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3 years ago

Is this a scientific model? Use complete sentences to explain why or why not.

Physics

1 answer:

grin007 [14]3 years ago

8 0

Answer:

A scientific model is an activity that is aimed at making a specific part or feature easier to understand and define.

Modeling is an essential scientific discipline and is used in the fields of education. Modeling seems to represent empirical objects, phenomena in a logical and objective way.

The oval-shaped world map showing vegetation in green color is a 2D model of the entire earth and demarcates us the world vegetation distribution and growth.

Hence this is a scientific model as its helps in visualizing the land surface of the earth on paper.

Explanation:

Yes because a scientific models doesn't have to be complex, just like this one

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A lens of focal length 10.0 cm

fgiga [73]

Answer:

30cm

Explanation:

7 0

3 years ago

A mine car (mass=390 kg) rolls at a speed of 0.50 m/s on a horizontal track, as the drawing shows. A 250-kg chunk of coal has a

Nady [450]

Answer:

v=0.60 m/s

Explanation:

Given that

m ₁= 390 kg ,u ₁= 0.5 m/s

m₂ = 250 kg ,u₂ = 0.76 m/s

As we know that if there is no any external force on the system the total linear momentum of the system will be conserve.

Pi = Pf

m ₁u ₁+m₂u₂ = (m₂ + m ₁ ) v

Now putting the values in the above equation

390 x 0.5 + 250 x 0.76 = (390 + 250 ) v

$v=\dfrac{390\times 0.5+250\times 0.76}{390+250}\ m/s$

v=0.60 m/s

Therefore the velocity of the system will be 0.6 m/s.

8 0

4 years ago

The water in a river flows uniformly at a constant speed of 2.50 m/s between parallel banks 80.0 m apart. You are to deliver a p

NISA [10]

Answer:

a) The swimmer should travel perpendicular to the bank to minimize the spent in getting to the other side.

b) 133.33 m

c) 53.13°

d) 106.67 m

Explanation:

a) The swimmer should travel perpendicular to the bank to minimize the spent in getting to the other side.

b) velocity = distance * time

Let the velocity of the swimmer be $v_{s}$ = 1.5 m/s

The separation of the two sides of the river, d = 80 m

The time taken by the swimmer to get to the other end of the river bank,

$t = \frac{d}{v_{s} }$

t = 80/1.5

t = 53.33 s

The swimmer will be carried downstream by the river through a distance, s

Let the velocity of the river be $v_{r}$ = 2.5 m/s

$S = v_{r} t$

S = 53.33 * 2.5

S = 133.33 m

c) To minimize the distance traveled by the swimmer, his resultant velocity must be perpendicular to the velocity of the swimmer relative to water

That is ,

$cos \theta = \frac{v_{s} }{v_{r} } \\cos \theta = 1.5/2.5\\cos \theta = 0.6\\\theta = cos^{-1} 0.6\\\theta = 53.13^{0}$

d) Downstream velocity of the swimmer, $v_{y} = v_{s} sin \theta\\$

$v_{y} = 1.5 sin 53.13\\v_{y} = 1.2 m/s$

The vertical displacement is given by, $y = v_{y} t$

80 = 1.2 t

t = 80/1.2

t = 66.67 s

the horizontal speed,

$v_{x} = 2.5 - 1.5cos53.13\\v_{x} = 1.6 m/s$

The downstream horizontal distance of the swimmer, $x = v_{x} t$

x = 1.6 * 66.67

x = 106.67 m

7 0

3 years ago

Which material is very strong and tough but shows very little elongation as it absorbs energy? A. spider silk B.rubber C.Kevlar®

Furkat [3]

C. Kevlar

:) ++ * ++ !

5 0

3 years ago

Read 2 more answers

A 3.15-kg object is moving in a plane, with its x and y coordinates given by x = 6t2 − 4 and y = 5t3 + 6, where x and y are in m

ArbitrLikvidat [17]

Answer:

Explanation:

The sum of force along both components x and y is expressed as;

$\sum Fx = ma_x \ and \ \sum Fy = ma_y$

The magnitude of the net force which is also known as the resultant will be expressed as $R =\sqrt{(\sum Fx)^2 + (\sum Fx )^2}$

To get the resultant, we need to get the sum of the forces along each components. But first lets get the acceleration along the components first.

Given the position of the object along the x-component to be x = 6t² − 4;

$a_x = \frac{d^2 x }{dt^2}$

$a_x = \frac{d}{dt}(\frac{dx}{dt} )\\ \\a_x = \frac{d}{dt}(6t^{2}-4 )\\\\a_x = \frac{d}{dt}(12t )\\\\a_x = 12m/s^{2}$

Similarly,

$a_y = \frac{d}{dt}(\frac{dy}{dt} )\\ \\a_y = \frac{d}{dt}(5t^{3} +6 )\\\\a_y = \frac{d}{dt}(15t^{2} )\\\\a_y = 30t\\a_y \ at \ t= 2.15s; a_y = 30(2.15)\\a_y = 64.5m/s^2$

$\sum F_x = 3.15 * 12 = 37.8N\\\sum F_y = 3.15 * 64.5 = 203.18N$

$R = \sqrt{37.8^2+203.18^2}\\ \\R = \sqrt{1428.84+41,282.11}\\ \\R = \sqrt{42.710.95}\\ \\R = 206.67N$

Hence, the magnitude of the net force acting on this object at t = 2.15 s is approximately 206.67N

7 0

3 years ago