Which of the following describes a seamount?

Which of the following is not a component of skill related fitness

MArishka [77]

What are the options / answers

4 0

3 years ago

An irregular object of mass 3 kg rotates about an axis, about which it has a radius of gyration of 0.2 m, with an angular accele

Artemon [7]

Answer:

0.06 Nm

Explanation:

mass of object, m = 3 kg

radius of gyration, k = 0.2 m

angular acceleration, α = 0.5 rad/s^2

Moment of inertia of the object

$I = mK^{2}$

I = 3 x 0.2 x 0.2 = 0.12 kg m^2

The relaton between the torque and teh moment off inertia is

τ = I α

Wheree, τ is torque and α be the angular acceleration and I be the moemnt of inertia

τ = 0.12 x 0.5 = 0.06 Nm

6 0

3 years ago

The Problems: 1. Xavier starts at a position of 0 m and moves with an average speed of 0.50 m/s for 3.0 seconds. He normally mov

NemiM [27]

Answer:

(1). His final position is 1.5 m.

(2). The final position of the hedgehog is 3 m.

(3). The final position of the tortoise

(4). Her race time is 80 sec.

(5). It take to finish in 5 hr.

Explanation:

(1). Given that,

Initial position = 0 m

Average speed = 0.50 m/s

Time = 3.0 s

We need to calculate the final position

Using formula of average speed

$v_{av}=\dfrac{x_{f}+x_{i}}{t}$

Where, $x_{f}$ = final position

$x_{i}$ = Initial position

t = total time

Put the value into the formula

$0.50=\dfrac{x_{f}+0}{3.0}$

$x_{f}=0.50\times3.0$

$x_{f}=1.5\ m$

(2). Given that,

Initial position = 0 m

Average speed = 0.75 m/s

Time = 4.0 s

We need to calculate the final position

Using formula of average speed

$v_{av}=\dfrac{x_{f}+x_{i}}{t}$

Put the value into the formula

$0.75=\dfrac{x_{f}+0}{4.0}$

$x_{f}=0.75\times4.0$

$x_{f}=3\ m$

(3). Given that,

Average speed = 1.25 m/s

Time = 3.0 sec

Initial position = 1.0 m

We need to calculate the final position

Using formula of average speed

$v=\dfrac{x_{f}+x_{i}}{t}$

Put the value into the formula

$1.25=\dfrac{x_{f}+1.0}{3.0}$

$x_{f}=1.25\times3.0-1.0$

$x_{f}=2.75\ m$

(4). Given that,

Average speed = 1.25 m/s

Distance = 100 m

We need to calculate the time

Using formula of time

$t=\dfrac{d}{v}$

Put the value into the formula

$t=\dfrac{100}{1.25}$

$t=80 sec$

(5). Given that,

Average speed = 5 miles/hr

Suppose, distance = 25 miles

We need to calculate the time

Using formula of time

$t=\dfrac{d}{v}$

Put the value into the formula

$t=\dfrac{25}{5}$

$t=5\ hr$

Hence, (1). His final position is 1.5 m.

(2). The final position of the hedgehog is 3 m.

(3). The final position of the tortoise

(4). Her race time is 80 sec.

(5). It take to finish in 5 hr.

5 0

4 years ago

An object's distance from a converging lens is 3.78 times the focal length. a) Determine the location of the image. Express the

Romashka-Z-Leto [24]

Answer:

a) v=1.36f

b)m=.36

c) real

d) Inverted

Explanation:

a)

In this question we have given,

object distance from converging lens,u=-3.78f

focal length of converging lens,=f

we have to find location of image,v=?

and magnification,m=?

we know that u, v and f are related by following formula

$\frac{1}{f} =\frac{1}{v}- \frac{1}{u}$ .............(1)

put values of f u in equation (1)

we got,

$\frac{1}{f} =\frac{1}{v}- \frac{1}{-3.78f}$

$\frac{1}{f}-\frac{1}{3.78f} =\frac{1}{v}$

$\frac{2.78}{3.78f} =\frac{1}{v}$

or

v=1.36f

b) Magnification, $m=\frac{v}{u} \\m=\frac{1.36f}{3.78f}\\m=.36$

c)

Here the object is located further away from the lens than the focal point therefore image is real image and inverted.

d) image is inverted

7 0

3 years ago

An object 2.7 cmtall is placed12 cmin front of a mirror. •What type of mirror and what radius of curvature are needed to createa

nasty-shy [4]

Answer:

a. Concave mirror, radius of curvature = 16 cm b. magnification = 2

Explanation:

a. Since the image is upright and larger than the object, we need a concave mirror.

We know image height, h'/object height, h = -image distance, d'/object distance, d

h'/h = -d'/d

Using the real is positive convention,

h'= + 5.4 cm, h = + 2.7 cm and d = + 12 cm.

So, + 5.4 cm/+ 2.7 cm = -d'/+ 12 cm

2 = -d'/12

d' = -2 × 12 cm

= -24 cm

Using the mirror formula 1/d + 1/d' = 2/r where r = radius of curvature of the mirror

1/+12 + 1/- 24 = 2/r

1/12(1 - 1/2) = 2/r

1/12(1/2) = 2/r

1/24 = 2/r

r/2 = 24

r = 2 × 24

r = 48 cm

b.

magnification = image height, h'/object height,h = + 5.4 cm/+ 2.7 cm = 2

3 0

3 years ago