The acceleration of the SRB and main engine during the first 2.0 minutes of flight is 52.16 m/s².
The given parameters;
initial velocity of the engine, u = 1341 m/sfinal velocity of the engine, v = 7600 m/stime of motion, t = 2 minutes = 2 x 60 s = 120 sThe acceleration of the SRB and main engine is calculated as follows;
[tex]a = \frac{\Delta v}{\Delta t } \\\\a = \frac{7600 - 1341}{2 \times 60 s} \\\\a = 52.16 \ m/s^2[/tex]
Thus, the acceleration of the SRB and main engine during the first 2.0 minutes of flight is 52.16 m/s².
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A friend has suggested that you go swimming in a pool having water of temperature 350 K. What would this temperature be on the Fahrenheit scale?
109°F
123°F
170°F
202°F
This temperature would be 170° F on the Fahrenheit scale. Hence, option (C) is correct.
What is temperature?The physical concept of temperature indicates in numerical form how hot or cold something is. A thermometer is used to determine temperature. Thermometers are calibrated using a variety of temperature scales, which historically defined distinct reference points and thermometric substances.
The most popular scales are the Celsius scale, sometimes known as centigrade, with the unit symbol °C, the Fahrenheit scale (°F), and the Kelvin scale (K), with the latter being mostly used for scientific purposes.
the relation between Kelvin scale and Fahrenheit scale is given by:
(F - 32)/180 = (K - 273)/100
F - 32 = (350 - 273)(9/5)
F = 32 + (350 - 273)(9/5)
F = 170
Hence, this temperature would be 170° F on the Fahrenheit scale.
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If a battery causes a wire to carry a current of 4 Amps how many coulombs of charge flow past any point in the wire in 3 seconds
[tex]▪▪▪▪▪▪▪▪▪▪▪▪▪ {\huge\mathfrak{Answer}}▪▪▪▪▪▪▪▪▪▪▪▪▪▪[/tex]
According to above question ~
Current (I) = 4 AmperesTime (t) = 3 seconds Charge (q) = ?Let's find the charge (q) by using formula ~
[tex]I = \dfrac{q}{t} [/tex][tex]4 = \dfrac{q}{3} [/tex][tex]q = 4 \times 3[/tex][tex]q = 12 \: \: coulombs[/tex]Hence, 12 coulombs of charge flow past any point in the wire in 3 seconds
A cubical box with sides of length 0.368 m contains 1.980 moles of neon gas at a temperature of 298 K. What is the average rate (in atoms/s) at which neon atoms collide with one side of the container? The mass of a single neon atom is 3.35x10-26 kg.
The average rate at which the neon atoms collide with one side of the container is [tex]3.31 \times 10^{26} \ atoms/s[/tex].
The given parameters;
length of the cubical box, L = 0.368number of moles of the gas, n = 1.98temperature, T = 298 Kmass of the gas, m = 3.35 x 10⁻²⁶ kgThe average kinetic energy of the gas molecules is calculated as follows;
[tex]K = \frac{3}{2} \frac{R}{N_a} T\\\\K = \frac{3 \times 8.314\times 298}{2\times 6.022 \times 10^{23}} \\\\K = 6.17\times 10^{-21} \ J/atoms[/tex]
The average speed of the gas molecules is calculated as follows;
[tex]K = \frac{1}{2}mv_{rms}^2\\\\v_{rms} = \sqrt{\frac{2K}{m} } \\\\v_{rms} = \sqrt{\frac{2\times 6.17 \times 10^{-21}}{3.35\times 10^{-26}} } \\\\v_{rms} = 607 \ m/s[/tex]
The time of collision of the gas molecules with the walls of the container is calculated as follows;
[tex]t = \frac{2d}{v} \\\\t = \frac{2\times 0.368}{607} \\\\t = 0.0012 \ s[/tex]
The average rate at which the gases collide with a single wall out of the 3 identical walls is calculated as follows;
[tex]rate =\frac{1}{3} \times \frac{n \times N_a}{t} \\\\rate = \frac{1.98 \times 6.02 \times 10^{23} \ atoms}{3 \times 0.0012 \ s} \\\\rate = 3.31 \times 10^{26} \ atoms/s[/tex]
Thus, the average rate at which the gases collide with one side of the container is [tex]3.31 \times 10^{26} \ atoms/s[/tex].
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How would you best define the word drug?
A: Something that makes you tired
B: Something that can kill you
C: Something that effects your body and mind
D: Stored for energy
someone help
Answer:
C
Explanation:
Definition of drug: a medicine or other substance which has a physiological effect when ingested or otherwise introduced into the body
a boat's engine can give it a velocity of 25m/s. if the boat heads east across a river which of the following due south with a velocity of 8.5m/s; what is the resultant velocity of the boat? (remember you must find both a magnitude and direction!)
Answer:
Explanation:
v = √(25² + 8.5²) = 26.40549... = 26 m/s
θ = arctan(8.5/25) = 18.77803... = 19° S of E
a student lifts a toy car from a bench and places the toy car at the top of a slope describe an energy transfer that occurs when the student lifts the toy car from the bench and places the toy car at the top of the slope.
Answer:
Assuming there are no energy losses due to friction or drag, the gravitational potential energy will change into kinetic potential energy as the car reaches the bottom of the slope.
G.P.E = m*g*h
K.E = (m*v^2)/2
where
m = mass of toy car (kg)
g = gravity (m/s^2)
h = heigh of your car from the bottom (m)
v = velocity of the toy car as it reaches the bottom (m/s)
Equate K.E to G.P.E
G.P.E = K.E
m*g*h = (m*v^2)/2
make v the subject of the formula
v = (2*g*h)^(1/2)
Substitute g = 9.81 m/s^2 and h = 2m into the equation to get v
v = (2*9.81*2)^(1/2)
v = 6.264 m/s
Children in a tree house lift a small dog in a basket 3.85 m up to their house. If it takes 201 J of work to do this, what is the combined mass of the dog and basket
Answer:
Explanation:
The work will equal the increase in potential energy.
PE = mgh
m = PE/gh = W/gh = 201/(9.81(3.85)) = 5.32 kg
I NEED THE ANSWER ASAPP
Answer:
Explanation:
a) The spring force will equal the weight.
b) If up is positive
kx - mg = 0
mg = kx kx = 25 N
c) m = kx/g = 25/10 = 2.5 kg
A man applies a force of 540 N to the barrow in a direction 75 from the horizontal. He moves the barrow 30 m along the level ground. Calculate the work he does against friction?
The work done by the man against friction is 4,192.86 J.
The given parameters;
force applied, F = 540 Nangle of inclination, θ = 75⁰horizontal distance, x = 30 mThe work done by the man against friction is calculated as follows;
[tex]W = F \times d \times cos(75)\\\\W = 540 \times 30 \times cos(75)\\\\W = 4,192.86 \ J[/tex]
Thus, the work done by the man against friction is 4,192.86 J.
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What is a list of all the states of matter?
Answer:
3
Explanation:
state of matter are solid
liquid and
gases
Answer:
3
Explanation:
state of a matter are solid liquid and gas
A gold doubloon 6.1 cm in diameter and 2.0mm thick is dropped over the side of a Pirate Ship. When it comes to rest on the ocean floor at a depth of 770m how much has its volume changed
The volume of a material is the total amount of matter that it can contain. The volume of the given coin has been determined to be 5.85 x [tex]10^{-6}[/tex] [tex]m^{3}[/tex]. Since the gold doubloon do not absorb water, then its volume remains constant at the ocean floor.
The volume of the gold doubloon can be determined by;
volume = [tex]\pi r^{2}[/tex] + h
where r is the radius of the coin and h is its thickness.
Such that; diameter = 6.1 cm (61 mm) and h = 2.0 mm
r = [tex]\frac{diameter}{2}[/tex]
= [tex]\frac{61}{2}[/tex]
r = 30.5 mm
Thus,
volume of the coin = [tex]\frac{22}{7}[/tex] x [tex](30.5)^{2}[/tex] x 2
= 5847.2857
Therefore, the volume of the gold doubloon is 5847.3 [tex]mm^{3}[/tex]. This can also be expressed as 5.85 x [tex]10^{-6}[/tex] [tex]m^{3}[/tex].
Since the gold doubloon is not miscible with water, thus its volume at a depth of 770 m at the ocean floor is the same as its initial volume.
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The second hand on a clock is 3.00 cm long. What is the speed of the outermost tip of that second hand
In 60 minutes or 3600 seconds, the tip of the minute hand traverses the circumference of a circle with radius 3.00 cm, so it moves with a tangential speed of
(3.00 cm)/(3600 s) ≈ 0.00083 cm/s = 8.3 μm/s
Which properties make a metal a good material to use for electrial wires
Answer:
Most importantly metals can pass an electric current without being affected and changed by the electricity. Electrical conductivity combined with ductility makes metals the most suitable materials for electrical transmission wires.
Find the dimension of the gravitational constant in this equation F=Gm1m2/r¹r²
The gravitational force acting between the two bodies is given by:
F=G
r
2
m
1
m
2
G=
m
1
m
2
Fr
2
The dimension of the force is [MLT
−2
]
=
[M][M]
[MLT
−2
][L
2
]
=M
−1
L
3
T
−2
A man is whirling a 0.25 kg ball on a 1.5 m long string at 3 m/s. Find the centripetal acceleration of this ball.
Question 2 options:
0.5 m/s2
13.5 m/s2
6 m/s2
2 m/s2
The centripetal acceleration of this ball is equal to 12 [tex]m/s^2[/tex]
Given the following data:
Diameter = 1.5 mSpeed, V = 3 m/s.Mass = 0.25 kgRadius = [tex]\frac{Diameter}{2} = \frac{1.5}{2} = 0.75 \;meters[/tex]
To find the centripetal acceleration of this ball:
The acceleration of an object along a circular track is referred to as centripetal acceleration.
Mathematically, the centripetal acceleration of an object is given by the formula:
[tex]A_c = \frac{V^2}{r}[/tex]
Where:
Ac is the centripetal acceleration.r is the radius of the circular track.V is the velocity of an object.
Substituting the given parameters into the formula, we have;
[tex]A_c = \frac{3^2}{0.75}\\\\A_c = \frac{9}{0.75}\\\\A_c = \frac{9}{0.75}[/tex]
Centripetal acceleration = 12 [tex]m/s^2[/tex]
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jshshwjs sbwiwiw910mw s x djjskskekwkq
Answer:
jsbdhdndmlsusgsbkaksudgnslsosufhbf ffb
An object is dropped from a vertical height of 1.89 m above the balcony level. What is the object’s speed when it is 2.20 m below the balcony level if 10.0% energy is lost due to the air resistance? Does it matter when to apply 10% loss before V calculations or after? [8.49m/s] [yes it does, 0.9Energy result in √0.9Velocity]
Answer:
The 10% must come off the energy, not the velocity
Explanation:
Change in height h = 1.89 + 2.20 = 4.09 m
Potential energy = mgh = m(9.81)(4.09) = 40.1m J
Friction losses 40.1m(0.10) = 4.01m J
Kinetic energy at end 40.1m - 4.01m = 36.1m J
KE = ½mv²
v = √(2KE/m)
v = √(2(36.1m)/m = √72.2
v = 8.4983...
v = 8.50 m/s
A 0.035-kg bullet is fired vertically at 214 m/s into a 0.15-kg baseball that is initially at rest. How high does the combined bullet and baseball rise after the collision, assuming the bullet embeds itself in the ball
Answer:
Explanation:
conservation of momentum during the collision
0.035(214) + 0.15(0) = 0.185v
v = 40.486 m/s
The kinetic energy after impact will convert to gravity potential energy
(ignoring air resistance)
mgh = ½mv²
h = v²/2g
h = 40.486² / (2(9.8))
h = 83.6303...
h = 84 m
electron and proton are projected with same velocity normal to the magnetic field which one will suffer greater deflection? why
Answer:
Explanation:
The deflection of a charged particle by a magnetic field is proportional to its electric charge and to its velocity. The deflection is also inversely proportional to its mass. So given a proton and an electron going at the same velocity in a magnetic field and having equal (but opposite) electric charge the electron will deflect much more since the ratio of the masses is 1836.
A 2.0 kg particle moving along the z-axis experiences the
force shown in (Figure 1). The particle's velocity is
3.0 m/s at x = 0 m.
A) At what point on the x axis does the particle have a turning point?
At point x = 0, the particle accelerates. Since there will be change of velocity at that point. The the force of the particle will change from negative sign to positive sign according to the given figure, we can therefore conclude that the particle will have a turning point at point x = 0.
Given that a 2.0 kg particle moving along the z-axis experiences the force shown in a given figure.
Force is the product of mass and acceleration. While acceleration is the rate of change of velocity. Both the force and acceleration are vector quantities. They have both magnitude and direction.
If the particle's velocity is 3.0 m/s at x = 0 m, that mean that the particle experience change of velocity at point x = 0. Since the the force of the particle will change from negative sign to positive sign according to the given figure, we can therefore conclude that the particle will have a turning point at point x = 0.
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Two airplanes taxi as they approach the terminal. Plane 1 taxies with a speed of 13 m/s due north. Plane 2 taxies with a speed of 8.5 m/s in a direction 20 ∘ north of west.
Part A
What is the magnitude of the velocity of plane 1 relative to plane 2?
Part B
What is the direction of the velocity of plane 1 relative to plane 2?
Part C
What are the magnitude of the velocity of plane 2 relative to plane 1?
Answer:
Explanation:
Plane 2 is moving north at
8.5sin20 = 2.9 m/s
Plane 2 is moving west at
8.5cos20 = 8.0 m/s
Part A
v = √((13 - 2.9)² + 8.0²) = 12.876... 13 m/s
Part B
θ = arctan((13 - 2.9) / 8.0) = 51.617... 52° N of E
Part C
13 m/s 52° S of W
relative velocity magnitude is independent of reference frame
Upton Chuck is riding the Giant Drop at Great America. If Upton free falls for 2.60 seconds, what will be his final velocity and how far will he fall?
Answer:
d = -33.1 m and Vf = -25.5 m/s
Explanation:
Given:a = -9.8 m
t = 2.6 s
Vᵢ = 0 m/s
To Find:d = ?
Vf = ?
Now,
d = Vᵢ × t + 0.5 × a × t²
d = (0 m/s) × (2.60 s) + 0.5 × (-9.8 m/s²) × (2.60 s)²
d = -33.1 m (- indicates direction)
Vf = Vᵢ + a × t
Vf = 0 + (-9.8 m/s²) × (2.60 s)
Vf = -25.5 m/s (- indicates direction)
Thus, d = -33.1 m and Vf = -25.5 m/s
-TheUnknownScientist 72
It is known that a general solution for the displacement from equilibrium of a harmonic oscillator is x(t)=Ccos(ωt)+Ssin(ωt), where C, S, and ω are constants.
A) Using the general equation for x(t) given in the problem introduction, express the initial position of the block xinit in terms of C, S, and ω (Greek letter omega).
b) Find the value of S using the given condition that the initial velocity of the block is zero: v(0)=0.
c)What is the equation x(t) for the block? Express your answer in terms of t, ω, and xinit.
d)Find the equation for the block's position xnew(t) in the new coordinate system.
Express your answer in terms of L, xinit, ω (Greek letter omega), and t.
The characteristics of the expression of the simple harmonic motion allows to find the results for the expression of the mass- block system are:
A) The constant Ces: C = xinit
B) The ocsntna S is: S = 0
C) The equation of the system is: x = xinit cos wt
D) If the reference system is at some extreme, the equation is:
[tex]L - x_{init} = x_{init} \ cos \ wt[/tex]
The simple harmonic movement is an oscillatory movement where the restoring force is proportional to the displacement, the general equation that describes this movement is indicated.
x = C cos wt + S sin wt
Where x is the displacement C and S are constants. W the angular velocity and t the time.
A) The initial position of the body occurs when the time is zero, t = 0
We substitute.
x = C cos 0 + S sin 0
[tex]x_{init}[/tex] = C
B) The velocity of the particle is defined.
[tex]v= \frac{dx}{dt} \\ v= C w \ sin \ wt - Sw \ cos \ wt[/tex]
The initial velocity occurred for time zero t = 0
v = - S w
It indicates that the initial velocity is zero, since the angular velocity must be different from zero, it implies that the constant is valid.
S = 0
C) The equation for the block remains.
x (t) = [tex]x_{init} \ cos \ wt[/tex]
D) In some cases it is measured with respect to another reference system, the most common are:
For maximum compression it is the zero of the system. The maximum extension is the zero of the system.
In these cases, the change that must be made is
x = [tex]L - x_{min}[/tex] t
we substitute
[tex]L - x_{init} = x_{init} \ cos \ wt[/tex]
L = [tex]x_{init}[/tex] (1 + cos wt)
In conclusion, using the characteristics of the expression of the simple harmonic motion we can find the results for the expression of the mass- block system are:
A) The constant Ces: C = xinit
B) The ocsntna S is: S = 0
C) The equation of the system is: x = xinit cos wt
D) If the reference system is at some extreme, the equation is:
[tex]L - x_{init} = x_{init} \ cos \ wt[/tex]
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Convection currents occur when _________ energy transfers between two parts of a fluid
Answer:
heat
Explanation:
Which of the vectors in the graph below is the negative of the vector v
A. a
B. d
C. c
D. b
Answer:
Option c.
Explanation:
Contrast the behavior of a water wave that travel by a stone barrier to a sound wave that travels through a door
calvin carter
Explanation:
here the file has everything
A mass undergoes SHM with amplitude of 4 cm. The energy is 8.0 J at this time. The mass is cut in half, and the system is again set in motion with amplitude 4.0 cm. What is the energy of the system now?
Hi there!
[tex]\large\boxed{E_{total} = 8.0 \text{ J}}[/tex]
For a mass undergoing SHM, the total energy of the system is given as:
[tex]ME = \frac{1}{2}kA^2[/tex]
Where:
k = Spring constant (N/m)
A = amplitude (m)
There is no quantity of mass in the equation, so the total mechanical energy of the system is NOT impacted by the object's mass.
Thus, the energy of the system will still be 8.0 J.
A 1-kg mass at the Earth's surface weighs how much
Answer:
the answer is weight=10N
Answer:
[tex]\boxed {\boxed {\sf 9.8 \ Newtons}}[/tex]
Explanation:
Weight is also called the force of gravity. This force acts on all objects at all times, pulling them down toward the center of the Earth.
It is calculated by multiplying the mass by the acceleration due to gravity.
[tex]F_g=mg[/tex]
The mass of the object is 1 kilogram. This scenario is occurring on Earth, so the acceleration due to gravity is 9.8 meters per second squared.
m= 1 kg g= 9.8 m/s²Substitute the values into the formula.
[tex]F_g= 1 \ kg *9.8 \ m/s^2[/tex]
Multiply.
[tex]F_g= 9.8 \ kg*m/s^2[/tex]
Convert the units. 1 kilogram meter per second squared is equal to 1 Newton, so our answer of 9.8 kilogram meters per second squared is equal to 9.8 Newtons.
[tex]F_g= 9.8 \ N[/tex]
A 1 kilogram mass at Earth's surface weighs 9.8 Newtons.
Please Help
A projectile fired over level ground has an initial total velocity of 41.3 m/s. It is in the air for 5.1 s. What is the x-component of the projectile's initial velocity?
Answer:
Explanation:
In the vertical analysis assuming launch from ground level.
0 = 0 + (41.3sinθ)(5.1) + ½(-9.8)5.1²
(41.3sinθ)(5.1) = ½(9.8)5.1²
(41.3sinθ) = ½(9.8)5.1
sinθ = ½(9.8)5.1/41.3
sinθ = 0.60508...
θ = 37.235°
vx = 41.3cos37.235
vx = 32.881452...
vx = 32.9 m/s
A small 1240-kg SUV has a wheelbase of 3.2 m. If 67% of its weight rests on the front wheels, how far behind the front wheels is the wagon's center of mass
Answer:
Explanation:
Let d be the distance to the center of mass from the front wheels
Sum moments about the front wheel contact point to zero
1240(9.8)[d] - 1240(9.8)(1 - 0.67)[3.2] = 0
1240(9.8)[d] = 1240(9.8)(1 - 0.67)[3.2]
d = (1 - 0.67)[3.2]
d = 1.056 m