Answer:
F1= 196 N
F2= 392 N
Explanation:
Given:
length of the plank = 2 m
mass of the plank = 20 kg
Weight of the plank = 20 x 9.8 =196 N
Torque due to the weight of the plank with respect to the pivoted end (i.e the end held by the hand) Counter clockwise torque = 196 x cog of wood
= 196 x 1 = 196 Nm
Clockwise torque = F2 x 0.5
for the balanced case
F2 x 0.5 = 196
F2 = 196/ 0.5
F2= 392 N
now,
the net force
Net downward force =Net upward force
F1 + weight of plank = F2
F1 + 196 = 392N
F1 = 392 – 196
F1= 196 N
To find the forces F1 and F2 exerted on the plank, we can use the principle of equilibrium. The sum of the forces acting on the plank must be zero, and the sum of the torques must also be zero. By solving the equations derived from these conditions, we can determine the values of F1 and F2.
Explanation:To find the forces F1 and F2, we can use the principle of equilibrium. According to the principle, the sum of the forces acting on an object in equilibrium must be zero, and the sum of the torques must also be zero.
For the vertical forces:
F1 + F2 - mg = 0
Where m is the mass of the plank and g is the acceleration due to gravity.
For the torques:
F1 * 2m = F2 * 0.5m
Solving these equations will give us the values for F1 and F2.
Learn more about the Equilibrium of Forces here:https://brainly.com/question/30916838
#SPJ3
Suppose the energy transferred to a dead battery during charging is W. The recharged battery is then used until fully discharged again. Is the total energy transferred out of the battery during use also W?
Answer:
No
Explanation:
The amount of energy transferred out of battery will not be the same as given during charging.
A battery has its internal resistance as well which will draw some energy, through this internal resistance energy is lost
P = I² R
where P is energy, I is current passing and R is resistance.
As per law of conservation energy will be saved as it is the sum of drawn energy and energy wasted in internal resistance. But the total energy transferred out will not be same.
BASEBALL IN SPACE baseball in space You are observing a baseball in space, with mass m = 0.145 kg. It is moving past your spacecraft with speed v = 22.9 m/sec, to the north. Your Acme Tractor Beam can exert 1.3 Newtons of pull force on that baseball. CALCULATE: How long will it take your Acme Tractor Beam to slow down the baseball to a stop? Enter the numeric part of your answer, to the nearest hundredth of a second, i.e., 0.01 sec. E.g., if your answer is 60.5329 sec, then type in 60.53 in the answer box.
Answer:
It takes 2.55 sec
Explanation:
First we need to find the aceleration with this equation:
F = m*a
[tex]a = \frac{F}{m}[/tex]
Where: F = 1.3 Newtons
m = 0.145 kg
Then [tex]a = \frac{1.3 Newtons}{0.145 kg}[/tex]
[tex]a = 8.9655 m/sec^{2}[/tex]
Now we are ready to calculate the time with the next equation:
[tex]V_{f} = V_{o} + a*t[/tex]
[tex]V_{f} - V_{o} = a*t[/tex]
[tex]t = \frac{V_{f} - V_{o}}{a}[/tex]
Where: [tex]V_{o} = 22.9 m/sec[/tex] (at the beginning)
[tex]V_{f} = 0 m/sec[/tex] (at the end because it stops)
We must use [tex]a = - 8.9655 m/sec^{2}[/tex] because in this case speed is decreasing
Finally: [tex]t = \frac{0 m/sec - 22.9 m/sec}{- 8.9655 m/sec^{2} }[/tex]
t = 2.55 sec
By applying Newton's second law of motion and the motion equation, we find that it will take approximately 2.55 seconds for the Acme Tractor Beam to slow down the baseball to a stop in space.
Explanation:This problem is a classic example of the application of Newton's second law of motion which states: F = ma, where F is the force, m is the mass, and a is the acceleration. Here we are given the force F exerted by your Acme Tractor Beam (1.3 N) and the mass of the baseball m (0.145 kg).
First, use F=ma to calculate acceleration a. The acceleration a = F/m = 1.3 N / 0.145 kg = 8.97 m/s². This is the rate at which the baseball would slow down under the influence of the tractor beam.
Next, use the formula v = u + at, where 'v' is the final velocity, 'u' is the initial velocity, 't' is the time, and 'a' is the acceleration we just calculated. We want to know the time 't' when the baseball comes to stop (v = 0), so we rearrange this equation to solve for 't': t = (v - u) / a.
Plugging in the given numbers: t = (0 - 22.9 m/s) / -8.97 m/s², we find t = 2.55 s.
So it will take approximately 2.55 seconds for your Acme Tractor Beam to slow down the baseball to a stop.
Learn more about Physics of Motion here:https://brainly.com/question/13966796
#SPJ3
The 3.6-lb rod AB is hanging in the vertical position. A 2.2-lb block, sliding on a smooth horizontal surface with a velocity of 12 ft/s, strikes the rod at its end B.
What is the velocity of the block immediately after the collision. The coefficient of restitution between the block and the rod at B is e = 0.85?
Answer:
v = 4.17 m/s
Explanation:
as we know that block is moving towards the end of the rod
so let say just after the collision the block is moving with speed v' and rod is rotating with speed "w"
so we have angular momentum conservation about the hinge point
[tex]mvL = mv'L + I\omega[/tex]
[tex](2.2)(12)L = (2.2)v'(L) + (\frac{3.6 L^2}{3}) \omega[/tex]
[tex]12 = v' + 0.545(\omega L)[/tex]
now by the equation of coefficient of restitution we can say
[tex]0.85 = \frac{(\omega L) - v'}{12}[/tex]
now we have
[tex]\omega L - v' = 10.2[/tex]
now we have
[tex]\omega L = 14.37 [/tex]
[tex]v' = 4.17 m/s[/tex]
so velocity of block just after collision is 4.17 m/s
To find the velocity of the block after the collision, we can use the principle of conservation of momentum and the coefficient of restitution.
Explanation:First, we need to calculate the velocity of the 3.6-lb rod AB after the collision with the 2.2-lb block.
To do this, we can use the principle of conservation of momentum. The momentum before the collision is given by the sum of the momentums of the rod and the block: (3.6 lb) x (0 ft/s) + (2.2 lb) x (12 ft/s) = 26.4 lb-ft/s.
The coefficient of restitution (e) is defined as the ratio of the relative velocity of separation to the relative velocity of approach. In this case, the relative velocity of approach is 12 ft/s, and the relative velocity of separation is the sought-after velocity of the block immediately after the collision.
An electron is released from rest in a uniform electric field. The electron accelerates vertically upward, traveling 4.50 m in the first [math]3.00 \mu s[/math] after it is released. (a) What are the magnitude and direction of the electric field? (b) Are we justified in ignoring the effects of gravity? Justify your answer quantitatively.
To determine the magnitude and direction of the electric field when an electron is released in a uniform electric field, we use the displacement formula and compare the acceleration due to the electric field with the acceleration due to gravity.
Explanation:(a) To determine the magnitude of the electric field, we can use the formula E = delta V / delta x. The electron experiences constant acceleration, so the displacement equation is given by x = (1/2)at^2, where x is the displacement, a is the acceleration, and t is the time. Plugging in the given values, we find that the acceleration is 5.00 x 10^11 m/s^2 (upward).
(b) We can justify ignoring the effects of gravity by comparing the acceleration due to the electric field with the acceleration due to gravity. For an electron in a uniform electric field, the acceleration is much greater than the acceleration due to gravity. The acceleration due to gravity is approximately 9.8 m/s^2. Therefore, the effects of gravity can be ignored in this scenario.
Learn more about Magnitude and direction of electric field here:https://brainly.com/question/2085280
#SPJ12
The magnitude and direction of the electric field can be determined by calculating the electric field strength. Gravity can be ignored in this situation because the electric field is much stronger.
Explanation:(a) To find the magnitude and direction of the electric field, we can use the equation:
Electric field strength (E) = Force (F) / Charge (q)
Given that the electron is accelerating upward, the direction of the electric field is downward. Using the formula, we can rearrange it to find the electric field strength:
Electric field strength = Force / Charge = mass x acceleration / charge = (9.11 x 10^-31 kg) x (9.8 m/s^2) / (1.6 x 10^-19 C) = 5.67 x 10^11 N/C
The magnitude of the electric field is 5.67 x 10^11 N/C, and the direction is downward.
(b) We can justify ignoring the effects of gravity by comparing the magnitude of the electric field to the gravitational field strength. The gravitational field strength near the surface of the Earth is approximately 9.8 N/kg. Comparing this to the electric field strength of 5.67 x 10^11 N/C, we can see that the electric field is much stronger than the gravitational field. Therefore, the effects of gravity can be ignored.
Learn more about Electric field here:https://brainly.com/question/8971780
#SPJ2
A toy truck has a speed of 4 m/s and kinetic energy of 48 J. What is its mass?
A. 8 kg
B. 4 kg
C. 6 kg
D. 10 kg
48/4= 12
12/2= 6
Answer: C. 6 kg
The pressure of a gas in a container is 1.85 atm and occupies a volume of 12.5 L. If the original volume is reduced by half at constant temperature, what would happen to the pressure?
Answer:
3.7 atm i.e., pressure doubles
Explanation:
P₁ = Initial pressure = 1.85 atm
P₂ = Final pressure
V₁ = Initial volume = 12.5 L
V₂ = Final volume = 0.5V₁
T = Temperature is constant
From ideal gas law
P₁V₁ = P₂V₂
[tex]\\\Rightarrow P_2=P_1\frac{V_1}{V_2}\\\Rightarrow P_2=1.85\frac{V_1}{0.5V_1}\\\Rightarrow P_2=1.85\frac{1}{0.5}\\\Rightarrow P_2=1.85\times 2\\\Rightarrow P_2=3.7\ atm[/tex]
∴ Final pressure is 3.7 atm i.e., pressure doubles
Final answer:
Boyle's law describes the relationship between pressure and volume of a gas at constant temperature. When the volume is reduced by half, the pressure of the gas would double.
Explanation:
The pressure of a gas in a container is directly proportional to its volume when the temperature is constant, as described by Boyle's law. In this case, when the original volume is reduced by half, the pressure would double, resulting in a pressure of 3.7 atm.
Generation of electricity in coal-burning power plants and nuclear power plants both involve _______.
Answer:
Heating water to produce steam which drives a turbine
Explanation:
Generation of electricity in coal-burning power plants and nuclear power plants both involve heating water to produce steam which drives a turbine.
The sum of potential and kinetic energies in the particles of a substance is called energy.
Explanation:
This sentence is the description of a specific type of energy: the mechanical energy.
To unerstand it better:
The mechanical energy of a body, a system or a substance is that which is obtained from the speed of its movement (kinetic energy) or its specific position (potential energy), in order to produce a mechanical work. This means mechanical energy involves both the kinetic energy and the potential energy (which can be elastic or gravitational, for example).
In addition, it should be noted that mechanical energy is conserved in conservative fields and is a scalar magnitude.
Therefore:
The sum of potential and kinetic energies in the particles of a substance is called Mechanical Energy
Answer:
Explanation:
internal!!!!!!!!
Review Problem. A light string with a mass per unit length of 8.20 g/m has its ends tied to two walls separated by a distance equal to three-fourths the length of the string (Fig. P16.30). An object of mass m is suspended from the center of the string, putting a tension in the string. (A) (mg)^1/2(B) 1/4(mg x 1000)^1/2 (C) 1/8(5mg x 1000)^1/2(D) None
Answer:
The answer is 'D' none
Explanation:
In the figure shown we have
[tex]2Tsin(\theta )=mg\\\\\therefore T=\frac{mg}{2sin(\theta )}[/tex]
From the figure we can see that
[tex]cos(\theta )=\frac{\frac{3L}{8}}{\frac{L}{2}}\\\\\therefore \theta =cos^{-1}(\frac{3}{4})\\\\sin(\theta )=\frac{\sqrt{7}}{4}[/tex]
Thus value of tension be will be
[tex]T=\frac{2mg}{\sqrt{7}}[/tex]
The tension in the string when an object of mass m is suspended from the center is given by T = (mg)/(2L), where T is the tension, m is the mass of the object, g is the acceleration due to gravity, and L is the length of the string. So, the correct answer is (A) (mg)^1/2.
Explanation:To find the tension in the string when an object of mass m is suspended from the center, we can use the formula for tension in a string:
T = (mg)/(2L)
Where T is the tension, m is the mass of the object, g is the acceleration due to gravity, and L is the length of the string.
Substituting the given values, the tension in the string is:
T = (m)(9.8)/(2L)
Therefore, the correct answer is (A) (mg)^1/2.
Learn more about tension in a string here:https://brainly.com/question/32999602
#SPJ3
On what factors capacitance of parallel plate capacitor depends?
Answer:
Separation between the plates, area of the plates and dielectric constant
Explanation:
The capacitance of a parallel plate capacitor is given by:
[tex]C=k \epsilon_0 \frac{A}{d}[/tex]
where
k is the dielectric constant
[tex]\epsilon_0[/tex] is the vacuum permittivity (which has a constant value)
A is the area of the plates
d is the separation between the plates
Therefore from the formula we see that the capacitance of a parallel plate capacitor depends on the following factors:
- Separation between the plates
- Area of the plates
- Dielectric constant
A car travelling 85km/h strikes a tree. The front end compresses and the driver comes to rest after travelling 0.80m. What was the average acceleration of the driver during the collision?
Answer:
The average acceleration of the driver during the collision is [tex]-348.4\frac{m}{s^{2}}[/tex]
Explanation:
Initial speed of car , [tex]u=85\frac{km}{h}=\frac{85\times 5}{18}\frac{m}{s}[/tex]
=>[tex]u=23.61\frac{m}{s}[/tex]
Finally the car comes to rest .
Therefore final speed of the car , [tex]v=0\frac{m}{s}[/tex]
Distance traveled while coming to rest , s = 0.80 m
Using equation of motion
[tex]v^{2}=u^{2}+2as[/tex]
=>[tex]0^{2}=23.61^{2}+2\times 0.80\times a[/tex]
=>[tex]a=-348.4\frac{m}{s^{2}}[/tex]
Thus the average acceleration of the driver during the collision is [tex]-348.4\frac{m}{s^{2}}[/tex]
A car travels from Boston to Hartford in 4 hours. The two cities are 240 kilometers apart. What was the average speed of the car during the trip? A. 4 km/hour B. 960 km/hour C. 60 km/hour
Answer:
Average speed is 60 km/hour
Explanation:
When we need to calculate average speed, we use this equation:
[tex]V = \frac{x_{f} - x_{o}}{t_{f} - t_{o}}[/tex]
Where: [tex]x_{o} = 0 km[/tex] position at the beginning
[tex]x_{f} = 240 km[/tex] at the end
[tex]t_{o} = 0 hours[/tex]
[tex]t_{f} = 4 hours[/tex]
Then: [tex]V = \frac{240 km - 0 km}{4 hours - 0 hours}[/tex]
[tex]V = \frac{240 km}{4 hours}[/tex]
Finally V = 60 km/hour
Two cylindrical resistors are made from same material and have the same length. When connected across the same battery, one dissipates twice as much power as the other. How do their diameters compare ?
Answer:
[tex]\frac{d_2}{d_1} = \sqrt2 = 1.41[/tex]
Explanation:
As resistor is connected to the battery of constant EMF then the power across the resistor is given as
[tex]P = \frac{E^2}{R}[/tex]
now if two resistors are made up of same material and of same length then due to different cross sectional area they both have different resistance
Due to different resistance they both will have different power
Since power is inversely depends on the resistance
So if the power is twice that of the other then the resistance must be half
so we have
[tex]R_1 = \rho \frac{L}{A_1}[/tex]
[tex]R_2 = \rho\frac{L}{A_2}[/tex]
since one resistance is half that of other resistance
So the area of one must be twice that of other
so we have
[tex]\frac{A_2}{A_1} = 2[/tex]
[tex]\frac{\pi d_2^2}{\pi d_1^2} = 2[/tex]
[tex]d_2 = 1.41 d_1[/tex]
A 10-cm-long spring is attached to the ceiling. When a 2.0 kg mass is hung from it, the spring stretches to a length of 15 cm.?A) What is the spring constant k? B) How long is the spring when a 3.0 kg mass is suspended from it?
A) 392 N/m
The spring constant can be found by applying Hooke's Law:
F = kx
where
F is the force applied to the spring
k is the spring constant
x is the stretching of the spring
Here we have:
F is the weight of the block hanging from the spring, which is
[tex]F=mg=(2.0 kg)(9.8 m/s^2)=19.6 N[/tex]
The stretching of the spring is
[tex]x=15 cm - 10 cm = 5 cm = 0.05 m[/tex]
Therefore its spring constant is
[tex]k=\frac{F}{x}=\frac{19.6 N}{0.05 m}=392 N/m[/tex]
B) 17.5 cm
Now that we know the value of the spring constant, we can calculate the new stretching of the spring when a mass of m=3.0 kg is applied to it. In this case, the force applied on the spring is
[tex]F=mg=(3.0 kg)(9.8 m/s^2)=29.4 N[/tex]
Therefore the stretching of the spring is
[tex]x=\frac{F}{k}=\frac{29.4 N}{392 N/m}=0.075 m = 7.5 cm[/tex]
And since the natural length of the spring is 10 cm, the new length will be
L = 10 cm + 7.5 cm = 17.5 cm
Listed following are three possible models for the long-term expansion (and possible contraction) of the universe in the absence of dark energy. Rank each model from left to right based on the ratio of its actual mass density to the critical density, from smallest ratio (mass density much smaller than critical density) to largest ratio (mass density much greater than critical density).
(A) coasting universe
(B) critical universe
(C) recollapsing universe
Answer:
1) Recollapsing universe
2) critical universe
3) Coasting universe
Explanation:
According to the smallest ration (ratio actual mass density to current density) to largest ration, rank of models for expansion of universe are
1) Recollapsing universe -in this, metric expansion of space is reverse and universe recollapses.
2) critical universe - in this, expansion of universe is very low.
3) Coasting universe - in this, expansion of universe is steady and uniform
______ is the energy released by the sun is caused by thermonuclear fusion.
Answer: Nuclear energy
Explanation:
Nuclear energy (also called atomic energy) is the energy found in the nucleus of an atom.
This energy is released spontaneously (within the stars and the sun) or artificially (in nuclear reactors built by humans) in nuclear reactions, which are divided into two types:
-Fission (separation of the components of the atom nucleous)
-Fusion (joining two light nuclei to form a heavier nucleous)
In the case of the Sun, the nuclear reactions that occur are due fusion, in which the hydrogen is converted into helium.
An object with mass 3.5 kg is attached to a spring with spring stiffness constant k = 270 N/m and is executing simple harmonic motion. When the object is 0.020 m from its equilibrium position, it is moving with a speed of 0.55 m/s.(a) Calculate the amplitude of the motion._____ m(b) Calculate the maximum velocity attained by the object. [Hint: Use conservation of energy.]______ m/s
Answer:
Part a)
A = 0.066 m
Part b)
maximum speed = 0.58 m/s
Explanation:
As we know that angular frequency of spring block system is given as
[tex]\omega = \sqrt{\frac{k}{m}}[/tex]
here we know
m = 3.5 kg
k = 270 N/m
now we have
[tex]\omega = \sqrt{\frac{270}{3.5}}[/tex]
[tex]\omega = 8.78 rad/s[/tex]
Part a)
Speed of SHM at distance x = 0.020 m from its equilibrium position is given as
[tex]v = \omega \sqrt{A^2 - x^2}[/tex]
[tex]0.55 = 8.78 \sqrt{A^2 - 0.020^2}[/tex]
[tex]A = 0.066 m[/tex]
Part b)
Maximum speed of SHM at its mean position is given as
[tex]v_{max} = A\omega[/tex]
[tex]v_{max} = 0.066(8.78) = 0.58 m/s[/tex]
A 1024 Hz tuning fork is used to obtain a series of resonance levels in a gas column of variable length, with one end closed and the other open. The length of the column changes by 20 cm from resonance to resonance. From this data, the speed of sound in this gas is:
Answer:
The speed of sound in this gas is 409.6 m/s.
Explanation:
The length of the column changes from 20 cm from resonance to resonance. Thus,
[tex]L=\frac {(2n+1)\lambda}{4}[/tex]
The length change from one resonance to resonance. so, there is 1 loop change. So,
ΔL = 1 loop = λ/2
ΔL = 20 cm (given)
Also, 1 cm = 0.01 m
So,
ΔL = 0.2 cm (given)
The wavelength is:
λ = ΔL×2
λ = 2x0.2 = 0.4 m
Given:
Frequency (ν) = 1024 Hz
Velocity of the sound in the gas = ν×λ = 1024×0.4 m/s = 409.6 m/s
You recently purchased a large plot of land in the Amazon jungle at an extremely low cost. You are quite pleased with yourself until you arrive there and find that the nearest source of electricity is 1500 miles away, a fact that your brother-in-law, the real estate agent, somehow forgot to mention. Since the local hardware store does not carry 1500-mile-long extension cords, you decide to build a small hydroelectric generator under a 35.0-m high waterfall located nearby. The flow rate of the waterfall is 0.150x10^2 m^3/h, and you anticipate needing 1750 kW h/wk to run you lights, air conditioner, and television.
What is the maximum power theoretically available from the waterfall? Is the power sufficient to meet your needs?
Answer:
The maximum power available from the water fall is 239.568 kWh/week.
Explanation:
Given:
Flow rate = 0.15 × 10² m³/h
now the mass of water flowing per hour will be = flow rate × mass density of water(i.e 1000 kg/m³)
mass of water flowing per hour will be = 0.15 × 10² m³/h × 1000 kg/m³ = 15000 kg/h
The gravitational potential energy of the falling water = mgh = 15000 × 9.8 × 35 = 5145000 J/h
or
5145000/3600 = 1426.166 J/s (as 1 h = 3600 seconds)
or
1426.166 W = 1.426 kW
Now, the number of hours in a week = 7 × 24 = 168 hours
Now the energy produce in a week = 1.426 kW × 168 hr = 239.568 kWh/week.
No it is not sufficient to meet the needs
The maximum hydraulic power generated by the waterfall is around 14.35 kW, which is more than the required energy for weekly consumption (2.28 kW). Therefore, the waterfall can especially serve as a power source.
Explanation:The problem involves the calculation of hydraulic power, which is given by the formula Power = p * g * h * Q, where p is the density of water (around 1000 kg/m³ in standard conditions), g is the gravitational acceleration (approximated as 9.8 m/s²), h is the waterfall height (in meters), and Q is the flow rate (in m³/s). First, we need to convert the flow rate from m³/h to m³/s. We get Q = 0.150 * 10² m³/h = 0.04167 m³/s.
After substituting the figures into the formula, we get: Power = 1000 kg/m³ * 9.8 m/s² * 35.0 m * 0.04167 m³/s = 14350.5 Watts, or about 14.35 kW.
Given the energy requirement per week is 1750 kWh, let's convert it to the energy required per second. That is 1750 kWh/wk = (1750 * 1000) / (7 * 24 * 60 * 60) = 2.28 kW. This means, the maximum power available from the waterfall is more than enough to meet your requirements.
Learn more about Hydraulic Power here:
https://brainly.com/question/35132579
#SPJ3
A campground official wants to measure the distance across an irregularly-shaped lake, but can't do so directly. He picks a point south of the lake. From that point to the left side of the lake, the distance is 105 meters. From that point to the right side of the lake is 119 meters. The angle between these two measurements is 83 degrees. How far is it across the lake?
Answer:
149 m
Explanation:
The distances across the lake is forming a triangle.
let the distance between the point and the left side be 'x'
and the distance between the point and the right be 'y'
and the distance across the lake be 'z' and the angle opposite to 'z' be 'Z' given:
∠Z = 83°
x = 105 m
y = 119 m
Now, applying the Law of Cosines, we get
z² = x² + y² - 2xycos(Z)
Substituting the values in the above equation, we get
z² = 105² + 119² - 2×105×119×cos(83°)
or
z = √22140.48
or
z = 148.796 m ≈ 149 m
The point is 149 m across the lake
A 75.0-kg person is riding in a car moving at 20.0 m/s when the car runs into a bridge abutment. (a) calculate the average force on the person if he is stopped by a padded dashboard that compresses an average of 1.00 cm. (b) calculate the average force on the person if he is stopped by an air bag that compresses an average of 15.0 c
(a) [tex]-1.5\cdot 10^6 N[/tex]
First of all, we need to calculate the acceleration of the person, by using the following SUVAT equation:
[tex]v^2 -u ^2 = 2ad[/tex]
where
v = 0 is the final velocity
u = 20.0 m/s is the initial velocity
a is the acceleration
d = 1.00 cm = 0.01 m is the displacement of the person
Solving for a,
[tex]a=\frac{v^2-u^2}{2d}=\frac{0-20.0^2}{2(0.01)}=-20000 m/s^2[/tex]
And the average force on the person is given by
[tex]F=ma[/tex]
with m = 75.0 kg being the mass of the person. Substituting,
[tex]F=(75)(-20000)=-1.5\cdot 10^6 N[/tex]
where the negative sign means the force is opposite to the direction of motion of the person.
b) [tex]-1.0\cdot 10^5 N[/tex]
In this case,
v = 0 is the final velocity
u = 20.0 m/s is the initial velocity
a is the acceleration
d = 15.00 cm = 0.15 m is the displacement of the person with the air bag
So the acceleration is
[tex]a=\frac{v^2-u^2}{2d}=\frac{0-20.0^2}{2(0.15)}=-1333 m/s^2[/tex]
So the average force on the person is
[tex]F=ma=(75)(-1333)=-1.0\cdot 10^5 N[/tex]
The average force on the person if he is stopped by a padded dashboard is -1.5 x 10⁶ N.
The average force on the person if he is stopped by an air bag is -10,000 N.
The given parameters;
mass of the person, m = 75 kgvelocity of the person, u = 20 m/sThe deceleration of the car when the padded dashboard compresses an average of 1.00 cm.
s = 0.01 m
[tex]v^2 = u^2 + 2as\\\\0 = 20^2 + (2\times 0.01)a\\\\0 = 400 + 0.02a\\\\0.02a = -400\\\\a = \frac{-400}{0.02} \\\\a = -20,000 \ m/s^2[/tex]
The average force is calculated as;
F = ma
F = 75 x (-20,000)
F = -1.5 x 10⁶ N.
The deceleration of the car when the air bag compresses an average of 15 cm.
s = 0.15 m
[tex]v^2 = u^2 + 2as\\\\0 = u^2 + 2as\\\\-2as = u^2\\\\a = \frac{u^2}{-2s} = \frac{20^2}{-2\times 0.15} = -1,333.33 \ m/s^2[/tex]
The average force is calculated as follows;
F = ma
F = 75 x (-1,333.33)
F = -10,000 N.
Learn more here:https://brainly.com/question/19887955
Find the angular acceleration produced given the mass lifted is 12 kg at a distance of 27.1 cm from the knee joint, the moment of inertia of the lower leg is 0.959 kg m2 kg m 2 , the muscle force is 1504 1504 N, and its effective perpendicular lever arm is 3.3 3.3 cm.
Answer:
Angular acceleration, α = 26.973 rad/s²
Explanation:
Given data:
Lifted mass, M = 12 kg
Distance of the lifted mass = 27.1 cm = 0.271 m
Effective lever arm, d = 3.3 cm = 0.033 m
Moment of inertia, I = 0.959 kg.m²
Applied force, F = 1504 N
Now,
the torque (T) is given as:
T = F × d
also,
T = I × α
where,
α is the angular acceleration
Now,
Total moment of inertia, I = 0.959 + 12×(0.271)² = 1.840 kg.m²
Now equation both the torque formula and substituting the respective values, we get
1504 × 0.033 = 1.840 × α
⇒ α = 26.973 rad/s²
Coherent light of wavelength 540 nm passes through a pair of thin slits that are 3.4 × 10-5 m apart. At what angle away from the centerline does the second bright fringe occur?
Answer: [tex]1.8\°[/tex]
Explanation:
The diffraction angles [tex]\theta_{n}[/tex] when we have a slit divided into [tex]n[/tex] parts are obtained by the following equation:
[tex]dsin\theta_{n}=n\lambda[/tex] (1)
Where:
[tex]d=3.4(10)^{-5}m[/tex] is the width of the slit
[tex]\lambda=540 nm=540(10)^{-9}m[/tex] is the wavelength of the light
[tex]n[/tex] is an integer different from zero.
Now, the second-order diffraction angle is given when [tex]n=2[/tex], hence equation (1) becomes:
[tex]dsin\theta_{2}=2\lambda[/tex] (2)
Now we have to find the value of [tex]\theta_{2}[/tex]:
[tex]sin\theta_{2}=\frac{2\lambda}{d}[/tex] (3)
Then:
[tex]\theta_{2}=arcsin(\frac{2\lambda}{d})[/tex] (4)
[tex]\theta_{2}=arcsin(\frac{2(540(10)^{-9}m)}{3.4(10)^{-5}m})[/tex] (5)
Finally:
[tex]\theta_{2}=1.8\°[/tex] (6)
All objects near the earths surface - regardless of size and weight - hhave the same force of gravityvacting on them.
A. True
B. False
Answer:
B. False
Explanation:
Not all objects near the earths surface - regardless of size and weight - have the same force of gravity on them.
A sloping surface separating air masses that differ in temperature and moisture content is called a _________.
Answer:
A sloping surface separating air masses that differ in temperature and moisture content is called a front.
A sloping surface separating air masses that differ in temperature and moisture content is called a front.
Explanation:A sloping surface separating air masses that differ in temperature and moisture content is called a front. Fronts occur when warm air and cold air meet, creating a boundary between them. The warm air is forced to rise over the cold air, resulting in changes in weather conditions.
There are several types of fronts, each associated with distinct weather patterns:
Cold Front: A cold front forms when a cold air mass advances and replaces a warmer air mass. As the cold air displaces the warm air, it forces the warm air to rise rapidly, leading to the formation of cumulonimbus clouds and potentially severe weather conditions such as thunderstorms.
Warm Front: In contrast, a warm front occurs when a warm air mass advances and overtakes a retreating cold air mass. As the warm air rises over the colder air, it produces widespread stratiform clouds and precipitation over an extended area.
Stationary Front: When two air masses meet but neither advances, a stationary front is formed. This results in a prolonged period of cloudy and wet weather along the boundary.
Occluded Front: An occluded front develops when a fast-moving cold front overtakes a slow-moving warm front. This complex type of front often leads to a mix of weather conditions, including precipitation and strong winds.
Learn more about front here:https://brainly.com/question/1387503
#SPJ2
What is meant by constructive and destructive interference
A frog jumps for 4.0 seconds at a maximum horizontal distance of 0.8m. what is its velocity along the road?
Answer:
The frog's horizontal velocity is 0.2 m/s.
Explanation:
To solve this problem, we must first remember what velocity is and how we solve for it. Velocity can be solved for using the formula x/t, where x represents horizontal distance and t represents time (in seconds), that it takes to travel this distance. If we plug in the given numbers for these variables and solve, we get the following:
v = x/t
v = 0.8m/4s
v = 0.2 m/s
Therefore, the correct answer is 0.2 m/s. We can verify that these units are correct because the formula calls for distance divided by time, so meters per second is a sensible answer.
Hope this helps!
On July 15, 2004, NASA launched the Aura spacecraft to study the earth's climate and atmosphere. This satellite was injected into an orbit 705 km above the earth's surface, and we shall assume a circular orbit.
(a) How many hours does it take this satellite to make one orbit?
h
(b) How fast (in km/s) is the Aura spacecraft moving?
km/s
Answer:
a). T = 1.64 hr
b). v = 7.503 km/s
Explanation:
Given :
NASA launched the Aura spacecraft to study the earth's climate and atmosphere.
Height of the satellite orbit from the earth's surface, h = 705 km
= 705000 m
Therefore we know that,
a).Time period of the space craft is
[tex]T = 2\pi \sqrt{\frac{(R+h)^{3}}{G\times M}}[/tex]
where, G = Universal Gravitational constant
= [tex]6.67 \times 10^{-11} N-m^{2}/kg^{2}[/tex]
M = Mass of the earth
= 5.98 x [tex]10^{24}[/tex] kg
R = Radius of the earth
= 6.38 x [tex]10^{6}[/tex] m
∴[tex]T = 2\pi \sqrt{\frac{(R+h)^{3}}{G\times M}}[/tex]
[tex]T = 2\pi \sqrt{\frac{((6.36\times 10^{6})+705000)^{3}}{6.67\times 10^{-11}\times 5398\times 10^{24}}}[/tex]
[tex]T = 5933[/tex] s
= 1.64 hr
Thus, the satellite will take 1.64 hr to make one orbit.
b). We know velocity of the spacecraft is given by
[tex]v=\sqrt{\frac{G\times M}{R+h}}[/tex]
[tex]v=\sqrt{\frac{6.67\times 10^{-11}\times 5.98\times 10^{24}}{(6.38\times 10^{6})+705000}}[/tex]
v = 7503 m/s
= 7.503 km/s
Thus, the Aura satellite is moving with velocity v = 7.503 km/s
A) The number of hours it takes for the satellite to make one orbit ( h ) = 1.64 hours
b) The speed of the Aura spacecraft = 7.5 Km/s
Given data
Height of satellite Orbit ( h ) = 705 km ≈ 705000 m
A) Determine the time taken in hours for the satellite to make a single orbit
T = [tex]2\pi \sqrt{\frac{(R+h)^3}{GM} }[/tex] -------- ( 1 )
where ; G = 6.67 * 10⁻¹¹ N-m²/kg², M ( mass of earth ) = 5.98 * 10²⁴,
R = 6.38 * 10⁶ m , h = 705000 m
Insert the values into equation ( 1 )
T = 5933 secs
= 1.64 hours ( time taken to complete one orbit )
B) Determine how fast Aura spacecraft is moving
V = [tex]\sqrt{\frac{GM}{R+h} }[/tex] -------- ( 2 )
where ; G = 6.67 * 10⁻¹¹ N-m²/kg², M ( mass of earth ) = 5.98 * 10²⁴,
R = 6.38 * 10⁶ m , h = 705000 m
Insert values into equation ( 2 )
∴ V = 7503 m/s
= 7.5 km/s
Hence we can conclude that the number of hours it takes for the satellite to make one orbit ( h ) = 1.64 hours and The speed of the Aura spacecraft = 7.5 Km/s
Learn more : https://brainly.com/question/25357037
easy A solid disk rotates in the horizontal plane at an angular velocity of 0.067 rad/s with respect to an axis perpendicular to the disk at its center. The moment of inertia of the disk is 0.10 kg · m2. From above, sand is dropped straight down onto this rotating disk, so that a thin uniform ring of sand is formed at a distance of 0.40 m from the axis. The sand in the ring has a mass of 0.50 kg. After all the sand is in place, what is the angular velocity of the disk?
Answer:
[tex]\omega_f = 0.0067\ rad/s[/tex]
Explanation:
Given:
Initial angular velocity of the disc, [tex]\omega_o[/tex] = 0.067 rad/s
Initial moment of inertia of the disc, [tex]I_o[/tex] = 0.10 kg.m²
Distance of sand ring from the axis, r = 0.40m
Mass of the sand ring = 0.50 kg
Now, no external torque is applied to the rotating disk.
Thus, the angular momentum of the system will remain conserved.
Also,from the properties of moment of inertia, the addition of the sand ring will increase the initial moment of inertia by an amount Mr²
thus, we have
Initial Angular Momentum = Final Angular Momentum
or
[tex]I_o\omega_o = I_f\times\omega_f[/tex]
[tex]I_o\omega_o = (I_o + Mr^2)\times\omega_f[/tex]
Where,
[tex]I_f[/tex] = Final moment of inertia
[tex]\omega_f[/tex] = Final angular velocity
substituting the values, we get
[tex]0.10\times0.067 = (0.10 + 0.50\times 0.40^2)\times\omega_f[/tex]
or
[tex]0.0067 = (0.18)\times\omega_f[/tex]
or
[tex]\omega_f = 0.0067\ rad/s[/tex]
The rate of change of angular displacement is defined as angular velocity. After all the sand is in place the angular velocity of the disk will be 0.067 rad/sec.
What is the definition of Angular velocity?The rate of change of angular displacement is defined as angular velocity, and it is stated as follows:
ω = θ t
Where,
θ is the angle of rotation,
t is the time
ω is the angular speed
ω₀ is the initial angular velocity of the disc = 0.067 rad/s
I₀ is the initial moment of inertia of the disc, = 0.10 kg.m²
r is the distance of sand ring from the axis = 0.40m
m is the mass of the sand ring = 0.50 kg
If the net external torque is applied to the rotating disk is zero
According to the angular momentum conservation principle;
Initial Angular Momentum = Final Angular Momentum
[tex]\rm I_0 \omega_0= I_f \omega_f \\\\[/tex]
According to parallel axis theorem ;
[tex]\\\\ I_f = I_0 + mr^2[/tex]
[tex]\rm I_0 \omega_0= (I_0 + mr^2 ) \omega_f[/tex]
[tex]\rm 0.10 \times 0.067= (0.10 + 1020 \times 0.40^2 ) \omega_f \\\\ \rm \omega_f=0.0067\ rad/sec[/tex]
Hence the angular velocity of the disk will be 0.067 rad/sec.
To learn more about the angular speed refer to the link;
https://brainly.com/question/9684874
A 10.0-g bullet is fired into, and embeds itself in, a 1.95-kg block attached to a spring with a force constant of 16.6 N/m and whose mass is negligible. How far is the spring compressed if the bullet has a speed of 300 m/s just before it strikes the block and the block slides on a frictionless surface? Note: You must use conservation of momentum in this problem because of the inelastic collision between the bullet and block.
Answer:
Distance the spring is compressed, x = 0.52 m
Explanation:
Given :
Mass of the bullet, m = 10 g = 0.01 kg
Mass of the block, M = 1.95 kg
spring force constant, k = 16.6 N/m
Distance the spring is compressed =x
Speed of the bullet, v = 300 m/s
Speed of the block = V
Therefore we know that according to the law of conservation of momentum,
m.v = ( m+M )V
or [tex]V = \frac{m\times v}{m+M}[/tex]
[tex]V = \frac{0.01\times 300}{0.01+1.95}[/tex]
= 1.53 m/s
Now according to the law of conservation of momentum,
[tex]\frac{1}{2}\times M\times V^{2} =\frac{1}{2}\times k\times x^{2}[/tex]
[tex]x = \sqrt{\frac{\left ( M+m \right ).V^{2}}{k}}[/tex]
[tex]x = \sqrt{\frac{\left ( M+m \right )}{k}}\times V[/tex]
[tex]x = \sqrt{\frac{\left ( 1.95+0.01 \right )}{16.6}}\times 1.53[/tex]
[tex]x = 0.52[/tex] m
Thus, distance the spring is compressed, x = 0.52 m