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3 years ago

Under metric prefixes, if a meter were divided into 100 equal-sized subunits, a single unit would be called a

Physics

1 answer:

joja [24]3 years ago

5 0

Answer:

In metric prefixes, if a meter were divided into 100 equal-sized subunits, a single unit would be called a centimeter.

Explanation:

Here we asked to divide 1 meter in to 100 equal parts.

Let us find out what is the length of piece when 1 m is divided in to 100 equal parts.

Length

$l = \frac{1}{100} = 0.01m$

That is length of 1 m divided into 100 equal parts is 0.01m.

We know that 0.01 m is 1 centimeter.

So, in metric prefixes, if a meter were divided into 100 equal-sized subunits, a single unit would be called a centimeter.

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An airplane is flying at a speed of 200 m/s in level flight at an altitude of 800 m. A package is to be dropped from the airplan

MArishka [77]

Answer:

2560m or 2.56km (rounded to 3 significant figures)

Explanation:

First, list all known and desired values/variables (initial vertical velocity is 0 as the plane is kept level and vertical acceleration is just gravity):

$Vertical \ velocity \ (\frac{m}{s} ) = u_{v} = 0 \\\\ Horizontal \ velocity \ (\frac{m}{s} ) = u_{h} = 200\\\\ Vertical \ acceleration \ (\frac{m}{s^{2} } ) = a_{v} = 9.8 \\\\ Horizontal \ acceleration \ (\frac{m}{s^{2} } ) = a_{h} = 0 \\\\ Vertical \ displacement \ (m) = s_{v} = 800 \\\\ Horizontal \ displacement \ (m) = s_{h}$

The horizontal displacement is going to be the distance travelled, horizontally of course, once the package is released;

First thing to understand is that the vertical and horizontal components are to be dealt with separately because they don't affect each other;

Since there is no horizontal acceleration (ignoring air resistance), we simply require a velocity and time to find the horizontal displacement, using the formula v = d/t (or speed = distance/time);

What we have is the horizontal velocity but we don't have the time taken;

One thing we know is that the time elapsed for the vertical fall of 800m and for the horizontal displacement must be the same;

What we do, therefore, is find the time taken for the vertical displacement using the formula, s = ut + ¹/₂·at², since we know the vertical velocity, height and acceleration:

800 = (0)t + ¹/₂·(9.8)t²

800 = 4.9t²

t² = 163.26...

t = 12.77...

We now have the time taken for the vertical fall and the horizontal displacement, we can use this with the horizontal velocity we know already and get the horizontal displacement:

$u_{h} = \frac{s_{h} }{t} \\\\ 200 = \frac{s_{h} }{12.77...} \\\\ s_{h} = 200(12.77...) \\\\ s_{h} = 2555.5...$

7 0

2 years ago

Consider three scenarios in which a particular box moves downward under the pull of gravity :

luda_lava [24]

Answer:

1. True WA > WB > WC

Explanation:

In this exercise they give work for several different configurations and ask that we show the relationship between them, the best way to do this is to calculate each work separately.

A) Work is the product of force by distance and the cosine of the angle between them

WA = W h cos 0

WA = mg h

B) On a ramp without rubbing

Sin30 = h / L

L = h / sin 30

WB = F d cos θ

WB = F L cos 30

WB = mf (h / sin30) cos 30

WB = mg h ctan 30

C) Ramp with rubbing

W sin 30 - fr = ma

N- Wcos30 = 0

W sin 30 - μ W cos 30 = ma

F = W (sin30 - μ cos30)

WC = mg (sin30 - μ cos30) h / sin30

Wc = mg (1 - μ ctan30) h

When we review the affirmation it is the work where there is rubbing is the smallest and the work where it comes in free fall at the maximum

Let's review the claims

1. True The work of gravity is the greatest and the work where there is friction is the least

2 False. The job where there is friction is the least

3 False work with rubbing is the least

4 False work with rubbing is the least

5 0

3 years ago

What conditions must be satisfied for a body to be in equilibrium <br>

dem82 [27]

For an object to be in equilibrium, it must be experiencing no acceleration. This means that both the net force and the net torque on the object must be zero.

5 0

3 years ago

An elevator car weighs 5500 N. If the car accelerates upwards at a rate of 4.0 m/s2, what is the tension in the support cable li

nadya68 [22]

Answer:

Explanation:

The equation for this, since we are talking about weight on an elevator, is Newton's 2nd Law adjusted to fit our needs:

$F_n=ma+w$ where the Normal Force needed to lift that elevator car is the tension. So the equation then becomes

T = ma + w where T is the tension in the cable to lift the elevator, m is the mass of the elevator (which we have to solve for), a is the acceleration of the elevator (positive since it's going up), and w is the weight of the elevator (which we have as 5500 N). Solving first for mass:

w = mg and

5500 =- m(10) so

m = 550 kg. Now we have what we need to solve for the tension:

T = 550(4.0) + 5500 and

T = 2200 + 5500 so

T = 7700 N

4 0

3 years ago

Which of the following is NOT a good example of a short-term goal?

Molodets [167]

Loose 10 pounds im pretty sure

5 0

3 years ago