In which switching technique, a dedicated

Where does most of the energy we use at home/our cars come from?

KengaRu [80]

It comes from the sun, and then it is converted to energy/electricity (by solar panels)

4 0

3 years ago

15 POINTS PLEASE HELP! NO GUESSING

torisob [31]

He answer is A. <span>encourage agricultural usage in the watershed

if you want to read it for yourself go to
www.nature.org/ourinitiatives/regions/northamerica/unitedstates/indiana/journeywithnature/watersheds...

hope this helps you!!</span>

7 0

3 years ago

Read 2 more answers

When a hammer thrower releases her ball, she is aiming to maximize the distance from the starting ring. Assume she releases the

Taya2010 [7]

Answer:

The angular velocity is 15.37 rad/s

Solution:

As per the question:

$\theta = 54.6^{\circ}$

Horizontal distance, x = 30.1 m

Distance of the ball from the rotation axis is its radius, R = 1.15 m

Now,

To calculate the angular velocity:

Linear velocity, v = $\sqrt{\frac{gx}{sin2\theta}}$

v = $\sqrt{\frac{9.8\times 30.1}{sin2\times 54.6}}$

v = $\sqrt{\frac{294.98}{sin109.2^{\circ}}} = 17.67\ m/s$

Now,

The angular velocity can be calculated as:

$v = \omega R$

Thus

$\omega = \frac{v}{R} = \frac{17.67}{1.15} = 15.37\ rad/s$

8 0

3 years ago

A top-fuel dragster starts from rest and has a constant acceleration of 42.0 m/s2. What are (a) the final velocity of the dragst

disa [49]

Answer:

a) Final velocity of the dragster at the end of 1.8 s = 75.6 m/s

b) Final velocity of the dragster at the end of 3.6 s = 151.2 m/s

c) The displacement of the dragster at the end of 1.8 s = 68.04 m

d) The displacement of the dragster at the end of 3.6 s = 272.16 m

Explanation:

a) We have equation of motion v = u + at

Initial velocity, u = 0 m/s

Acceleration , a = 42 m/s²

Time = 1.8 s

Substituting

v = u + at

v = 0 + 42 x 1.8 = 75.6 m/s

Final velocity of the dragster at the end of 1.8 s = 75.6 m/s

b) We have equation of motion v = u + at

Initial velocity, u = 0 m/s

Acceleration , a = 42 m/s²

Time = 3.6 s

Substituting

v = u + at

v = 0 + 42 x 3.6 = 75.6 m/s

Final velocity of the dragster at the end of 3.6 s = 151.2 m/s

c) We have equation of motion s= ut + 0.5 at²

Initial velocity, u = 0 m/s

Acceleration , a = 42 m/s²

Time = 1.8 s

Substituting

s= ut + 0.5 at²

s = 0 x 1.8 + 0.5 x 42 x 1.8²

s = 68.04 m

The displacement of the dragster at the end of 1.8 s = 68.04 m

d) We have equation of motion s= ut + 0.5 at²

Initial velocity, u = 0 m/s

Acceleration , a = 42 m/s²

Time = 3.6 s

Substituting

s= ut + 0.5 at²

s = 0 x 3.6 + 0.5 x 42 x 3.6²

s = 272.16 m

The displacement of the dragster at the end of 3.6 s = 272.16 m

3 0

3 years ago

A rock thrown straight up takes 4.2 s to reach its maximum height what was its initial velocity

liubo4ka [24]

Consider the upward direction of motion as positive and downward direction of motion as negative.

a = acceleration due to gravity in downward direction = - 9.8 $\frac{m }{s^{2}}$

v₀ = initial velocity of rock in upward direction = ?

v = final velocity of rock at the highest point = 0 $\frac{m }{s}$

t = time to reach the maximum height = 4.2 sec

Using the kinematics equation

v = v₀ + a t

inserting the values

0 = v₀ + (- 9.8) (4.2)

v₀ = 41.2 $\frac{m }{s}$

8 0

3 years ago