A flashlight bulb is connected to a dry cell of voltage 2.25 V. It draws 35.0 mA (1000 mA = 1 A). What is its resistance?

List the forms of renewable energy that are curently in use

aliya0001 [1]

Here are the ones that I know about
and can think of just now:

-- wind
-- solar
-- nuclear
-- tidal
-- hydro
-- geothermal
-- biomass

4 0

3 years ago

What is the kinetic energy of a 1.40 kg discus with a speed of 22.5 m/s?

goldenfox [79]

Apply E=(mv^2)/2 . It will give the answer.

6 0

4 years ago

A car has a kinetic energy of 1.9 × 103 joules. If the velocity of the car is decreased by half, what is its kinetic energy? 2.2

Harman [31]

If the velocity is decreased by half, then Kinetic energy will reduced to it's 4 times,
= 1.9 * 10³ / 4
= 4.75 * 10²

In short, Your Answer would be Option B

Hope this helps!

8 0

3 years ago

Read 2 more answers

The threshold of hearing is defined as the minimum discernible intensity of the sound. It is approximately 10âˆ’12w/m2. Find the

Monica [59]

Answer:

The distance d from the car at which the sound from the stereo can still be discerned = 97720.5 m

Explanation:

Sound intensity heard at distance is related to the distance with the relation = (power of sound at the source)/(surface area of the wall of an imaginary sphere at the distance in question)

I = P/4πd²

Assuming the car has 2 speakers,

P = 0.06 W × 2 = 0.12 W

d = ?

For the intensity of the least discernible sound,

I = 10⁻¹² W/m²

10⁻¹² = 0.12/4πd²

d = 97720.5 m

8 0

3 years ago

A rookie quarterback throws a football with an initial upward velocity component of 17.0 m/s and a horizontal velocity component

vitfil [10]

Answer:

(a) 1.73 s

(b) 14.75 m

(c) 3.36 s

(d) double

(e) 63.32 m

Explanation:

Vertical component of initial velocity, uy = 17 m/s

Horizontal component of initial velocity, ux = 18.3 m/s

(A) At highest point of trajectory, the vertical component of velocity is zero. Let the time taken is t.

Use first equation of motion in vertical direction

vy = uy - gt

0 = 17 - 9.8 t

t = 1.73 seconds

(B) Let the highest point is at height h.

Use III equation of motion in vertical direction

$v^{2}=u^{2}-2gh$

0 = 17 x 17 - 2 x 9.8 x h

h = 14.75 m

(C) The time taken by the ball to return to original level is T.

Use second equation of motion i vertical direction.

$h = ut + 0.5at^2$

h = 0 , u = 17 m/s

0 = 17 t - 0.5 x 9.8 t^2

t = 3.46 second

(D) It is the double of time calculated in part A

(E) Horizontal distance = horizontal velocity x total time

d = 18.3 x 3.46 = 63.32 m

8 0

3 years ago