(SCIENCE) 22 POINTS AND CROWN ASAP

Let v1, , vk be vectors, and suppose that a point mass of m1, , mk is located at the tip of each vector. The center of mass for

g100num [7]

Answer:

Explanation:

Center of mass is give as

Xcm = (Σmi•xi) / M

Where i= 1,2,3,4.....

M = m1+m2+m3 +....

x is the position of the mass (x, y)

Now,

Given that,

u1 = (−1, 0, 2) (mass 3 kg),

m1 = 3kg and it position x1 = (-1,0,2)

u2 = (2, 1, −3) (mass 1 kg),

m2 = 1kg and it position x2 = (2,1,-3)

u3 = (0, 4, 3) (mass 2 kg),

m3 = 2kg and it position x3 = (0,4,3)

u4 = (5, 2, 0) (mass 5 kg)

m4 = 5kg and it position x4 = (5,2,0)

Now, applying center of mass formula

Xcm = (Σmi•xi) / M

Xcm = (m1•x1+m2•x2+m3•x3+m4•x4) / (m1+m2+m3+m4)

Xcm = [3(-1, 0, 2) +1(2, 1, -3)+2(0, 4, 3)+ 5(5, 2, 0)]/(3 + 1 + 2 + 5)

Xcm = [(-3, 0, 6)+(2, 1, -3)+(0, 8, 6)+(25, 10, 0)] / 11

Xcm = (-3+2+0+25, 0+1+8+10, 6-3+6+0) / 11

Xcm = (24, 19, 9) / 11

Xcm = (2.2, 1.7, 0.8) m

This is the required center of mass

6 0

3 years ago

When atoms are split, they release energy. this concept applies to

balu736 [363]

This applies to nuclear reactions, specifically nuclear fission.

This huge release of energy has been used in atomic bombs and in the nuclear reactors that generate electricity.

6 0

3 years ago

Q 1 . How many significant figures are in the following measurement? 0.0009(1 point)

Crazy boy [7]

Here we have some questions about experimental errors.

Q1) We want to see how many significant figures have the measure:

0.0009

The number of significant figures is the number of known digits that are not the leading zeros.

Here we can see four leading zeros, and a single-digit different than zero, which is a 9.

Then we have only one significant figure, the 9.

Q2) Here we will use the measure that is the less exact, as the error of that measure may be larger than the smaller significant figures of the other measures.

Then:

31.2 lb + 38.02lb + 45 lb

The worst measure is 45lb, so the smallest significant figure that we should use is the first one at the left of the decimal point, then we need to round the other two measures to the next whole number, we will get:

31 lb + 38 lb + 45 lb = 114lbs

Q3) We know that the measure is 11.5 seconds and the uncertainty of 1.7%, then the uncertainty will be the 1.7% of the above measure:

(1.7%/100%)*11.5 s = 0.1955 s

Notice that our measure has one significant figure after the decimal point, so we need to round the error to the same significant figure.

0.1955 s ≈ 0.2s

Then the measure is:

11.5 s ± 0.20 s

Q4) We have the measure:

312.0 mph ± 3.9 mph.

The percent uncertainty will be the quotient between the error and the measure times 100%, or:

(3.9 mph/312.0 mph)*100% = 1.25%

This is a percent error, we do not need to round this.

If you want to learn more, you can read:

brainly.com/question/17339020

5 0

2 years ago

A pickup truck is traveling down the highway at a steady speed of 30.1 m/s. The truck has a drag coefficient of 0.45 and a cross

Sav [38]

Answer:

The energy that the truck lose to air resistance per hour is 87.47MJ

Explanation:

To solve this exercise it is necessary to compile the concepts of kinetic energy because of the drag force given in aerodynamic bodies. According to the theory we know that the drag force is defined by

$F_D=\frac{1}{2}\rhoC_dAV^2$