Which fact is true during a fission chain reaction?

What will happen if a car experiences a 300 N force to the right from the engine and a separate 150 N force due to friction and

gayaneshka [121]

Force applied on the car due to engine is given as

$F_1 = 300 N$ towards right

Also there is a force on the car towards left due to air drag

$F_2 = 150 N$ towards left

now the net force on the car will be given as

$\vec F_{net} = \vec F_1 + \vec F_2$

now we can say that since the two forces are here opposite in direction so here the vector sum of two forces will be the algebraic difference of the two forces.

So we can say

$F_{net} = F_1 - F_2$

$F_{net} = 300 - 150$

$F_[net} = 150 N$

So here net force on the car will be 150 N towards right and hence it will accelerate due to same force.

5 0

2 years ago

A man attempts to pick up his suitcase of weight ws by pulling straight up on the handle. However, he is unable to lift the suit

kvasek [131]

Answer:

b)

Explanation:

Normal force, is always directed upward the surface over which is placed the object, and can adopt any value, as required to meet Newton's 2nd Law.

In this case, as the external force on the suitcase pulls upward, in order to counteract the influence of gravity, normal force is less than the weight of the suitcase, as follows:

F + Fn = m*g

⇒ Fn = m*g - F

So, the normal force is equal to the magnitude of the weight of the suitcase (m*g) minus the magnitude of the force of the pull (F) which is the same expressed by the statement b.

4 0

2 years ago

Which of the following is not an advantage that attractive people enjoy in interpersonal relationships?

Licemer1 [7]

Psychology on Egenuity Oct 5th 2018 says answer is C

5 0

3 years ago

Read 2 more answers

Find the mass of an object if a 40 N of force causes the object to accelerate at 5.5 m/s/s

Murrr4er [49]

Answer:

<h3>The answer is 8 kg</h3>

Explanation:

The mass of the object can be found by using the formula

$m = \frac{f}{a} \\$

f is the force

a is the acceleration

From the question we have

$m = \frac{40}{5} \\$

We have the final answer as

Hope this helps you

7 0

2 years ago

A traffic light is weighing 200N hangs from a vertical cable tied to two other cables that are fastened to a support. The upper

Levart [38]

Answer:

T₁ = 93.6 N , T₂ = 155.6 N , T₃ = 200 N

Explanation:

This is a balance exercise where we must apply the expressions for translational balance in the two axes

∑ F = 0

Suppose that cable t1 goes to the left and the angles are 41º with respect to the horizontal and cable t2 goes to the right with angles of 63º

decompose the tension of the two upper cables

cos 41 = T₁ₓ / T1

sin 41 = T₁y / T1

T₁ₓ = T₁ cos 41

T₁y= T₁ sin 41

for cable gold

cos 63 = T₂ / T₂

sin 63 = $T_{2y}$ / T₂

We apply the two-point equilibrium equation: The junction point of the three cables and the point where the traffic light joins the vertical cable.

Let's start by analyzing the point where the traffic light meets the vertical cable

T₃ - W = 0

T₃ = W

T₃ = 200 N

now let's write the equations for the single point of the three wires

X axis

- T₁ₓ + T₂ₓ = 0

T₁ₓ = T₂ₓ

T1 cos 41 = T2 cos 63

T1 = T2 cos 63 / cos 41 (1)

y Axis

$T_{1y}$ + T_{2y} - T3 = 0

T₁ sin 41 + T₂ sin 63 = T₃ (2)

to solve the system we substitute equation 1 in 2

T₂ cos 63 / cos 41 sin 41 + T₂ sin 63 = W

T₂ (cos 63 tan 41 + sin 63) = W

T₂ = W / (cos 63 tan 41 + sin 63)

We calculate

T₂ = 200 / (cos 63 tan 41 + sin 63)

T₂ = 200 / 1,2856

T₂ = 155.6 N

we substitute in 1

T₁ = T₂ cos 63 / cos 41

T₁ = 155.6 cos63 / cos 41

T₁ = 93.6 N

therefore the tension in each cable is

T₁ = 93.6 N

T₂ = 155.6 N

T₃ = 200 N

6 0

2 years ago