The location where an earthquake begins

Which workout schedule would be BEST suited for a weight-lifting workout that builds both strength and endurance

VLD [36.1K]

My workout schedule is day on - day off

This helps the body rest after the work out and let it gain/build muscle mass

Good luck

4 0

4 years ago

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A record is spinning at the rate of 25rpm. If a ladybug is sitting 10cm from the center of the record.

marin [14]

A) Angular speed: 0.42 rev/s

B) Frequency: 0.42 Hz

C) Tangential speed: 26.4 cm/s

D) Distance travelled: 528 cm

Explanation:

A)

In this problem, the ladybug is rotating together with the record.

The angular velocity of the ladybug, which is defined as the rate of change of the angular position of the ladybug, in this problem is

$\omega = 25 rpm$

where here it is measured in revolutions per minute.

Keeping in mind that

1 minute = 60 seconds

We can rewrite the angular speed in revolutions per second:

$\omega = 25 \frac{rev}{min} \cdot \frac{1}{60 s/min}=0.42 rev/s$

B)

The relationship between angular speed and frequency of revolution for a rotational motion is given by the equation

$\omega = 2 \pi f$ (1)

where

$\omega$ is the angular speed

f is the frequency of revolution

For the ladybug in this problem,

$\omega=0.42 rev/s$

Keeping in mind that $1 rev = 2\pi rad$ , the angular speed can be rewritten as

$\omega = 0.42 \frac{rev}{s} \cdot 2\pi = 2\pi \cdot 0.42$

And re-arranginf eq.(1), we can find the frequency:

$f=\frac{\omega}{2\pi}=\frac{(2\pi)0.42}{2\pi}=0.42 Hz$

And the frequency is the number of complete revolutions made per second.

C)

For an object in circular motion, the tangential speed is related to the angular speed by the equation

$v=\omega r$

where

$\omega$ is the angular speed

v is the tangential speed

r is the distance of the object from the axis of rotation

For the ladybug here,

$\omega = 2\pi \cdot 0.42 rad/s$ is the angular speed

r = 10 cm = 0.10 m is the distance from the center of the record

So, its tangential speed is

$v=(2\pi \cdot 0.42)(0.10)=0.264 m/s = 26.4 cm/s$

D)

The tangential speed of the ladybug in this motion is constant (because the angular speed is also constant), so we can find the distance travelled using the equation for uniform motion:

$d=vt$

where

v is the tangential speed

t is the time elapsed

Here we have:

v = 26.4 cm/s (tangential speed)

t = 20 s

Therefoe, the distance covered by the ladybug is

$d=(26.4)(20)=528 cm$

Learn more about circular motion:

brainly.com/question/9575487

brainly.com/question/9329700

brainly.com/question/2506028

#LearnwithBrainly

7 0

4 years ago

Read 2 more answers

Determine the value of the force exerted by the surface (normal force) on a

Advocard [28]

Answer:

53.5 N

Explanation:

Vertical component of the F force 50 sin30 = 25 N upward

force of gravity = m g = 8 * 9.81 =78.5 N Downward

NET downward force by block on table = net upward force exerted by table = 78.5 -25 =53.5 N

8 0

2 years ago

A 2cm length of wire centered on the origin carries a 20A current directed in the positive y direction. Determine the magnetic f

skad [1K]

Answer:

The magnetic field at a distance x = 5 m is 1.59 nT

Explanation:

Length of the wire, L = 2 cm = 0.02 m

Current, I = 20 A

x = 5 m

Magnetic field at a distance x = 5 m due to an infinitely long wire is given by:

$B = \frac{\mu_{o}IL}{4\pi x\sqrt{x^{2} + L^{2}}}$

$B = \frac{4\pi\times 10^{- 7}\times 20\times 0.02}{4\pi \times 5\sqrt{5^{2} + 0.02^{2}}} = 1.59\times 10^{- 9}\ T$

3 0

3 years ago

Briefly outline the caloric theory about the nature of heat

Stells [14]

Briefly outline the caloric theory about the nature of heat

4 0

3 years ago