A force or an event that changes the earth's surface is known as

At the local swimming hole, a favorite trick is to run horizontally off a cliff that is 8.0 m above the water. One diver runs of

Alika [10]

Answer:

Number of revolutions=1.532 revolutions

Explanation:

Given data

Distance s=8.0 m

Angular speed a=1.2 rev/s

To find

Number of revolutions

Solution

From the equation of simple motion we not that

$S=ut+1/2gt^{2}\\ where\\u=0\\So\\8.0m=0+(1/2)(9.8m/s^{2} )t^{2}\\ t^{2}=\frac{8.0m}{0.5*9.8m/s^{2} } \\ t^{2}=1.63\\t=\sqrt{1.63} \\t=1.28s$

So for the number of revolutions she makes is given as

$n=a*t\\n=(1.2rev/s)(1.28s)\\n=1.532revolutions$

8 0

3 years ago

Apakah yang dimaksud dengan gaya

tangare [24]

Nah gaya seperti titik fashion seperti apa yang Anda mana dan hal-hal seperti itu. (tell me if you cant understand)

8 0

3 years ago

Read 2 more answers

A very weak pressure wave, i.e., a sound wave, across which the pressure rise is 30 Pa moves through air which has a temperature

Fofino [41]

Answer:

Density change, Δρ = 2.4 × 10⁻⁴ kg/m³

Temperature Change, ΔT = 0.0258 K

Velocity Change, Δc = 0.0148 m/s

Explanation:

For sound waves moving through the air,

Pressure and Temperature varies thus

(P₀/P) = (T₀/T)^(k/(k-1))

Where P₀ = initial pressure of air = 101KPa = 101000 Pa

P = final pressure of air due to the change brought about by the moving sound wave = 101000+30 = 101030 Pa

T₀ = initial temperature of air = 30°C = 303.15 K

T = final temperature of air = ?

k = ratio of specific heats = Cp/Cv = 1.4

(101000/101030) = (303.15/T)^(1.4/(1.4-1))

0.9990703 =(303.15/T)^(3.5)

Solving This,

T = 303.1758 K

ΔT = T - T₀ = 303.1758 - 303.15 = 0.0258 K

Density can be calculate in two ways,

First method

Δρ = ρ - ρ₀

P₀ = ρ₀RT₀

ρ₀ = P₀/RT₀

R = gas constant for air = 287 J/kg.k

where all of these are values for air before the wave propagates

P₀ = 101000 Pa, R = 287 J/kg.K, T₀ = 303.15K

ρ₀ = 101000/(287 × 303.15) = 1.1608655 kg/m³

ρ = P/RT

P = 101030 Pa, T = 303.1758K

ρ = 101030/(287×303.1758) = 1.1611115 kg/m³

Δρ = ρ - ρ₀ = 1.1611115 - 1.1608655 = 0.00024 kg/m³ = 2.4 × 10⁻⁴ kg/m³

Second method

(ρ₀/ρ) = (T₀/T)^(1/(k-1))

Where ρ₀ is initially calculated from ρ₀ = P₀/RT₀, then ρ is then computed and the diff taken.

Velocity Change

c₀ = √(kRT₀) = √(1.4 × 287 × 303.15) = 349.00669 m/s

c = √(kRT) = √(1.4 × 287 × 303.1758) = 349.0215415 m/s

Δc = c₀ - c = 349.0215415 - 349.00669 = 0.0148 m/s

QED!

5 0

3 years ago

S GP A projectile of mass m moves to the right with a speed vi (Fig. P11.51a). The projectile strikes and sticks to the end of a

Andrews [41]

What is the kinetic energy of the system after the collision?

$K_f=\frac{3}{2} \frac{m^{2}v_i^{2} }{(M+3m)}$

How this is calculated?

Given:

Initial speed= $v_i$

mass of rod=M

Let, Initial kinetic energy = $K_i$

Final kinetic energy= $K_f$

Moment of inertia =I

What is the moment of inertia?

$I=(I_p)_0+(I_{rod})_0\\I=m(\frac{d}{2})^{2} +\frac{Md^{2} }{12} \\I=\frac{(M+3m)d^{2} }{12}$

What is the angular momentum?

By conservation of angular momentum,

$L_i=L_f$

$mv_i\frac{d}{2}=\frac{(M+3m)d^{2}\omega }{12} \\\omega=\frac{6mv_i^{2} }{d(M+3m)}$

We know that, the final kinetic energy is given by,

$K_f=I\omega^{2}\\K_f=\frac{1}{2} *\frac{(M+3m)d^{2} }{12} *\frac{36m^{2}v_i^{2}}{d^{2}(M+3m)^{2}}\\ K_f=\frac{3}{2} \frac{m^{2}v_i^{2} }{(M+3m)}$

What is the kinetic energy?

In physics, the kinetic energy of an object is the energy that it possesses due to its motion.
It is defined as the work needed to accelerate a body of a given mass from rest to its stated velocity.
Having gained this energy during its acceleration, the body maintains this kinetic energy unless its speed changes.

To know more about kinetic energy, refer:

brainly.com/question/114210

#SPJ4

8 0

1 year ago

A spring has a spring constant of 90N/m.How much potential energy does it store when stretched by 2 cm?

il63 [147K]

Answer:

The potential energy stored in the spring is 0.018 J.

Explanation:

Given;

spring constant, k = 90 N/m

extension of the spring, x = 2 cm = 0.02 m

The potential energy stored in the spring is calculated as;

U = ¹/₂kx²

where;

U is the potential energy stored in the spring

Substitute the given values in the equation above;

U = ¹/₂ x 90 N/m x (0.02 m)²

U = 0.018 J

Therefore, the potential energy stored in the spring is 0.018 J.

5 0

3 years ago