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3 years ago

What do you notice from the picture?

Physics

2 answers:

Irina18 [472]3 years ago

7 0

What I notice in the picture is some type of scientists examining a animal and making guesses on what it could be

Schach [20]3 years ago

5 0

Answer:there talking about the house

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David is driving a steady 30.0 m/s when he passes Tina, who is sitting in her car at rest. Tina begins to accelerate at a steady

SpyIntel [72]

Answer:

Explanation:

Let after time t , Tina catches up David .

Distance travelled by them are equal ,

Distance travelled by Tina

s = ut + 1/2 a t²

= .5 x 2.10 t²

= 1.05 t²

Distance travelled by David

= 30 t ( because of uniform velocity )

1.05 t² = 30t

t = 28.57 s

Distance travelled by Tina

= 1/2 a t²

= .5 x 2.10 x 28.57²

= 857 m approx.

7 0

3 years ago

Read 2 more answers

Mass of block A above is 25-kg. Calculate the force of gravity on block A.

stiv31 [10]

Answer:w+mg

Explanation:

4 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:

Given:

$T_1$ = 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 = $1.01325\times 10^5\ Pa$

Assume:

$P_2$ = 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

An eight-turn coil encloses an elliptical area having a major axis of 40.0 cm and a minor axis of 30.0 cm. The coil lies in the

Darina [25.2K]

Answer:

9.25 x 10^-4 Nm

Explanation:

number of turns, N = 8

major axis = 40 cm

semi major axis, a = 20 cm = 0.2 m

minor axis = 30 cm

semi minor axis, b = 15 cm = 0.15 m

current, i = 6.2 A

Magnetic field, B = 1.98 x 10^-4 T

Angle between the normal and the magnetic field is 90°.

Torque is given by

τ = N i A B SinФ

Where, A be the area of the coil.

Area of ellipse, A = π ab = 3.14 x 0.20 x 0.15 = 0.0942 m²

τ = 8 x 6.20 x 0.0942 x 1.98 x 10^-4 x Sin 90°

τ = 9.25 x 10^-4 Nm

thus, the torque is 9.25 x 10^-4 Nm.

5 0

3 years ago

Which measurement is a potential difference? A.115J B.115V C.115N D.115C

borishaifa [10]

Answer:

Explanation:

Potential difference has a SI Unit of Volt and its symbol is V. Hence answer is B.

A is wrong as it has the unit Joule (J) which is the SI unit for energy.

C is wrong as it has the unit Newton (N) which is the SI unit for force.

D is wrong as it has the unit Coulomb (C) which is the SI unit of charge.

5 0

4 years ago

What do you notice from the picture?​

What do you notice from the picture?