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katen-ka-za [31]
3 years ago
11

A football is kicked with a velocity of 31 meters per second at an angle of 41 degrees what is the balls acceleration in the ver

tical direction as it flies through the air
Physics
2 answers:
IgorC [24]3 years ago
8 0

Answer:

The vertical component of the acceleration is equal to the acceleration due to gravity g.

Explanation:

Given data,

The initial velocity of the football, u = 31 m/s

The angle of direction of the football with the horizontal, Ф = 41°

In free fall, the vertical component of acceleration always remains the same.

Since the ball is clicked in a particular direction that has a vertical component,  then the vertical component of the velocity of the ball is always acted upon by the gravitational force of the earth.

The horizontal component of the velocity of the projectile remains the same because the gravitational force doesn't act in the horizontal direction.

Since the only force is the gravitational force acts on the projectile, the vertical component of the acceleration is equal to the acceleration due to gravity g.

BaLLatris [955]3 years ago
8 0

Answer:

-9.8 m/s^2

Explanation:

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3 years ago
A 8.00g sample of substance (substance, molar mass = 152.0 g/mol) was combusted in a bomb calorimeter with a heat capacity of 6.
aleksandrvk [35]

Answer:

ΔH°comb=-5899.5 kJ/mol

Explanation:

First, consider the energy balance:

m_{c} *Cp*(T_{2}-T_{1})=-n_{s} *H_{c} Where m_{c} is the calorimeter mass and n_{s} is the number of moles of the samples; H_{c} is the combustion enthalpy. The energy balance says that the energy that the reaction release is employed in rise the temperature of the calorimeter, which is designed to be adiabatic, so it is suppose that the total energy is employed rising the calorimeter temperature.

The product m_{c} *Cp is the heat capacity, so the balance equation is:

6.21\frac{kJ}{K}*(75-25)=-8.00g*\frac{mol}{152.0g}*H_{c}

So, the enthalpy of combustion can be calculated:

H_{c}=-5899.5\frac{kJ}{mol}

I will be happy to solve any doubt you have.

4 0
4 years ago
Why must one use a reference point to determine whether or not an object is in motion
GaryK [48]
Because there's no such thing as "really" moving. 
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Here's an example:
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comfortable, your eyes are getting so heavy, finally the book slips
out of your hand, falls into your lap, and you are fast asleep.

-- Relative to you, the book is not moving at all.
-- Relative to the seat, you are not moving at all.
-- Relative to the wall and the window, the seat is not moving at all.
-- But your seat is in a passenger airliner.  Relative to people on the
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-- Relative to the center of the Earth, the people on the ground are moving
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How fast are they REALLY moving ?
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7 0
3 years ago
The hydraulic lift in a car repair shop has an output diameter of 30 cm and is to lift cars up to 2000 kg. determine the fluid g
leonid [27]
Given Data: Diameter 'd' = 30 cm = 0.3 m Lifting Weight 'W' = mg = 2000*9.81 N = 19,620 N Calculations: Area of the lift 'A' = <span>pi\over4*d^2=pi\over4*0.3^2=0.07 m^2

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4 0
3 years ago
Read 2 more answers
An apple weighs 1.00 N. When you hang it from the end of a long spring of force constant 1.50 N/m and negligible mass, it bounce
Ierofanga [76]

Answer:

2.67 m

Explanation:

k = Spring constant = 1.5 N/m

g = Acceleration due to gravity = 9.81 m/s²

l = Unstretched length

Frequency of SHM motion is given by

f_s=\dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}}

Frequency of pendulum is given by

f_p=\dfrac{1}{2\pi}\sqrt{\dfrac{g}{l}}

Given in the question

f_p=\dfrac{1}{2}f_s

\dfrac{1}{2\pi}\sqrt{\dfrac{g}{l}}=\dfrac{1}{2}\dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}}\\\Rightarrow \sqrt{\dfrac{g}{l}}=\dfrac{1}{2}\sqrt{\dfrac{k}{m}}\\\Rightarrow \dfrac{g}{l}=\dfrac{1}{4}\dfrac{k}{m}\\\Rightarrow l=\dfrac{4gm}{k}\\\Rightarrow l=\dfrac{4\times 9.81\times \dfrac{1}{9.81}}{1.5}\\\Rightarrow l=2.67\ m

The unstretched length of the spring is 2.67 m

6 0
3 years ago
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