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3 years ago

In your own words, what was the purpose of the field study?

Physics

1 answer:

DanielleElmas [232]3 years ago

6 0

Answer:

Hi there! I have only the procedure and the scientific names of the creatures. Hope this helps!

Procedure

I studied the physical features of ten creatures and classified them using the key. I completed each section below and recorded the creature's scientific name in the data section.

The Creatures Scientific Names:

Fuzzus tallywag

Fuzzus pointilus

Silkus duosquirmus

Fuzzus chompilus

Silkus stretchilus

Silkus tallyhas

Fuzzus feelzalot

Silkus monosquirmus

Fuzzus squarilus

Silkus monowrestle

Have a thrilling Thursday!

~Lola

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Which one of the following accurately pairs the device with its function?

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Its A. Bourdon gage - indicate steam pressure in a furnace.

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4 years ago

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A bullet is fired in a horizontal direction with a muzzle velocity of 300m/s.In the absence of air resistance,how far will it ha

Kruka [31]

Answer:

304.9m

Explanation:

Given

Velocity = 300m/s

Tim = 1sec

Required

Horizontal distance S

Using the formula

S= ut+1/2gt²

S = 300(1)+1/2(9.8)(1)²

S = 300+4.9

S = 304.9m

Hence the distance travelled is 304.9m

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3 years ago

on the surface of theearth ,the weight of a boy is 400N but on a mountainpeak his weight is 360N,Calculatethe value of ''g' on t

andreyandreev [35.5K]

The value of g at sea level is 9.81 ms^-2.

The boy's mass is constant wherever he is in the universe but his weight will depend on the strength gravity where he is.

By proportion its value on the mountain peak is (360 /400) * 9.81

= 0.9 * 9.81 = 8.83 ms^-2 to nearest hundredth, (answer).

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3 years ago

An 80- quarterback jumps straight up in the air right before throwing a 0.43- football horizontally at 15 . How fast will he be

lord [1]

Answer:

the quarterback will be moving back at speed of 0.080625 m/s

the distance moved horizontally by the quarterback is 0.0241875 m or 2.41875 cm

Explanation:

Given the data in the question;

How fast will he be moving backward just after releasing the ball?

using conservation of momentum;

m₁v₁ = m₂v₂

v₂ = m₁v₁ / m₂

where m₁ is initial mass ( 0.43 kg )

m₂ is the final mass ( 80 kg )

v₁ is the initial velocity ( 15 m/s )

v₂ is the final velocity

so we substitute

v₂ = ( 0.43 × 15 ) / 80

v₂ = 6.45 / 80

v₂ = 0.080625 m/s

Therefore, the quarterback will be moving back at speed of 0.080625 m/s

b) Suppose that the quarterback takes 0.30 to return to the ground after throwing the ball. How far d will he move horizontally, assuming his speed is constant?

we make use of the relation between time, distance and speed;

s = d/t

d = st

where s is the speed ( 0.080625 m/s )

t is time ( 0.30 s )

so we substitute

d = 0.080625 × 0.30

d = 0.0241875 m or 2.41875 cm

Therefore, the distance moved horizontally by the quarterback is 0.0241875 m or 2.41875 cm

5 0

3 years ago

Big Ben, a large artifact in England, has a mass of 1x10^8 kilograms and the Empire State Building 1x10^9 kilograms. The distanc

TiliK225 [7]

Answer:

The force, exerted by Big Ben on the Empire State Building is 2.66972 × 10⁻⁷ N

Explanation:

The question relates to the force of gravity experienced between two bodies

The given parameters are;

The mass of Big Ben, M₁ = 1 × 10⁸ kg

The mass of the Empire State Building, M₂ = 1 × 10⁹ kg

The distance between the two Big Ben and the Empire State Building, r = 5,000,000 meters

By Newton's Law of gravitation, we have;

$F=G \times \dfrac{M_{1} \times M_{2}}{r^{2}}$

Where;

F = The force exerted by Big Ben on the Empire State Building and vice versa

G = The universal gravitational constant = 6.67430 × 10⁻¹¹ N·m²/kg²

M₁, M₂, and r are the given parameters

By plugging in the values of the parameters and the constant into the equation for Newton's Law of gravitation, we have;

$F=6.67430 \times 10^{-11} \times \dfrac{1 \times 10^8 \times 1 \times 10^9}{(5,000,000)^{2}} = 2.66972 \times 10^{-7}$

The force, 'F', exerted by Big Ben on the Empire State Building is F = 2.66972 × 10⁻⁷ N.

3 0

3 years ago