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4 years ago

A snail moves 18 millimeters every 24 seconds what is the snails rate of speed

Physics

1 answer:

Nataly_w [17]4 years ago

3 0

The speed of the snail is given by:

$v=\frac{S}{t}$

where S is the distance covered by the snail and t the time taken.

The snail in the problem moves by S=18 mm in t=24 s, therefore its speed is

$v=\frac{S}{t}=\frac{18 mm}{24 s}=0.75 mm/s$

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If the wavelength of one wave is 0.25 meters, how many waves will be formed on a rope that is one meter long?

ikadub [295]

<h3><u>Answer;</u></h3>

4 waves

<h3><u>Explanation;</u></h3>

Wavelength of one wave = 0.25 meters

Length of the rope = 1 meter

Therefore;

Number of waves = Length of the rope/wavelength of one wave

= 1/0.25

= 4 waves

6 0

3 years ago

When changing a tire sized 195/75 R15 to a 5 percent lower profile, the correct tire size would be _____

valentinak56 [21]

Answer and explanation:

When you are changing a car tire, the most important thing is to keep the total diameter as equal as possible.

The total car tire diameter can be calculated as:

$D_{tot orig}=\frac{195 \cdot 75}{2540 \cdot 2}+15''=26.5''$

The profile of this tire is 75 (the higher/taller relation), therefore a 5 percent lower profile would be:

pr=0.95·75=71.25

The problem is that the profiles are normalized and the nearest profile available is 70.

If we take a theorical tire with a profile of 71.25:

$D_{tot orig}=D_{new}\\\frac{195 \cdot 75}{2540 \cdot 2}+15''=\frac{X \cdot 71.25}{2540 \cdot 2}+15''\\X=205.26$

The theorical tire size should be 205/71 R15.

If we look for a real tire size, we should look for a tire with a diameter nearest to 26.5'' and a profile of 70.

The best option for real tire size is: Tire 225/70 R14 (wheel diameter of 26.4'') or 205/70 R15 (wheel diameter of 26.3'').

3 0

4 years ago

A 500 lb steel beam is lifted up by a crane to a height of 100 ft and is held there.

Nadya [2.5K]

A. How much work is being done to hold the beam in place?

Work is the product of Force and Displacement. Since there is no Displacement involved in just holding the beam in place, hence the work is zero.

B. How much work was done to lift the beam?

In this case, force is simply equal to weight or mass times gravity. Hence the work is:

Work = weight * displacement

Work = 500 lbf * 100 ft

Work = 50,000 lbf * ft

C. How much work would it take if the steel beam were raised from 100 ft to 200ft?

The displacement is still 100 ft since 200 – 100 = 100 ft, hence the work done is still similar in B which is:

5 0

3 years ago

A solid ball is released from rest and slides down a hillside that slopes downward at an angle 51.0 ∘ from the horizontal. what

Romashka [77]

In order for the object not to slip, the component of the weight parallel to the surface must be equal to the frictional force (which acts in the opposite direction):
$F_{//}= F_a$

The parallel component of the weight is:
$F_{//} = mg \sin \alpha$
where m is the object mass and $\alpha$ is the angle of the inclined plane.

The frictional force is
$F_a = \mu m g \cos \alpha$
where $\mu$ is the coefficient of static friction.

Equalizing the two forces, we have
$mg \sin \alpha = \mu m g \cos \alpha$
from which we find
$\mu = \tan \alpha$

and so, in our problem the coefficient of static friction must be
$\mu=\tan 51^{\circ} =1.23$

4 0

3 years ago

Show the relationship between the quantity of matter and the mass of objects

IRISSAK [1]

Answer:

actually quantity of matter present in a body is its mass.

3 0

3 years ago