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3 years ago

Ill mark brainliest!!!

Physics

1 answer:

navik [9.2K]3 years ago

6 0

Answer: A

si, di, are expressed as negative and f is expressed as positive values.

Explanation:

When the object is located at a location in front of the focal point of a convex len, the image will always be located somewhere on the same side of the lens as the object. The image is located behind the object. In this case, the image will be an upright image and the image is enlarged

In this question, the relationship between the focal length, image distance di and object distance is

I/f = 1/di + 1/do that is

I/f - 1/di - 1/do = 0

The image size si is also negative since the image is a virtual image.

Therefore, si, di, are expressed as negative, f is expressed as positive values.

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A projectile is launched with an initial velocity of 25 m/s at an angle of 30° above the horizontal. The projectile reaches maxi

Alchen [17]

Answer:

x ≈ 56 m

Explanation:

vertical initial velocity = $v_{0y}(t)$ = 25 m/s* sin(30°)= 12.5 m/s

height = h

$h =v_{0y}t+\frac{at^{2}}{2} \\\\65m = 12.5m/s*t + \frac{9.8m/s^{2}*t^{2}}{2} \\\\t=2.584 s$

t- time is found solving quadratic equation.

horizontal velocity = $v_{0x}=25m/s*cos(30^{o})=21.65 m/s$

Horizontal velocity is constant, so distance $x=v_{0x}*t =21.65 m/s *2.584 s=55.9 = 56 m$

6 0

3 years ago

What is the average velocity of a car that travels 450 km north in 9.0 hours?

Hatshy [7]

The average velocity of a car that travels 450 km north in 9.0 h is 5.0 x 10 km/h

3 0

3 years ago

URGENT!!!!! If a 6 Ω resistor is in a circuit connected to a voltage source, and the current through the circuit is 2 amps, what

Eduardwww [97]

Answer:

Explanation:

according to ohms law we know that

v=IR

given current =2 amps

given resistance =6Ω

so voltage is

v=2*6 =12 V

3 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

Can someone tell me the answers??? thanks i need it asap!! i will give brainlist!!

Sloan [31]

Increases then decreases

4 0

4 years ago

Read 2 more answers