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3 years ago

13

What causes seasons on Earth?

2 answers:

Bezzdna [24]3 years ago

6 0

Explanation:

A. answer is the correct answer

Naily [24]3 years ago

4 0

A is the correct answer

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A simple pendulum consists of a mass M attached to a string oflength L andnegligible mass. For this system, when undergoing smal

PilotLPTM [1.2K]

The frequency of the pendulum is independent of the mass on the end. (c)

This means that it doesn't matter if you hang a piece of spaghetti or a school bus from the bottom end. If there is no air resistance, and no friction at the top end, and the string has no mass, then the time it takes the pendulum to swing from one side to the other <u><em>only</em></u> depends on the <u><em>length</em></u> of the string.

8 0

3 years ago

classic physics problem states that if a projectile is shot vertically up into the air with an initial velocity of 128 feet per

ddd [48]

Answer:

$t_1 = 0.28 s$

$t_2 = 7.72 s$

Explanation:

Given that height of the projectile as a function of time is

$h = -16 t^2 + 128 t + 112$

here we know that

h = 147 ft

so from above equation

$147 = -16 t^2 + 128 t + 112$

$16 t^2 - 128 t + 35 = 0$

now by solving above quadratic equation we know that

$t_1 = 0.28 s$

$t_2 = 7.72 s$

6 0

3 years ago

An object travels in a circular path of radius 5.0 meters at a uniform speed of 10. m/s. What is the magnitude of the object's a

vladimir1956 [14]

I think F= mv²/r
And F=ma
So, ma = mv²/r
a = v²/r
a = 100/5
a = 20 m/s

8 0

3 years ago

Read 2 more answers

Shanti is riding on a train that is moving at a speed of 90 km/h. He is carrying a power cord for his phone that is 1.2 m long.

faltersainse [42]

Answer:equal to 1.2 m

Explanation: i took the test

3 0

3 years ago

Read 2 more answers

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