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3 years ago

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A zebra drinks from a watering hole and is ambushed by a crocodile. What's more important and why: the zebra having a high maxim

um speed or a high acceleration?

1 answer:

elena55 [62]3 years ago

5 0

Answer:

before this type of attack, high acceleration is the most important thing.

Explanation:

As the zebra is ambushed by the crocodile the most important thing is a quick reaction, in this attack the most likely is that the crocodile is in the water so it cannot run after the zebra.

Consequently, before this type of attack, high acceleration is the most important thing.

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ludmilkaskok [199]

Answer:

1) ironing a shirt 2) writing on surfaces 3) working of an eraser

5 0

3 years ago

A large refrigerator (mass 80kg) sits at rest inside a house. The homeowner wants to move the refrigerator across the room, so s

Liono4ka [1.6K]

Answer:

400 N

Explanation:

By the law of friction,

$F=\mu R$

$F$ is the maximum frictional force, $\mu$ is the coefficient of friction and $R$ is the reaction on the refrigerator. On a horizontal surface, the reaction is equal to the weight of the refrigerator.

$R=mg$

$F=\mu mg$

While not moving, the fricition on the refrigerator is static friction. So, $\mu=0.65$

$F=0.65 \times80\times9.8=509.6 \text{ N}$

This is the maximum frictional force and is more than the applied horizontal force of 400 N. Frictional force cannot be more than the applied force, else it would actually pull the refrigerator backwards (a strange thing, if it were to happen). It is equal to the extent of the applied force because the applied force is not enough to overcome the maximum.

Hence the frictional force is 400 N.

PS: Note that we do not use the coefficient of kinetic friction because applied force could not overcome the static friction.

6 0

4 years ago

Where is the near point of an normal eye when accidentally wear a contact lens with a power of +2.0 diopters?

Lerok [7]

Answer:

The near point of an eye with power of +2 dopters, u' = - 50 cm

Given:

Power of a contact lens, P = +2.0 diopters

Solution:

To calculate the near point, we need to find the focal length of the lens which is given by:

Power, P = $\frac{1}{f}$

where

f = focal length

Thus

f = $\frac{1}{P}$

f = $\frac{1}{2}$ = + 0.5 m

The near point of the eye is the point distant such that the image formed at this point can be seen clearly by the eye.

Now, by using lens maker formula:

$\frac{1}{f} = \frac{1}{u} + \frac{1}{u'}$

where

u = object distance = 25 cm = 0.25 m = near point of a normal eye

u' = image distance

Now,

$\frac{1}{u'} = \frac{1}{f} - \frac{1}{u}$

$\frac{1}{u'} = \frac{1}{0.5} - \frac{1}{0.25}$

$\frac{1}{u'} = \frac{1}{f} - \frac{1}{u}$

Solving the above eqn, we get:

u' = - 0.5 m = - 50 cm

7 0

3 years ago

A car is traveling at 50 mi/h when the brakes are fully applied, producing a constant deceleration of 38 ft/s2. what is the dist

e-lub [12.9K]

Convert 38 ft/s^2 to mi/h^2. Then we se the conversion factor > 1 mile = 5280 feet and 1 hour = 3600 seconds.

So now we show it > $38 \frac{ft}{s^2} x \frac{1mi}{5280ft} x \frac{(3600s)^2}{(1h)^2} = 93272.27 \frac{mi}{h^2}$

Then we have to use the formula of constant acceleration to determine the distance traveled by the car before it ended up stopping.

Which the formula for constant acceleration would be > $v_2^2=v_1^2 + 2as$

The initial velocity is 50mi/h $(v_1=50)$

When it stops the final velocity is $(v_2=0)$

Since the given is deceleration it means the number we had gotten earlier would be a negative so a = -93272.27

Then we substitute the values in....

$0^2 = 50^2 + 2(-93272.27)s

0 = 2500 - 186544.54s

Isolate S next.

185644.54s= 2500

s = 2500/(185644.54)

s=0.0134
$

So we can say the car stopped at 0.0134 miles before it came to a stop but to express the distance traveled in feet we need to use the conversion factor of 1 mile = 5280 feet in otherwards > $0.0134 mi * \frac{5280ft}{1mi} = 70.8 ft$
So this means that the car traveled in feet 70.8 ft before it came to a stop.

4 0

3 years ago

A 40.0 kg wagon is towed up a hill inclined at 18.5 degrees with respect to the horizontal. The tow rope is parallel to the incl

sp2606 [1]

Answer:

7.9m/s

Explanation:

We are given that

Mass of wagon=40 kg

$\theta=18.5^{\circ}$

Tension= $140 N$

Initial velocity of wagon= $u=0$

Displacement=s=80 m

Net force applied on wagon= $F=T-mgsin\theta=140-40(9.8)sin18.5=15.7 N$

By using $g=9.8m/s^2$

$a=\frac{F}{a}=\frac{15.7}{40}=0.39m/s^2$

We know that

$v^2-u^2=2as$

Using the formula

$v^2=2\times 0.39\times 80$

$v=\sqrt{2\times 0.39\times 80}=7.9m/s$

5 0

3 years ago