What is the current if 4C of charge passes in 2 s?

Questions 8 out of 20

MrMuchimi

Answer:

potential energy

Explanation:

8 0

3 years ago

In a novel from 1866 the author describes a spaceship that is blasted out of a cannon with a speed of about 11.000 m/s. The spac

Elan Coil [88]

Answer:

$a=0.284\ m/s^2$

Explanation:

Given that,

Initially, the spaceship was at rest, u = 0

Final velocity of the spaceship, v = 11 m/s

Distance accelerated by the spaceship, d = 213 m

We need to find the acceleration experienced by the occupants of the spaceship during the launch. It is a concept based on the equation of kinematics. Using the third equation of motion to find acceleration.

$v^2-u^2=2ad\\\\a=\dfrac{v^2-u^2}{2d}\\\\a=\dfrac{(11)^2-(0)^2}{2\times 213}\\\\a=0.284\ m/s^2$

So, the acceleration experienced by the occupants of the spaceship is $0.284\ m/s^2$ .

5 0

3 years ago

in order to qualify for the finals in a racing event, a race car must achieve an average speed of 278 km/h on a track with a tot

Pachacha [2.7K]

The minimum average speed it must have in the second half of the event in order to qualify is 414.7 km/h.

<h3>What is average speed?</h3>

The average speed of an object is the ratio of total distance traveled by the object to the total time of motion of the object.

<h3>Total time taken by the car during the entire race</h3>

time = distance/average speed

time = (1.41 km) / (278 km/h)

time = 0.0051 hr

The car travels the first half of the race, d (¹/₂ x 1410 m) at 210 km/h;

d = 705 m = 0.705 km

t1 = 0.705/210

t1 = 0.0034 hr

<h3>time for the second half</h3>

t2 = 0.0051 - 0.0034 hr

t2 = 0.0017 hr

<h3>minimum average speed of the second half</h3>

v = d/t

v = 0.705 km / 0.0017 hr

v = 414.7 km/hr

Thus, the minimum average speed it must have in the second half of the event in order to qualify is 414.7 km/h.

Learn more about average speed here: brainly.com/question/4931057

#SPJ1

6 0

1 year ago

Some enterprising physics students working on a catapult decide to have a water balloon fight in the school hallway. The ceiling

sergejj [24]

Answer:

$\alpha =54.7$ º

Explanation:

From the exercise we have our initial information

$y=3.4m\\v_{o}=10m/s\\g=-9.8m/s^2$

When the balloon gets to the ceiling its velocity at that moment is 0 m/s. Being said that we can calculate velocity at the vertical direction

$v_{y}^2=v_{oy}^2+ag(y-y_{o})$

Since $v_{y}=0$ and $y_{o}=0$

$0=v_{oy}^2-2(9.8m/s^2)(3.4m)$

$v_{oy}=\sqrt{2(9.8m/s^2)(3.4m)}=8.16m/s$

Knowing that

$v_{oy}=v_{o}sin\alpha$

$sin\alpha =\frac{v_{oy} }{v_{o} }$

$\alpha =sin^{-1}(\frac{v_{oy}}{v_{o}})=sin^{-1}(\frac{8.16m/s}{10m/s})=54.7$ º

8 0

3 years ago

The eyes of amphibians such as frogs have a much flatter cornea but a more strongly curved (almost spherical) lens than do the e

Lapatulllka [165]

Answer:

0.2cm towards the retina.

Explanation:

the focal length of the frog eye is

(1/f) = (1/10) + (1/0.8)

f = 0.74cm

Since the distance of the object is 15cm Hence

(1/0.74) = (1/15) + (1/V)

V = 0.78cm

Therefore the distance the retina is to move is

0.78cm - 0.8cm = 0.02cm towards the retina.

3 0

3 years ago