All categories

3 years ago

Identify the object or objects in motion in each example of kinetic energy.

2 answers:

7 0

A: a speeding train

Reasoning)
The train is getting faster making it in motion, unlike the bowl of hot soup which is an object that is not in motion.

Leto [7]3 years ago

5 0

Answer:

A speedy train

Explanation:

You might be interested in

In a playground, there is a small merry-go-round of radius 1.20 m and mass 160 kg. Its radius of gyration is 91.0 cm. (Radius of

aksik [14]

Answer:

a) 145.6kgm^2

b) 158.4kg-m^2/s

c) 0.76rads/s

Explanation:

Complete qestion: a) the rotational inertia of the merry-go-round about its axis of rotation

(b) the magnitude of the angular momentum of the child, while running, about the axis of rotation of the merry-go-round and

a) From I = MK^2

I = (160Kg)(0.91m)^2

I = 145.6kgm^2

b) The magnitude of the angular momentum is given by:

L= r × p The raduis and momentum are perpendicular.

L = r × mc

L = (1.20m)(44.0kg)(3.0m/s)

L = 158.4kg-m^2/s

c) The total moment of inertia comprises of the merry- go - round and the child. the angular speed is given by:

L = Iw

158.4kgm^2/s = [145kgm^2 + ( 44.0kg)(1.20)^2]

w = 158.6/208.96

w = 0.76rad/s

7 0

3 years ago

What’s the difference between the resistance of 1,000 feet of aluminum No. 14 wire and 1,000 feet of No. 14 copper wire? A. The

Lapatulllka [165]

the answer is A. The aluminum has 0.84 ohms more resistance.

3 0

3 years ago

An electron moving parallel to a uniform electric field increases its speed from 2.0 × 107 m/s to 4.0 × 107 m/s over a distance

jeka94

Answer:

$1.8\times 105 N/C$

Explanation:

We are given that

$u=2\times 10^7 m/s$

$v=4\times 10^7 m/s$

d=1.9 cm= $\frac{1.9}{100}=0.019 m$

Using 1m=100 cm

We have to find the electric field strength.

$v^2-u^2=2as$

Using the formula

$(4\times 10^7)^2-(2\times 10^7)^2=2a(0.019)$

$16\times 10^{14}-4\times 10^{14}=0.038a$

$0.038a=12\times 10^{14}$

$a=\frac{12}{0.038}\times 10^{14}=3.16\times 10^{16}m/s^2$

$q=1.6\times 10^{-19} C$

Mass of electron,m $=9.1\times 10^{-31} kg$

$E=\frac{ma}{q}$

Substitute the values

$E=\frac{9.1\times 10^{-31}\times 3.16\times 10^{16}}{1.6\times 10^{-19}}$

$E=1.8\times 105 N/C$

7 0

4 years ago

Match the major muscle group with its functional role.

Naily [24]

Answer:

Abdominal

Sitting up, postural alignment

Biceps

Lifting, pulling

Deltoids

Overhead lifting

Erector Spinae

Postural alignment

Gastronemius & Soleus

Push off for walking, standing on tiptoes

Gluteus

Climbing stairs, walking, standing up

Hamstrings

Walking

Latissimus Dorsi & Rhomboids

Postural alignment, pulling open a door

Obliques

Rotation and side flexion of body

Pectoralis

Push up, pull up, bench press

Quadriceps

Climbing stairs, walking, standing up

Trapezius

Moves head sideways

Triceps

Pushing

God bless you. Because my soul almost left my body when i had to do this.

7 0

3 years ago

A steel cable with Cross Sectional Area 3.00cm² has an elastic limit of 2.40 x 10^8pascals. Find the maximum upward acceleration

bazaltina [42]

Answer:

Stress = F / A force per unit area

A = 3.00 cm^2 = 3 E-4 m^2

F = 2.4E8 N/m^2 * 3E-4 m^2 = 7.2E4 N max force applied

F/3 = 2.4E4 N if force not to exceed limit (= f)

f = M a

a = 2.4 E4 N / 1.2 E3 kg = 20 m / s^2 about 2 g

3 0

3 years ago