The distance traveled by car A when they pass each other is 3278.2 ft.
What is the time taken for the two cars to meet each other?The time taken for the two cars to meet each other is calculated by applying the concept of relative velocity.
(Va + Vb)t = d
where;
Va is the velocity of car AVb is the velocity of car Bt is the time of motion of the two carsd is the distance between the cars(74 + 56)t = 5760
130t = 5760
t = 5760/130
t = 44.3 s
The distance traveled by car A when they pass each other is calculated as follows;
da = 74 ft/s x 44.3 s
da = 3,278.2 ft
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. gun sights are adjusted to aim high to compensate for the effect of gravity, effectively making the gun accurate only for a specific range. (a) if a gun is sighted to hit targets that are at the same height as the gun and 100.0 m away, how low will the bullet hit if aimed directly at a target 150.0 m away?
The problem describes a situation where a bullet is launched at a certain angle respect to the horizontal to hit in the target, given the launching conditions and the distance to the target. The formulas to compute the magnitudes involved in the projectile motion are
x=[tex]v0[/tex] cos0 t
Y=[tex]V0[/tex] sin0 t
xm=[tex]v0[/tex]² sin 20/g
[tex]Xm[/tex] is the maximum horizontal distance the object can reach when it returns to the ground (or launching height).
let's check if X is computed correctly .
The set of data is correct. We can now proceed to the required distance. We know the bullet is shot directly to the target, so the angle is 0, keeping the same speed. The new maximum distance is 150 m, so we can compute the time needed to complete the flight
since
Let's find the time from the formula
X=[tex]V0[/tex] cos0 t
t=[tex]Xm[/tex]/[tex]V0[/tex] cos0
t=150/100=1.5
Y=(9.8)(0.66)²÷2
Y=-2.134
The bullet will hit 2.134 meters below the target
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What type of energy is the sum of kinetic and potential energy in an object that is used to do work?.
Mechanical energy type of energy is the sum of kinetic and potential energy in an object that is used to do work.
What is an example mechanical energy?Potential energy plus kinetic energy are combined to form mechanical energy. It refers to the power underlying an object's location and motion. A compressed spring, for instance, has mechanical energy in the form of potential energy, while a moving vehicle has mechanical energy in the form of kinetic energy.
How does mechanical energy become electrical energy?A sequence of blades installed on a rotor shaft in a turbine generator are pushed by a flowing fluid, including such air, water, steam, combustion gases, or water. A generator's rotor shaft rotates or spins due to the fluid's force on the blades. In turn, the generator converts the rotor's physical (kinetic) energy into electricity.
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A meteorite is found to contain two isotopes of element x, x-12, and x-14. Experimental analysis determined 95% of the element is isotope x-12, whereas 5% of the element is isotope x-14. What is the average atomic mass of element x?.
Considering the definition isotopes and atomic mass of an element, the average atomic mass of element x is 12.1
Definition of isotopeThe same chemical element can have the same atomic numbers, but the number of neutrons is different. These atoms are called isotopes of the element.
Definition of atomic massThe atomic masses of chemical elements are calculated as the weighted average of the masses of the different isotopes of each element, taking into account the relative abundance of each of them.
Atomic mass of the element in this caseIn this case, you know:
Isotope x-12 is 95% abundant (the atomic mass is 12).Isotope x-14 is 5% abundant (the atomic mass is 14).The average mass of the element x can be calculated as:
average mass of element x=12 amu×0.95 + 14 amu×0.05
average mass of element x= 12.1 amu
Finally, the atomic mass of the element x is 12.1.
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a long ladder 20 feet long is leaning against a wall with one end on the floor. the bottom end of the the ladder is being pulled along the floor away from the wall at a constant rate of 1 2 feet/sec. assume that the top of the ladder is always in contact with the wall. what are the velocity and acceleration of the top of the ladder when the bottom of the ladder is 12 feet from the wall?
When a ladder with 20 ft. long is pulled along the floor away with a constant rate 1.2 ft/s, then the top of the ladder is moving toward the floor with velocity - 0.9 ft/s and there is no acceleration.
Let:
x = the distance between the bottom of the ladder from the wall
y = the distance between the top of the ladder from the floor
When x = 12 ft, then using the Pythagorean Theorem:
x² + y² = 20² (Equation 1)
12² + y² = 20²
y² = 20² - 12² = 16²
y = 16 ft.
Take the derivative of equation 1 with respect to t
2x . dx/dt + 2y. dy/dt = 0
Substitute x = 12, y = 16, dx/dt = 1.2 ft/s
2 . 12 . (1.2) + 2 . 16 . dy/dt = 0
dy/dt = - 0.9 ft/s
Hence, the velocity is - 0.9 ft/s
Since the floor is moving with a constant rate, the acceleration is zero.
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A force of 30 N to the right is applied to an object. An opposite force of 20 N to the left is applied to the same object. What is the net force applied to the object?
A. 10 N to the left
B. 50 N to the right
C. 50 N to the left
D. 10 N to the right
The net force on an object if a force of 30 N to the right is applied to an object. An opposite force of 20 N to the left is applied to the same object is 10 N to the right. And the right option is D.
What is net force?The net force is the vector sum of all the forces that act upon an object.
To calculate the net force applied to the object, we use the formula below.
Formula:
F = Fa-Fb................. Equation 1Where:
F = Net force applied to the object.Fa = Force applied to the rightFb = Force applied to the leftFrom the question,
Given:
Fa = 30 N to the rightFb = 20 N to the leftSubstitute these values into equation 1
F = 30-20F = 10 N to the right.Hence the right option is D. 10 N to the right.
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a woman stands near the edge of a cliff and drops a stone over the edge. exactly one second later she drops another stone. one second after that, how fast is the distance between the two stones changing?
The two stones will fall at the different time. But the difference between the time is just of one second.
The distance of the cliff is also considered. Here, the speed or velocity is asked. To calculate this we have to use the equation of motion.
There are three equations of motion.
We have;
The kinematic equation of motion that can be used is s = ut - (1/2)·gt²
For the first stone, we have, s₁ = ut₁ - (1/2)·gt₁²
For the second stone, we get; s₂ = ut₂ - (1/2)·gt₂²
g = The acceleration due to gravity ≈ 9.81 m/s²
s = The distance below the cliff top
The initial velocity of the stones, u = 0
To calculate the change of distance between the two stones,
we will subtract the two equation of motions are
s₁-s₂
With all the values kept together in the equation of motion, we can easily find the change of distance between the two stones.
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Given a block which is at rest on a perfectly flat smooth surface with no friction. You push with your hand horizontally on the block with some force. If the force on the block= 4.00 N and the block moves a distance of 2.00 m on the table, then what is the work done on the block? A) 1.29 JB) 8.00 JC) 9.31 JD) 3.65 JE) 4.57 J
ANSWER
B) 8.00 J
EXPLANATION
Given:
• The force applied to move the block, F = 4.00N
,• The distance the block is moved, d = 2.00 m
Unknown:
• The work done on the block, W
If there is no friction between the block and the table, the only force acting upon the block in the direction of movement is the applied force F. Since the force is applied in the direction of movement, then the angle between the force and the displacement is 0, so the work done on the block is,
[tex]W=F\times d=4.00N\times2.00m=8.00J[/tex]Hence, the work done on the block is 8.00 J.
a person threw a small bundle toward their friend on a balcony 10 meters above them. how fast did they throw it up?
The speed with which they threw the bundle is 14 m/s.
What is the speed of the bundle?We know that the speed of the bundle can be obtained from the use of the equations of kinematics. Now we can use the equation of the upward motion under gravity here.
Using;
v^2 = u^2 - 2gh
v = final velocity
u = initial velocity
g = acceleration due to gravity
h = height
Now;
v = 0 m/s at the maximum height
u^2 = 2gh
We now have;
u = √2gh
u = √ 2 * 9.8 * 10
u = 14 m/s
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What is the weight of a boy with a mass of 50 kg on Earth?
Use g= 9.81 m/s² and do not include units in your answer.
Answer:
490.5
Explanation:
Force = Mass × Acceleration
Since, Weight is a force and 'g' is the acceleration due to the earth's gravity,
Weight = Mass × g
Weight = 50 kg × 9.81 m/s² = 490.5 kg-m/s² or 490.5 N
suppose we want to place a weather satellite into a circular orbit 300 kmkm above the earth's surface. what speed, period, and radial acceleration must it have? the earth's radius is 6380km
a. The speed of the satellite is 244 m/s
b. The period of the satellite is 1.72 × 10⁷ s
c. The radial acceleration of the satellite is 8913 rad/s
What is the speed of the satellite?The speed of the satellite is given by v = √(GM/R) where
G = universal gravitational constant = 6.67 × 10⁻¹¹ Nm²/kg², M = mass of earth = 5.972 × 10²⁴ kg and R = radius of orbit = r + r' where r = radius of earth = 6380 km and r' = radius of orbit = 300 kmSo, substituting the values of the variables into the equation, we have that
v = √(GM/R)
v = √(6.67 × 10⁻¹¹ Nm²/kg² × 5.972 × 10²⁴ kg /6380 km + 300 km)
v = √(6.67 × 10⁻¹¹ Nm²/kg² × 5.972 × 10²⁴ kg /6680 km)
v = √(39.83324 × 10¹³ Nm²/kg ÷ 6.680 × 10⁶ m)
v = √(5.963 × 10⁻³ × 10⁷ Nm²/kg)
v = √(5.963 × 10⁴ Nm/kg)
v = 2.44 × 10² m/s
V = 244 m/s
So, the speed of the satellite is 244 m/s
b. What is the period of the orbit?The period of the orbit is given by T = 2πR/v where
R = radius of orbit = 6680 km and v = speed of orbit = 244 m/sSo, substituting the values of the variables into the equation, we have
T = 2πR/v
T = 2π × 6680 km/244 m/s
T = 2π × 6680 × 10³ m/244 × 10⁵ m/s
T = 13660π × 10³ m/244 × 10⁵ m/s
T = 41971.68 × 10³ m/244 × 10⁵ m/s
T = 17201.5 × 10³ s
T = 1.72015 × 10⁷ s
T ≅ 1.72 × 10⁷ s
So, the period of the satellite is 1.72 × 10⁷ s
What is the radial acceleration of the satellite?The radial acceleration of the satellite is given by a = v²/R where
v = speed of satellite = 244 m/s and R = radius of orbit = 6680 kmSubstituting the values of the variables into the equation, we have
a = v²/R
a = (244 m/s)²/6680 km
a = 59536 m²/s²/6680 × 10³ m
a = 8.913 × 10³ rad/s
a = 8913 rad/s
So, the radial acceleration of the satellite is 8913 rad/s
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a cliff diver steps off a cliff and hits the water 2.2 seconds later. How high was the cliff?
The height of the cliff is 24.2 m.
Let:
'h' be the height of the cliff'u' be the initial velocity of diver; u = 0m/s'+'ve been the annotation for the Downward directiona = 10m/s^2 be the approximation of acceleration due to gravityUsing the Second Equation of Motion:
[tex]h = ut + 1/2 at^{2} \\h = 0 + 1/2 (10)(2.2)^{2} \\h = 0 + 5(2.2)^{2} \\h = 5(4.84)\\h = 24.2m[/tex]
The Second Equation of Motion establishes a relationship between displacement and time. It can be used to calculate the displacement and time, in different conditions.
The cliff was 24.2m high above the water surface, into which the cliff diver dived.
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A car is driving with a constant velocity around a circular track. The track’s radius is 156.1 m, and the car returns to its starting position in 49.3 s. What is the car’s centripetal acceleration?
A car is driving with a constant velocity around a circular track. The track’s radius is 156.1 m, and the car returns to its starting position in
49.3 s the car’s centripetal acceleration is 0.327 m/sec².
What is acceleration?Rate of change of velocity with respect to time is called acceleration.
when the magnitude of velocity is constant and direction is changing
than the acceleration is centripetal acceleration.
Given that is question radius of track is 156.1 m and car returns to its starting position in 49.3 s.
Velocity of car = Circumference of Circle / total time
= 352.308/ 49.30
v = 7.14 m/sec
Centripetal acceleration = v²/r
= 51.06/156.1
= 0.327 m/sec²
The car’s centripetal acceleration is 0.327 m/sec².
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a 68.1-kg boy is surfing and catches a wave which gives him an initial speed of 1.60 m/s. he then drops through a height of 1.56 m, and ends with a speed of 8.51 m/s. how much nonconservative work (in kj) was done on the boy?
The non-conservative work done on the boy is 1.337 kJ
Given that,
Mass of boy, m = 68.1 kg
Initial speed of boy, u = 1,60 m/s
The boy then drops through a height of 1.56 m
Final speed of boy, v = 8.51 m/s
Now, we have to find the non-conservative work was done on the boy.
The Work done by the non-conservative forces is equal to the sum of total change in kinetic energy and total change in potential energy.
[tex]W=\frac{1}{2}m (v^{2}-u^{2})+(0-mgh)[/tex]
[tex]W=\frac{1}{2}m (v^{2}-u^{2})-mgh[/tex]
[tex]W=\frac{1}{2}(68.1) ((8.51)^{2}-(1.60)^{2})-68.1*9.8*1.56[/tex]
W = 1337.62 Joules
or
W = 1.337 kJ
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The longest shish kebab ever made was 881.0m long. Suppose the meat and veggies need to be delivered in a cart from one end of this shish kebab skewer to the other end. A cook pulls the cart by applying a force of friction acting on the cart is 28.00N, what is the net work done on the cart and it’s content during the delivery.
Answer:
Explanation:
Given:
L = 881.0 m
F = 28.0 N
___________
A - ?
The work :
A = L*F = 881.0 * 28.0 ≈ 24 670 J
A horizontal rope is attached from a truck to a 1475-kg car. As the truck tows the car on a horizontal straight road, the rope will break if the tension is greater than 2551 N. Ignoring friction, what is the maximum possible acceleration of the truck if the rope does not break?
Given data
*The given mass of the car is m = 1475 kg
*The maximum tension is T = 2551 N
The formula for the maximum possible acceleration of the truck is given by Newton's second law as
[tex]\begin{gathered} T=ma_{\max } \\ a_{\max }=\frac{T}{m} \end{gathered}[/tex]Substitute the known values in the above expression as
[tex]\begin{gathered} a_{\max }=\frac{2551}{1475} \\ =1.72m/s^2 \end{gathered}[/tex]Hence, the maximum possible acceleration of the truck is a_max = 1.72 m/s^2
the barrel of a shotgun: group of answer choices is generally shorter than that of a rifle. is wider at the end to concentrate shot. is smooth without the grooves and lands found in rifles. is indistinguishable from that of a rifle.
Measures the diameter of a shotgun's barrel in terms of the quantity of lead balls needed to make the bore weigh one pound (12 gauge is the diameter of a lead ball weighing one-twelfth of a pound).
What differentiates a shotgun from a rifle?A shotgun has many projectile properties while a rifle only comes with one. Shotguns are helpful at close ranges when rifles are useful at a distance. Shotguns only have a front sight, whereas rifles have both front and rear sights.
Rifles are typically a little noisier than shotguns. Analyzing test results reveals that shotguns and rifles have decibel levels that are comparable. The typical decibel range
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the radius of the planet mercury is 2.43*10^6m and its mass is 3.2*10^23 kg. find the period of a satellite orbiting mercury 265,000 m above its surface?
The period of a satellite orbiting mercury 265,000 m above its surface is 100 minutes if the radius of the planet mercury is 2.43*10^6m and its mass is 3.2*10^23 kg.
v = √ G M / r
r = R + d
v = Orbital velocity
G = Gravitational constant
M = Mass of planet
r = Radius of the orbit
R = Radius of planet ( mercury )
d = Distance from surface to satellite
m = 3.2 * [tex]10^{23}[/tex] kg
R = 2.43 * [tex]10^{6}[/tex] m
d = 265000 m
G = 6.67 * [tex]10^{-11}[/tex] N m² / kg²
r = R + d
r = 2.43 * [tex]10^{6}[/tex] + 265000
r = 2.695 * [tex]10^{6}[/tex] m
v = √ ( 6.67 * [tex]10^{-11}[/tex] * 3.2 * [tex]10^{23}[/tex] ) / 2.695 * [tex]10^{6}[/tex]
v = √ 7.92 * [tex]10^{6}[/tex]
v = 2.81 * 10³ m / s
v = 2.81 km / s
v = 2 π r / T
T = Time period
T = 2 * 3.14 * 2.695 * [tex]10^{6}[/tex] / 2.81 * 10³
T = 6.02 * 10³ s
T = 100 minutes
Therefore, the period of a satellite orbiting mercury is 100 minutes
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you have two solid metal sphere that are identical except thatr one has a radius exactly twice the other(with the same density). what is the ratio of their moment of inertia?
The elasticity and Poisson's ratio for this metal is 0.232.
What is ratio?
The realation between two numbers which shows how much bigger one quantity is than another.
Sol-
As per the given question
P=190KN
d=16 mm
Lo=50mm
X=50.1349-50=0.1349mm
Y=15.99-16=-0.01mm
The formula-
E=ó/€
Ó=P/A
A=r/4 d^2 =π/4(16)^2=201.062 mm
ó={190(1000)}201.062=944.982 Mpa
E=944.982/0.002698=350.253 GPa
€y=-0.000625
v=0.232(answer)
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Air pressure increases as you travel higher above sea level. This is the reason that cabins in commercial airliners require pressurization.
True
False
The answer is true about the cabins in commercial airliners that require pressurization.
Why are the cabins of commercial airplanes pressurized?Airplanes are pressurized because the air is very thin at the high altitude where they fly. The passenger jet has a cruising altitude of about 30,000 - 40,000 feet. At this altitude or height, humans can't breathe very well and our body gets less amount of oxygen. Most aircraft cabins are pressurized to an altitude about 8,000 feet. This is called cabin altitude. Aircraft pilots have access to the control's mode of a cabin pressure control system and if needed it can command the cabin to depressurize.
So we can conclude that cabins in commercial airliners require pressurization because of the greater pressure of the surrounding environment.
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an electron moves from one point a to point b. while moving, the electron remains on an equipotential surface created by nearby charges. what can you say about the speed of the electron as it moved?
At initial speed must the quarter back through the ball for it to reach the reciver.
What is initial speed
The initial speed is the velocity at which motion start.
Sol-
The horizontal velocity is Vi*cos38.6and the vertical velocity is Vi*sin38.6
horizontal
28.4=Vi*cos38.6 time find time in terms of Vi.
Vertical
hf=hi+Vi*sin38.6*t -1/2 g t^2
now hf=hi=0
so
0=Vi*sin 38.6-1/2 g t and put the expression you got for t, then solve for Vi. You will encounter the need for one trig identity.
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a clod of dirt falls from the bed of a moving truck. it strikes the ground directly below the end of the truck. what is the direction of its velocity relative to the truck just before it hits? is this the same as the direction of its velocity relative to ground just before it hits? explain your answers.
The direction of the velocity of the clod of dirt relative to the truck just before it hits is in the same direction as that of the truck.
The direction of the velocity of the clod of dirt relative to the ground just before it hits is in the opposite direction as that of the ground.
What is relative velocity?The relative velocity of an object in motion is the velocity of that object when viewed from a reference point.
The direction of the velocity of the clod of dirt before it hits the ground is in the same direction as that of the truck.
However, compared to the ground, the direction of the relative velocity is in the opposite direction.
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6. What is the acceleration of a 15 kg object being pushed with 200 N of force?
The acceleration of the object will be 13.33 m/s^2.
According to Newton's second law of motion,
The magnitude of Force is equal to the product of the mass of the body and the acceleration it is given ( or achieves ).
The units of force, mass and acceleration are Newton, kg and m/s^2 respectively.
In numerical terms,
Force = mass × acceleration
i.e. F = m × a
Given,
mass of the body = 15 kg
Force applied to the body = 200 N.
Putting these values in the equation,
200 = 15 × a
=> a = 200/15
=> a = 13.33 m/s^2.
Therefore, the acceleration of the given body will be 13.33 m/s^2.
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A 10 kg box is pulled across a level floor, where the coefficient of kinetic friction is
0.35. What horizontal force is required for an acceleration of 2.0 m/s2?
A 10 kg box is pulled across a level floor, where the coefficient of kinetic friction is 0.35. The horizontal force is required for an acceleration of 2.0 m/s2 will be 54.3 N
Newton’s second law states that the acceleration of an object depends upon two variables – the net force acting on the object and the mass of the object. The acceleration of the body is directly proportional to the net force acting on the body and inversely proportional to the mass of the body.
equation of motion will be
F(net) = mass * acceleration
F - fr = m * a
where
F = force by which the box is moving
fr = force due to friction
m = mass of the object
a = acceleration by the object is moving
F = fr + m * a
= (mu * N) + ma where N = mg
= (mu * mg) + ma
= (0.35 * 10 * 9.8) + 10 * 2
= 54.3 N
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this si a 2 part question84) A shock absorber is designed to quickly damp out the oscillations that a car would otherwise make because it is suspended on springs. (a) Find the period of oscillation of a 1610-kg car that is suspended by springs that make an effective force constant of 5.75×104 N/m. (b) Find the damping constant b that will reduce the amplitude of oscillations of this car by a factor of 5.00 within a time equal to half the period of oscillation.
Given data
*The given mass of the car is m = 1610 kg
*The given effective force constant is k = 5.75 × 10^4 N/m
(a)
The formula for the period of oscillation of a 1610 kg car is given as
[tex]T=2\pi\sqrt[]{\frac{m}{k}}[/tex]Substitute the known values in the above expression as
[tex]\begin{gathered} T=2\times3.14\times\sqrt[]{\frac{1610}{5.75\times10^4}} \\ =1.05\text{ s} \end{gathered}[/tex]Hence, the time period of oscillation of a 1610 kg car is T = 1.05 s
(b)
As from the given data, the amplitude of the oscillation of the car decreases by a factor of 5.00. Then, the expression for the amplitude of the oscillation, and the damping constant (b) is calculated as
[tex]A=A_0e^{-\frac{bt}{2m}}[/tex]Substitute the known values in the above expression as
[tex]\begin{gathered} \frac{A_0}{5.0}=A_0e^{-\frac{bt}{2m}} \\ bt=2m\ln (5.0)_{} \\ b(\frac{T}{2})=2m\ln (5.0) \\ b=\frac{4m\ln (5.0)}{T} \\ =\frac{4\times1610\times\ln (5.0)}{1.05} \\ =9871.2\text{ kg/s} \end{gathered}[/tex]Hence, the damping constant is b = 9871.2 kg/s
A freight train moves due north with a speed of 8 m/s. The mass of the train is 6 x 106 kg. How fast would a 1500 kg automobile have to be moving due north to have the same momentum as the train
The freight train has a momentum of mv = (4.5 × 105 kg)(1.4 m/s) = 6.3 ×105 kg-m/s.
An automobile with a mass of 1800 kg would need to go at a speed of 6.3 × 105 kg-m/s ÷ 1800 kg = 350 m/s in order to have a momentum of the same magnitude. (Roughly 800 mph)
A moving body's amount of motion is referred to as momentum. It has units of kg m/s and is theoretically represented as p = m * v.
The momentum equation connects the total force exerted on a fluid element to its acceleration or rate of change of momentum and is a formulation of Newton's Second Law. You're certainly familiar with the formula F = ma, which links applied force and acceleration in solid mechanics analysis.
Momentum is the result of a particle's mass and velocity. Being a vector quantity, momentum possesses both magnitude and direction. According to Isaac Newton's second equation of motion, the force applied on a particle is equal to the time rate of change of momentum.
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a cubical box of mass 2kg with each side 2 cm is lying on the ground. Calculate the pressure exerted by the block on the ground
Answer:
25000Nm⁻² using rounded g figure
Explanation:
The box is in cm. It needs to be converted into SI metres.
1m = 100cm
1cm = ( 1 / 100 )m
2cm = ( 1 / 50 )m
A = ( 1 / 2500 )m²
Assume that g = 10ms⁻¹. Use your own definition.
F = 10N
P = F / A
P = 10N( 2500 )m⁻²
25000Nm⁻²
Answer:
Explanation:
Given:
m = 2 kg
a = 2 cm = 0.02 m
g =9.8 m/s²
_______________
P - ?
Area of the cube:
S = a² = 0.02² = 0,0004 m²
Gravity:
F = m·g = 2·9.8 = 19.6 N
Pressure:
p = F / S = 19.6 / 0,0004 = 49 000 N/m²
Jose applies a force of 20 N North, Johnny applies a force of 10 N West and Janie applies a force of 25 N south on a heavy dresser. Find the resultant force on the dresser.
Answer:
Resultant force of the given forces: approximately [tex]11.2\; {\rm N}[/tex] at approximately [tex]53^{\circ}[/tex] west of south.
Explanation:
The resultant force of these forces will be:
[tex]25\; {\rm N} - 20\; {\rm N} = 5\; {\rm N}[/tex] to the south (the [tex]20\; {\rm N}[/tex] force to the north partially balances the [tex]25\; {\rm N}[/tex] force to the south), and[tex]10\; {\rm N}[/tex] to the west.Refer to the diagram attached. The resultant [tex]5\; {\rm N}[/tex] force to the south and the [tex]10\; {\rm N}[/tex] force to the west are perpendicular to each other, forming the two legs of a right triangle. The hypotenuse of this right triangle will be the net effect of these two forces.
Apply Pythagorean's Theorem on this triangle to find the magnitude of this net effect:
[tex]\begin{aligned}(\text{length of hypotenuse}) &= \sqrt{10^{2} + 5^{2}} \approx 11.2\end{aligned}[/tex].
Hence, the magnitude of this net effect will be approximately [tex]11.2\; {\rm N}[/tex].
Let [tex]\theta[/tex] denote the angle between this resultant force and west. In this right triangle:
[tex]\begin{aligned} \tan(\theta) &= \frac{(\text{opposite})}{(\text{adjacent})} \\ &= \frac{5\; {\rm N}}{10\; {\rm N}} \\ &= \frac{1}{2}\end{aligned}[/tex].
[tex]\begin{aligned} \theta &= \arctan\left(\frac{1}{2}\right) \approx 27^{\circ}\end{aligned}[/tex].
Hence, the angle between this resultant force and south will be approximately [tex](90^{\circ} - 27^{\circ}) = 63^{\circ}[/tex]. This resultant force will be at approximately [tex]63^{\circ}[/tex] west of south.
Isaac is practicing his volleyball skills by volleying a ball straight up and down, over and over again. His teammate
Marie notices that after one volley, the ball rises 3.6 m above Isaac's hands. What is the speed with which the ball left
Isaac's hand? (8.4 m/s)
Answer:
the velocity of a ball at highest point is 8.5m/s
Explanation:
: The Freefall Problem
Xue,is standing on the roof of a building. Emily is standing below and tosses a ball straight upwards to Xue. It travels up past
Xue, comes back down and Xue reaches out and catches it. Xue catches the ball 6.0 m above Emily’s hands. The ball was
travelling at 12.0 m/s upwards the moment it left Emily’s hand. We would like to know how much time this trip takes.
1. Represent. Complete part A below. Indicate the y-origin for position measurements and draw a sign convention where
upwards is positive. Label the important events.
2. Represent. Complete part C below. Make sure the two graphs line-up vertically. Draw a single dotted vertical line
through the graphs indicating the moment when the ball is at its highest.
A: Pictorial Representation
Sketch, coordinate system, label givens, conversions, describe events
Event 1:
Event 2:
Event 3:
C: Physics Representation
Motion diagram, motion graphs, key events
The total length of the path traveled by an object is the distance. The change in position, from one event to another is the
displacement. Distance is a scalar quantity and displacement is a vector quantity.
3. Reason. Explain why this in this example it is relatively easy to find the displacement, but harder to find the distance.
4. Reason. The BIG 5 equations are valid for any interval of motion where the acceleration is uniform. Does the ball
accelerate uniformly during events 1 and 3? Explain D: Mathematical Representation
Describe steps, complete equations, substitutions with final statement isThere are multiple ways of solving…one
will require the use of the quadratic formula.
The other way will take additional steps.
x = -b+-under root b^2-4ac/2a
how would an increase in force affect the mechanical advantage of a simple machine
Mechanical advantage is a measure of the ratio of the output force produced to the input force applied in a system, mainly in the case of simple machines, such as levers and pulleys. It is a dimensionless quantity.
From the definition, the mechanical advantage of a simple machine is inversely proportional to the input force applied on the system. So, an increase in this force decreases the mechanical advantage of a simple machine.
For example, the mechanical advantage of a lever can be decreased by increasing the force on load arm or by reducing the force on the effort arm.
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assume that you weight yourself on a bathroom scale. would the springs inside a bathroom scale be more compressed or less compressed if you weighed yourself in an elevator that was accelerating upward?
The springs inside a bathroom scale will be more compressed.
What is acceleration due to gravity?
In physics, gravitational acceleration is the acceleration of an object in free fall in a vacuum (hence no resistance). This is a steady increase in velocity caused solely by the gravitational pull. All objects accelerate at the same speed in a vacuum, regardless of their mass or composition . The measurement and analysis of these velocities is known as gravimetry.
At a fixed point on the surface, the magnitude of the Earth's gravity is the combined effect of gravity and the centrifugal force due to the rotation of the Earth. Free-fall acceleration at various points on the Earth's surface ranges from 9.764 to 9.834 m/s2 (32.03 to 32.26 ft /s2) depending on altitude, latitude and longitude . The old standard is precisely defined as 9.80665 m/s2 (32.1740 ft/s2). Locations that deviate significantly from this value are called gravity anomalies. Other effects such as lift and air resistance are not taken into account.
When the elevator accelerates and shows a higher weight reading. As the lift accelerates and descends, the springs in the bathroom scale are compressed less and the displayed weight value is reduced.
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