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3 years ago

In order for combustion to occur what needs to happen

2 answers:

5 0

Pretty much combustion is the reaction of oxygen to a compound that contains carbon and hydrogen. That's why it creates CO2 and H2O

larisa86 [58]3 years ago

5 0

<span>The ignition point for a given fuel is met.</span>

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A bowling ball has a mass of 5 kg. A student rolls the bowling ball one time at 1 m/s and a second time and 2 m/s. Compare the m

sasho [114]

The answer is 5kg m/s with the second momentum being 10 kg m/s. i took the test and it was right. hope this helps yalls

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3 years ago

The benefits of jumping rope include: (which of these things)

Svetach [21]

Answer:

E. All of the Above

Explanation:

By doing any kind of exercise or physical activity, you are increasing your overall health. Your muscle strengthen because of the jumping and movement of arms. You are more alert because you have to time each jump right in order to keep going. By breathing evenly while jumping, you do help your Cardiorespiratory fitness as well. And of course, you increase your athletic ablility over all with much endurance and practice.

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3 years ago

1. A 100-kg crate is pulled across a warehouse floor using a rope with a force of 250 N at an angle of 45o from the horizontal.

harkovskaia [24]

Answer:

(a) The net force is 80.394 N

The acceleration of the crate is 0.804 m/s²

(b) the final velocity of the crate is 5.02 m/s

Explanation:

Given;

mass of the crate, m = 100 kg

applied force, F = 250 N

angle of inclination, θ = 45°

coefficient of friction, μ = 0.12

Applied force in y-direction, $F_y = Fsin \theta = 250sin45 = 176.78 \ N$

Applied force in x-direction, $F_x = Fcos \theta = 250cos45 = 176.78 \ N$

The normal force is calculated as;

N + Fy -W = 0

N = W - Fy

N = (100 x 9.8) - 176.78

N = 980 - 176.78 = 803.22 N

The frictional force is given by;

Fk = μN

Fk = 0.12 x 803.22

Fk = 96.386 N

(a) The net force is given by;

$F_{net} = F_x - F_k\\\\F_{net} = 176.78-96.386\\\\F_{net} = 80.394 \ N$

Apply Newton's second law of motion;

F = ma

$a = \frac{F_{net}}{m}\\\\ a = \frac{80.394}{100}\\\\ a = 0.804 \ m/s^2$

(b) the velocity of the crate after 5.0 s

$F = ma= \frac{m(v-u)}{t} \\\\Ft =m(v-u)\\\\v-u = \frac{Ft}{m}\\\\ v = \frac{Ft}{m} + u\\\\v = \frac{F_{net}*t}{m} + u\\\\v = \frac{80.394*5}{100} + 1\\\\v = 5.02 \ m/s$