Brainliest if right

While running these tests, crall notices a similarity in the velocity measured at the ground independent of whether the tennis b

lara [203]

At the ground the ball will always have velocity along the direction of gravity. If upward motion is taken positive it will always have negative velocity at the ground because, if the ball was given an initial upward velocity then gravity will decelerate it and bring it down with a negative final velocity. If the ball is given an initial downward velocity then the ball will be further accelerated by gravity in the downward direction only, again maintaining negative direction. The magnitude however in both cases will be different. the final velocity at the ground will have higher magnitude in case of elevator moving downwards.

6 0

3 years ago

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A wave traveling in a string has a wavelength of 35 cm, an amplitude of 8. 4 cm, and a period of 1. 2 s. What is the speed of th

AlekseyPX

0.29 m/s (wave velocity = wavelength (lamda)/period (T) in metres)

35 / 1.2 = 29.16

29.16 ÷ 100 = 0.29

Wave velocity in string:

The properties of the medium affect the wave's velocity in a string. For instance, if a thin guitar string is vibrated while a thick rope is not, the guitar string's waves will move more quickly. As a result, the linear densities of the two strings affect the string's velocity. Linear density is defined as the mass per unit length.

Instead of the sinusoidal wave, a single symmetrical pulse is taken into consideration in order to comprehend how the linear mass density and tension will affect the wave's speed on the string.

Learn more about density here:

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3 0

2 years ago

The total resistance of a parallel circuit is 25 ohms. If the total current is 100mA, how much current is through a 220 ohm resi

gulaghasi [49]

Answer:

The current across the resistance is 0.011 A.

Explanation:

Total resistance, R = 25 ohms

Total current, I = 100 mA = 0.1 A

Let the voltage is V.

By the Ohm's law

V = I R

V = 0.1 x 25 = 2.5 V

Now the resistance is R' = 220 ohm

As they are in parallel so the voltage is same. Let the current is I'.

V = I' x R'

2.5 = I' x 220

I' = 0.011 A

7 0

3 years ago

A major strike-slip earthquake on the San Andreas fault in California will cause a catastrophic tsunami affecting residents of S

ZanzabumX [31]

Answer:

a) true

Explanation:

The san andres is a transform fault that forms boundary between the Pacific and the North Atlantic plate and this slip strike is characterized by the latex motion the fault runs in the length of the California state.

This plate is widely estimated for the high magnitude of earthquakes and varies from 7.7 to 8.3 magnitude. They are capable of producing a deadly tsunami that can devastate the pacific northwest.

6 0

3 years ago

An automobile tire is inflated with air originally at 10.0°C and normal atmospheric pressure. During the process, the air is com

solong [7]

Answer:

(a) $3.81\times 10^5\ Pa$

(b) $4.19\times 1065\ Pa$

Explanation:

<u>Given:</u>

$T_1$ = The first temperature of air inside the tire = $10^\circ C =(273+10)\ K =283\ K$

$T_2$ = The second temperature of air inside the tire = $46^\circ C =(273+46)\ K= 319\ K$

$T_3$ = The third temperature of air inside the tire = $85^\circ C =(273+85)\ K=358 \ K$

$V_1$ = The first volume of air inside the tire

$V_2$ = The second volume of air inside the tire = $30\% V_1 = 0.3V_1$

$V_3$ = The third volume of air inside the tire = $2\%V_2+V_2= 102\%V_2=1.02V_2$

$P_1$ = The first pressure of air inside the tire = $1.01325\times 10^5\ Pa$

<u>Assume:</u>

$P_2$ = The second pressure of air inside the tire

$P_3$ = The third pressure of air inside the tire
n = number of moles of air

Since the amount pof air inside the tire remains the same, this means the number of moles of air in the tire will remain constant.

Using ideal gas equation, we have

$PV = nRT\\\Rightarrow \dfrac{PV}{T}=nR = constant\,\,\,(\because n,\ R\ are\ constants)$

Part (a):

Using the above equation for this part of compression in the air, we have

$\therefore \dfrac{P_1V_1}{T_1}=\dfrac{P_2V_2}{T_2}\\\Rightarrow P_2 = \dfrac{V_1}{V_2}\times \dfrac{T_2}{T_1}\times P_1\\\Rightarrow P_2 = \dfrac{V_1}{0.3V_1}\times \dfrac{319}{283}\times 1.01325\times 10^5\\\Rightarrow P_2 =3.81\times 10^5\ Pa$

Hence, the pressure in the tire after the compression is $3.81\times 10^5\ Pa$ .

Part (b):

Again using the equation for this part for the air, we have

$\therefore \dfrac{P_2V_2}{T_2}=\dfrac{P_3V_3}{T_3}\\\Rightarrow P_3 = \dfrac{V_2}{V_3}\times \dfrac{T_3}{T_2}\times P_2\\\Rightarrow P_3 = \dfrac{V_2}{1.02V_2}\times \dfrac{358}{319}\times 3.81\times 10^5\\\Rightarrow P_3 =4.19\times 10^5\ Pa$

Hence, the pressure in the tire after the car i driven at high speed is $4.19\times 10^5\ Pa$ .

8 0

3 years ago