Answer: So, if the forces are in the same direction, it'll be 50N+30N= 80N!
If the forces are in opprsite direction it'll be 50N-30N= 20N (or visa-versa)
Explanation: This is because if the forces are in the same direction, like 2 people are pushing a body with 0 mass (I know, weird but imagine it) and both the people are applying the force in the same direction, let's say East, so the total fore exerted on the body will add up to become 50N+30N=80N.
And if the forces are phusing the body in oposite direction, then, let's say the bigger guy is pushing on the West Direction with 50N of force and the weaker guy is pushing on the opposite direction with 30N of force, so this time their forces cancel out and it becomes 50N-30N= 20N (or 30N-50N= -20N).
A goalie kicks a soccer ball straight vertically into the air. It takes 5.00 s for the ball to reach its maximum height and come back down to the level of the crossbar. Assume the crossbar of a soccer goal is 2.44 m above the ground. (a) How fast was the ball originally moving when it was kicked. (b) How much longer would it take the ball to reach the ground?
Answer:
(a) vo = 24.98m/s
(b) t = 5.09 s
Explanation:
(a) In order to calculate the the initial speed of the ball, you use the following formula:
[tex]y=y_o+v_ot-\frac{1}{2}gt^2[/tex] (1)
y: vertical position of the ball = 2.44m
yo: initial vertical position = 0m
vo: initial speed of the ball = ?
g: gravitational acceleration = 9.8m/s²
t: time on which the ball is at 2.44m above the ground = 5.00s
You solve the equation (1) for vo and replace the values of the other parameters:
[tex]v_o=\frac{y-y_o+1/2gt^2}{t}[/tex]
[tex]v_o=\frac{2.44m-0.00m+1/2(9.8m/s^2)(5.00s)^2}{5.00s}\\\\v_o=24.98\frac{m}{s}[/tex]
The initial speed of the ball is 24.98m/s
(b) To find the time the ball takes to arrive to the ground you use the equation (1) for y = 0m (ground) and solve for t:
[tex]0=24.98t-\frac{1}{2}(9.8)t^2\\\\t=5.09s[/tex]
The time that the ball takes to arrive to the ground is 5.09s
We have that for the Question, it can be said that the speed of ball and How much longer would it take the ball to reach the ground is
u=25.13m/sX=0.095sec
From the question we are told
A goalie kicks a soccer ball straight vertically into the air. It takes 5.00 s for the ball to reach its maximum height and come back down to the level of the crossbar. Assume the crossbar of a soccer goal is 2.44 m above the ground.
(a) How fast was the ball originally moving when it was kicked.
(b) How much longer would it take the ball to reach the ground?
a)
Generally the Newton equation for the Motion is mathematically given as
[tex]S=ut+1/2at^2\\\\Therefore\\\\2.44=ut+1/2(9.8)(5)^2\\\\u=25.13m/s\\\\[/tex]
b)
Generally the Newton equation for the Motion is mathematically given as
[tex]S=ut+1/2at^2\\\\Therefore\\\\t=\frac{-24}{a}\\\\t=\frac{-2*25.013}{9.81}\\\\t=5.095sec\\\\[/tex]
Therefore
[tex]X=5.095-5[/tex]
X=0.095sec
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1) A net force of 75.5 N is applied horizontally to slide a 225 kg crate across the floor.
a. Compute the acceleration of the crate?
Answer:
The acceleration of the crate is [tex]0.3356\,\frac{m}{s^2}[/tex]
Explanation:
Recall the formula that relates force,mass and acceleration from newton's second law;
[tex]F=m\,a[/tex]
Then in our case, we know the force applied and we know the mass of the crate, so we can solve for the acceleration as shown below:
[tex]F=m\,a\\75.5\,N=225\,\,kg\,\,a\\a=\frac{75.5}{225} \,\frac{m}{s^2} \\a=0.3356\,\,\frac{m}{s^2}[/tex]
Two small identical speakers are connected (in phase) to the same source. The speakers are 3 m apart and at ear level. An observer stands at X, 4 m in front of one speaker. If the amplitudes are not changed, the sound he hears will be least intense if the wavelength is:
a. 1 m
b. 2 m
c. 3 m
d. 4 m
e. 5 m
Answer:
b. 2 m
Explanation:
Given that:
the identical speakers are connected in phases ;
Let assume ; we have speaker A and speaker B which are = 3 meter apart
An observer stands at X = 4m in front of one speaker.
If the amplitudes are not changed, the sound he hears will be least intense if the wavelength is:
From above; the distance between speaker A and speaker B can be expressed as:
[tex]\sqrt{3^2 + 4^2 } \\ \\ = \sqrt{9+16 } \\ \\ = \sqrt{25} \\ \\ = 5 \ m[/tex]
The path length difference will now be:
= 5 m - 4 m
= 1 m
Since , we are to determine the least intense sound; the destructive interference for that path length will be half the wavelength; which is
= [tex]\dfrac{1}{2}*4 \ m[/tex]
= 2 m
The sound will be heard with least intensity if the wavelength is 2 m. Hence, option (b) is correct.
Given data:
The distance between the speakers is, d = 3 m.
The distance between the observer and speaker is, s = 4 m.
The amplitude of sound wave is the vertical distance from the base to peak of wave. Since sound amplitudes are not changed in the given problem. Then the distance between speaker A and speaker B can be expressed as:
[tex]=\sqrt{3^{2}+4^{2}}\\\\=\sqrt{25}\\\\=5\;\rm m[/tex]
And the path length difference is,
= 5 m - 4 m
= 1 m
Since , we are to determine the least intense sound; the destructive
interference for that path length will be half the wavelength; which is
[tex]=\dfrac{1}{2} \times s\\\\=\dfrac{1}{2} \times 4[/tex]
= 2 m
Thus, we can conclude that the sound will be heard with least intensity if the wavelength is 2 m.
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Determine the minimum gauge pressure needed in the water pipe leading into a building if water is to come out of a faucet on the fourteenth floor, 43
Answer:
The gauge pressure is [tex]P = 4.2*10^{5} \ N/m^2[/tex]
Explanation:
From the question we are told that
The height of the 14th floor from the point where the water entered the building is h = 43 m
The gauge pressure is mathematically represented as
[tex]P = mgh[/tex]
Where the m is the mass of the water which is mathematically represented as
[tex]m = \frac{\rho}{V}[/tex]
Where [tex]\rho[/tex] is the density of the water which has a constant value of [tex]\rho = 1000 \ kg/m^3[/tex] and this standard value of density the volume is [tex]1 m^3[/tex] so
[tex]m = \frac{1000}{1}[/tex]
[tex]m = 1000 \ kg[/tex]
Thus
[tex]P = 1000 * 9.8 * 43[/tex]
[tex]P = 4.2*10^{5} \ N/m^2[/tex]
Water is boiled at 1 atm pressure in a 20-cm-internal diameter polished copper pan on an electric range. If it is observed that the water level in the pan drops by 8.00 cm in 15 minutes, determine the inner surface temperature of the pan.
Answer:
11.3298W
Explanation:
The rate of heat transfer is determined from the enthalpy of vaporization at the give pressure obtained and the mass flow rate. The mass flow rate is determined from the volume of the boiled water, the given time interval and the specific volume of the saturated liquid.
Given that
1atm as the atmospheric pressure
Internal diameter = 20cm = 0.2m
Time = 15mins = (15×60)secs
Latent heat of vaporization (hevap) = 2256.6
Q = mh(evap)
= m/∆t . hevap
= V/αliq∆t ×h(evap)
D^2π∆h/4αliq ∆t × hevap
= 0.2^2 ×π×0.8×2256.5/4×0.001043×15×60
=0.04×3.142×0.08×2256.6/2.00256
= 22.68876/2.00256
Q = 11.3298W
Two cannonballs are dropped from a second-floor physics lab at height h above the ground. Ball B has four times the mass of ball A. When the balls pass the bottom of a first-floor window at height above the ground, the relation between their kinetic energies, KA and KB, is
Answer:
1:4
Explanation:
The formula for calculating kinetic energy is:
[tex]KE=\dfrac{1}{2}mv^2[/tex]
If the mass is multiplied by 4, then, the kinetic energy must be increased by 4 as well. Since they will be travelling at the same speed when they are at the same point, the relation between KA and KB must be 1:4 or 1/4. Hope this helps!
The relation between the kinetic energies of the freely falling balls A and B is obtained as [tex]\frac{KE_{A}}{KE_{B}} =\frac{1}{4}[/tex].
Kinetic EnergyThe kinetic energy of an object depends on the mass and velocity with which it moves.
While under free-fall, the mass of an object does not affect the velocity with which it falls.
So, the velocities of both the balls are the same.
Let the mass of ball A is 'm'
So, the mass of ball B is '4m'
The kinetic energy of ball A is given by;
[tex]KE_{A}=\frac{1}{2} mv^2[/tex]
The kinetic energy of ball B is given by;
[tex]KE_{B}=\frac{1}{2} 4mv^2 = 2mv^2[/tex]
Therefore, the ratio of kinetic energies of A and B is,
[tex]\frac{KE_{A}}{KE_{B}} =\frac{1}{4}[/tex]
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A cowboy fires a silver bullet with a muzzle speed of 200 m/s into the pine wall of a saloon. Assume all the internal energy generated by the impact remains with the bullet. What is the temperature change of the bullet?
Explanation:
KE = q
½ mv² = mCΔT
ΔT = v² / (2C)
ΔT = (200 m/s)² / (2 × 236 J/kg/°C)
ΔT = 84.7°C
This question involves the concepts of the law of conservation of energy.
The temperature change of the bullet is "84.38°C".
What is the Law of Conservation of Energy?According to the law of conservation of energy, total energy of the system must remain constant. Therefore, in this situation.
[tex]Kinetic\ energy\ of\ bullet\ before\ impact=heat\ absorbed\ in\ bullet\\\\\frac{1}{2}mv^2=mC\Delta T\\\\\Delta T = \frac{v^2}{2C}[/tex]
where,
ΔT = change in temperature of the bullet = ?C = specific heat capacity of silver = 237 J/kg°Cv = speed of bullet = 200 m/sTherefore,
[tex]\Delta T = \frac{(200\ m/s)^2}{2(237\ J/kg.^oC)}[/tex]
ΔT = 84.38°C
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If 2 balls had the same volume but ball a has twice as much mass as babil which one will have the greater density
The electron density in copper is 8.49 × 1028 electrons/m3. When a 1.00 A current is present in a copper wire with a 0.40 cm2 cross-section, the electron drift velocity, in m/s, with direction defined relative to the current density, is
Answer:
Drift velocity in m/s is [tex]1.84 \times 10^{-8}\ m/s[/tex].
Explanation:
Formula for Drift Velocity is:
[tex]u = \dfrac{I}{nAq}[/tex]
Where I is the current
n is the number of electrons in 1 [tex]m^3[/tex] or the electron density
A is the area of cross section and
q is the charge of one electron.
We are given the following:
n = [tex]8.49 \times 10^{28}\ electrons/m^3[/tex]
I = 1 A
A = 0.40 [tex]cm^2[/tex] = 40 [tex]\times 10^{-4}[/tex] [tex]m^{2}[/tex]
We know that q = [tex]1.6\times10^{-19} C[/tex]
Putting all the values to find drift velocity:
[tex]u = \dfrac{1}{8.49 \times 10 ^{28} \times 40 \times 10^{-4}\times 1.6 \times 10^{-19}}\\u = \dfrac{1}{543.36 \times 10 ^{5} }\\u = 1.84 \times 10^{-8}\ m/s[/tex]
So, drift velocity in m/s is [tex]1.84 \times 10^{-8}\ m/s[/tex].
According to the model in which active galactic nuclei are powered by supermassive black holes, the high luminosity of an active galactic nucleus primarily consists of
Answer:
the high luminosity of an active galactic nucleus primarily consists of light emitted by hot gas in an accretion disk that swirls around the black hole
Doubling the potential across a given capacitor causes the energy stored in that capacitor to reduce to:_______
a. one-half.
b. double.
c. reduce to one-fourth.
d. quadruple.
Answer:
D. quadrupleExplanation:
The stored energy varies with the square of the electric charge stored in the capacitor. If you double the charge, the stored energy in the capacitor will quadruple or increase by a factor of 4.
Doubling the potential across a given capacitor causes the energy stored in that capacitor to reduce to :
D. Quadruple
"Energy"Doubling the potential across a given capacitor causes the energy stored in that capacitor to reduce to Quadruple.
The stored energy shifts with the square of the electric charge put away within the capacitor.
In case you twofold the charge, the put away vitality within the capacitor will fourfold or increment by a calculate of 4.
Thus, the correct answer is D.
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The back wall of an auditorium is 30.0 m from the stage. If you are seated in the middle row, how much time elapses between a sound from the stage reaching your ear directly and the same sound reaching your ear after reflecting from the back wall?
Answer:
0.0875 secExplanation:
Using the relationship to calculate the time that elapse 2x = vt
x = distance between the source and the reflector
v is the velocity of sound in air
t = time elapsed
Given x = 30.0m and v = 343m/s
substituting the values into the formula above to get the total time elapsed t;
t = 2x/v
t = 2(30)/343
t= 60/343
t = 0.175sec
If you are seated in the middle row, the time that will elapse between the sound from the stage reaching your ear directly will be;
0.175/2 = 0.0875 secs
which of the following has the lowest density? A. Water B. Air C. Mineral Water D. Salt Water
Air has lower density than water, mineral Water, or salt water. (B)
5. (10 points) Which of the following statements is(are) correct: A. Resistivity purely depends on internal properties of the conductor; B. Resistance purely depends on internal properties of the conductor; C. Resistivity depends on the size and shape of the conductor; D. Resistance depends on the size and shape of the conductor; E. A and D; F. B and C.
Answer:
B and D
Explanation:
Because
R= resistivity xlenght/ Area
Where R= resistance
what are all the rays that come from the sun called List all?
Explanation:
what are all the rays that come from the sun called List all?
cosmetic rays, gamma rays, x-rays, ultraviolet rays, infrared rays, microwaves, short radio waves and long radio waves.
Two people play tug of war. The 100-kg person on the left pulls with 1,000 N, and the 70-kg person on the right pulls with 830 N. Assume that neither person releases their grip on the rope with either hand at any time, assume that the rope is always taut, and assume that the rope does not stretch. What is the magnitude of the tension in the rope in Newtons
Answer:
The tension on the rope is T = 900 N
Explanation:
From the question we are told that
The mass of the person on the left is [tex]m_l = 100 \ kg[/tex]
The force of the person on the left is [tex]F_l = 1000 \ N[/tex]
The mass of the person on the right is [tex]m_r = 70 \ kg[/tex]
The force of the person on the right is [tex]F_r = 830 \ N[/tex]
Generally the net force is mathematically represented as
[tex]F_{Net} = F_l - F_r[/tex]
substituting values
[tex]F_{Net} = 1000-830[/tex]
[tex]F_{Net} = 170 \ N[/tex]
Now the acceleration net acceleration of the rope is mathematically evaluated as
[tex]a = \frac{F_{net}}{m_I + m_r }[/tex]
substituting values
[tex]a = \frac{170}{100 + 70 }[/tex]
[tex]a = 1 \ m/s ^2[/tex]
The force [tex]m_i * a[/tex]) of the person on the left that caused the rope to accelerate by a is mathematically represented as
[tex]m_l * a = F_r -T[/tex]
Where T is the tension on the rope
substituting values
[tex]100 * 1 = 1000 - T[/tex]
=> T = 900 N
An inquisitive physics student and mountain climber climbs a 47.0-m-high cliff that overhangs a calm pool of water. He throws two stones vertically downward, 1.00 s apart, and observes that they cause a single splash. The first stone has an initial speed of 2.12 m/s.
(a) How long after release of the first stone do the two stones hit the water?
(b) What initial velocity must the second stone have if the two stones are to hit the water simultaneously?
magnitude =
(c) What is the speed of each stone at the instant the two stones hit the water?
first stone =
second stone =
Answer:
a) Only the first root is physically reasonable. Therefore, both stones hit the water in 2.866 seconds, b) The initial velocity of the second stone is -16.038 meters per second, c) The speed of the first stone is 30.227 meters per second and the speed of the second stone is 34.338 meters per second.
Explanation:
a) The time after the release after the release of the first stone can be get from the following kinematic formula for the first rock:
[tex]y_{1} = y_{1,o} + v_{1,o} \cdot t +\frac{1}{2}\cdot g \cdot t^{2}[/tex]
Where:
[tex]y_{1}[/tex] - Final height of the first stone, measured in meters.
[tex]y_{1,o}[/tex] - Initial height of the first stone, measured in meters.
[tex]v_{1,o}[/tex] - Initial speed of the first stone, measured in meters per second.
[tex]t[/tex] - Time, measured in seconds.
[tex]g[/tex] - Gravity constant, measured in meters per square second.
Given that [tex]y_{1,o} = 47\,m[/tex], [tex]y_{1} = 0\,m[/tex], [tex]v_{1,o} = -2.12\,\frac{m}{s}[/tex] and [tex]g = -9.807\,\frac{m}{s^{2}}[/tex], the following second-order polynomial is built:
[tex]-4.984\cdot t^{2} - 2.12\cdot t + 47 = 0[/tex]
Roots of the polynomial are, respectively:
[tex]t_{1} \approx 2.866\,s[/tex] and [tex]t_{2}\approx -3.291\,s[/tex]
Only the first root is physically reasonable. Therefore, both stones hit the water in 2.866 seconds.
b) As the second stone is thrown a second later than first one, its height is represented by the following kinematic expression:
[tex]y_{2} = y_{2,o} + v_{2,o}\cdot (t-t_{o}) + \frac{1}{2}\cdot g \cdot (t-t_{o})^{2}[/tex]
[tex]y_{2}[/tex] - Final height of the second stone, measured in meters.
[tex]y_{2,o}[/tex] - Initial height of the second stone, measured in meters.
[tex]v_{2,o}[/tex] - Initial speed of the second stone, measured in meters per second.
[tex]t[/tex] - Time, measured in seconds.
[tex]t_{o}[/tex] - Initial absolute time, measured in seconds.
[tex]g[/tex] - Gravity constant, measured in meters per square second.
Given that [tex]y_{2,o} = 47\,m[/tex], [tex]y_{2} = 0\,m[/tex], [tex]t_{o} = 1\,s[/tex], [tex]t = 2.866\,s[/tex] and [tex]g = -9.807\,\frac{m}{s^{2}}[/tex], the following expression is constructed and the initial speed of the second stone is:
[tex]1.866\cdot v_{2,o}+29.926 = 0[/tex]
[tex]v_{2,o} = -16.038\,\frac{m}{s}[/tex]
The initial velocity of the second stone is -16.038 meters per second.
c) The final speed of each stone is determined by the following expressions:
First stone
[tex]v_{1} = v_{1,o} + g \cdot t[/tex]
Second stone
[tex]v_{2} = v_{2,o} + g\cdot (t-t_{o})[/tex]
Where:
[tex]v_{1,o}, v_{1}[/tex] - Initial and final velocities of the first stone, measured in meters per second.
[tex]v_{2,o}, v_{2}[/tex] - Initial and final velocities of the second stone, measured in meters per second.
If [tex]v_{1,o} = -2.12\,\frac{m}{s}[/tex] and [tex]v_{2,o} = -16.038\,\frac{m}{s}[/tex], the final speeds of both stones are:
First stone
[tex]v_{1} = -2.12\,\frac{m}{s} + \left(-9.807\,\frac{m}{s^{2}} \right)\cdot (2.866\,s)[/tex]
[tex]v_{1} = -30.227\,\frac{m}{s}[/tex]
Second stone
[tex]v_{2} = -16.038\,\frac{m}{s} + \left(-9.807\,\frac{m}{s^{2}} \right) \cdot (2.866\,s-1\,s)[/tex]
[tex]v_{2} = -34.338\,\frac{m}{s}[/tex]
The speed of the first stone is 30.227 meters per second and the speed of the second stone is 34.338 meters per second.
Isaac drops a rubber ball drom height of 2.0m and it bounces to a height of 1.5m. a) What fraction of it's initial energy is lost during the bounce? b)What is the rubber ball's speed just before and just after it bounces? c)Where did the energy go?
Answer:
a) ΔE = 25 %
b) v = 8,85 m/s
c) The energy was used against air resistance
Explanation:
In any situation total energy of a body is equal to potential energy plus
kinetic energy, then, just at the moment when Isaac dop the ball the situation is:
Ei = Ep + Ek where Ep = m*g*h and Ek = 1/2*m*v²
As v = 0 (Isaac drops the ball)
Ei = Ep = m*g*h = 2*m*g
At the end (when the ball bounced to 1,5 m
E₂ = Ep₂ + Ek₂ again at that point v =0 and
E₂ = 1,5*m*g*
Ei = E₂ + E(lost)
E(lost) = Ei - E₂
E(lost) = 2*m*g* - 1,5*m*g and the fraction of energy lost is
E(lost)/Ei
ΔE = (2*m*g* - 1,5*m*g )/ 2*m*g
ΔE = 0,5*m*g / 2*m*g
ΔE = 0,5/2
ΔE = 0,25 or ΔE = 25 %
b) The speed of the ball is
Potential energy is converted in kinetic energy just when the ball is touching the ground, then
m*g*h = 1/2*m*v²
2*h*g = 1/2 *v²
v² = 4*g*h
v² = 4*2*9,8
v² = 78,4
v = 8,85 m/s
If the impact is an elastic collision, then Ek before and after the impact is the same.
When a certain capacitor carries charge of magnitude Q on each of its plates, it stores energy Ep. In order to store twice as much energy, how much charge should it have on its plates
2Q
Explanation:
When a capacitor carries some certain charge, the energy stored in the capacitor is its electric potential energy E. The magnitude of this potential energy is given by;
E = [tex]\frac{1}{2}qV[/tex] ------------(i)
Where;
q = charge between the plates of the capacitor
V = potential difference between the plates of the capacitor
From the question;
q = Q
E = Ep
Therefore, equation (i) becomes;
Ep = [tex]\frac{1}{2} QV[/tex] ----------------(ii)
Make V subject of the formula in equation (ii)
V = [tex]\frac{2E_{p}}{Q}[/tex]
Now, when the energy is doubled i.e E = 2Ep, equation (i) becomes;
2Ep = [tex]\frac{1}{2}qV[/tex]
Substitute the value of V into the equation above;
2Ep = [tex]\frac{1}{2}[/tex]([tex]q *\frac{2E_{p}}{Q}[/tex])
Solve for q;
[tex]2E_{p}[/tex] = [tex]\frac{2qE_p}{2Q}[/tex]
[tex]2E_{p}[/tex] = [tex]\frac{qE_p}{Q}[/tex]
[tex]q = 2Q[/tex]
Therefore, the charge, when the energy stored is twice the originally stored energy, is twice the original charge. i.e 2Q
Light in vacuum is incident on the surface of a glass slab. In the vacuum the beam makes an angle of 38.0° with the normal to the surface, while in the glass it makes an angle of 26.0° with the normal. What is the index of refraction of the glass?
Answer:
n_glass = 1.404
Explanation:
In order to calculate the index of refraction of the light you use the Snell's law, which is given by the following formula:
[tex]n_1sin\theta_1=n_2sin\theta_2[/tex] (1)
n1: index of refraction of vacuum = 1.00
θ1: angle of the incident light respect to normal of the surface = 38.0°
n2: index of refraction of glass = ?
θ2: angle of the refracted light in the glass respect to normal = 26.0°
You solve the equation (1) for n2 and replace the values of all parameters:
[tex]n_2=n_1\frac{sin\theta_1}{sin\theta_2}=(1.00)\frac{sin(38.0\°)}{sin(26.0\°)}\\\\n_2=1.404[/tex]
The index of refraction of the glass is 1.404
The air is partially removed at the upper part of the tube and the valve is closed. The height of the water above its pool surface is 0.7 m.The height of the liquid above its pool surface is 0.33 m.The difference in the heights of the pool surfaces is 0.22 m.
Required:
Find the density of the unknown liquid.
Answer:
2121.2kg/m^3 is the density of the test liquid on the left
Explanation:
See attached file
A particle with charge q is to be brought from far away to a point near an electric dipole. Net nonzero work is done if the final position of the particle is on:__________
A) any point on the line through the charges of the dipole, excluding the midpoint between the two charges.
B) any point on a line that is a perpendicular bisector to the line that separates the two charges.
C) a line that makes an angle of 30 ∘ with the dipole moment.
D) a line that makes an angle of 45 ∘with the dipole moment.
Answer:
Net nonzero work is done if the final position of the particle is on options A, C and D
Explanation:
non zero work is done if following will be the final position of the charges :
A) Any point on the line through the charges of the dipole , excluding the midpoint between the two charges.
C) A line that makes an angle 30° with the dipole moment.
D) A line that makes an angle 45° with the dipole moment.
A charging bull elephant with a mass of 5500 kg comes directly toward you with a speed of 4.70 m/s . You toss a 0.160-kg rubber ball at the elephant with a speed of 7.50 m/s(a) When the ball bounces back toward you, what is its speed? (b) How do you account for the fact that the ball's kinetic energy has increased?
Answer:
v2 = - 16.899 m/s
velocity of ball increases so that the kinetic energy of the ball increases.
Explanation:
given data
mass of elephant, m1 = 5500 kg
mass of ball, m2 = 0.160 kg
initial velocity of elephant, u1 = - 4.70 m/s
initial velocity of ball, u2 = 7.50 m/s
solution
we consider here final velocity of ball = v2
so collision formula is express as for v2
[tex]v_{2}=\left ( \frac{2m_{1}}{m_{1}+m_{2}} \right )u_{1}+\left ( \frac{m_{2}-m_{1}}{m_{1}+m_{2}} \right )u_{2}[/tex] .................1
put here value and we get
[tex]v_{2}=\left ( \frac{2\times 5500}{5500+0.160} \right )(-4.70)+\left ( \frac{0.16-5500}{5500+0.160} \right )(7.50)[/tex]
solve it we get
v2 = - 16.899 m/s
here negative sign shows that the ball bounces back towards you
and
here we know the velocity of ball increases so that the kinetic energy of the ball increases.
and due to this effect, it will gain in energy is due to the energy from the elephant mass
How much work will it take to lift a 2-kg pair of hiking boots 2 meters off the
ground and onto a shelf in your closet?
O A. 2.45 J
OB. 4J
C. 39.2 J
D. 20 J
Answer:
Option C - 39.2 J
Explanation:
We are given that;
Mass; m = 2 kg.
Distance moved off the floor;d = 10 m.
Acceleration due to gravity;g = 9.8 m/s².
We want to find the work done.
Now, the Formula for work done is given by;
Work = Force × displacement.
In this case, it's force of gravity to lift up the boots, thus;
Formula for this force is;
Force = mass x acceleration due to gravity
Force = 2 × 9.8 = 19.2 N
∴ Work done = 19.6 × 2
Work done = 39.2 J.
Hence, the Work done to life the boot of 2 kg to a height of 2 m is 39.2 J.
Answer:39.2J
Explanation: I just answered this question and this was the correct answer. 4J is the wrong answer.
In separate experiments, a large number of particles (all with the same charge, but with a wide variety of masses and speeds) are fired into a region containing a uniform magnetic field. The velocities of the particles are all perpendicular to the field. What do the particles that move in orbits of the same radius have in common
Answer:
they have the same mass
Explanation:
The force applied by the field is a function of the charge and velocity, so the acceleration experienced by a particle will be dependent upon its mass. Particles in orbits with the same radius are exhibiting the same acceleration, so must have the same mass.
A 2.0-kg stone is tied to a 0.50 m long string and swung around a circle at a constant angular velocity of 12 rad/s. The circle is parallel to the xy plane and is centered on the z axis, 0.75 m from the origin. The magnitude of the torque about the origin is
Answer:
108 NmExplanation:
Given data
mass m= 2 kg
radius r= 0.5 m
angular velocity ω= 12 rad/s
distance d= 0.75 m
we know that
[tex]Force= mass * acceleration\\\ acceleration= w^2r\\\\ Force= m*w^2r\\\\Force =2*12^2*0.5= 144 N[/tex]
we know that torque = F*d= 144*0.75= 108 Nm
The ball tends to come back to the centerline of the flow when it is pushed by an external disturbance. Explain this phenomenon using the curvature of streamlines.
Answer is given below
Explanation:
given data
we will consider here
Ping-Pong ball weighs = 3.1 g
diameter = 4.2 cm
solution
Whenever the ball is pushed, the length of the airflow along the outer edge increases and it accelerates. According to Bernoulli's equation. As the speed increases, the pressure decreases, so the pressure at the outer end is reduced. As the pressure at the outer edge is low, the extra air jet pushes it back to the center line.
Suppose your friend claims to have discovered a mysterious force in nature that acts on all particles in some region of space. He tells you that the force is always pointed away a definite point in space, which we can call the force center. The magnitude of the force turns out to be proportional to B/r3, where r is the distance from the force center to any other point. Your friend says that it has been determined that the constant of proportionality has been determined to be B= 2 (in units to be determined later), so that the magnitude of the force on a particle (in newtons) can be written as 2r^3, when the particle is at a distance r from the force center.
Required:
Write an expression of potential energy.
Answer:
U = 1 / r²
Explanation:
In this exercise they do not ask for potential energy giving the expression of force, since these two quantities are related
F = - dU / dr
this derivative is a gradient, that is, a directional derivative, so we must have
dU = - F. dr
the esxresion for strength is
F = B / r³
let's replace
∫ dU = - ∫ B / r³ dr
in this case the force and the displacement are parallel, therefore the scalar product is reduced to the algebraic product
let's evaluate the integrals
U - Uo = -B (- / 2r² + 1 / 2r₀²)
To complete the calculation we must fix the energy at a point, in general the most common choice is to make the potential energy zero (Uo = 0) for when the distance is infinite (r = ∞)
U = B / 2r²
we substitute the value of B = 2
U = 1 / r²
Four point charges have the same magnitude of 2.4×10^−12C and are fixed to the corners of a square that is 4.0 cm on a side. Three of the charges are positive and one is negative. Determine the magnitude of the net electric field that exists at the center of the square.
Answer:
7.2N/C
Explanation:
Pls see attached file
Find the potential energy associated with a 79-kg hiker atop New Hampshire's Mount Washington, 1900 m above sea level. Take the zero of potential energy at sea level.
Answer:
P = 1470980 J
Explanation:
We have,
Mass of the hiker is 79 kg
It is required to find the potential energy associated with a 79-kg hiker atop New Hampshire's Mount Washington, 1900 m above sea level.
It is given by :
[tex]P=mgh\\\\P=79\times 9.8\times 1900\\\\P=1470980\ J[/tex]
So, the potential energy of 1470980 J is associated with a hiker.