in seconds?b) At time zero, the particle is at the origin. At what later time t does it pass the origin again?c) Derive an expression for the x-component of velocity.d) At what time t is the particle momentarily at rest?e) Derive an expression for the x-component of the particles acceleration, ax
Answer:
We are given x= bt +ct²
So
A. bxt= m
Because m/s*s= m
So b= m/s and c= m/s²
B.
x= bt-ct²
So at x=0 t=0
x=0 t= 2
We have
bt = ct² so t = b/c at x= 0
So b-2ct= 0
B. To find velocity we use
dx / dt = b - 2 Ct
C. At rest wen V= 0
We have t= b/2c
D. To find acceleration we use
dv / dt = - 2C
When We are given x= bt +ct²
Then A. bx.t = mJust Because of the m/s*s= mSo that, b= m/s and c= m/s²Now B. x= bt-ct²the So at x=0 t=0After that x=0 t= 2Now We have bt = ct² so t = b/c at x= 0After that So b-2ct= 0Then B. To find velocity we use that isThen dx / dt = b - 2 CtThen C. At rest wen V= 0We have t= b/2cNow D. To find acceleration we useNow the answer is dv / dt = - 2CLearn more about:
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Which is NOT one of the 11 organ systems of the body?
Integumentary system
Immune system
Skeletal system
Digestive system
Answer:
Immune
Explanation:
because I remember taking that test and I I specifically remember that question and that's the answer
A toy car that is 0.12 m long is used to model the actions of an actual car that
is 6 m long. Which ratio shows the relationship between the sizes of the
model and the actual car?
A: 5:1
B: 1:50
C: 50:1
D: 1:5
Answer:
it is B 1:50
Explanation:
just did it on apex
Which of the following are properties of a conductor in electrostatic equilibrium? The force on a charge near the conductor is zero. Electric field lines just outside the conductor are perpendicular to the surface. All net charge in the conductor is on the surface. The electric field inside the conductor is zero.
Answer:
The electric field inside the conductor is zero.
Explanation:
In electrostatic equilibrium charges are assumed to be in rest. Therefore no current.
I = 0
From the ohm's law
ΔV= IR = 0
⇒ΔV =0
Therefore, potential V is same through the conductor.
Also, E = -ΔV d = 0
Hence, The electric field inside the conductor is zero.
If a car is driving at 10 m/s and is driving for 2 minutes, what is the total distance it will
cover?
Distance = (speed) x (time)
Distance = (10 meter/second) x (2 minutes)
Distance = (10 meter/second) x (2 minutes) x (60 second/minute)
Distance = (10 x 2 x 60) (meter-minute-second / second-minute)
Distance = 1,200 meters
A car moves in a straight line at a speed of 72.1 km/h.
How far (in km) will the car move in 7.74 minutes at this speed?
Speed=distance/time.
7.74mins=464.4secs.
72.1=distance/464.4.
distance=464.4×7.21.
=3348.324m.
A 2250 kg truck has a velocity of 25 m/s to the east. What is the momentum of the truck? p = mv
The momentum of the truck is "56250 kgm/s".
According to the question,
Mass,
m = 2250 kgVelocity,
v = 25 m/sAs we know,
→ [tex]p = mv[/tex]
By putting the values, we get
→ [tex]= 2250\times 25[/tex]
→ [tex]= 56250 \ kgm/s[/tex]
Thus the answer is right.
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A projectile is launched at some angle to the horizontal with some initial speed vi, and air resistance is negligible.(a) Is the projectile a freely falling body? (b) What is its acceleration in the vertical direction? (c) What is its acceleration in the horizontal direction?
Answer:
Explanation:
When a projectile is launched at some angle to the horizontal with some speed vi , and air resistance is negligible , it is definitely a freely falling body .
It is so because it is free to accelerate towards the earth with acceleration of g . Air has no resistance , hence no force is acting on it except the gravitational force . Hence it is a freely falling body .
b )
The acceleration in the vertical direction is due to force exerted by the earth that is gravitational force on it . Hence its acceleration is equal to g in vertically downward direction .
c )
It has zero acceleration in horizontal direction . It is so because no force is acting on it in horizontal direction . So no acceleration will be present in horizontal direction . It will move in horizontal direction with constant speed of vi cos θ where θ is the angle vi make with the horizontal .