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Vedmedyk [2.9K]

3 years ago

12

A football is kicked at ground level with the speed of 32 m/s at an angle of 40 degrees to the horizontal. how much later does i

t hit the ground?

1 answer:

LUCKY_DIMON [66]3 years ago

4 0

the question was a football is kicked at ground level with a speed of 32 m/s the an angle of 40 degrees to the horizontal how much later does it hit the ground probably like 30 more or 40 sense it was around 32

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At a rock concert, the sound intensity 1.0 m in front of the bank of loudspeakers is 0.10 W/m². A fan is 30 m from the loudspeak

Klio2033 [76]

To solve this problem we will apply the concepts related to the Area, the power and the proportionality relationships between intensity and distance.

The expression for sound power is,

$P = AI$

Here,

A = Area

I = Intensity

P = Power

At the same time the area can be written as,

$A = \frac{\pi d^2}{4}$

Now the intensity is inversely proportional to the square of the distance from the source, then

$I \propto \frac{1}{r^2}$

The expression for the intensity at different distance is

$\frac{I_1}{I_2}= \frac{r^2_2}{r_1^2}$

Here,

$I_1$ = Intensity at distance 1

$I_2$ = Intensity at distance 2

$r_1$ = Distance 1 from light source

$r_2$ = Distance 2 from the light source

If we rearrange the expression to find the intensity at second position we have,

$I_2 = I_1 (\frac{r_1^2}{r_2^2})$

If we replace with our values at this equation we have,

$I_2 = (0.10W/m^2)(\frac{1.0m^2}{30.0m^2})$

$I_2 = 1.11*10^{-4} W/m^2$

Now using the equation to find the area we have that

$A = \frac{\pi (8.4*10^{-3}m)^2}{4}$

$A = 5.5*10^{-5}m^2$

Finally with the intensity and the area we can find the sound power, which is

$P = AI$

$P = (5.5*10^{-5}m^2)(1.11*10^{-4}W/m^2)$

$P = 6.1*10^{-9}J/s$

Power is defined as the quantity of Energy per second, then

$E = 6.1*10^{-9}J$

8 0

3 years ago

A planet exerts a gravitational force of magnitude 9e22 N on a star. If the planet were 2 times closer to the star (that is, if

Dmitrij [34]

To solve this problem we will use the related concepts in Newtonian laws that describe the force of gravitational attraction. We will use the given value and then we will obtain the proportion of the new force depending on the Radius. From there we will observe how much the force of attraction increases in the new distance.

Planet gravitational force

$F_p = 6*10^{22}N$

$F_p = \frac{GMm}{R^2}$

$F_p = 9*10^{22}N$

Distance between planet and star

$r = \frac{R}{2}$

Gravitational force is

$F = \frac{GMm}{r^2}$

Applying the new distance,

$F = \frac{GMm}{(\frac{R}{2})^2}$

$F = 4\frac{GMm}{R^2}$

Replacing with the previous force,

$F = 4F_p$

Replacing our values

$F= 4(9*10^{22}N)$

$F = 36*10^{22}N$

Therefore the magnitude of the force on the star due to the planet is $36*10^{22}N$

5 0

3 years ago

A pickup truck is being driven down the highway carrying eight 100 gallon fish tank full of water. A hole is punctured in each t

irakobra [83]

Answer:

Explanation:

The velocity of the vehicle would increase because the the tanks (when filled with water) must have exerted a force which would reduce the velocity of the vehicle at a certain pressure on the gas pedal. Note that force equals mass multiplied by acceleration; as the mass decreases, so the force decreases. Thus, when the mass exerted by this tanks (on the vehicle) decrease as a result of the hole punctured in them, the force exerted by the tanks would also decrease causing an increase in velocity of the pick up truck when the same pressure is applied on the gas pedal throughout (before and after the puncture).

The conservation law that applied here is the law of conservation of energy which states that energy can neither be created nor destroyed but can be transformed from one form to another. This is because the energy the vehicle used in carrying the load (the tanks) was transformed to the energy that resulted in increasing it's velocity (no new energy was formed as the pressure on the gas pedal remained the same).

6 0

3 years ago

What is the moment of inertia of the object starting from rest if it has a final velocity of 5.9 m/s? Express the moment of iner

Bogdan [553]

Answer:

The moment of inertia is I = 0.126*R^2*M

Explanation:

We can calculate the moment of inertia of an object that starts from rest and has a final velocity using the energy conservation equation, as follows:

Ek1 + Ep1 = Ek2 + Ep2, where

Ek1 = kinetic energy of the object before to roll down

Ep1 = potential energy of the object

Ek2 = kinetic energy when the object comes down

Ep2 = potential energy of the object at the bottom

We have the follow:

Ek1 = 0

Ep1 = M*g*h

Ek2 = ((I*w)/2) + ((M*v^2)/2)

Ep2 = 0

Replacing values:

0 + M*g*h = ((I*w)/2) + ((M*v^2)/2) + 0

where:

M = mass of the object

g = gravitational acceleration

I = moment of the inertia

w = angular velocity = v/R

h = height

M*g*h = ((1/2) * I * (v^2/R^2)) + ((M*v^2)/2)

M*9.8*2 = (I*(5.9^2)/(2*R^2)) + ((5.9^2 * M)/2)

19.6 * M = ((17.4*I)/R^2) + 17.4*M

Clearing I, we have:

I = 0.126*R^2*M

5 0

3 years ago

If the time for 20 vibrations is 100 second then find

Lilit [14]

hey guy

the vibration per second is frequency

then for 20 vibration time taken is 100 second

in 1 second 20/100 vibration took place

it means 0.2 is frequency

for time period , we have

t=1/f

=1/0.2

=5

8 0

3 years ago

Read 2 more answers