When you put a book on a table the table pushes back. which newton law

What are the body parts to this figure? Any of the body parts. (Please)

vlabodo [156]

1 - Skull
2 - Mandible
3 - Scapula
4 - Sternum
5 - Ulna
6 - Radius
7 - Pelvis
8 - Femur
9 - Patella
10 - Tibia
11 - Fibula
12 - Metatarsals
13 - Clavicle
14 - Ribs (rib cage)
15 - Humerus
16 - Spinal column
17 - Carpals
18 - Metacarpals
19 - Phalanges
20 - Tarsals
21 - Phalanges

3 0

3 years ago

Suppose that an asteroid traveling straight toward the center of the earth were to collide with our planet at the equator and bu

vlada-n [284]

Answer:

$\frac{1}{10}$ M

Explanation:

To apply the concept of angular momentum conservation, there should be no external torque before and after

As the asteroid is travelling directly towards the center of the Earth, after impact ,it does not impose any torque on earth's rotation, So angular momentum of earth is conserved

⇒ $I_{1} \times W_{1} =I_{2} \times W_{2}$

$I_{1}$ is the moment of interia of earth before impact
$W_{1}$ is the angular velocity of earth about an axis passing through the center of earth before impact
$I_{2}$ is moment of interia of earth and asteroid system
$W_{2}$ is the angular velocity of earth and asteroid system about the same axis

let $W_{1}=W$

since $\text{Time period of rotation}∝$ $\frac{1}{\text{Angular velocity}}$

⇒ if time period is to increase by 25%, which is $\frac{5}{4}$ times, the angular velocity decreases 25% which is $\frac{4}{5}$ times

therefore $W_{1}$ $=$ $\frac{4}{5} \times W_{1}$

$I_{1}=\frac{2}{5} \times M\times R^{2}$ (moment of inertia of solid sphere)

where M is mass of earth

R is radius of earth

$I_{2}=\frac{2}{5} \times M\times R^{2}+M_{1}\times R^{2}$

(As given asteroid is very small compared to earth, we assume it be a particle compared to earth, therefore by parallel axis theorem we find its moment of inertia with respect to axis)

where $M_{1}$ is mass of asteroid

⇒ $\frac{2}{5} \times M\times R^{2} \times W_{1}=}(\frac{2}{5} \times M\times R^{2}$ $+$ $M_{1}\times R^{2})\times(\frac{4}{5} \times W_{1})$

$\frac{1}{2} \times M\times R^{2}$ = $(\frac{2}{5} \times M\times R^{2}$ + $M_{1}\times R^{2})$

$M_{1}\times R^{2}=$ $\frac{1}{10} \times M\times R^{2}$

⇒ $M_{1}=}$ $\frac{1}{10} \times M$

3 0

3 years ago

Pls help Asapppppppppp.It is due in a little bit.

bija089 [108]

C I’m pretty sure !

7 0

3 years ago

Read 2 more answers

A 6.0 kg object undergoes an acceleration of 2.0 m/s2 .

Alenkinab [10]

A. Fnet=ma
6*2=12N of force acting on the object in the direction it is accelerating

B. Fnet=ma
4*2=8N of force action on the object in the direction it is accelerating

6 0

3 years ago

A 100-g toy car moves along a curved frictionless track. At first, the car runs along a flat horizontal segment with an initial

aleksandrvk [35]

Answer:

The final velocity of the car is 2.02 m/s

Explanation:

Hi there!

The kinetic energy of the car as it runs along the first flat horizontal segment can be calculated using the following equation:

KE = 1/2 · m · v²

Where:

KE = kinetic energy

m = mass

v = velocity

Then, the initial kinetic energy will be:

KE = 1/2 · 0.100 kg · (2.77 m/s)²

KE = 0.384 J

When the car gains altitude, it gains potential energy. The amount of gained potential energy will be equal to the loss of kinetic energy. So let´s calculate the potential energy of the car as it reaches the top:

PE = m · g · h

Where:

PE = potential energy.

m = mass

g = acceleration due to gravity.

h = height.

PE = 0.100 kg · 9.8 m/s² · 0.184 m

PE = 0.180 J

Then, the final kinetic energy will be (0.384 J - 0.180 J) 0.204 J

Using the equation of kinetice energy, we can obtain the velocity of the car:

KE = 1/2 · m · v²

0.204 J = 1/2 · 0.100 kg · v²

2 · 0.204 J / 0.100 kg = v²

v = 2.02 m/s

The final velocity of the car is 2.02 m/s

7 0

3 years ago