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3 years ago

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Consider the expression below. Assume m is an integer. 6m(2m + 18) Enter an expression in the box that uses the variable m and m

akes the equation true. (Simplify your answer completely. If no expression exists, enter DNE.)

1 answer:

Anika [276]3 years ago

8 0

6x2=12m
6x18=108
12m+108
Simplified: m+9 bc 12/12 and 108/12

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A proton moving at 3.0 × 10^4 m/s is projected at an angle of 30° above a horizontal plane. If an electric field of 400 N/C is a

GuDViN [60]

Answer:

The time it takes the proton to return to the horizontal plane is 7.83 X10⁻⁷ s

Explanation:

From Newton's second law, F = mg and also from coulomb's law F= Eq

Dividing both equations by mass;

F/m = Eq/m = mg/m, then

g = Eq/m --------equation 1

Again, in a projectile motion, the time of flight (T) is given as

T = (2usinθ/g) ---------equation 2

Substitute in the value of g into equation 2

$T = \frac{2usin \theta}{\frac{Eq}{m}} =\frac{m* 2usin \theta}{Eq}$

Charge of proton = 1.6 X 10⁻¹⁹ C

Mass of proton = 1.67 X 10⁻²⁷ kg

E is given as 400 N/C, u = 3.0 × 10⁴ m/s and θ = 30°

Solving for T;

$T = \frac{(1.67X10^{-27}* 2*3X10^4sin 30}{400*1.6X10^{-19}}$

T = 7.83 X10⁻⁷ s

6 0

3 years ago

The ice cap at the North Pole may be 1000 m thick, with a density of 920 kg/m3. Find the pressure at the bottom and the correspo

bija089 [108]

<span>Pice=920kg/m^3 deltaP=PgH=920kg/m^3 X 9.80665m/s^2 X 1000m = 9022118 Pa P=Po + deltaP=101.325 + 9022 = 9123kPa</span>

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3 years ago

Examine figure 26-1 which area is the convection zone<br><br>A. A<br>B. B<br>C. C<br>D. D

kykrilka [37]

I think the answer is B

4 0

3 years ago

Read 2 more answers

kherson [118]

The friction factor and head loss when velocity is 1m/s is 0.289 and 1.80 × 10^8 respectively. Also, the friction factor and head loss when velocity is 3m/s is 0.096 and 5.3 × 10^8 respectively.

<h3>How to determine the friction factor</h3>

Using the formula

μ = viscosity = 0. 06 Pas

d = diameter = 120mm = 0. 12m

V = velocity = 1m/s and 3m/s

ρ = density = 0.9

a. Velocity = 1m/s

friction factor = 0. 52 × $\frac{0. 06}{0. 12* 1* 0. 9}$

friction factor = 0. 52 × $\frac{0. 06}{0. 108}$

friction factor = 0. 52 × 0. 55

friction factor $= 0. 289$

b. When V = 3mls

Friction factor = 0. 52 × $\frac{0. 06}{0. 12 * 3* 0. 9}$

Friction factor = 0. 52 × $\frac{0. 06}{0. 324}$

Friction factor = 0. 52 × 0. 185

Friction factor $= 0.096$

Loss When V = 1m/s

Head loss/ length = friction factor × 1/ 2g × velocity^2/ diameter

Head loss = 0. 289 × $\frac{1}{2*6. 6743 * 10^-11}$ × $\frac{1^2}{0. 120}$ × $\frac{1}{100}$

Head loss = 1. 80 × 10^8

Head loss When V = 3m/s

Head loss = $0. 096$ × $\frac{1}{1. 334 *10^-10}$ × $\frac{3^2}{0. 120}$ × $\frac{1}{100}$

Head loss = 5. 3× 10^8

Thus, the friction factor and head loss when velocity is 1m/s is 0.289 and 1.80 ×10^8 respectively also, the friction factor and head loss when velocity is 3m/s is 0.096 and 5.3 ×10^8 respectively.

Learn more about friction here:

brainly.com/question/24338873

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4 0

1 year ago

A block of aluminum measures 4.0 cm x 5.0 cm x 2.0 cm is completely immersed in a tank of water.

Aneli [31]

Answer:

upthrust or BUOYANT FORCE =Vdg

volume=LWH

upthrust=(4cm×5cm×2cm)×1g/cm²×g

upthrust=40cm³×1g/cm³×g

upthrust=40gf or 0.04kg×10m/s²=0.4N

weight of the displaced liquid is upthrust.

so mass=40g or 0.04kg

upthrust=40gf or 0.4Nand mass of the displaced liquid=40g or 0.04kg

please mark brainliest, hope it helped

7 0

2 years ago