It takes 1 minute for 45 c to pass a point in a circuit, what is the current flowing through the circuit?

A ball is ejected to the right with an unknown horizontal velocity from the top of a pillar that is 50 meters in height. At the

dimulka [17.4K]

Answer:

15.67 m/s

Explanation:

The ball has a projectile motion, with a horizontal uniform motion with constant speed and a vertical accelerated motion with constant acceleration g=9.8 m/s^2 downward.

Let's consider the vertical motion only first: the vertical distance covered by the ball, which is S=50 m, is given by

$S=\frac{1}{2}gt^2$

where t is the time of the fall. Substituting S=50 m and re-arranging the equation, we can find t:

$t=\sqrt{\frac{2S}{g}}=\sqrt{\frac{2(50 m)}{9.8 m/s^2}}=3.19 s$

Now we now that the ball must cover a distance of 50 meters horizontally during this time, in order to fall inside the carriage; therefore, the velocity of the carriage should be:

$v=\frac{d}{t}=\frac{50 m}{3.19 s}=15.67 m/s$

8 0

3 years ago

During a field experiment about speed, a scientist created the chart above. The chart shows distance and time measurements for a

BaLLatris [955]

Answer:

yes

Explanation:

its not a good thing for the rest of your life but you have a PS4

6 0

2 years ago

What molecule determines the characteristics a child will inherit from its parents?

Korvikt [17]

Answer:

A molecule called DNA (deoxyribonucleic acid) is passed from adult organisms to their offspring during reproduction. This molecule containsthe instructions for an organism to develop, grow, survive and reproduce.

8 0

2 years ago

10. According to Newton's First Lawy , if a box is pushed with no external resistance, what will the action of the box be ?

kati45 [8]

Explanation:

According to Newton's First Law of motion, if a box is pushed with no external resistance, the box will keep on moving due to the absence of external force. It might gets stopped due to frictional force that is acting between the surface and the ball. The first law of motion is also known as law of inertia. the magnitude of force acting on the object is given by second law of motion.

4 0

3 years ago

Read 2 more answers

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