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3 years ago

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A man claims that he can hold onto a 13.0-kg child in a head-on collision as long as he has his seat belt on. Consider this man

in a collision in which he is in one of two identical cars each traveling toward the other at 59.0 mi/h relative to the ground. The car in which he rides is brought to rest in 0.11 s.

1 answer:

Softa [21]3 years ago

4 0

Answer:

F = 3116.49 N

Explanation:

given,

mass of the child = 15 kg

speed = 59 mi/hr

time taken to come to rest = 0.11 s

force on the boy = ?

1 mi/hr = 0.447 m/s

59 mi/hr = 59 × 0.447 = 26.37 m/s

$a = \dfrac{dv}{dt}$

$a = \dfrac{26.37}{0.11}$

a=239.73 m/s²

the calculation of force

Force = mass × acceleration

F = m × a

F = 13 × 239.73

F = 3116.49 N

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An elevated water tank holds 1 million gallons of water. Determine the force (in lbf) that the tower structure must be able to w

bixtya [17]

Answer:

8,345,925 lb-f

Explanation:

We know pressure P = F/A = ρgh ⇒ F = ρghA = ρgV

F = ρgV where F = force, ρ = density of water = 1000 kg/m³ , g = acceleration due to gravity = 9.8 m/s² and V = 1 × 10⁶ gallons = 1 × 10⁶ gallons × 1m³/264.2 gallons = 3785 m³

So, F = ρgV = 1000 kg/m³ × 9.8 m/s² × 3785 m³ = 37,093,000 N

We now convert Newtons to pound-force.

1 N = 0.225 lb-f

So, F = 37,093,000 N = 37,093,000 N × 0.225 lb-f/1 N = 8,345,925 lb-f

So, the tower must be able to withstand 8,345,925 lb-f

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3 years ago

A 1232 kg car moving north at 25.6 m/s collides with a 2028 kg car moving north at 17.5 m/s . They stick together. In what direc

Citrus2011 [14]

Answer:

I. Angle = 41.7° Northeast.

II. Vr = 7.08m/s

Explanation:

Let the two cars be denoted by A and B

<u>Given the following data;</u>

Mass of car A = 1232 Kg

Velocity of car A = 25.6 m/s

Mass of car B = 2028 Kg

Velocity of car B = 17.5m/s

First of all, we would solve for momentum;

Momentum = mass × velocity

Momentum, M1 = 1232 × 25.6

Momentum, M1 = 31539.2 Kgm/s

Momentum, M2 = 2028 × 17.5

Momentum, M2 = 35490 Kgm/s

Now, let's find the resultant momentum using the Pythagoras theorem;

R² = M1² + M2²

R² = 31539.2² + 35490²

R² = 994721136.6 + 1259540100

R² = 2254261237

Taking the square root of both sides, we have

Resultant momentum, R = 47479.06 Kgm/s

To find the direction;

Angle = tan¯¹(M1/M2)

Angle = tan¯¹(31539.2/35490)

Angle = tan¯¹(0.89)

<em>Angle = 41.7° Northeast.</em>

To find the speed;

R = (M1 + M2)Vr

47479.06 = (31539.2 + 35490)Vr

47479.06 = 67029.2Vr

Vr = 47479.06/67029.2

<em>Vr = 7.08m/s</em>

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3 years ago

A homeowner is trying to move a stubborn rock from his yard. By using a a metal rod as a lever arm and a fulcrum (or pivot point

pogonyaev

Answer:

L = 1.545 m

Explanation:

Let the total length of the rod is L

now the torque must applied on the other end of the rod so that it will balance the torque due to weight of rock on other side of fulcrum

so we will have

$mg \times d = F(L - d)$

so we have

$325\times 9.8 \times 0.266 = F(L - 0.266)$

F = 663 N

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3 years ago

Distance in scientific notation

n200080 [17]

Answer:

(See explanation for further information)

Explanation:

a) The distances of each planet with respect to the Sun is:

Mercury - $57.909 \times 10^{6}\,mi$

Venus - $67.240\times 10^{6}\,mi$

Earth - $92.960\times 10^{6}\,mi$

Mars - $141.600\times 10^{6}\,mi$

Jupiter - $483.800\times 10^{6}\,mi$

Saturn - $888.200\times 10^{6}\,mi$

Uranus - $1,787\times 10^{6}\,mi$

Neptune - $2,795\times 10^{6}\,mi$

b) The solutions are presented below:

A. The distance between Venus and Jupiter is:

$\Delta s = 483.800\times 10^{6}\,mi-67.240\times 10^{6}\,mi$

$\Delta s = 416.560\times 10^{6}\,mi$

B. The combined distance from the Sun is:

$\Delta s = 57.909\times 10^{6}\,mi + 67.240\times 10^{6}\,mi +92.960\times 10^{6}\,mi$

$\Delta s = 218.109\times 10^{6}\,mi$

Which is less than the distance from the Sun to Neptune ( $2,795\times 10^{6}\,mi$ ).

C. The new distance of Earth is $929.60 \times 10^{6}\,mi$ ( $929,600,000\,mi$ ). Saturn would be the closest planet to Earth, whose distance:

Scientific notation:

$\Delta s = 929.600\times 10^{6}\,mi-888.200\,\times 10^{6}\,mi$

$\Delta s = 41.4\times 10^{6}\,mi$

Standard notation:

$\Delta s = 929,600,000\,mi - 888,200,000\,mi$

$\Delta s = 41,400.000\,mi$

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4 years ago

Which pairs of angles in the figure below are verticals angles ? check all that apply.

Dmitriy789 [7]

Answer:

mark two options:

B. TSN and ISW

D. ISN and TSW

Explanation:

Recall the definition of vertical angles as: <em>those angles opposed by the vertex, and which are the result of two lines that intersect</em>.

In the figure given there are two pairs such angles, one pair I depicted in red, and the other one in blue, while the intersecting lines are shown in green. Notice that when two lines intersect, there are always two pairs of vertical angles being generated.

Therefore, according to the notation given, two boxes need to be checked in the set of options:

1) the one corresponding to : TSN and ISW (blue ones in the attached image)

and

2) the one corresponding to ISN and TSW (red ones in the attached image)

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