Answer:
Physical properties are properties that can be measured or observed without changing the chemical nature of the substance. Some examples of physical properties are:
color (intensive)
density (intensive)
volume (extensive)
mass (extensive)
boiling point (intensive): the temperature at which a substance boils
melting point (intensive): the temperature at which a substance melts
Explanation:
Question 9 of 10
According to the law of conservation of momentum, the total initial
momentum equals the total final momentum in a(n)
A. Interacting system
B. System interacting with one other system
C. Isolated system
D. System of balanced forces
Answer:
The answer is C. Isolated System
Answer:
C. Isolated system
Explanation :
∵According to law of conservation of momentum ,In an isolated system ,the total momentum remains conserved.
The block in the drawing has dimensions L0×2L0×3L0,where L0 =0.2 m. The block has a thermal conductivity of 150 J/(s·m·C˚). In drawings A, B, and C, heat is conducted through the block in three different directions; in each case the temperature of the warmer surface is 35 ˚C and that of the cooler surface is 16 ˚C Determine the heat that flows in 6 s for each case.
Answer:
1140 J, 6840 J, 10260 J
Explanation:
Lo x 2 Lo x 3 Lo, Lo = 0.2 m, K = 150 J/(s · m · C˚) , T = 35 ˚C, T' = 16 ˚C,
time, t = 6 s
The heat conducted is
[tex]H = \frac{K A (T - T') t}{d}\\\\H = \frac{150\times 3\times 0.2\times 0.2\times (35-16) \times 6}{3\times 0.2}\\\\H = 1140 J[/tex]
The heat conducted is
[tex]H = \frac{K A (T - T') t}{d}\\\\H = \frac{150\times 3\times 0.2\times 2\times0.2\times (35-16) \times 6}{3\times 0.2}\\\\H = 6840 J[/tex]
The heat conducted is
[tex]H = \frac{K A (T - T') t}{d}\\\\H = \frac{150\times 3\times 0.2\times 2\times0.2\times (35-16) \times 6}{2\times 0.2}\\\\H = 10260 J[/tex]
prove mathematically :
1. v = u + at
2. s = ut+1*2 at
Answer:
a.v=u+v/2
a.v=s/t
combining two equation we get,
u+v/2=s/t
(u+v)t/2=s
(u+v)t/2=s
{u+(u+at)}t/2=s
(u+u+at)t/2=s
(2u+at)t/2=s
2ut+at^2/2=s
2ut/2+at^2/2=s
UT +1/2at^2=s
proved
a=v-u/t
at=v-u
u+at=v
A transverse sine wave with an amplitude of 2.50 mm and a wavelength of 1.80 m travels, from left to right along a long, horizontal stretched string with a speed of 36.0 m s. I Take the origin at the left end of the undisturbed string. At time t = 0 the left end of the string has its maximum upward displacement,
(a) What is the frequency of the wave?
(b) What is the angular frequency of the wave?
(c) What is the wave number of the wave?
(d) What is the function y(x,t) that describes the wave?
(e) What is y(t) for a particle at the left end of the string?
(f) What is y(t) for a particle 1.35 m to the right of the origin?
(g) What is the maximum magnitude of transverse velocity of any particle of the string?
(h) Find the transverse displacement of a particle 1.35 m to the right of the origin at time t = 0.0625 s.
(i) Find the transverse velocity of a particle 1.35 m to the right of the origin at time t = 0.0625 s.
Explanation:
Given that,
Amplitude, A = 2.5 nm
Wavelength,[tex]\lambda=1.8\ m[/tex]
The speed of the wave, v = 36 m/s
At time t = 0 the left end of the string has its maximum upward displacement.
(a) Let f is the frequency. So,
[tex]f=\dfrac{v}{\lambda}\\\\f=\dfrac{36}{1.8}\\\\f=20\ Hz[/tex]
(b) Angular frequency of the wave,
[tex]\omega=2\pi f\\\\=2\pi \times 20\\\\=125.7\ rad/s[/tex]
(c) The wave number of the wave[tex]=\dfrac{1}{\lambda}[/tex]
[tex]=\dfrac{1}{1.8}\\\\=0.56\ m^{-1}[/tex]
When an apple falls towards the earth,the earth moves up to meet the apple. Is this true?If yes, why is the earth's motion not noticeable?
Answer:
because the mass of the apple is very less compared to the mass of earth. Due to less mass the apple cannot produce noticable acceleration in the earth but the earth which has more mass produces noticable acceleration in the apple. thus we can see apple falling on towards the earth but we cannot see earth moving towards the apple.
I need help with this problem can anybody help me please , it’s physics 2 course
Answer:
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Explanation:
The 2 capacitors in the middle are connected in parallel so simply add their capacitance together:
[tex]5.0\:\mu\text{F} + 8.0\:\mu\text{F} = 13.0\:\mu \text{F}[/tex]
Now we have 3 capacitors connected in series so their equivalent capacitance [tex]C_{eq}[/tex] is
[tex]\dfrac{1}{C_{eq}} = \dfrac{1}{10.0\:\mu \text{F}} + \dfrac{1}{13.0\:\mu \text{F}} + \dfrac{1}{9.0\:\ mu \text{F}} [/tex]
or
[tex]C_{eq} = 3.5\:\mu \text{F}[/tex]
1.- Que distancia recorrió una carga de 2,5x10-6 coul, generando así un campo eléctrico de 55new/coul.
Answer:
r = 20.22 m
Explanation:
Given that,
Charge,[tex]q=2.5\times 10^{-6}\ C[/tex]
Electric field, [tex]E=55\ N/C[/tex]
We need to find the distance. We know that, the electric field a distance r is as follows :
[tex]E=\dfrac{kq}{r^2}\\\\r=\sqrt{\dfrac{kq}{E}}\\\\r=\sqrt{\dfrac{9\times 10^9\times 2.5\times 10^{-6}}{55}}\\\\r=20.22\ m[/tex]
So, the required distance is 20.22 m.
If you and a friend are standing side-by-side watching a soccer game, would you both view the motion from the same reference frame?
a. Yes, we would both view the motion from the same reference point because both of us are at rest in Earth’s frame of reference.
b. Yes, we would both view the motion from the same reference point because both of us are observing the motion from two points on the same straight line.
c. No, we would both view the motion from different reference points because motion is viewed from two different points; the reference frames are similar but not the same.
d. No, we would both view the motion from different reference points because response times may be different; so, the motion observed by both of us would be different.
Answer:
the correct is C
Explanation:
The concept of a frame of reference is of crucial importance in physics, because it is the system from which measurements are made. Therefore, the relationships between the different reference frames must be clear so that the measurements made can be compared correctly.
In this case, the first observed sees the movement of the ball, suppose it moves a distance r, the second observed is next to me, separated by a distance x, therefore a frame of reference located in the movement of the ball. ball r '.
Consequently, the measurement carried out is related by
r = r’ + x
where the bold letters indicate wind blowers.
With these explanations we review the different answers, the correct one is C
a vessel with mass 10kg intially moving withthe velocicity 12m s along the x axis explodes into three exactly identical pieces Just after the explosion one piece moves with speed 10 m s along the x axis and asecond piece moves with speed 10 m s along the y axis What iis the magnitude of the component of velocity of the third piece along the y axiss
Answer:
Explanation:
Apply law of conservation of momentum along y-axis.
Initially there was no momentum along y-axis. So there will be nil momentum along y-axis again finally.
Let the mass of each piece after breaking be m .
Momentum of piece moving along positive y-axis
= m x 10 = 10m .
Let the component of velocity of third piece along y-axis be v .
Its momentum along the same direction = m v .
Total momentum along y -axis = 10 m + m v
According to law of conservation of momentum
10 m + mv = 0
v = - 10 m/s .
Component of velocity of the third piece along y-axis will be - 10 m/s .
In other words it will be along negative y-axis with speed of 10 m/s.
ACCORDING TO NEWTON'S THIRD LAW EVERY ACTION HAS EQUAL AND OPPOSITE REACTION BUT THEN WHY DON'T WE FLY WHEN WE FART??
Answer:
Your fart only has so much force, not nearly enough to launch you into oblivion. Your fart and you still exert a force onto each other, so I guess, hypothetically, you could fly if you really, really try hard enough. Just make sure you don't try too hard and prolapse as a result :)
What is the energy equivalent of an object with a mass of 2.5 kg? 5.5 × 108 J 7.5 × 108 J 3.6 × 1016 J 2.25 × 1017 J
Answer:
E = m c^2 = 2.5 * (3 * 10E8)^2 = 2.25 * 10E17 Joules
Answer:
The answer is D. 2.25 × 1017 J
Explanation:
got it right on edge 2021
The relation of mass m, angular velocity o and radius of the circular path r of an object with the centripetal force is-
a. F = m²wr
b. F = mwr²
c. F = mw²r
d. F = mwr.
Answer:
Correct option not indicated
Explanation:
There are few mistakes in the question. The angular velocity ought to have been denoted with "ω" and not "o" (as also suggested in the options).
The formula to calculate a centripetal force (F) is
F = mv²/r
Where m is mass, v is velocity and r is radius
where
While the formula to calculate a centrifugal force (F) is
F = mω²r
where m is mass, ω is angular velocity and r is radius of the circular path.
From the above, it can be denoted that the relationship been referred to in the question is that of a centrifugal force and not centripetal force, thus the correct option should be C.
NOTE: Centripetal force is the force required to keep an object moving in a circular path/motion and acts inward towards the centre of rotation while centrifugal force is the force felt by an object in circular motion which acts outward away from the centre of rotation.
A 1,200kg roller coaster car starts rolling up a slope at a speed of 15m/s. What is the highest point it could reach
Answer: 11.36 m
Explanation:
Given
Mass of roller coaster is m=1200 kg
Initial speed of roller coaster is v=15 m/s
Energy at bottom and at the top is same i.e.
[tex]\Rightarrow \dfrac{1}{2}mv^2=mgh\\\\\Rightarrow \dfrac{1}{2}\times 1200\times 15^2=1200\times 9.8\times h\\\\\Rightarrow h=\dfrac{15^2}{2\times 9.8}\\\\\Rightarrow h=11.36\ m[/tex]
Thus, the highest point reach by the roller coaster is 11.36 m
Answer:
11.36m
Explanation:
A rigid tank contains 10 lbm of air at 30 psia and 60 F. Find the volume of the tank in ft3. The tank is now heated until the pressure doubles. Find the heat transfer in Btu.
Answer:
Hence the amount of heat transfer is 918.75 Btu.
Explanation:
Now,
Assume that I = E/(R + r), prove that 1/1 = R/E + r/E
[tex]\implies {\blue {\boxed {\boxed {\purple {\sf { \frac{1}{I} = \frac{R}{E} + \frac{r}{E} }}}}}}[/tex]
[tex]\large\mathfrak{{\pmb{\underline{\orange{Step-by-step\:explanation}}{\orange{:}}}}}[/tex]
[tex]I = \frac{ E}{ R + r} \\[/tex]
[tex] ➺\:\frac{I}{1} = \frac{E}{R + r} \\[/tex]
Since [tex]\frac{a}{b} = \frac{c}{d} [/tex] can be written as [tex]ad = bc[/tex], we have
[tex]➺ \: I \: (R + r) = E \times 1[/tex]
[tex]➺ \: \frac{1}{I} = \frac{R + r}{E} \\ [/tex]
[tex]➺ \: \frac{1}{I} = \frac{R}{E} + \frac{r}{E} \\ [/tex]
[tex]\boxed{ Hence\:proved. }[/tex]
[tex]\red{\large\qquad \qquad \underline{ \pmb{{ \mathbb{ \maltese \: \: Mystique35ヅ}}}}}[/tex]
Find the volume of cuboid of side 4cm. Convert it in SI form
Answer:
0.000064 cubic meters.
Explanation:
Given the following data;
Length of side = 4 centimeters
Conversion:
100 centimeters = 1 meters
4 cm = 4/100 = 0.04 meters
To find the volume of cuboid;
Mathematically, the volume of a cuboid is given by the formula;
Volume of cuboid = length * width * height
However, when all the sides are equal the formula is;
Volume of cuboid = L³
Volume of cuboid = 0.04³
Volume of cuboid = 0.000064 cubic meters.
You need to calculate the volume of berm that has a starting cross-sectional area of 118 SF, and an ending cross-sectional area of 245 SF. The berm is 300 ft long and is assumed to taper evenly between the two cross-sectional areas, what is the calculated volume of the berm in cubic feet
A ball has a mass of 4.65kg and approximates a ping pong ball of mass 0.060kg that is at rest by striking it in an elastic collision. The initial velocity of the bowling ball is 5.00m/s, determine the final velocities of both masses after the collision.
Answer:
Look at work
Explanation:
Elastic Collision: Ki=Kf
M1=4.65kg
M2: 0.060kg
v1=5m/s
v2=0m/s
4.65*5+0.060*0=4.65*v1'+0.060*v2'
23.25+0=4.65v1'+0.060v2'
Also since it is an elastic collision we can use
v1+v1'=v2+v2'
4.65+v1'=v2'
4.65+v1'=v2'
Substitute into the earlier equation
23.25=4.65v1'+0.060(4.65+v1')
Expand
23.25=4.65v1'+0.279+0.06v1'
Solve for v1'
22.971=4.71v1'
v1'=4.88m/s
v2'=4.65+4.88=9.53m/s
A grade 12 Physics student shoots a basketball
from the ground at a hoop which is 2.0 m above
her release. The shot was at a velocity of 10 m/s
and at an angle of 80° to the ground.
a. Determine the vertical velocity of the ball
when it is at the level of the net. You
should get two answers.
Please show ALL steps
Answer:
7.84 m/s
Explanation:
Height, h = 2 m
Initial velocity, u = 10 m/s
Angle, A = 80°
(a) Let the time taken to go to the net is t.
Use second equation of motion
[tex]h = u t + 0.5 at^2\\\\- 2 = - 10 sin 80 t - 4.9 t^2\\\\4.9 t^2 + 9.8 t - 2 = 0 \\\\t= \frac{- 9.8\pm\sqrt{9.8^2 + 4\times 4.9\times 2}}{9.8}\\\\t = \frac{- 9.8 \pm 11.6}{9.8}\\\\t = - 2.2 s , 0.2 s[/tex]
Time cannot be negative.
So, t = 0.2 s
The vertical velocity at t = 0.2 s is
v = u + at
v = 10 sin 80 - 9.8 x0.2
v = 9.8 - 1.96 = 7.84 m/s
During a practice shot put throw, the 7.9-kg shot left world champion C. J. Hunter's hand at speed 16 m/s. While making the throw, his hand pushed the shot a distance of 1.4 m. Assume the acceleration was constant during the throw.
Required:
a. Determine the acceleration of the shot.
b. Determine the time it takes to accelerate the shot.
c, Determine the horizontal component of the force exerted on the shot by hand.
Answer:
a) a = 91.4 m / s², b) t = 0.175 s, c)
Explanation:
a) This is a kinematics exercise
v² = vox ² + 2a (x-xo)
a = v² - 0/2 (x-0)
let's calculate
a = 16² / 2 1.4
a = 91.4 m / s²
b) the shooting time
v = vox + a t
t = v-vox / a
t = 16 / 91.4
t = 0.175 s
c) let's use Newton's second law
F = ma
F = 7.9 91.4
F = 733 N
The working substance of a certain Carnot engine is 1.50 mol of an ideal monatomic gas. During the isothermal expansion portion of this engine's cycle, the volume of the gas doubles, while during the adiabatic expansion the volume increases by a factor of 5.7. The work output of the engine is 940 J in each cycle. Compute the temperatures of the two reservoirs between which this engine
operates.
Answer:
The hot temperature is 157.5 K
The cold temperature is 48.8 K
Explanation:
Step 1: Data given
The working substance of a certain Carnot engine is 1.50 mol of an ideal monatomic gas.
The volume increases by a factor of 5.7
The work output of the engine is 940 J in each cycle.
During the isothermal expansion portion of this engine's cycle, the volume of the gas doubles. This means V2 = 2*V1 (and V4 = 2*V3)
Step 2:For a carnot engine:
V2/V1 = V4/V3
Work = nR((T1)ln(V2/V1) - (T2)ln(V4/V3))
⇒with Work = the work done in the cycle = 940J
⇒with n = the number of moles = 1.50 moles
⇒with R = the gas constant = 8.314 J/mol*K
⇒with T1 = the hot temperature
⇒With T2⇒ the cold temperature
where R = 8.31 J/mol K Gas Constant
940J = 1.5moles * 8.314 J/mol*K * (T1*ln(2) - T2*ln(2)))
940 = 1.5 * 8.314 ln(2) * (T1-T2)
(T1-T2) = 940 / (1.5*8.314*ln(2))
(T1-T2) = 108.7K
For the reversible adiabatic expansion: T2 = T1*(V1/V2)^(R/Cv). Where V2/V1 = 5.7 (Because during the adiabatic expansion the volume increases by a factor of 5.7)
For a monatomic ideal gas, Cv = 3/2R
When we combine both, we'll have:
T2 = T1*(1/5.7)^(R/3/2R)
T2 = T1*(1/5.7)^(2/3)
T2= T1 * 0.31
Since we know that (T1-T2) = 108.7K
we have:
T1 - 0.31T1= 108.7K
0.69T1 = 108.7K
T1 = 157.5K
T2 = 157.5*0.31 = 48.8K
A 31 kg block is initially at rest on a horizontal surface. A horizontal force of 83 N is required to set the block in motion. After it is in motion, a horizontal force of 55 N i required to keep it moving with constant speed. From this information, find the coefficients of static and kinetic friction
Answer:
The static and kinetic coefficients of friction are 0.273 and 0.181, respectively.
Explanation:
By Newton's Laws of Motion and definition of maximum friction force, we derive the following two formulas for the static and kinetic coefficients of friction:
[tex]\mu_{s} = \frac{f_{s}}{m\cdot g}[/tex] (1)
[tex]\mu_{k} = \frac{f_{k}}{m\cdot g}[/tex] (2)
Where:
[tex]\mu_{s}[/tex] - Static coefficient of friction, no unit.
[tex]\mu_{k}[/tex] - Kinetic coefficient of friction, no unit.
[tex]f_{s}[/tex] - Static friction force, in newtons.
[tex]f_{k}[/tex] - Kinetic friction force, in newtons.
[tex]m[/tex] - Mass, in kilograms.
[tex]g[/tex] - Gravitational constant, in meters per square second.
If we know that [tex]f_{s} = 83\,N[/tex], [tex]f_{k} = 55\,N[/tex], [tex]m = 31\,kg[/tex] and [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], then the coefficients of friction are, respectively:
[tex]\mu_{s} = \frac{83\,N}{(31\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)}[/tex]
[tex]\mu_{s} = 0.273[/tex]
[tex]\mu_{k} = \frac{55\,N}{(31\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)}[/tex]
[tex]\mu_{k} = 0.181[/tex]
The static and kinetic coefficients of friction are 0.273 and 0.181, respectively.
The force an ideal spring exerts on an object is given by , where measures the displacement of the object from its equilibrium position. If , how much work is done by this force as the object moves from to
Answer:
The correct answer is "1.2 J".
Explanation:
Seems that the given question is incomplete. Find the attachment of the complete query.
According to the question,
x₁ = -0.20 mx₂ = 0 mk = 60 N/mNow,
⇒ [tex]W=\int_{x_1}^{x_2}F \ dx[/tex]
⇒ [tex]=\int_{x_1}^{x_2}-kx \ dx[/tex]
⇒ [tex]=-k \int_{-0.20}^{0}x \ dx[/tex]
By putting the values, we get
⇒ [tex]=-(60)[\frac{x^2}{2} ]^0_{-0.20}[/tex]
⇒ [tex]=-60[\frac{0}{2}-\frac{0.04}{2} ][/tex]
⇒ [tex]=1.2 \ J[/tex]
A mass is attached to the end of a spring and set into oscillation on a horizontal frictionless surface by releasing it from a compressed position. The record of time is started when the oscillating mass first passes through the equilibrium position, and the position of the mass at any time is described by
The question is incomplete. The complete question is :
A mass is attached to the end of a spring and set into oscillation on a horizontal frictionless surface by releasing it from a compressed position. The record of time is started when the oscillating mass first passes through the equilibrium position, and the position of the mass at any time is described by x = (4.7 cm)sin[(7.9 rad/s)πt].
Determine the following:
(a) frequency of the motion
(b) period of the motion
(c) amplitude of the motion
(d) first time after t = 0 that the object reaches the position x = 2.6 cm
Solution :
Given equation : x = (4.7 cm)sin[(7.9 rad/s)πt].
Comparing it with the general equation of simple harmonic motion,
x = A sin (ωt + Φ)
A = 4.7 cm
ω = 7.9 π
a). Therefore, frequency, [tex]$f=\frac{\omega}{2 \pi}$[/tex]
[tex]$=\frac{7.9 \pi}{2 \pi}$[/tex]
= 3.95 Hz
b). The period, [tex]$T=\frac{1}{f}$[/tex]
[tex]$T=\frac{1}{3.95}[/tex]
= 0.253 seconds
c). Amplitude is A = 4.7 cm
d). We have,
x = A sin (ωt + Φ)
[tex]$x_t=4.7 \sin (7.9 \pi t)$[/tex]
[tex]$2.6 = 4.7 \sin (7.9 \pi t)$[/tex]
[tex]$\sin (7.9 \pi t) = \frac{26}{47}$[/tex]
[tex]$7.9 \pi t = \sin^{-1}\left(\frac{26}{47}\right)$[/tex]
Hence, t = 0.0236 seconds.
In the figure, particle A moves along the line y = 31 m with a constant velocity v with arrow of magnitude 2.8 m/s and parallel to the x axis. At the instant particle A passes the y axis, particle B leaves the origin with zero initial speed and constant acceleration a with arrow of magnitude 0.35 m/s2. What angle between a with arrow and the positive direction of the y axis would result in a collision?
Answer:
59.26°
Explanation:
Since a is the acceleration of the particle B, the horizontal component of acceleration is a" = asinθ and the vertical component is a' = acosθ where θ angle between a with arrow and the positive direction of the y axis.
Now, for particle B to collide with particle A, it must move vertically the distance between A and B which is y = 31 m in time, t.
Using y = ut + 1/2a't² where u = initial velocity of particle B = 0 m/s, t = time taken for collision, a' = vertical component of particle B's acceleration = acosθ.
So, y = ut + 1/2a't²
y = 0 × t + 1/2(acosθ)t²
y = 0 + 1/2(acosθ)t²
y = 1/2(acosθ)t² (1)
Also, both particles must move the same horizontal distance to collide in time, t.
Let x be the horizontal distance,
x = vt (2)where v = velocity of particle A = 2.8 m/s and t = time for collision
Also, using x = ut + 1/2a"t² where u = initial velocity of particle B = 0 m/s, t = time taken for collision, a" = horizontal component of particle B's acceleration = asinθ.
So, x = ut + 1/2a"t²
x = 0 × t + 1/2(ainsθ)t²
x = 0 + 1/2(asinθ)t²
x = 1/2(asinθ)t² (3)
Equating (2) and (3), we have
vt = 1/2(asinθ)t² (4)
From (1) t = √[2y/(acosθ)]
Substituting t into (4), we have
v√[2y/(acosθ)] = 1/2(asinθ)(√[2y/(acosθ)])²
v√[2y/(acosθ)] = 1/2(asinθ)(2y/(acosθ)
v√[2y/(acosθ)] = ytanθ
√[2y/(acosθ)] = ytanθ/v
squaring both sides, we have
(√[2y/(acosθ)])² = (ytanθ/v)²
2y/acosθ = (ytanθ/v)²
2y/acosθ = y²tan²θ/v²
2/acosθ = ytan²θ/v²
1/cosθ = aytan²θ/2v²
Since 1/cosθ = secθ = √(1 + tan²θ) ⇒ sec²θ = 1 + tan²θ ⇒ tan²θ = sec²θ - 1
secθ = ay(sec²θ - 1)/2v²
2v²secθ = aysec²θ - ay
aysec²θ - 2v²secθ - ay = 0
Let secθ = p
ayp² - 2v²p - ay = 0
Substituting the values of a = 0.35 m/s, y = 31 m and v = 2.8 m/s into the equation, we have
ayp² - 2v²p - ay = 0
0.35 × 31p² - 2 × 2.8²p - 0.35 × 31 = 0
10.85p² - 15.68p - 10.85 = 0
dividing through by 10.85, we have
p² - 1.445p - 1 = 0
Using the quadratic formula to find p,
[tex]p = \frac{-(-1.445) +/- \sqrt{(-1.445)^{2} - 4 X 1 X (-1)}}{2 X 1} \\p = \frac{1.445 +/- \sqrt{2.088 + 4}}{2} \\p = \frac{1.445 +/- \sqrt{6.088}}{2} \\p = \frac{1.445 +/- 2.4675}{2} \\p = \frac{1.445 + 2.4675}{2} or p = \frac{1.445 - 2.4675}{2} \\p = \frac{3.9125}{2} or p = \frac{-1.0225}{2} \\p = 1.95625 or -0.51125[/tex]
Since p = secθ
secθ = 1.95625 or secθ = -0.51125
cosθ = 1/1.95625 or cosθ = 1/-0.51125
cosθ = 0.5112 or cosθ = -1.9956
Since -1 ≤ cosθ ≤ 1 we ignore the second value since it is less than -1.
So, cosθ = 0.5112
θ = cos⁻¹(0.5112)
θ = 59.26°
So, the angle between a with arrow and the positive direction of the y axis would result in a collision is 59.26°.
The cannon on a battleship can fire a shell a maximum distance of 33.0 km.
(a) Calculate the initial velocity of the shell.
Answer:
v = 804.23 m/s
Explanation:
Given that,
The maximum distance covered by a cannon, d = 33 km = 33000 m
We need to find the initial velocity of the shell. Let it is v. It can be calculated using the conservation of energy such that,
[tex]v=\sqrt{2gh} \\\\v=\sqrt{2\times 9.8\times 33000} \\\\v=804.23\ m/s[/tex]
So, the initial velocity of the shell is 804.23 m/s.
A child is outside his home playing with a metal hoop and stick. He uses the stick to keep the hoop of radius 45.0 cm rotating along the road surface. At one point the hoop coasts downhill and picks up speed. (a) If the hoop starts from rest at the top of the hill and reaches a linear speed of 6.35 m/s in 11.0 s, what is the angular acceleration, in rad/s2, of the hoop? rad/s2 (b) If the radius of the hoop were smaller, how would this affect the angular acceleration of the hoop? i. The angular acceleration would decrease. ii. The angular acceleration would increase. iii. There would be no change to the angular acceleration.
Answer:
a) [tex] \alpha = 1.28 rad/s^{2} [/tex]
b) Option ii. The angular acceleration would increase
Explanation:
a) The angular acceleration is given by:
[tex] \omega_{f} = \omega_{0} + \alpha t [/tex]
Where:
[tex] \omega_{f} [/tex]: is the final angular speed = v/r
v: is the tangential speed = 6.35 m/s
r: is the radius = 45.0 cm = 0.45 m
[tex]\omega_{0}[/tex]: is the initial angular speed = 0 (the hoop starts from rest)
t: is the time = 11.0 s
α: is the angular acceleration
Hence, the angular acceleration is:
[tex] \alpha = \frac{\omega}{t} = \frac{v}{r*t} = \frac{6.35 m/s}{0.45 m*11.0 s} = 1.28 rad/s^{2} [/tex]
b) If the radius were smaller, the angular acceleration would increase since we can see in the equation that the radius is in the denominator ([tex] \alpha = \frac{v}{r*t} [/tex]).
Therefore, the correct option is ii. The angular acceleration would increase.
I hope it helps you!
A cylindrical tank with radius 7 m is being filled with water at a rate of 2 m3/min. How fast is the height of the water increasing (in m/min)?
Answer:
0.013 m/min
Explanation:
Applying,
dV/dt = (dh/dt)(dV/dh)............. Equation 1
Where
V = πr²h................ Equation 2
Where V = volume of the tank, r = radius, h = height.
dV/dh = πr²............ Equation 3
Substitute equation 3 into equation 1
dV/dt = πr²(dh/dt)
From the question,
Given: dV/dt = 2 m³/min, r = 7 m, π = 3.14
Substitute these values into equation 3
2 = (3.14)(7²)(dh/dt)
dh/dt = 2/(3.14×7²)
dh/dt = 0.013 m/min
A system gains 1500J of heat and 2200J of work is done by the system on its surroundings. Determine the change in internal energy of the system
Answer:
-700
formula is heat gained - work done
The change in internal energy if A system gains 1500J of heat and 2200J of work is done by the system on its surroundings, is 700 joules.
What is Energy?Energy is the ability to perform work in physics. It could exist in several different forms, such as potential, kinetic, thermal, electrical, chemical, radioactive, etc.
Additionally, there is heat and work, which is energy being transferred from one body to another. Energy is always assigned based on its nature once it has been transmitted. Thus, heat transmitted may manifest as thermal energy while work performed may result in mechanical energy.
Given:
A system gains 1500J of heat and 2200J of work is done by the system on its surroundings,
Calculate the change in internal energy as shown below,
The change in internal energy = heat gained - work done
The change in internal energy = 1500 - 2200
The change in internal energy = -700 J
Thus, the change in internal energy is 700 joules.
To know more about Energy:
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The 1 kg box is sliding along a frictionless surface. It collides with and sticks to the 2 kg box. Afterward, the speed of the two boxes is:__________.
A) 0 m/s
B) 1 m/s
C) 2 m/s
D) 3 m/s
E) Not enough info
Answer:
The correct option is (E).
Explanation:
Given that,
Mass of object 1, m₁ = 1 kg
Mass of object 2, m₂ = 2 kg
They collides after the collision. We need to find the speed of the two boxes after the collision.
The initial speeds of both boxes is not given. So, we can't put the values of their speeds in the momentum conservation equation.
So, the information is not enough.
