If the air pressure is doubled, the speed of sound

During low-intensity activity, your body obtains most of its energy from

Gemiola [76]

Carbohydrates, in cellular respiration.

3 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

How could you dissolve more solid solute in a saturated solution in a liquid solvent

tankabanditka [31]

If you saturated the solid it will turn into liquid and soon become an air

3 0

3 years ago

The height of a typical playground slide is about 1.8 m and it rises at an angle of 30 ∘ above the horizontal.

Salsk061 [2.6K]

Answer:

$5.94\ \text{m/s}$

$1.7$

$0.577$

Explanation:

g = Acceleration due to gravity = $9.81\ \text{m/s}^2$

$\theta$ = Angle of slope = $30^{\circ}$

v = Velocity of child at the bottom of the slide

$\mu_k$ = Coefficient of kinetic friction

$\mu_s$ = Coefficient of static friction

h = Height of slope = 1.8 m

The energy balance of the system is given by

$mgh=\dfrac{1}{2}mv^2\\\Rightarrow v=\sqrt{2gh}\\\Rightarrow v=\sqrt{2\times 9.81\times 1.8}\\\Rightarrow v=5.94\ \text{m/s}$

The speed of the child at the bottom of the slide is $5.94\ \text{m/s}$

Length of the slide is given by

$l=h\sin\theta\\\Rightarrow l=1.8\sin30^{\circ}\\\Rightarrow l=0.9\ \text{m}$

$v=\dfrac{1}{2}\times5.94\\\Rightarrow v=2.97\ \text{m/s}$

The force energy balance of the system is given by

$mgh=\dfrac{1}{2}mv^2+\mu_kmg\cos\theta l\\\Rightarrow \mu_k=\dfrac{gh-\dfrac{1}{2}v^2}{gl\cos\theta}\\\Rightarrow \mu_k=\dfrac{9.81\times 1.8-\dfrac{1}{2}\times 2.97^2}{9.81\times 0.9\cos30^{\circ}}\\\Rightarrow \mu_k=1.73$

The coefficient of kinetic friction is $1.7$ .

For static friction

$\mu_s\geq\tan30^{\circ}\\\Rightarrow \mu_s\geq0.577$

So, the minimum possible value for the coefficient of static friction is $0.577$ .

8 0

3 years ago

Systems distribute heat energy by the circulation of air.

Ulleksa [173]

That seems like a statement more than a question. Where's the question?

6 0

4 years ago

Read 2 more answers