Pleaseeee help got this homework tomorrow ; if an object vibrates five times every second it has a frequency of ...?

Richardson pulls a toy 3.0 m across the floor by a string, applying a force of0.50 N. During the first meter, the string is para

Anastasy [175]

Answer:

Total Work done =0.65 joule

Explanation:

Work done is given Mathematically as

W=F *d

Where w=work done in joules

F=applied force

d= distance moved

The work done to move the toy accros the first meter is

W1=0.5*1

W1=0.5joule

The work done to move the toy across the next 2m at an angle of 30° is

.W2=0.5*2cos30

W2=0.5*2*0.154

W2=0.154joule

Hence total work done is

W1+W2=0.5+0.154

Total Work done =0.65 joule

7 0

3 years ago

In the simulation above, as the projectile travels upward, how does the vertical velocity change? Question 9 options:

enot [183]

Answer:

Vertical velocity decreases.

Explanation:

The motion of the ball is a projectile ball, which consists of two independent motions:

- a horizontal motion, with constant velocity

- a vertical motion, with constant acceleration g=9.8 m/s^2 towards the ground

In the vertical motion, there is a constant acceleration directed downward: this means that the vertical velocity decreases as the ball goes higher. In fact, it decreases following the equation

$v(t)=v_0 -gt$

And it decreases until the ball reaches its maximum height, then it starts increasing again.

4 0

3 years ago

Two identical satellites orbit the earth in stable orbits. One satellite orbits with a speed v at a distance r from the center o

jolli1 [7]

Answer:

c)At a distance greater than r

Explanation:

For a satellite in orbit around the Earth, the gravitational force provides the centripetal force that keeps the satellite in motion:

$G\frac{Mm}{r^2}=m\frac{v^2}{r}$

where

G is the gravitational constant

M is the Earth's mass

m is the satellite's mass

r is the distance between the satellite and the Earth's centre

v is the speed of the satellite

Re-arranging the equation, we write

$r = \frac{GM}{v^2}$

so we see from the equation that when the speed is higher, the distance from the Earth's centre is smaller, and when the speed is lower, the distance from the Earth's centre is larger.

Here, the second satellite orbit the Earth at a speed less than v: this means that its orbit will have a larger radius than the first satellite, so the correct answer is

c)At a distance greater than r

7 0

3 years ago

The Reynolds number, rho VD/mu, is a very important parameter in fluid mechanics. Verify that the Reynolds number is dimensionle

Agata [3.3K]

Answer:

Re = 1 10⁴

Explanation:

Reynolds number is

Re = ρ v D /μ

The units of each term are

ρ = [kg / m³]

v = [m / s]

D = [m]

μ = [Pa s]

The pressure

Pa = [N / m²] = [Kg m / s²] 1 / [m²] = [kg / m s²]

μ = [Pa s] = [kg / m s²] [s] = [kg / m s]

We substitute the units in the equation

Re = [kg / m³] [m / s] [m] / [kg / m s]

Re = [kg / m s] / [m s / kg]

RE = [ ]

Reynolds number is a scalar

Let's evaluate for the given point

Where the data for methane are:

viscosity μ = 11.2 10⁻⁶ Pa s

the density ρ = 0.656 kg / m³

D = 2 in (2.54 10⁻² m / 1 in) = 5.08 10⁻² m

Re = 0.656 4 2 5.08 10⁻² /11.2 10⁻⁶

Re = 1.19 10⁴

4 0

3 years ago

You have two windlasses (winches) named X and Y. Windlass X drum diameter is 50cm and windlass Y drum diameter is 75cm. Both are

Blababa [14]

You have two windlasses named X and Y. Windlass X drum diameter is 50cm and windlass Y drum diameter is 75cm. Both are hoisting a 50kg weight, Hence, X will lift faster and easier than Y

<h3>What is windlasses (winches) will lift faster?</h3>

Generally, the equation for the Torque is mathematically given as

Torque(X)=weight x radius of drum

T=50x9.81x0.25

T=122.625 Nm

For, Torque (Y)

T=50x9.81x0.375

T=183.93 Nm

Generally, the equation for the Equation of Power is mathematically given as

P= Torque * Angular velocity

Angular velocity (w)= 2 x pi x N/60

As a consequence, torque has an inverse relationship to speed. Because of this, the torque will decrease as the speed rises.

According to the previous statement, windlass X has a speed of lift that is at its maximum, but windlass Y has a speed that is very low.

Additionally, power is exactly proportional to torque, which means that the highest possible torque required the highest possible amount of energy and vice versa.

Therefore, X should raise at the highest possible speed while using the least amount of energy.

In conclusion, X will lift faster and easier than Y