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3 years ago

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an object of mass 6000 kg rests on the flatbed of a truck. it is held in place by metal brackets that can exert a maximum horizo

ntal force of 9000 N. when the truck is traveling 15 m/s, what is the minimum stopping distance if the load is not to slide forward into the cab?

1 answer:

Otrada [13]3 years ago

4 0

Answer:

minimum stopping distance will be d = 75 m

Explanation:

Maximum force exerted by the bracket is given as

F = 9000 N

now we know that mass of the object is

m = 6000 kg

so the maximum acceleration that truck can have is given as

$F = ma$

$9000 = 6000 a$

$a = 1.5 m/s^2$

now for finding minimum stopping distance of the truck

$v_f^2 - v_i^2 = 2a d$

$0^2 - 15^2 = 2(-1.5) d$

$d = 75 m$

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Answer:

The value is $t = 689.029 \ hours$

Explanation:

From the question we are told that

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The initial temperature is $T_i = -186 ^o F = (-186-32) *\frac{5}{9} +273.15 = 152\ K$

The final temperature is $T_f = 78 ^o F = (78-32) *\frac{5}{9} +273.15 = 298.7 \ K$

The specific heat capacity is $c_h = 0.099 [kJ/(mol K)] = 0.099 *10^3 J/(mol/K)$

The power available is $P = 300 \ W$

The mass of the fuel is $m = 1640 \ kg$

Generally the number of moles of hydrazine present is

$n = \frac{m}{Z}$

=> $n = \frac{1640}{= 0.032}$

=> $n = 51250 \ mol$

Generally the quantity of heat energy needed is mathematically represented as

$Q = n * c_h * (T_f -T_i)$

=> $Q = 51250 * 0.099 *10^3 * (298.7 - 152)$

=> $Q = 7.441516913 * 10^{8} \ J$

Generally the time taken is mathematically represented as

$t = \frac{Q}{P}$

=> $t = \frac{7.441516913 * 10^{8} }{300}$

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6 0

3 years ago

Please help me with this question

valkas [14]

Answer: m∠P ≈ 46,42°

because using the law of sines in ΔPQR

=> sin 75°/ 4 = sin P/3

so ur friend is wrong due to confusion between edges

+) we have: sin 75°/4 = sin P/3

=> sin P = sin 75°/4 . 3 = (3√6 + 3√2)/16

=> m∠P ≈ 46,42°

Explanation:

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3 years ago

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soldi70 [24.7K]

Answer:

2 charges of electron (2C)

Explanation:

I = Q/t

2 = Q/1

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Q = 2 charge of electron

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Which type of waves from the electromagnetic spectrum are used in heat lamps and heat sensing devices?

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Infrared waves are used in heat lamps and other heat sensing devices. Infrared waves or commonly known as Infrared radiations (IR) is the type of electromagnetic radiation we encounter most in our everyday life. Heat lamps are electrical devices which emit infrared radiation.

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Read 2 more answers

A 4 kg textbook sits on a desk. It is pushed horizontally with a 50 N applied force against a 15 N frictional force.

GarryVolchara [31]

a) See free-body diagram in attachment

b) The book is stationary in the vertical direction

c) The net horizontal force is 35 N in the forward direction

d) The net force on the book is 35 N in the forward horizontal direction

e) The acceleration is $8.75 m/s^2$ in the forward direction

Explanation:

a)

The free-body diagram of a body represents all the forces acting on the body using arrows, where the length of each arrow is proportional to the magnitude of the force and points in the same direction.

From the diagram of this book, we see there are 4 forces acting on the book:

- The applied force, F = 50 N, pushing forward in the horizontal direction

- The frictional force, $F_f = 15 N$ , pulling backward in the horizontal direction (the frictional force always acts in the direction opposite to the motion)

- The weight of the book, $W=mg$ , where m is the mass of the book and $g=9.8 m/s^2$ is the acceleration of gravity, acting downward. We can calculate its magnitude using the mass of the book, m = 4 kg:

$W=(4)(9.8)=39.2 N$

- The normal reaction exerted by the desk on the book, N, acting upward, and balancing the weight of the book

b)

The book is in equilibrium in the vertical direction, therefore there is no motion.

In fact, the magnitude of the normal reaction (N) exerted by the desk on the book is exactly equal to the weight of the book (W), so the equation of motion along the vertical direction is

$N-W=ma$

where a is the acceleration; however, since N = W, this becomes

$a=0$

And since the book is initially at rest on the desk, this means that there is no motion.

c)

We said there are two forces acting in the horizontal direction:

- The applied force, F = 50 N, forward

- The frictional force, $F_f = 15 N$ , backward

Since they act along the same line, we can calculate their resultant as

$\sum F = F - F_f = 50 - 15 = 35 N$

and therefore the net force is 35 N in the forward direction.

d)

The net force is obtained as the resultant of the net forces in the horizontal and vertical direction. However, we have:

- The net force in the horizontal direction is 35 N

- The net force in the vertical direction is zero, because the weight is balanced by the normal reaction

Therefore, this means that the total net force acting on the book is just the net force acting on the horizontal direction, so 35 N forward.

e)

The acceleration of the book can be calculated by using Newton's second law:

$\sum F = ma$

where

$\sum F$ is the net force

m is the mass

a is the acceleration

Here we have:

$\sum F = 35 N$ (in the forward direction)

m = 4 kg

Therefore, the acceleration is

$a=\frac{\sum F}{m}=\frac{35}{4}=8.75 m/s^2$ (forward)

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