Why do people eat bo oty

B)

Vlad1618 [11]

Answer:

a) 16m/s b) 192m

Explanation:

v1=32m/s a=-2m/s^2 t=8s v2=? d=??

a) I will use this equation v2= v1 + a*t

v2= 32m/s + -2m/s^2 * 8s

v2= 32m/s + -16m/s

v2= 16m/s

b) v2^2=v1^2 + 2ad

rearranging

v2^2-v1^2=2ad

v2^2-v1^2/2= a d

v2^2-v1^2/2a=d

16m/s^2 - 32m/s^2/ 2 x-2m/s^2 =d

d=192m

5 0

3 years ago

How does the endurance enhance performance in physical activities

White raven [17]

it's Also called aerobic exercise, endurance exercise includes activities that increase your breathing and heart rate such as walking, jogging, swimming, and biking. Endurance activity keeps your heart, lungs and circulatory system healthy and improves your overall fitness.

5 0

3 years ago

1. If an object that stands 3 centimeters high is placed 12 centimeters in front of a plane

igor_vitrenko [27]

Answer:

1. 12 cm

2. 0.133 m

3. 0.03 m

4. Plane mirror

Virtual image

Upright

Behind the mirror

The same size as the object

Concave mirror when the object is located a distance greater than the focal length from the mirror's surface

Real image

Inverted image

In front of the the mirror

Diminished when the object is beyond the center of curvature

Same size as object when the object is placed at the center of curvature

Enlarged when the object is placed between the center of curvature of the mirror and the focus of the mirror

Concave mirror when the object is located a distance less than the focal length from the mirror's surface

Virtual image

Upright image

Behind the the mirror

Enlarged

Convex mirror

Type = Virtual image

Appearance = Upright image

Placement = Behind the mirror

Size = Smaller than the object

Explanation:

1. For plane mirror, since there is no magnification, the virtual image distance from the mirror = object distance from the mirror = 12 cm behind the mirror

2. The height of the object = 0.3 m

The distance of the object from the mirror = 0.4 meters

Height of image formed = 0.1 meter

We have;

$Magnification, \ m = \dfrac{Image \ height }{Object \ height } = \dfrac{Image \ distance \ from \ mirror }{Object\ distance \ from \ mirror }$

$m = \dfrac{0.1}{0.3 } = \dfrac{Image \ distance \ from \ mirror }{0.4 }$

Image distance from the mirror = 0.1/0.3×0.4 = 2/15 = 0.133 m

Image distance from the mirror = 0.133 m

3. $m = \dfrac{Image \ height}{0.10 } = \dfrac{0.06 }{0.20 }$

The image height = 0.06/0.2×0.1 = 3/100 = 0.03 meter

The image height = 0.03 meter

4. Plane mirror

Type = Virtual image

Appearance = Upright image with the left transformed to right

Placement = Behind the mirror

Size = The same size as the object

Concave mirror when the object is located a distance greater than the focal length from the mirror's surface

Type = Real image

Appearance = Inverted image

Placement = In front of the the mirror

Size = Diminished when the object is beyond the center of curvature

Same size as object when the object is placed at the center of curvature

Enlarged when the object is placed between the center of curvature of the mirror and the focus of the mirror

Concave mirror when the object is located a distance less than the focal length from the mirror's surface

Type = Virtual image

Appearance = Upright image

Placement = Behind the the mirror

Size = Enlarged

Convex mirror

Type = Virtual image

Appearance = Upright image

Placement = Behind the mirror

Size = Smaller than the object.

3 0

4 years ago

What’s the answer to the summary ?

Talja [164]

Because of the different speeds..

7 0

3 years ago

An electron enters a region of space containing a uniform 2.71 × 10 − 5 2.71×10−5 T magnetic field. Its speed is 197 197 m/s and

Andrews [41]

Answer:

r = 0.0414mm

F = 757,692.3Hertz

Explanation:

If the body enters space with uniform magnetic field B, the force experienced by the object is expressed as

F = qvBsintheta... 1

Also, if the body undergoes a circular motion, the force experienced by the body in a circular path is given as

Fc = mv²/r... 2

Equating both forces

F = Fc

qvBsin theta = mv²/r

Since the body enters perpendicular to the field, theta = 90°

The equality becomes;

qvB sin90° = mv²/r

qvB = mv²/r

qB = mv/r

r = mv/qB

Given mass of the electron m = 9.11×10^-31kg

Velocity of the object v = 197m/s

Charge on the electron q = 1.6×10^-19C

Magnetic field B = 2.71×10^-5T

Substituting this value into the equation to get the radius r we have;

r = 9.11×10^-31 × 197/1.6×10^-19 × 2.71×10^-5

r = 1794.67×19^-31/4.336×10^-24

r = 413.89×10^-7

r = 0.0000414m

r = 0.0414mm

b) Frequency of the motion F = w/2π where w is the angular velocity

Since w = v/r

F = (v/r)/2π

F = v/2πr

F = 197/2π(0.0000414)

F = 757,692.3Hertz

6 0

4 years ago