A chemistry student needs of 40mL diethylamine for an experiment. By consulting the CRC Handbook of Chemistry and Physics, the student discovers that the density of diethylamine is 0.7 . Calculate the mass of diethylamine the student should weigh out. Round your answer to significant digits.
Explanation:
Mass = density × volume
m = (0.7 g/mL) × (40 mL)
m = 28 g
Rounding to 1 significant figure, the student should weigh out 30 grams of diethylamine.
By using the formula Mass = Density x Volume, we find that the student needs to weigh out 28 grams of diethylamine for 40mL volume, as the density of diethylamine is 0.7g/mL.
Explanation:The question asks for the mass of diethylamine the student should weigh out to obtain 40mL of diethylamine. Diethylamine has a density of 0.7 g/mL, according to the CRC Handbook of Chemistry and Physics. The formula to find mass when you have volume and density is:Mass = Density x Volume. Therefore, to find the mass of the diethylamine, we multiply the volume (40 mL) by the density (0.7 g/mL), which equals 28 grams. So, the student should weigh out 28 grams of diethylamine.
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Starting from rest and height of 7 m, a 3kg object slides down a 30degrees incline, reaching the bottom with a speed of 10m/s . what is the work done by friction? please elaborate.
Answer:
The work done by the friction is 55.8 J.
Explanation:
Given that,
Height h = 7 m
Mass of the object = 3 kg
Angle = 30°
Speed = 10 m/s
The initial potential energy of the object is converted in to the kinetic energy at the bottom of the incline and the work done by the friction
[tex]mgh=\dfrac{1}{2}mv^2+W[/tex]
Where, m = mass of the object
v = final velocity
g = acceleration due to gravity
h = height
Put the value in the equation
[tex]3\times9.8\times7=\dfrac{1}{2}\times3\times(10)^2+W[/tex]
[tex]W = (205.8-150)\ J[/tex]
[tex]W=55.8\ J[/tex]
Hence, The work done by the friction is 55.8 J.
A woman walks 237 m in the direction 27.2 east of north, then 335 m directly east. Find the difference between the distance the woman walks and the magnitude of her displacement. (Answer in m.)
Answer:
81.1 m
Explanation:
X = 237 m 27.2 east of north 237 (Sin27.2 i + Cos 27.2 j)
X = 108.33 i + 210.8 j
Y = 335 east = 335 i
Displacement = X + Y = 108.33 i + 210.8 j + 335 i = 443.33 i + 210.8 j
(magnitude of displacement)^2 = 443.33^2 + 210.8^2
magnitude of displacement = 490.9 m
Distance traveled = 237 + 335 = 572 m
Difference in the magnitude of displacement and distance = 572 - 490.9
= 81.1 m
A small sphere with a mass of 441 g is moving upward along the vertical +y-axis when it encounters an electric field of 5.00 N/C iˆ. If, due to this field, the sphere suddenly acquires a horizontal acceleration of 13.0 m/s2 iˆ, what is the charge that it carries?
Answer:
Charge, q = 1.15 C
Explanation:
It is given that,
Mass of sphere, m = 441 g = 0.441 kg
Electric field, E = 5 N/C
Due to this field, the sphere suddenly acquires a horizontal acceleration of 13.0 m/s² such that,
[tex]ma=qE[/tex]
[tex]q=\dfrac{ma}{E}[/tex]
[tex]q=\dfrac{0.441\ kg\times 13\ m/s^2}{5\ N/C}[/tex]
q = 1.146 C
or
q = 1.15 C
So, the charge carried by the small sphere is 1.15 C. Hence, this is the required solution.
The electric field accounts for the unbalanced motion of the sphere. The electrostatic forces balance the linear force, for which the charge acquired by the sphere is 1.146 C.
What is an electric field?The region or space where a charged particle experiences the electrostatic force of attraction or repulsion, is known as an electric field.
Given data:
The mass of the small sphere is, m = 441 g = 0.441 kg.
The magnitude of the electric field is, E = 5.00 N/C.
The magnitude of horizontal acceleration is, a = 13.0 m/s².
Clearly, under the influence of an electric field, the electrostatic force balances the linear force. Then,
Fe = FL
[tex]qE = ma[/tex]
Here,
q is the magnitude of charge.
Solving as,
[tex]q \times 5.0 = 0.441 \times 13\\\\q = \dfrac{0.441 \times 13}{5.0}\\\\q = 1.146 \;\rm C[/tex]
Thus, we can conclude that the magnitude of charge carried by the sphere is 1.146 C.
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A tension force of 165 N inclined at 35.0° above the horizontal is used to pull a 31.0 kg storage crate a distance of 5.60 m on a rough surface. If the crate moves at a constant speed, find (a) the work done by the tension force and (b) the coefficient of kinetic friction between the crate and surface. HINT (a) the work done by the tension force (in J) J (b) the coefficient of kinetic friction between the crate and surface
(a) 756.9 J
The work done by a force when moving an object is given by:
[tex]W=Fd cos \theta[/tex]
where
F is the magnitude of the force
d is the displacement of the object
[tex]\theta[/tex] is the angle between the direction of the force and of the displacement
In this situation,
F = 165 N
d = 5.60 m
[tex]\theta=35.0^{\circ}[/tex]
so the work done by the tension is
[tex]W=(165 N)(5.60 m)cos 35.0^{\circ}=756.9 J[/tex]
(b) 0.646
The crate is moving at constant speed: this means that the acceleration of the crate is zero, so the net force on the crate is also zero.
There are only two forces acting on the crate along the horizontal direction:
- The horizontal component of the tension, [tex]Tcos \theta[/tex], forward
- The frictional force, [tex]-\mu N[/tex], backward, with [tex]\mu[/tex] being the coefficient of kinetic friction, N being the normal reaction of the surface on the crate
The normal reaction is the sum of the weight of the crate + the vertical component of the tension, so
[tex]N=mg-T sin \theta = (31.0 kg)(9.8 m/s^2) - (165 N)sin 35.0^{\circ}=209.2 N[/tex]
Since the net force is zero, we have
[tex]T cos \theta-\mu N =0[/tex]
where
T = 165.0 N
[tex]\theta=35.0^{\circ}[/tex]
N = 209.2 N
Solving for [tex]\mu[/tex], we find the coefficient of friction:
[tex]\mu = \frac{T cos \theta}{N}=\frac{(165.0 N)(cos 35^{\circ})}{209.2 N}=0.646[/tex]
Final answer:
In this physics problem, we determine the work done by a tension force inclining at an angle above the horizontal and find the coefficient of kinetic friction when a crate moves at a constant speed.
Explanation:
A tension force of 165 N inclined at 35.0° above the horizontal is used to pull a 31.0 kg storage crate at a distance of 5.60 m on a rough surface.
(a) Work done by the tension force: To calculate work, use the formula W = Fd cos(θ), where F is the force, d is the distance moved, and θ is the angle between the force and displacement. Here, W = 165 N * 5.60 m * cos(35.0°).
(b) Coefficient of kinetic friction: Since the crate moves at a constant speed, the work done by the tension force equals the work done against friction. Find the frictional force using the work done formula and then the coefficient of kinetic friction.
A satellite orbits Earth at an altitude of 350 km. What is its orbital period? A) 1.1 min B) It can not be determined without the satellite's mass. C) 65 min D) 91 min E) 1440 min
The orbital period of a satellite at an altitude of 350 km above Earth's surface is approximately 1.93 hours or around 115.8 minutes. Given the options, the closest one is option D, with a period of 91 minutes.
Explanation:To answer the question about the satellite orbiting Earth at an altitude of 350 km, we can utilize some physics concepts related to orbital mechanics. Note that a significant factor in determining the orbital period of a satellite (time required for a satellite to complete one orbit) is not the satellite's mass but rather its altitude or, more accurately, the total distance from the center of the Earth, considering the Earth's radius plus the satellite's altitude.
In general, the closer a satellite is to Earth, the shorter its orbital period. The information given indicates that a satellite at an altitude of 350 km will likely have an orbital period around 1.93 hours, which is roughly equivalent to 115.8 minutes. Given that, none of the available options exactly match this result. But the value closest to it would be 91 minutes, which is represented by option D.
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An overhang hollow shaft carries a 900 mm diameter pulley, whose centre is 250 mm from the centre of the nearest bearing. The weight of the pulley is 600 N and the angle of lap is 180°. The pulley is driven by a motor vertically below it. If permissible tension in the belt is 2650 N and if coefficient of friction between the belt and pulley surface is 0.3, estimate, diameters of shaft, when the internal diameter is 0.6 of the external. Neglect centrifugal tension and assume permissible tensile and shear stresses in the shaft as 84 MPa and 68 MPa respectively.
Answer:yu
Explanation:
A charge q = 2.00 μC is placed at the origin in a region where there is already a uniform electric field E⃗ = (100 N/C) iˆ . Calculate the flux of the net electric field through a Gaussian sphere of radius R = 10.0 cm centered at the origin. (ε 0 = 8.85 × 10-12 C2/N · m2)
Answer:
The flux of the net electric field is [tex]2.26\times10^{5}\ Nm^2/C[/tex]
Explanation:
Given that,
Charge [tex]q= 2.00 \mu C[/tex]
Electric field E = 100 N/C i
Radius R = 10.0 cm
We need to find the flux. It can be calculate using Gauss's law
The flux of the net electric field
[tex]\phi=\dfrac{q}{\epsilon_{0}}[/tex]
[tex]\phi=\dfrac{2.00\times10^{-6}}{8.85\times10^{-12}}[/tex]
[tex]\phi=2.26\times10^{5}\ Nm^2/C[/tex]
Hence, The flux of the net electric field is [tex]2.26\times10^{5}\ Nm^2/C[/tex]
An engine is designed to obtain energy from the temperature gradient of the ocean. What is the thermodynamic efficiency of such an engine if the temperature of the surface of the water is 59°F (15°C) and the temperature well below the surface is 41°F (5°C)
Answer:
0.035 (3.5 %)
Explanation:
The thermodynamic efficiency is given by:
[tex]\eta = 1 - \frac{T_C}{T_H}[/tex]
where
[tex]T_C[/tex] is the cold temperature
[tex]T_H[/tex] is the hot temperature
In this problem we have
[tex]T_C = 5 ^{\circ}C+ 273 = 278 KT_H = 15^{\circ}C+273 = 288 K[/tex]
So the efficiency is
[tex]\eta = 1 - \frac{278 K}{288 K}=0.035[/tex]
The maximum thermodynamic efficiency of an engine operating with a hot reservoir at 288 K and a cold reservoir at 278 K, according to the Carnot efficiency formula, is approximately 3.47%.
Explanation:The student has asked about the thermodynamic efficiency of an engine that uses the temperature gradient of the ocean. To calculate this, we can use the Carnot efficiency formula:
Carnot Efficiency Formula
η = 1 - Tc/Th, where η represents the efficiency, Tc is the cold temperature, and Th is the hot temperature. It's important to remember these temperatures need to be in Kelvin.
First, convert the given temperatures from degrees Fahrenheit to Kelvin:
59°F (15°C) = 288 K (Th)41°F (5°C) = 278 K (Tc)Now, apply the Carnot efficiency formula:
η = 1 - (278 K / 288 K) = 1 - 0.9653 = 0.0347 or 3.47%
Therefore, the maximum thermodynamic efficiency of this oceanic temperature gradient engine would be approximately 3.47%.
A cyclotron designed to accelerate protons has a magnetic field of magnitude 0.15 T over a region of radius 7.4 m. The charge on a proton is 1.60218×10−19 C and its mass is 1.67262 × 10−27 kg. What is the cyclotron frequency? Answer in units of rad/s.
Explanation:
It is given that,
Magnetic field, B = 0.15 T
Charge on a proton, [tex]q=1.60218\times 10^{-19}\ C[/tex]
Mass of a proton, [tex]m=1.67262 \times 10^{-27}\ kg[/tex]
The cyclotron frequency is given by :
[tex]f=\dfrac{qB}{2\pi m}[/tex]
[tex]f=\dfrac{1.60218\times 10^{-19}\ C\times 0.15\ T}{2\pi \times 1.67262 \times 10^{-27}\ kg}[/tex]
f = 2286785.40 Hz
or
[tex]\omega=14368296.44\ rad/s[/tex]
[tex]\omega=1.43\times 10^7 rad/s[/tex]
Hence, this is the required solution.
Two balls with masses of 2.10 kg and 6.50 kg travel toward each other at speeds of 13.0 m/s and 4.10 m/s, respectively. If the balls have a head-on inelastic collision and the 2.10-kilogram ball recoils with a speed of 8.20 m/s, how much kinetic energy is lost in the collision?
Answer:
137 J
Explanation:
Momentum is conserved, so:
(2.10)(13.0) + (6.50)(-4.10) = (2.10)(-8.20) + (6.50) v
v = 2.75 m/s
Energy before collision:
E = 1/2 (2.10) (13.0)² + 1/2 (6.50) (4.10)²
E = 232 J
Energy after collision:
E = 1/2 (2.10) (8.20)² + 1/2 (6.50) (2.75)²
E = 95.2 J
Energy lost in collision:
232 J - 95.2 J = 137 J
Final answer:
A total of 161.488 Joules of kinetic energy is lost in this collision.
Explanation:
The subject question involves a head-on inelastic collision between two balls of different masses traveling towards each other at different speeds and asks for the amount of kinetic energy lost in the collision. To find the kinetic energy lost, we first calculate the initial and final kinetic energies and then take their difference.
Initial kinetic energy (Eki) is the sum of the kinetic energies of the two balls before the collision:
Kinetic energy of the 2.10 kg ball: [tex](1/2) * 2.10 kg * (13.0 m/s)^2[/tex]
Kinetic energy of the 6.50 kg ball: [tex](1/2) * 6.50 kg * (4.1 m/s)^2[/tex]
Final kinetic energy (Ekf) involves only the kinetic energy of the 2.10 kg ball, as the final velocity of the 6.50 kg ball is not given:
Kinetic energy of the 2.10 kg ball: (1/2) * 2.10 kg * [tex](8.20 m/s)^2[/tex]
Kinetic energy lost during the collision is calculated as:
Elost = Eki - Ekf
After calculation:
Initial kinetic energy, Eki = (1/2) * 2.10 kg * [tex](13.0 m/s)^2[/tex] + (1/2) * 6.50 kg * [tex](4.1 m/s)^2[/tex]
Final kinetic energy, Ekf = (1/2) * 2.10 kg * [tex](8.20 m/s)^2[/tex]
Kinetic energy lost, Elost = Eki - Ekf
Plugging the numbers in:
Eki = 0.5 * 2.10 kg * [tex](13.0 m/s)^2[/tex] + 0.5 * 6.50 kg * [tex](4.1 m/s)^2[/tex]
= (0.5 * 2.10 * 169) + (0.5 * 6.50 * 16.81)
= 177.45 J + 54.64 J
= 232.09 J
Ekf = 0.5 * 2.10 kg * [tex](8.20 m/s)^2[/tex]
= (0.5 * 2.10 * 67.24)
= 70.602 J
Therefore, kinetic energy lost, Elost = 232.09 J - 70.602 J = 161.488 J
A total of 161.488 Joules of kinetic energy is lost in this collision.
. A hot-air balloon is drifting straight downward with a constant speed of 2.40 m/s. When the balloon is 7.63 m above the ground, the balloonist decides to drop one of the ballast sandbags to the ground below. How much time elapses before the sandbag hits the ground?
Answer:
1.03 seconds
Explanation:
x = x₀ + v₀ t + ½ at²
0 = 7.63 + (-2.40) t + ½ (-9.8) t²
0 = 7.63 - 2.40 t - 4.9 t²
Solve with quadratic formula:
t = [ -b ± √(b² - 4ac) ] / 2a
t = [ 2.40 ± √(2.40² - 4(-4.9)(7.63)) ] / -9.8
t = -1.52, 1.03
Since t can't be negative here, the sandbag hits the ground after 1.03 seconds.
The sandbag is dropped from the hot air balloon with the balloon's existing speed. Using the second equation of motion and the known values for initial velocity, acceleration due to gravity, and distance, we can solve for the time taken for the sandbag to hit the ground.
Explanation:In the given question, a hot-Air balloon is drifting straight downward with a constant speed of 2.40 m/s. When the balloon is 7.63 m above the ground, a ballast sandbag is dropped.
Given the bag is dropped from rest relative to the balloon (i.e., the initial velocity (u) of the sandbag is -2.4 m/s, negative because the direction is downward), and the acceleration due to gravity (a) is -9.8 m/s² (negative due to the downward direction), we can use the second equation of motion (s = ut + 0.5at²) where s is the distance covered, which is 7.63 m, to find time (t).
Setting this up, -7.63 = -2.4t + 0.5(-9.8)t². Solving this quadratic equation for time should give us the answer. The result will be the time taken by the sandbag to hit the ground after it is dropped from the hot air balloon.
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You connect a voltmeter to an unknown resistor in a circuit in order to measure the potential difference across that resistor. How should you connect it to the resistor and what would the voltmeter's ideal internal resistance be? A. In series Fint B. In parallel; int- O C. In series; rint- 0 D. In parallel; rint →
Answer:
option (a)
Explanation:
To make a galvanometer into voltmeter, we have to connect a high resistance in series combination.
The voltmeter is connected in parallel combination with teh resistor to find the voltage drop across it.
An ideal voltmeter has very high resistance that means it has a resistance as infinity.
Answer:
A. In series Fint
Explanation:
If you connect it to the resistor, the voltmeter's ideal internal resistance be In series Fint.
Which of the following is NOT an example of retaliation?
Getting A Sudden Shift Change
Being Fired
Getting A Promotion
Getting A Bad Performance Review
Getting a promotion is NOT an example of retaliation.
C. Getting A Promotion
What is retaliation?the action of harming someone because they have harmed oneself; revenge.
An example of to retaliate is for a person to punch someone who has hit him. (intransitive) To do something harmful or negative to get revenge for some harm; to fight back or respond in kind to damage or affront. Johny insulted Peter to retaliate for Peter's acid remark earlier.
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A person takes a trip, driving with a constant speed of 92.5 km/h, except for a 28.0-min rest stop. The person's average speed is 72.2 km/h. How much time is spent on the trip?
Answer:
40.3 min
Explanation:
First of all, let's convert every quantity into SI units:
[tex]v_1 = 92.5 km/h = 25.7 m/s[/tex] (speed in the first part of the trip)
[tex]t_2 = 28.0 min = 1680 s[/tex] time during which the person has stopped
[tex]v=72.2 km/h = 20.1 m/s[/tex] (average speed of the whole trip)
The average speed is the ratio between the total distance covered, d, and the total time taken, t:
[tex]v=\frac{d}{t}[/tex] (1)
The total distance covered is simply
[tex]d = v_1 t_1[/tex]
where [tex]t_1[/tex] is the time during which the person has moved at 92.5 km/h.
The total time taken is
[tex]t= t_1 + t_2[/tex]
So (1) becomes
[tex]v=\frac{v_1 t_1}{t_1 + t_2}[/tex]
Solving for [tex]t_1[/tex]:
[tex]v t_1 + v t_2 = v_1 t_1\\vt_2 = (v_1+v)t_1\\t_1 = \frac{v t_2}{v_1+v}=\frac{(20.1 m/s)(1680 s)}{25.7 m/s + 20.1 m/s}=737.3 s[/tex]
which corresponds to
[tex]t_2 = 737.3 s = 12.3 min[/tex]
So the total time of the trip is
[tex]t = 28.0 min + 12.3 min = 40.3 min[/tex]
bullet of mass 0.02070 �� collides inelastically with a wooden block of mass 28.00 ��, initially at rest. After the collision the, system has speed of 1.180 �/�. (a) What was initial speed of the bullet
Explanation:
It is given that,
Mass of bullet, m₁ = 0.0207 kg
Mass of wooden block, m₂ = 28 kg
Wooden block is initially at rest, u₂ = 0 m/s
After the collision the, system has speed of, v = 1.180 m/s
We need to find the initial speed (u₁) of the bullet. It can be calculated using the conservation of linear momentum as :
[tex]m_1u_1+m_2u_2=(m_1+m_2)v[/tex]
[tex]0.0207\ kg\times u_1+0=(0.0207\ kg+28\ kg)\times 1.180\ m/s[/tex]
[tex]0.0207\times u_1=33.06[/tex]
[tex]u_1=1597.10\ m/s[/tex]
So, the initial speed of the bullet is 1597.10 m/s. Hence, this is the required solution.
You are working out on a rowing machine. Each time you pull the rowing bar (which simulates the oars) toward you, it moves a distance of 1.4 m in a time of 1.1 s. The readout on the display indicates that the average power you are producing is 84 W. What is the magnitude of the force that you exert on the handle?
Answer:
Magnitude of the force exerted on the handle = 66 N
Explanation:
Power is the ratio of work and time.
[tex]P=\frac{W}{t}\\\\W=Pt=84\times 1.1=92.4J[/tex]
We have work done by rower = 92.4 J
We also have
Work = Force x Displacement
92.4 = Force x 1.4
Force = 66 N
Magnitude of the force exerted on the handle = 66 N
The cylinder rotates about the fixed z-axis in the direction indicated. If the speed of point A is vA = 2.7 ft/sec and the magnitude of its acceleration is aA = 20.6 ft/sec2, determine the magnitudes of the angular velocity and angular acceleration of the cylinder. Is knowledge of the angle θ necessary?
To solve this problem, we need to utilize the relationship between linear and angular motion.
Firstly, we know the speed of point A is vA = 2.7 ft/sec. In a rotating cylinder, the linear speed v of a point at distance r from the center of rotation is given by v = ω * r, where ω is the angular speed. This would allow us to find ω by dividing the speed at point A by its distance from the center.
However, in this problem, we don't know the exact distance of point A from the center. So let's denote the unknown distance as r ft. Then, the angular speed ω = vA / r.
Secondly, it's given that the magnitude of the acceleration of the point A is aA = 20.6 ft/sec². Angular acceleration α is related to linear acceleration a by the factor of the radius, i.e., a = α * r.
We don't know r itself, but we know vA and aA, so we are able to compute the ratio of α to ω or α/ω = aA/vA, which equals to 20.6/2.7. This calculation enables us to conclude that the angular velocity and angular acceleration are proportional to each other, regardless of the values of radius r and angle θ
Taken together, from this reasoning, we can conclude that the ratio of angular to linear velocity/acceleration is approximately 7.63.
Finally, considering the issue of whether the knowledge of the angle θ is necessary, we can see that the angle θ does not appear in our ratio, and it does not affect the ratios of angular to linear velocities and accelerations. Therefore, the knowledge of the angle θ is not necessary for this calculation.
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A beam of white light from a flashlight passes through a red piece of plastic. a. What is the color of the light that emerges from the plastic? b. Is the emerging light as intense as, more intense than, or less intense than the white light? Explain. c. The light then passes through a blue piece of plastic. Describe the color and intensity of the light that emerges.
The light emerging from a red filter is red and less intense than the original white light. When passed through blue plastic, it would be very dim or blocked, as red light cannot pass through a blue filter.
When a beam of white light passes through a red piece of plastic, the color of the light that emerges is red. This color is seen because the red plastic acts as a filter, absorbing other colors and only allowing red light to pass through. The emerging light is less intense than the white light because some of the light energy has been absorbed by the plastic.
If the light that has already passed through a red filter is then passed through a blue piece of plastic, the emerging light would likely be very dim or completely blocked. This is because the red light, lacking blue components, will be mostly absorbed by the blue plastic, which only allows blue light to pass through.
A 51.0 kg cheetah accelerates from rest to its top speed of 31.7 m/s. HINT (a) How much net work (in J) is required for the cheetah to reach its top speed? J (b) One food Calorie equals 4186 J. How many Calories of net work are required for the cheetah to reach its top speed? Note: Due to inefficiencies in converting chemical energy to mechanical energy, the amount calculated here is only a fraction of the energy that must be produced by the cheetah's body. Cal
(a) [tex]2.56\cdot 10^4 J[/tex]
The work-energy theorem states that the work done on the cheetah is equal to its change in kinetic energy:
[tex]W= \Delta K = \frac{1}{2}mv^2 - \frac{1}{2}mu^2[/tex]
where
m = 51.0 kg is the mass of the cheetah
u = 0 is the initial speed of the cheetah (zero because it starts from rest)
u = 31.7 m/s is the final speed
Substituting, we find
[tex]W=\frac{1}{2}(51.0 kg)(31.7 m/s)^2 - \frac{1}{2}(51.0 kg)(0)^2=2.56\cdot 10^4 J[/tex]
(b) 6.1 cal
The conversion between calories and Joules is
1 cal = 4186 J
Here the energy the cheetah needs is
[tex]E=2.56\cdot 10^4 J[/tex]
Therefore we can set up a simple proportion
[tex]1 cal : 4186 J = x : 2.56\cdot 10^4 J[/tex]
to find the equivalent energy in calories:
[tex]x=\frac{(1 cal)(2.56\cdot 10^4 J)}{4186 J}=6.1 cal[/tex]
The cheetah needs 25,972.35 Joules or 6.2 Calories of work to reach its top speed of 31.7 m/s.
Explanation:The cheetah accelerates from a state of rest to its top speed. This involves work done on the cheetah which we can compute using the formula for kinetic energy, as work done is equal to change in kinetic energy. The initial speed of the cheetah is 0, so initial kinetic energy is also 0. The final kinetic energy when the cheetah is at its top speed is ½ m v^2, where m is the mass and v is the speed.
(a) So, the work done, W = ½ * 51.0kg * (31.7 m/s)^2 = 25972.35 J,
(b) To convert Joules into Calories, we use 1 Calorie = 4186 J, so the Calories of work done = 25972.35 J / 4186 = 6.2 Calories. Note that due to inefficiencies in the energy conversion process in bodies, the actual energy produced by the cheetah's body will be more than 6.2 Calories to reach this speed.
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A catapult is tested by Roman legionnaires. They tabulate the results in a papyrus and 2000 years later the archaeological team reads (distances translated into modern units): Range = 0.4 km; angle of launch = π/5 rad; landing height = launch height. What is the initial velocity of launch of the boulders if air resistance is negligible?
Answer:
64.2 m/s
Explanation:
In the x direction:
x = x₀ + v₀ₓ t + ½ at²
400 m = 0 m + v₀ cos (π/5) t + ½ (0 m/s²) t²
t = 400 / (v₀ cos (π/5))
In the y direction:
y = y₀ + v₀ᵧ t + ½ gt²
0 m = 0 m + v₀ sin (π/5) t + ½ (-9.8 m/s²) t²
0 = v₀ sin (π/5) - 4.9 t
t = v₀ sin (π/5) / 4.9
Therefore:
400 / (v₀ cos (π/5)) = v₀ sin (π/5) / 4.9
1960 = v₀² sin (π/5) cos(π/5)
1960 = ½ v₀² sin(2π/5)
3920 / sin(2π/5) = v₀²
v₀ = 64.2 m/s
Answer:
Initial velocity = 423.08m/s
Explanation:
Using formular for Range of projectile
R =V^2 sin2theta/g
Given:
Range=0.4km= 400m
Theta=3.142/5
g= 9.8m/s^2
400= V^2×sin(2×3.142/5)/9.8
400×9.8= V^2Sin 1.26
3920= 0.0219V^2
V^2= 3920/0.0219
V^2= 178995.43
V=sqrt 178995.43
V= 423.08m/s
A 11 kg mass on a frictionless inclined surface is connected to a 2.1 kg mass. The pulley is massless and frictionless, and the connecting string is massless and does not stretch. The 2.1 kg mass is acted upon by an upward force of 6.6 N, and thus has a downward acceleration of only 3.6 m/s 2 . The acceleration of gravity is 9.8 m/s 2 . What is the tension in the connecting string?
Answer:
6.4 N
Explanation:
Draw free body diagrams for each mass.
For the mass on the inclined surface, the sum of forces parallel to the incline are:
∑F = ma
T - Mg sin θ = Ma
For the hanging mass, the sum of the forces in the y direction are:
∑F = ma
T - mg + F = m(-a)
We're given that g = 9.8 m/s², M = 11 kg, m = 2.1 kg, F = 6.6 N, and a = 3.6 m/s².
From the second equation, we have everything we need to find the tension force.
T - (2.1)(9.8) + 6.6 = (2.1)(-3.6)
T = 6.42 N
Rounded to 2 sig figs, the tension is 6.4 N.
Assume your car reaches a speed of 21.7 m/s at a steady rate for 5.05 s after the light turns green. (a) What distance have you traveled during this time? (b) What is your average speed during this time?
Answer:
The distance and average speed are 54.79 m and 10.85 m.
Explanation:
Given that,
Speed = 21.7 m/s
Time = 5.05 s
(a). We need to calculate the distance
Firstly we will find the acceleration
Using equation of motion
[tex]v = u+at[/tex]
[tex]a = \dfrac{v-u}{t}[/tex]
Where, v = final velocity
u = initial velocity
t = time
Put the value in the equation
[tex]a = \dfrac{21.7-0}{5.05}[/tex]
[tex]a = 4.297 m/s^2[/tex]
Now, using equation of motion again
For distance,
[tex]s = ut+\dfrac{1}{2}at^2[/tex]
[tex]s = 0+\dfrac{1}{2}\times4.3\times(5.05)^2[/tex]
[tex]s=54.79\ m[/tex]
The distance is 54.79 m.
(b). We need to calculate the average speed during this time
[tex]v_{avg}=\dfrac{D}{T}[/tex]
Where, D = total distance
T = time
Put the value into the formula
[tex]v_{avg}=\dfrac{54.79}{5.05}[/tex]
[tex]v_{avg}=10.85\ m/s[/tex]
Hence, The distance and average speed are 54.79 m and 10.85 m.
The specific heat of copper is 385 J/kg·K. If a 2.6 kg block of copper is heated from 340 K to 450 K, how much thermal energy is absorbed?
Answer:
Heat absorbed, Q = 110110 J
Explanation:
It is given that,
The specific heat of copper is, [tex]c=385\ J/kg.K[/tex]
Mass of block, m = 2.6 kg
Initial temperature, [tex]T_1=340\ K[/tex]
Final temperature, [tex]T_2=450\ K[/tex]
Thermal energy is given by :
[tex]Q=mc\Delta T[/tex]
[tex]Q=mc(T_2-T_1)[/tex]
[tex]Q=2.6\times 385\times (450-340)[/tex]
Q = 110110 J
So, the thermal heat of 110110 J is absorbed. Hence, this is the required solution.
A person is in a room whose walls are maintained at a temperature of 17 °C. The temperature of the person’s skin is 32 °C. The surface area of the person’s skin is 1.4 m2. Determine the netheat transfer rate leaving the person.
Answer:
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Explanation:
The electric field between square plates of a parallel-plate capacitor has magnitude E. The potential across the plates is maintained with constant voltage by a battery as they are pulled apart to twice their original separation, which is small compared to the dimensions of the plates. The magnitude of the electric field between the plates is now equal to a)E b)E/4 c)E/2 d)4E e)2E
Answer:
the Answer is c: E/2
Explanation:
see attachment
The critical angle for diamond (n-2.42) surrounded by air is approximately 24 35 48 66
Answer:
sin(C) = 1/n = 1/2.42
C = 24.4 deg
The critical angle for the diamond-air interface is 24 degrees.
What is the critical angle for the diamond-air interface?If the ray of light in the diamond medium strikes the separating surface between the diamond and air making an angle more than 24 degrees from the normal to the point of incidence then that ray of light will undergo a total internal reflection and comes back to the diamond medium. This is the reason behind the sparkles of diamonds.Hence, the critical angle for the diamond-air interface is 24 degrees.To learn more about the critical angle for the diamond-air interface, refer to:https://brainly.com/question/1304781
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A particle moves with constant acceleration along a straight line starting from rest. The percentage increase in its displacement during the 4th second compared to its displacement in the 3rd second (1) 33 % . is (2) 40 % (4) 77 % / (3) 66 % A particle is thrown upward from groun
Answer:
Option 2 is the correct answer.
Explanation:
We have equation of motion s = ut + 0.5at²
Here u = 0 m/s
So, s = 0.5at²
Distance traveled in first second = 0.5 x a x 1² = 0.5 a
Distance traveled in second second = 0.5 x a x 2² - 0.5 x a x 1²= 1.5 a
Distance traveled in third second = 0.5 x a x 3² - 0.5 x a x 2²= 2.5 a
Distance traveled in fourth second = 0.5 x a x 4² - 0.5 x a x 3²= 3.5 a
The percentage increase in its displacement during the 4th second compared to its displacement in the 3rd second
[tex]=\frac{3.5a-2.5a}{2.5a}=0.4=40\%[/tex]
Option 2 is the correct answer.
An eagle is flying horizontally at a speed of 3.00 m/s when the fish in her talons wiggles loose and falls into the lake 5.00 m below. Calculate the velocity of the fish relative to the water when it hits the water.
The velocity of the fish relative to water is approximately 10.44 m/s when it hits the water. The fish inherits the eagle's horizontal velocity at the start and then accelerates downward due to gravity.
Explanation:The eagle's speed is given as 3.00 m/s horizontally. Therefore, when the fish falls, it also initially has a horizontal velocity of 3.00 m/s because it was brought to that speed by the eagle. The vertical component of the fish's velocity can be calculated using the second equation of motion: vf = vi + a*t, which simplifies to vf = g*t in the absence of initial vertical velocity. The time 't' it takes for the fish to hit the water can be calculated using the equation d = 0.5*a*t2, which gives t = √(2d/g). Substituting 5.00 m for 'd' and 9.81 m/s2 (approx gravitational acceleration) for 'g', we get t ≈ 1.017 seconds. Replacing 't' in the velocity equation with 1.017 seconds, which results in a vertical velocity of 9.965 m/s. The resultant velocity of the fish relative to the water when it hits would then be calculated using Pythagoras theorem since the horizontal and vertical movements are independent and at right angles to each other, giving a result of √((3.00 m/s)2 + (9.965 m/s)2) = 10.44 m/s.
Learn more about Velocity here:https://brainly.com/question/39711173
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The velocity of the fish relative to the water when it hits the water is approximately 10.36 m/s at an angle of about 73.74° below the horizontal.
To calculate the velocity of the fish relative to the water when it hits the water, we need to analyze both the vertical and horizontal components of its motion separately.
Given Data:
Horizontal speed of the eagle (and the fish when dropped): [tex]v_{horizontal} = 3.00 \, \text{m/s}[/tex]
Height from which the fish falls: [tex]h = 5.00 \, \text{m}[/tex]
Step 1: Calculate the time it takes for the fish to fall 5m.
To find the time of fall, we can use the kinematic equation for vertical motion:
[tex]h = \frac{1}{2} g t^2[/tex]
where:
[tex]h[/tex] is the height (5.00 m),
[tex]g[/tex] is the acceleration due to gravity (approximately [tex]9.81 \, \text{m/s}^2[/tex]),
[tex]t[/tex] is the time in seconds.
Rearranging the equation to solve for [tex]t[/tex], we have:
[tex]t = \sqrt{\frac{2h}{g}} = \sqrt{\frac{2 \times 5.00}{9.81}} \approx \sqrt{1.02} \approx 1.01 \, \text{s}[/tex]
Step 2: Calculate the vertical velocity of the fish just before hitting the water.
The vertical velocity can be calculated using:
[tex]v_{vertical} = g t[/tex]
Substituting the time we found:
[tex]v_{vertical} = 9.81 \, \text{m/s}^2 \times 1.01 \, \text{s} \approx 9.91 \, \text{m/s}[/tex]
Step 3: Combine the horizontal and vertical velocities to find the resultant velocity.
The horizontal velocity is constant at 3.00 m/s, and the vertical velocity just before impact is approximately 9.91 m/s. We can use the Pythagorean theorem to find the magnitude of the resultant velocity:
[tex]v_{resultant} = \sqrt{v_{horizontal}^2 + v_{vertical}^2}[/tex]
Substituting the values:
[tex]v_{resultant} = \sqrt{(3.00)^2 + (9.91)^2} = \sqrt{9 + 98.20} = \sqrt{107.20} \approx 10.36 \, \text{m/s}[/tex]
Step 4: Find the direction of the velocity relative to the horizontal using tangent.
The angle [tex]\theta[/tex] can be calculated using:
[tex]\theta = \tan^{-1}\left(\frac{v_{vertical}}{v_{horizontal}}\right)[/tex]
Substituting our values:
[tex]\theta = \tan^{-1}\left(\frac{9.91}{3.00}\right) \approx \tan^{-1}(3.30) \approx 73.74^\circ[/tex]
Two simple pendulums are in two different places. The length of the second pendulum is 0.4 times the length of the first pendulum, and the acceleration of gravity experienced by the second pendulum is 0.9 times the acceleration of gravity experienced by the first pendulum. Determine the comparison of the frequency of the first pendulum to the second pendulum. a) 2/3. b) 3/2 (5 marks) c) 4/9. d) 9/4
Answer:
[tex]\sqrt{\frac{4}{9}}[/tex]
Explanation:
The frequency of a simple pendulum is given by:
[tex]f=\frac{1}{2\pi}\sqrt{\frac{g}{L}}[/tex]
where
g is the acceleration of gravity
L is the length of the pendulum
Calling [tex]L_1[/tex] the length of the first pendulum and [tex]g_1[/tex] the acceleration of gravity at the location of the first pendulum, the frequency of the first pendulum is
[tex]f_1=\frac{1}{2\pi}\sqrt{\frac{g_1}{L_1}}[/tex]
The length of the second pendulum is 0.4 times the length of the first pendulum, so
[tex]L_2 = 0.4 L_1[/tex]
while the acceleration of gravity experienced by the second pendulum is 0.9 times the acceleration of gravity experienced by the first pendulum, so
[tex]g_2 = 0.9 g_1[/tex]
So the frequency of the second pendulum is
[tex]f_2=\frac{1}{2\pi}\sqrt{\frac{g_2}{L_2}}=\frac{1}{2\pi} \sqrt{\frac{0.9 g_1}{0.4 L_1}}[/tex]
Therefore the ratio between the two frequencies is
[tex]\frac{f_1}{f_2}=\frac{\frac{1}{2\pi}\sqrt{\frac{g_1}{L_1}}}{\frac{1}{2\pi} \sqrt{\frac{0.9 g_1}{0.4 L_1}}}=\sqrt{\frac{0.4}{0.9}}=\sqrt{\frac{4}{9}}[/tex]
Final answer:
The frequency of the first pendulum compared to the second pendulum is in the ratio of 2/3, since the length of the second pendulum is 0.4 times the first and gravity on the second is 0.9 times the first, impacting their respective periods and inversely their frequencies.
Explanation:
The frequency of a simple pendulum is dependent on its length and the acceleration due to gravity it experiences. The formula to find the frequency (f) of a pendulum is given by:
f = 1 / T
Where T is the period of the pendulum, which can be calculated by:
T = 2π√(L/g)
Let's denote L1 and g1 as the length and the acceleration due to gravity for the first pendulum, and L2 and g2 for the second pendulum. Since the length of the second pendulum is 0.4 times that of the first, we have L2 = 0.4L1. And because the acceleration of gravity is 0.9 times for the second pendulum, g2 = 0.9g1.
Using these proportions:
T1 = 2π√(L1/g1)
T2 = 2π√(L2/g2) = 2π√((0.4L1)/(0.9g1))
We can simplify T2 by putting it in terms of T1:
T2 = T1√(0.4/0.9)
Now, since frequency is the inverse of period:
f1 = 1/T1
f2 = 1/T2
So the ratio of their frequencies (f1/f2) will be the inverse of the ratio of their periods (T2/T1):
f1/f2 = T2/T1
Plugging in the ratio:
f1/f2 = √(0.4/0.9) = √(4/9) = 2/3
Therefore, the correct answer is (a) 2/3.