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3 years ago

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Your car is traveling at 70/mi/h (1000ft/s) your sneeze and your eyes close for 0.33 seconds how far did you travel during your

sneeze?

1 answer:

Novay_Z [31]3 years ago

6 0

If you're moving at 70 mi/hr, then you cover 33.88 feet in 0.33 sec. If you're moving at 1,000 ft/sec, then you cover 33 feet in 0.33 sec. 70 mph and 1,000 fps are not equivalent. 1,000 fps is about 682 mph, whereas 70 mph is about 103 fps.

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A child bounces a 52 g superball on the sidewalk. The velocity change of the superball is from 20 m/s downward to 14 m/s upward.

marusya05 [52]

Answer:

Force on superball will be $=9.8222\times 10^{-4}N$

Explanation:

We have given mass of superball m = 52 gram = 0.052 kg

Velocity change from 20 m/sec downward to 14 m/sec upward

Let downward velocity is positive then upward velocity is negative

So downward velocity is + 20 m/sec and upward velocity is -14 m/sec

Time is given as 1800 sec

We know that acceleration is rate of change of velocity

So $a=\frac{20-(-14)}{1800}=0.0188m/sec^2$

According to newton second law

Force = ma = 0.052×0.0188 $=9.8222\times 10^{-4}N$

6 0

3 years ago

Read 2 more answers

You may have noticed runaway truck lanes while driving in the mountains. These gravel-filled lanes are designed to stop trucks t

Sladkaya [172]

Answer:

0.767

Explanation:

The work done on the truck by the frictional drag force is given by

$W=-Fd$

where

F is the magnitude of the frictional force

d = 38.0 m is the maximum displacement allowed for the truck

The negative sign is due to the fact that the force of friction is opposite to the motion of the truck

The force of friction can also be written as:

$F=\mu mg$

where

$\mu$ is the coefficient of kinetic friction between the truck and the lane

m is the mass of the truck

g is the acceleration of gravity

So we can rewrite the work done as

$W=-\mu mg d$ (1)

According to the work-energy theorem, the work done by friction is equal to the change in kinetic energy of the truck:

$W=K_f - K_i = \frac{1}{2}mv^2-\frac{1}{2}mu^2$ (2)

where

v = 0 is the final velocity of the truck

u = 23.9 m/s is the initial velocity of the truck

By combining (1) and (2) we get

$-\frac{1}{2}mu^2 = -\mu mg d$

And solving for $\mu$ , we find the minimum coefficient of kinetic friction able to stop the truck in a distance d:

$\mu = \frac{u^2}{2gd}=\frac{23.9^2}{2(9.8)(38.0)}=0.767$

7 0

3 years ago

A person walks due south from point A for 500 yards and then due west for 300 yards, arriving at point B. Answer the following q

Pepsi [2]

The total displacement of the person walking from point A to point B is 300 yards.

As shown in the figure we can conclude that the required method to calculate the total displacement is the Pythagoras theorem.

<h3>Pythagoras theorem in brief :</h3>

According to the Pythagorean Theorem, the square that represents the hypotenuse, or side of a right triangle that faces the right angle, is equal to the total of the squares on the triangle's legs.(or, in popular algebraic notation, $a^2 + b^2 = c^2$ ).

<h3>Calculation: </h3>

Let,

a = 500

b= 300

Hence by using Pythagoras' theorem

Total displacement of the person = $\sqrt{500^{2} + 300^{2} }$ = $\sqrt{900000}$ = $300$

Thus the total displacement of the person from starting point is 300 yards.

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2 years ago

Lourdes mixes several ingredients in a bowl, creating a cake batter. She holds the bowl up and turns it upside down, causing the

sasho [114]

friction and gravity ....................

4 0

3 years ago

The center of gravity of a loaded truck depends on how the truck is packed. If it is 4.0 m high and 2.4 m wide, and its CG is 2.

Vitek1552 [10]

The slope of the road can be given as the ratio of the change in vertical

distance per unit change in horizontal distance.

The maximum steepness of the slope where the truck can be parked without tipping over is approximately <u>54.55 %</u>.

Reasons:

Width of the truck = 2.4 meters

Height of the truck = 4.0 meters

Height of the center of gravity = 2.2 meters

Required:

The allowable steepness of the slope the truck can be parked without tipping over.

Solution:

Let, <em>C</em> represent the Center of Gravity, CG

At the tipping point, the angle of elevation of the slope = θ

Where;

$tan\left(\theta \right) = \dfrac{\overline{AM}}{\overline{CM}}$

The steepness of the slope is therefore;

$\mathrm{The \ steepness \ of \ the \ slope}= \dfrac{\overline{AM}}{\overline{CM}} \times 100$

Where;

$\overline{AM}$ = Half the width of the truck = $\dfrac{2.4 \, m}{2}$ = 1.2 m

$\overline{CM}$ = The elevation of the center of gravity above the ground = 2.2 m

$\mathrm{The \ steepness \ of \ the \ slope}= \dfrac{1.2}{2.2} \times 100 \approx 54.55\%$

$tan\left(\theta \right) = \mathbf{\dfrac{2.2}{1.2}} = \dfrac{11}{6}$

$Elevation \ of \ the \ road \ \theta = arctan\left( \dfrac{6}{11} \right) \approx 28.6 ^{\circ}$

The maximum steepness of the slope where the truck can be parked is <u>54.55 %</u>.

Learn more here:

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3 0

2 years ago