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posledela
3 years ago
9

A projectile is launched in the horizontal direction. It travels 2.050 m horizontally while it falls 0.450 m vertically, and it

then strikes the floor. a. How long is the projectile in the air?b. What was the original velocity of the projectile?c. What is the velocity in the horizontal direction of the projectile when it strikes the floor? d. What is its velocity in the vertical direction at this time? e. What is the magnitude of its velocity as it strikes the floor?
Physics
1 answer:
kompoz [17]3 years ago
8 0

Answer:

a) 0.303 s b) 6.77 m/s c) - 2.97  m/s d) 7.39 m/s

Explanation:

horizontal distance travel = 2.050 m, vertical distance travel = 0.45 m

Using equation of linear motion

Sy = Uy t + 1/2 gt² Uy is the inital vertical component of the velocity, t is the time taken for the vertical motion in seconds, and S is the vertical distance traveled, taken downward vertical motion as negative

-0.45 = 0 - 0.5 × 9.81×t²

0.45 / (0.5 × 9.81) = t²

t = √0.0917 = 0.303 s

b) horizontal distance traveled = horizontal velocity (Ux) × t

Ux = horizontal distance traveled / t = 2.050 / 0.303 = 6.77 m/s

c) Velocity in the vertical direction can be calculated using

Vy = Uy - gt where g is negative since initial U is zero

Vy = - 9.81 × 0.303 = - 2.97  m/s

d) the magnitude of the velocity = resultant of the the two velocity

using Pythagoras theorem

the magnitude of the velocity = √ ( 6.77² + 2.97²) = 7.39 m/s

at angle = tan^-1 ( 2.97 / 6.77) =  23.69⁰

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Calculate the force of gravity between planet X and planet y if both planets are 3.75 X 10^11 m apart, planet X has a mass of 1.
GenaCL600 [577]

So, the force of gravity that the asteroid and the planet have on each other approximately \boxed{\sf{2.9 \times 10^{17} \: N}}

<h3>Introduction</h3>

Hi ! Now, I will help to discuss about the gravitational force between two objects. The force of gravity is not affected by the radius of an object, but radius between two object. Moreover, if the object is a planet, the radius of the planet is only to calculate the "gravitational acceleration" on the planet itself,does not determine the gravitational force between the two planets. For the gravitational force between two objects, it can be calculated using the following formula :

\boxed{\sf{\bold{F = G \times \frac{m_1 \times m_2}{r^2}}}}

With the following condition :

  • F = gravitational force (N)
  • G = gravity constant ≈ \sf{6.67 \times 10^{-11}} N.m²/kg²
  • \sf{m_1} = mass of the first object (kg)
  • \sf{m_2} = mass of the second object (kg)
  • r = distance between two objects (m)

<h3>Problem Solving</h3>

We know that :

  • G = gravity constant ≈ \sf{6.67 \times 10^{-11}} N.m²/kg²
  • \sf{m_X} = mass of the planet X = \sf{1.55 \times 10^{22}} kg.
  • \sf{m_Y} = mass of the planet Y = \sf{3.95 \times 10^{28}} kg.
  • r = distance between two objects = \sf{3.75 \times 10^{11}} m.

What was asked :

  • F = gravitational force = ... N

Step by step :

\sf{F = G \times \frac{m_X \times m_Y}{r^2}}

\sf{F = 6.67 \cdot 10^{-11} \times \frac{1.55 \cdot 10^{22} \cdot 3.95 \times 10^{28}}{(3.75 \times 10^{11})^2}}

\sf{F \approx \frac{40.84 \times 10^{-11 + 22 + 28}}{14.0625 \times 10^{22}}}

\sf{F \approx 2.9 \times 10^{39 - 22}}

\sf{F \approx 2.9 \times 10^{17} \: N}

<h3>Conclusion</h3>

So, the force of gravity that the asteroid and the planet have on each other approximately

\boxed{\sf{2.9 \times 10^{17} \: N}}

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2 years ago
How much is the tension number 2
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This is what I got:

Net force in the Y direction:

ΣFy = T1 - T2

F = ma

ma = T1 - T2

Isolate for T2

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Multiply by -1

T1 - ma = T2

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3 years ago
An experimenter adds 970 J of heat to 1.75 mol of an ideal gas to heat it from 10.0∘C to 25.0∘C at constant pressure. The gas do
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Answer:

(a) ΔU=747J

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Substitute the given values

ΔU=970J-223J

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We can calculate γ by ratio of heat capacities of the gas

γ=Cp/Cv

Where Cp is the molar heat capacity at constant pressure

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To calculate γ we first need to find Cp and Cv

So

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As we know

Q=nCpΔT

Cp=(Q/nΔT)

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From relation of Cv and Cp we know that

Cp=Cv+R

Where R is gas constant equals to 8.314J/mol.K

So

C_{v}=C_{p}-R\\C_{v}=37-8.314\\C_{v}=28.687J/mol.K\\

So

γ=Cp/Cv

γ=[(37J/mol.K) / (28.687J/mol.K)]

γ=1.3

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Answer:

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