What is matter? explain and give example

Which option correctly describes a model of methanol. URGENT

STALIN [3.7K]

The last one.

One grey sphere, Four white spheres, and One red spheres

4 0

3 years ago

A proton (mass=1.67x10^-27 kg, charge= 1.60x10^-19 C) moves from point A to point under the influence of an electrostatic force

Tom [10]

Answer:

$VB - VA = - 33.4$

Explanation:

Generally the workdone in moving the proton is mathematically represented as

$W = KE_f - KE_i$

Where $KE_i \ and \ KE_f \ are\ the\ initial \ and \ final \ kinetic \ energy$

So

$KE_i = \frac{1}{2} m v_a^2$

Here $v_a$ is the velocity at A with value 50 m/s

So

$KE_i = \frac{1}{2} (1.67*10^{-27}) * 50^2$

$KE_i = 2.09 *10^{-24} \ J$

Also

$KE_f = \frac{1}{2} m v_b^2$

Here $v_a$ is the velocity at A with value $80 km/s = 80000 m/s$

=> $KE_f = \frac{1}{2} (1.67*10^{-27}) * 80000^2$

=> $KE_f = 5.34 *10^{-18} \ J$

So

$W = 5.34 *10^{-18} - 2.09 *10^{-24}$

$W = 5.34 *10^{-18} m/s$

Now this workdone is also mathematically represented as

$W = q * V$

So

$q * V = 5.34 *10^{-18}$

Here $q = 1.60*10^{-19} C$

So

$V = \frac{5.34 *10^{-18} }{1.60*10^{-19}}$

$V = 33.4 \ V$

Generally proton movement is in the direction of the electric field it means that $VA>VB$

So

$VB - VA = - 33.4$

8 0

3 years ago

A large airplane typically has three sets of wheels: one at the front and two farther back, one on each side under the wings. Co

Tems11 [23]

(a) The force the ground exerts on each set of rear wheels when the plane is at rest on the runway is 0.743 MN.

(b) The force the ground exerts on the front set of wheels is 0.239 MN.

<h3>Center mass of the airplane</h3>

The concept of center mass of an object can be used to dtermine the mass distribution of the airplane along the line through the center.

<h3>Some assumptions</h3>

The wheels under the wind do not pass through the center line.
The position of the front wheel is constant and it is zero mark (origin).
The rear wheels are at 21.7 m mark

Position of the center mass of the plane is calculated as follows;

Let the position of the center mass, Xcm = y

the center mass is 3 m in front of rear wheels, that is

21.7 - y = 3

y = 21.7 - 3

y = 18.7 m

Xcm = 18.7 m

<h3>Mass of the plane at the position of the rear wheels</h3>

Let the mass of the plane at front wheels = M1

Let the mass of the plane at rear wheels = M2

$X_{cm} = \frac{M_1x_1 + M_2x_2}{M_1 + M_2}$

$18.7 = \frac{M_1(0) + M_2(21.7)}{177000} \\\\3,309,900 = 21.7M_2\\\\M_2 = \frac{3,309,900}{21.7} \\\\M_2 = 152,529.95 \ kg$

<h3>Force exerted by the ground on each rear wheel</h3>

There are two rear wheels, and the force exerted on each wheel due to mass of the airplane at this position is calculated as follows;

$W = mg\\\\W_1 = W_2 = \frac{1}{2} (mg) = \frac{1}{2} (152,529.95 \times 9.8) = 743,396.76 \ N= 0.743 \ MN$

<h3>Mass of the plane at the position of the front wheel</h3>

M1 + M2 = 177,000

M1 = 177,000 - M2

M1 = 177,000 - 152,529.95

M1 = 24,470.05 kg

<h3>Force exerted by the ground on the front wheel</h3>

W = mg

W = 24,470.05 x 9.8

W = 239,806.5 N = 0.239 MN

Learn more about center mass here: brainly.com/question/13499822

7 0

2 years ago

if the time is taken to bring a ball to rest from a certain velocity 'v'. is reduced to half, what will be the change in the val

777dan777 [17]

The answer is C) rate of change of momentum. The answer is not initial or final momentum as the start and end points are not changing. On the other hand, the time it takes for the ball to change velocity is. This change relates to the change of momentum. Hope this helped :))

6 0

3 years ago

a spring is compressed 15 centimeters when a 8.5 kilogram weight is set upon it. how much power would be needed to stretch out t

o-na [289]

Answer:

159.38 Watts

Explanation:

Initially;

Mass on the spring is 8.5 kg
Therefore, compression force is 85 N
Compression distance is 15 cm or 0.15 m

But;

F = kx

where F is the force of compression, k is the spring constant and x is the compression distance.

Thus;

k = F/x

= 85 N ÷0.15

= 566.67 N/m

We are required to determine the power needed to stretch the same spring for 1.5 m in 4 secs.

Power = Work done ÷ time

Work done is given by 0.5kx²

Therefore;

Power = 0.5kx²÷ t

= (0.5×566.67 N/m × 1.5² ) ÷ 4 seconds

= 159.38 Watts

Thus, the power needed is 159.38 watts

5 0

3 years ago