The cannon on a battleship can fire a shell a maximum distance of 36.0 km.
(a) Calculate the initial velocity of the shell.
m/s

(b) What maximum height does it reach? (At its highest, the shell is above a substantial part of the atmosphere--but air resistance is not really negligible as assumed to make this problem easier.)
m

(c) The ocean is not flat, since the earth is curved. How many meters lower will its surface be 36.0 km from the ship along a horizontal line parallel to the surface at the ship?
mDoes your answer imply that error introduced by the assumption of a flat earth in projectile motion is significant here? (Select all that apply.)
The error is insignificant compared to the distance of travel.
The error is significant compared to the distance of travel.
The error could be significant compared to the size of a target.
The error is insignificant compared to the size of a target.

Answers

Answer 1

(a)The initial velocity of the shell will be 594.27 m/sec

(b)The maximum height it reaches will be 9000 m.

c)101.249 m meters lower will its surface be 36.0 km from the ship along a horizontal line parallel to the surface of the ship.

d)The error could be significant compared to the size of a target. Option C is correct.

What is velocity?

The change of distance with respect to time is defined as speed. Speed is a scalar quantity. It is a time-based component. Its unit is m/sec.

The given data in the problem is

m is the mass of the block = Kg.

u is the initial velocity of fall = m/sec

h is the distance of fall =  m

g is the acceleration of free fall = m/sec²

v is the hitting velocity of =?

a)

The range of the projectile is;

[tex]\rm R = \frac{u^2 sin 2 \theta }{g} \\\\ 36 \times 10^ 3 = \frac{U^2 sin 45^0}{9.81} \\\\ U= 594.27 \ m/sec[/tex]

b)

The maximum height of the projectile is;

[tex]\rm H = \frac{u^2 sin 2 \theta }{2g} \\\\ H = \frac{(594.27)^2\times (sin 45)^2}{2 \times 9.81 } \\\\ H = 9000 \ m[/tex]

c)

The distance between its surface and the ship, measured in a horizontal arc parallel to the surface, will be 36.0 kilometers. The distance from the lower surface is found as;

[tex]\rm( R_e + h)^2 = R_e^2+(36)^2 \\\\ (R_e)^2 = h^2+2R_e h= R_e^2 + 12196 \\\\ h^2 + 12800 h - 1296 = 0 \\\\ h = 101.249 \ m[/tex]

d)

An error is a mistaken or erroneous action. In some contexts, an error is interchangeable with a mistake.

The difference between the calculated value and the original value is known as the error. The inaccuracy may be large in comparison to the target's size. Option C is correct.

Hence the initial velocity of the shell, maximum height, and the distance from the lower surface will be 594.27 m/sec,9000 m, and 101.249 m and option c for question d are correct respectively.

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The Cannon On A Battleship Can Fire A Shell A Maximum Distance Of 36.0 Km.(a) Calculate The Initial Velocity

Related Questions

The cost of gas is $1.27 per liter. How much does a gallon of gas costs?

Answers

it will be at least 30.00

The cost of gas is $1.27 per liter and the gallon of gas costs $4.80. where 1 gallon is equal to 3.78541 liters.

To convert the cost of gas from dollars per liter to dollars per gallon, The conversion factor between liters and gallons.

1 gallon is equal to 3.78541 liters (the exact conversion factor is 3.785411784).

Given the cost of gas is $1.27 per liter, let's calculate the cost per gallon:

Cost per gallon = Cost per liter × Liters per gallon

Cost per gallon = $1.27 × 3.78541

Cost per gallon = $4.80

So, a gallon of gas costs $4.80.

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A long uniform board weighs 52.8 N (10.6 lbs) rests on a support at its mid point. Two children weighing 206.0 N (41.2 lbs) and 272.0 N (54.4 lbs) stand on the board so that the board is balanced.
What is the upward force exerted on the board by the support?

Answers

The upward force exerted on the board by the support is 530.8 N.

Upward force exerted on the board by the support

The sum of the upward forces is equal to sum of downward forces;

total downward forces = 52.8 N + 206 N + 272 N = 530.8 N

downward force = upward force = 530.8 N

Thus, the upward force exerted on the board by the support is 530.8 N.

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The radius of curvature of a highway exit is r = 93.5 m. The surface of the exit road is horizontal, not banked. (See figure.)
If the coefficient of static friction between the tires of the car and the surface of the road is μ[tex]_{s}[/tex] = 0.402, then what is the maximum speed at which the car can safely exit the highway without sliding?

Answers

Static friction keeps the car from skidding off the road and points toward the center of the curve. By Newton's second law, the car experiences

• net vertical force

F [normal] - F [weight] = 0

• net horizontal force

F [friction] = ma = mv²/r

where v is the tangential speed of the car.

It follows that

F [normal] = F [weight] = mg

and when static friction is maximized at the car's maximum speed,

F [friction] = µ F[normal] = 0.402 mg

Solve for v :

0.402 mg = mv²/r   ⇒   v = √(0.402 g (93.5 m)) ≈ 19.2 m/s

In which part of the ear is the sound wave converted into an electrical impulse?

Answers

The Cochlea is filled with a fluid that moves in response to the vibrations from the oval window. As the fluid moves, 25,000 nerve endings are set into motion. These nerve endings transform the vibrations into electrical impulses that then travel along the eighth cranial nerve (auditory nerve) to the brain.

16. 1 800 cm³ of fresh water of density
1 000 kgm is mixed with
2 200 cm³ of sea water of density
1 025 kgm. Calculate the density of
the mixture.

Answers

The density of the mixture is determined as 1,013.75 kg/m³.

Density of the mixture

The density of the mixture is calculated as follows;

density = total mass of the mixture / total volume of the mixture

mass of the fresh water = density x volume

                                        = 1000 kg/m³ x 0.0018 m³

                                        = 1.8 kg

mass of the sea water = 1025 kg/m³ x 0.0022 m³

                                     = 2.255 kg

Total mass = 1.8 kg + 2.255 kg = 4.055 kg

Total volume = 0.0018 m³  + 0.0022 m³ = 0.004 m³

Density = 4.055 kg/0.004 m³ = 1,013.75 kg/m³

Thus, the density of the mixture is determined as 1,013.75 kg/m³.

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A long uniform board weighs 52.8 N (10.6 lbs) rests on a support at its mid point. Two children weighing 206.0 N (41.2 lbs) and 272.0 N (54.4 lbs) stand on the board so that the board is balanced.
What is the upward force exerted on the board by the support?
764.8 N is incorrect.

Answers

The upward force exerted on the board by the support is 530.8 N.

Upward force exerted on the board by the support

The sum of the upward forces is equal to sum of downward forces;

total downward forces = 52.8 N + 206 N + 272 N = 530.8 N

downward force = upward force = 530.8 N

Thus, the upward force exerted on the board by the support is 530.8 N.

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Many immigrants lived in​

Answers

California I think not sure

A car moves with an average speed of 75 kmh^-1 from town P to town Q in 2 hours. By using information, you may calculate the distance between two towns. convert the value 75kmh ^-1 to S.I unit​

Answers

Distance = Speed * Time,
D = (75)(2),
D = 150 km.

The distance car moves with an average speed of 75 km/hr from town P to town Q in 2 hours is 149,760 m.

Given:
Speed = 75 km/hr

Time = 2 hr

The distance can be calculated from the product of velocity and time.

The standard unit of distance is a meter.

Convert speed and time into meters per second and seconds:

v = 75×5/18

v = 20.8 m/s

t = 3600×2

t = 7200 s

The distance is computed as:

v = d/t

d = vt

d = 20.8×7200

d = 149,760 m

Hence, the distance car moves with an average speed of 75 km/hr from town P to town Q in 2 hours is 149,760 m.

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A
В
C
D
At which point will an image be formed?
ОА
Ов
Ос
OD

Answers

Answer:

oa

Explanation:

it may be oa is the right answer for this question

but I don't know properly

A cannonball is shot horizontally off a
10.5 m high castle wall at 47.4 m/s. How
far from the base of the wall does the
cannonball land?

Answers

Answer:it's 132.2m

Explanation:

Hello !

Because first recall parabolic peak height equation and

given that v=47.4m/s , h=10.5m

v²sin²(θ⁰)=2ℎ₀(47.4)²²(θ⁰)=2(9.81)(10.5) θ=17.63

secondly recall parabolic total distance equation

d₀=²(2θ⁰)/gd₀=(47.4)²(2(17.63)⁰)/9.81d₀=132.2m

The distance from the base of the wall that the cannonball land will be 138.7 m.

What is the peak height equation?

Consider that a body of mass 'm' is projected with a velocity of projection 'v' and angle of projection 'θ'. Assume that 'h' be the maximum height attained by the projectile.

2gh = v²sin²θ

This expression can be used to calculate the maximum height of the projectile.

Given, the height of the castle wall, h = 10.5 m

The velocity of the cannonball, v = 47.4m/s

The peak height equation can be used to find the angle θ:

v²sin²θ = 2gh

(47.4)² sin²θ = 2(9.81)(10.5)

Sinθ = 0.3028

θ = 17.63°

Therefore, the distance (d) from the base of the wall that the cannonball land:

d =  (47.4)² × 2 sin(17.63) / 9.81

d = 138.7m

Therefore the distance from the base of the wall that the cannonball land is 138.7m.

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Two football players collide head-on in midair while chasing a pass. The first player has a 101.0 kg mass and an initial velocity of 4.10 m/s, while the second player has a 117 kg mass and initial velocity of -6 m/s. What is their velocity (in m/s) just after impact if they cling together? (Indicate the direction with the sign of your answer.)

Answers

Assuming players stick together immediately after the collision, their speed will be 1.3206 m/sec, and - ve indicates their direction.

What is the law of conservation of momentum?

According to the law of conservation of momentum, the momentum of the body before the collision is always equal to the momentum of the body after the collision.

The given data in the problem is;

(m₁) is the mass of 1st player= 101.0 kg

(u₁) is the initial velocity of 1st player (u₁) = 4.10 m/s

(m₂) is the mass of 2nd player = 117 kg

(u₂) is the initial velocity of 2nd player= -6  m/s

(v) is the velocity after collision =.?

According to the law of conservation of momentum;

Momentum before collision =Momentum after collision

m₁u₁+m₂u₂=(m₁+m₂)V

101.0 kg × 4.10 m/sec + 117 kg × (-6 m/sec) = (101.0 kg + 117 kg) × V

V= - 1.3206 m/sec

Hence, the velocity of players just after impact, if they cling together, will be - 1.3206 m/sec, and - ve shows the direction.

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An AC voltage source is connected to a resistor R = 1.90 102 Ω. The output from an AC voltage source is given by the expression V = (2.20 102 V) sin 2ft. (a) What is the rms voltage across the resistor? V (b) What is the rms current flowing through the resistor? A

Answers

Answer:

(a) The rms voltage across the resistor is given by V = (2.20 102 V) sin 2ft.

V = (2.20 102 V) sin 2ft

V = (2.20 102 V) (0.707)

V = 1.56 102 V

(b) The rms current flowing through the resistor is given by I = V/R.

I = V/R

I = (1.56 102 V)/(1.90 102 Ω)

I = 8.21 A

Review procedures and confirm results to avoid possible errors.

Explanation:

At a circus, a human cannonball is shot from a cannon at 15m/s at an angle of 40 degrees above horizontal. She leaves the cannon at 2m above ground and lands in a net 1m above the ground. At what height does a human cannonball reach?

Answers

The maximum height reached by the human cannonball is 4.74 m.

Height reached by the human cannonball

The height reached is calculated as follows;

H = u²sin²θ/2g

where;

u is the initial velocityθ is the angle of projection

H = (15² x [sin(40)]² )/(2 x 9.8)

H = 4.74 m

Thus, the maximum height reached by the human cannonball is 4.74 m.

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A wooden baseball bat and an aluminum baseball bat have the exact same size, shape, and mass. Aluminum is much denser than wood. Explain how the two bats could be the same size, shape, and mass.

Answers

Due to density, the two bats have the same size, shape, and mass.

How the two bats could be the same size, shape, and mass?

The two bats could be the same size, shape, and mass if they have the same density because density is directly proportional to mass so if the density same, the mass will also be the same.

So we can conclude that due to density, the two bats have the same size, shape, and mass.

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Lesson 2 History of Physical Science
Write an expository essay explaining how science builds on itself. Use at least two specific examples from this lesson.

Answers

The prompt above requires an explanatory essay. The focus of an explanatory essay is to give clarity to a concept. See the sample below.

What is the explanatory essay as per the prompt above?

From history, it is clear that Science builds on itself. For avoidance of doubt, this means that improvements in science are incremental with each improvement feeding off the last.

From the lesson we have examples of this phenomena such as that of Galileo. Recall that the text indicates that He improved the telescope; and while carrying out the same research that his predecessor Copernicus had done in relation to the stars and planets, Galileo used math to justify his position.

It is also on record that Johannes Kepler who lived in the same period as Galileo took interest in Astronomy. In this case, science built on itself because, Kepler's work served as the foundation for Sir Isaac Newton's work. This is also another example of how science builds on itself.

From the above textual evidence, it is given beyond reasonable double that science indeed builds on itself with the discovery of each generation serving as the foundation for the advancement of the next.

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What net external force is exerted on a 1350-kg artillery shell fired from a battleship if the shell is accelerated at 2.00 ✕ 104 m/s2? (Enter the magnitude.)
N

What is the magnitude of the force exerted on the ship by the artillery shell?
N

Answers

The magnitude of the force exerted on the ship by the artillery shell will be 27× 10⁶ N.

What is force?

Force is defined as the push or pulls applied to the body. Sometimes it is used to change the shape, size, and direction of the body.

Force is defined as the product of mass and acceleration. Its unit is Newton.

Given data;

Force,F = ?

Mass,m = 1350-kg

Acceleration,a = 2.00 ✕ 10⁴ m/s²

The magnitude of the force exerted on the ship by the artillery shell is found as;

F=ma

F=1350-kg × 2.00 ✕ 10⁴ m/s²

F= 10,000 N

The magnitude of the force exerted on the ship by the artillery shell will be 27× 10⁶ N.

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a= 3t^3 + 2t^2 find
1) velocity at t= 2s.
2) displacement at 4s
please answer this question ASAP ​

Answers

Use the fundamental theorem of calculus.

[tex]f(t) = f(0) + \displaystyle \int_0^t f'(u) \, du[/tex]

In order to find velocity and position exactly, you need to know the initial velocity and position.

1) Integrate the acceleration function to get the velocity function. By the FTC,

[tex]v(t) = v(0) + \displaystyle \int_0^t a(u) \, du[/tex]

[tex]v(t) = v(0) + \displaystyle \int_0^t (3u^3+2u^2) \, du[/tex]

[tex]v(t) = \dfrac34 t^4 + \dfrac23 t^3 + v(0)[/tex]

At [tex]t=2\,\rm s[/tex], the velocity is

[tex]v(2) = \dfrac34 2^4 + \dfrac23 2^3 + v(0) = \boxed{\dfrac{52}3 + v(0)}[/tex]

2) Integrate the velocity function to the get the position function. By the FTC again,

[tex]x(t) = x(0) + \displaystyle \int_0^t v(u) \, du[/tex]

[tex]x(t) = x(0) + \displaystyle \int_0^t \left(\frac34 u^4 + \frac23 u^3 + v(0)\right) \, du[/tex]

[tex]x(t) = \displaystyle \frac3{20} t^5 + \frac16 t^4 + v(0) t + x(0)[/tex]

At [tex]t=4\,\rm s[/tex], the object's position is

[tex]x(4) = \dfrac3{20}4^5 + \dfrac16 4^4 + 4v(0) + x(0) = \dfrac{2944}{15} + 4v(0) + x(0)[/tex]

Fortunately, you don't need the initial position to find the displacement, since [tex]x(0)[/tex] cancels out in the end.

[tex]\Delta x = x(4) - x(0) = \boxed{\dfrac{2944}{15} + 4v(0)}[/tex]

M, a solid cylinder (M=2.15 kg, R=0.111 m) pivots on a thin, fixed, frictionless bearing. A string wrapped around the cylinder pulls downward with a force F which equals the weight of a 0.730 kg mass, i.e., F = 7.161 N. If instead of the force F an actual mass m = 0.730 kg is hung from the string, find the angular acceleration of the cylinder.

Answers

The angular acceleration of the cylinder is mathematically given as

[tex]\alpha = 60.2175 rad/s^2[/tex]

What is the angular acceleration of the cylinder.?

Question parameters

M=2.15kgR = R=0mW= 7.161 N

Generally, the equation for Torque generated in the cylinder is  mathematically given as

[tex]\sigma[/tex] = w×r

for circular motion torque is

[tex]\sigma = × \alpha[/tex]

t[tex]=0.5 × m ×r^2 × \alpha\\\\ =0.5 × 2.15 ×0.111 ^2 × \alpha[/tex]

t = 0.0132 ×[tex]\alpha[/tex]

Therefore

0.0132 × [tex]\alpha[/tex] = w × r

[tex]\\\\\alpha =\frac{ 7.161 0.111}{(0.0132)}[/tex]

[tex]\alpha = 60.2175 rad/s^2[/tex]

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Abdou was explaining to a classmate that graphite is a good lubricant
because it is bonded in layers that easily slip over each other. Which image
shows this quality?

Answers

The fact that the layers of graphite are held together by only weak Van der Walls forces implies that they can slide over each other.

Why is graphite a solid lubricant?

We know that graphite is composed of layers. These hexagonal layers are held together by weak Van Der Walls forces and as such are able to slide over each other. The carbon atom in each layer are held together by strong covalent bonds.

The fact that the layers of graphite are held together by only weak Van der Walls forces implies that they can slide over each other and as such make the graphite fluid.

Thus, the image that shows these layers of graphite is attached to this an answer

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Answer:it’s B

Explanation:

A school bus is moving 26.8 m/s on
flat ground when it begins to
accelerate at 4.73 m/s². How much
time does it take to travel 95.1 m?
(Unit=s)
Watch your minus signs!
time (s)
Enter

Answers

The time taken for the bus to travel 95.1 m is 2.8 s

How to determine the final velocityInitial velocity (u) = 26.8 m/sAcceleration (a) = 4.73 m/s²distance (s) = 95.1 mFinal velocity (v) = ?

v² = u² + 2as

v² = 26.8² + (2 × 4.73 × 95.1)

v² = 1617.886

Take the square root of both sides

v² = √1617.886

v = 40.2 m/s

How to determine the timeInitial velocity (u) = 26.8 m/sAcceleration (a) = 4.73 m/s²Final velocity (v) = 40.2 m/sTime (t) = ?

v = u + at

40.2 = 26.8 + (4.73 × t)

Collect like terms

4.73 × t = 40.2 - 26.8

4.73 × t = 13.4

Divide both side by 4.73

t = 13.4 / 4.73

t = 2.8 s

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which of the following examples is not considered a pure substance?
Answer asap

Answers

Answer:

Fruit Punch

Explanation:

A pure substance is whereby there is only one type of element or a compound in the periodic table in the substance.

Carbon: Well its just C in the periodic table, so this is definitely pure.

Water Molecule: What makes water? H2O right? Contains Hydrogen and Oxygen, and as we all know H2O is a compound, therefore this is a pure substance.

Fruit Punch: What makes fruit punch, water and fruits. Fruits may contain citric acid(a compound itself), and is mixed with water with already has a compound, so having 2 different compounds will result in a mixture and therefore it will not be pure.

Discuss the causes, characteristics, and major features of black holes. Explain why the concept of a black hole does or doesn't seem reasonable to you and provide some rationale for your views?

Answers

The concept of a black hole seem reasonable to me because of its valid and authentic evidences.

What are the causes, characteristics, and major features of black holes?

A black hole is a region in space where light is unable to escape due to strong gravitation force. The strong gravity is formed because matter has been pressed into a tiny space. This compression occurs at the end of a star's life. Some black holes are formed when a stars die.

So we can conclude that the concept of a black hole seem reasonable to me because of its valid and authentic evidences.

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21. An AC generator has an rms output voltage of 274 V at a frequency of 57.0 Hz.
The generator is connected across a 0.900-H inductor.
(a) Determine the inductive reactance.

(b) Determine the maximum voltage across the inductor.

(c) Determine the rms and maximum currents in the inductor

Answers

The answers are as follows;

a) the inductive reactance is 322 ohm

b) The maximum voltage is 387.5 V

c) The rms and maximum currents in the inductor are 1.2 A and 0.85 A.

What is the reactance?

The reactance is obtained from;

XL = 2πfL

XL = 2 * 3.14 * 57.0 * 0.900

XL = 322 ohm

The maximum voltage is obtained as;

Vo = Vrms * √2

Vo =  274 V * √2

Vo = 387.5 V

Io = Vo/XL

Io =  387.5 V/ 322 ohm

Io = 1.2 A

Irms = 274 V/322 ohm

Irms = 0.85 A

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M, a solid cylinder (M=2.15 kg, R=0.111 m) pivots on a thin, fixed, frictionless bearing. A string wrapped around the cylinder pulls downward with a force F which equals the weight of a 0.730 kg mass, i.e., F = 7.161 N.
1. Calculate the angular acceleration of the cylinder.
2. If instead of the force F an actual mass m = 0.730 kg is hung from the string, find the angular acceleration of the cylinder.
3. How far does m travel downward between 0.390 s and 0.590 s after the motion begins?
4. The cylinder is changed to one with the same mass and radius, but a different moment of inertia. Starting from rest, the mass now moves a distance 0.450 m in a time of 0.490 s. Find I[tex]_{cm}[/tex] of the new cylinder.

Answers

Hi there!

1.

Let's use the rotational equivalent of Newton's Second Law.

[tex]\Sigma \tau = I\alpha[/tex]

τ = Torque (Nm)

I = Moment of Inertia (1/2mR² for solid cylinder)

α = Angular acceleration (rad/sec²)

The torque is equivalent to:
[tex]\tau = r \times F[/tex]

r = distance from lever arm to pivot point, aka the radius for this cylinder (0.111 m)
F = applied force (7.161 N)

We are already given various values, so let's create a working equation and solve.

[tex]rF = I\alpha \\\\\alpha = \frac{rF}{I}[/tex]
[tex]\alpha = \frac{(0.111)(7.161)}{\frac{1}{2}(2.15)(0.111^2)} = \boxed{60.013 \frac{rad}{sec^2}}[/tex]

2.

So, let's begin by doing a summation of forces acting on the block.

We have its force of gravity downward(+), and the tension from the string upward. (-), assigning signs based on the path of the falling block.

[tex]\Sigma F = F_g - T\\\\\Sigma F = m_b g - T\\\\m_b a = m_b g - T[/tex]

Solving for 'T':

[tex]T = m_b g - m_b a[/tex]

Let's now do a summation of torques on the cylinder. We only have the torque caused by the tension in the string connected to the block.

[tex]\Sigma \tau = rT\\\\I\alpha = r(T = m_b g - m_b a)[/tex]

Using the relationship between angular and translational acceleration:

[tex]a = \alpha r\\\\alpha = \frac{a}{r}[/tex]

And the equation for the moment of inertia:
[tex]I = \frac{1}{2}Mr^2[/tex]

Simplify the expression.

[tex]\frac{1}{2}Mr^2(\frac{a}{r}) = r( m_b g - m_b a)\\\\\frac{1}{2}Ma = ( m_b g - m_b a)[/tex]

Solve for 'a':
[tex]\frac{1}{2}ma + m_b a= m_b g \\\\a = \frac{m_b g}{\frac{1}{2}M + m_b} = \frac{0.730(9.81)}{\frac{1}{2}(2.15) + 0.730} = \boxed{3.97 \frac{m}{s^2}}[/tex]

3.

This is just a case of kinematics since we have a constant acceleration. To solve for this instance, however, let's use calculus.

We know that:
[tex]a(t) = 3.97[/tex]

v(t) is the integral. There is no initial velocity (block starts from rest), so:
[tex]v(t) = 3.97t + C\\C = 0 m/s \\\\v(t) = 3.97t[/tex]

Now, we can take the integral of the velocity/time equation to find the displacement between the interval.

[tex]\int\limits^{0.590}_{0.390} {3.97t} \, dt = \boxed{0.389 m}[/tex]

4.

So, let's now find the translational acceleration produced by the new cylinder. Use the kinematic equation:
[tex]\Delta x = v_0 t + \frac{1}{2}at^2[/tex]

Since the block starts from rest, we can do some simplifying and rearranging:
[tex]2\Delta x = at^2\\\\a = \frac{2\Delta x}{t^2} = \frac{2(0.450)}{0.490^2} = 3.748 \frac{m}{s}^2[/tex]

So, let's go back to our summation of torques equation. This time, however, we cannot simplify 'I'.

[tex]I(\frac{a}{r}) = r( m_b g - m_b a)[/tex]

Solving for 'I':
[tex]I = \frac{r^2( m_b g - m_b a)}{a} = \frac{(0.111^2)(7.161 - .730(3.748))}{3.748}\\\\ = \boxed{0.0145 kgm^2}[/tex]

A Skier starting from rest rolls down a 35.0 m ramp. When he arrives at the bottom of the ramp his
speed is 10.43 m/s. (a) Determine the magnitude of his acceleration, assumed to be constant, (b) If the
ramp is inclined at 60.0 degrees with respect to the ground, what is the component of his acceleration
that is parallel to the ground. (c) Provide me with a free body diagram, one for the 35 m ramp situation.
Also, provide a nice sketch of the skier.

Answers

The acceleration is  1.55 m/s^2 while is component that is parallel to the ground is 0.775  m/s^2.

What is acceleration?

The term acceleration is the rate of change of speed. Given that the distance covered is 35.0 m and the final velocity is 10.43 m/s it the follows that;

v^2 = u^2 + 2as

u = 0 because he started from rest

v^2 = 2as

a = v^2/2s

a = (10.43 m/s)^2/2 * 35.0 m

a = 1.55 m/s^2

The component of this acceleration which is parallel to the ground is;

ax = a sin θ

ax =  1.55 m/s^2 cos 60

ax = 0.775  m/s^2

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What is your design for an efficient commercial vehicle chassis that incorporates as many commercial truck components into its structure as realistically possible? ​

Answers

The design that would be used to make an efficient commercial vehicle chassis that has many realistic truck components is:

StrengthAdequate bending stiffnessMaximum stressDeflection, etc

What is a Chassis Frame?

This refers to the main vehicle structure of a car that bears all the main stress that other components are attached to.

Hence, we can see that to make a good design for an efficient commercial vehicle chassis that has many realistic truck components, we would have to consider various factors such as handling maximum stress, maximum equilateral stress, and deflection.

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In 450 BCE, Greek philosopher Democritus first thought of the existence of tiny particles that compose everything around us. He named them 'atomos', meaning _________.

A. None of these
B. Indivisible
C. Invisible
D. Particle

Answers

B. The meaning of 'atomos' according to Democritus in 450 BCE is indivisible.

The building block of every matter

A Greek philosopher  known as Democritus first thought of the existence of tiny particles that compose everything around us.

Democritus of Abdera, named the building blocks of matter atomos, meaning literally “indivisible.

Thus, the meaning of 'atomos' according to Democritus in 450 BCE is indivisible.

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Three strings, attached to the sides of a rectangular frame, are tied together by a knot as shown in the figure. The magnitude of the tension in the string labeled C is 56.3 N.
Calculate the magnitude of the tension in the string marked A. (You'll need to get the various positions from the graph. The ends of the strings are exactly on one of the tic marks.)

Answers

The magnitude of the tension in the string marked A and B is mathematically given as

A = 52.5 NB = 39.5 N

What is the magnitude of the tension in the string marked A?

Generally, the equation for is mathematically given as

The angle at A is...

[tex]tan\theta = 3/8[/tex]

When below the negative x.

B= tan\phi

[tex]tan\phi = 5/4[/tex]

When below the negative x

C=

[tex]tan \rho = 1/6[/tex]

Hence, considering the Horizontal components

74.9cos(9.46) = A*cos(20.6) + B*cos(51.3)

A = 78.9 - 0.668B

Hence, considering the Vertical components

74.9*sin(9.46) + Asin(20.6) = Bsin(51.3)

40.07 = 1.015B

B = 39.5 N

In conclusion, Sub the value of B is the equation of A

A = 78.9 - 0.668B

Sub

A = 78.9 - 0.668( 39.5 N)

A = 52.5 N

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A stone is dropped from rest into a well. The sound of the splash is heard exactly 2.70 s later. Find the depth of the well if the air temperature is 11.0°C.

Answers

The depth of the well at the given air temperature is 456.3 m.

Depth of the well

The depth of the well is calculated from the principle of echo.

v = 2d/t

2d =  vt

d = vt/2

where;

v is speed of sound in air at  11.0°C = 338 m/s

d = (338 x 2.7)/2

d = 456.3 m

Thus, the depth of the well at the given air temperature is 456.3 m.

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I need help with my homework

Answers

Answer:

A. the center of mass

Explanation:

[tex]C_m[/tex] is usually used to designate the center of mass of an object or system of objects.

B. is incorrect because the combined mass of a system is usually denoted by ∑m.

C. is incorrect because the momentum of a system is denoted by the letter p.

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