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4 years ago

Which of these statements are true about scientific theories and laws?

2 answers:

7 0

<span>The best response would be that theories are not opinions. While laws and theories are similar, laws tend to be a collection of theories about the same phenomenon. Laws can be expressed as math formulae, but more often they're in the form of "if-then" statements. Laws and theories both have to be rigorously tested before they reach these classiffications.</span>

V125BC [204]4 years ago

5 0

Answer:

Theories are not opinions.

Explanation:

Theories are all of them tested, with the scientific method in order to be accepted or refuted, then Laws are tested and are observable by any means all around the globe, the only option that is actually correct would be that Theories are not opinions because they are scientific statements that can be tested.

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A rocket in deep space has an empty mass of 220 kg and exhausts the hot gases of burned fuel at 2500 m/s. What mass of fuel is n

Scilla [17]

Answer:

Explanation:

Let fuel is released at the rate of dm / dt where m is mass of the fuel

thrust created on rocket

= d ( mv ) / dt

= v dm / dt

this is equal to force created on the rocket

= 220 dv / dt

so applying newton's law

v dm / dt = 220 dv / dt

v dm = 220 dv

dv / v = dm / 220

integrating on both sides

∫ dv / v = ∫ dm / 220

lnv = ( m₂ - m₁ ) / 220

ln4000 - ln 2500 = ( m₂ - m₁ ) / 220

( m₂ - m₁ ) = 220 x ( ln4000 - ln 2500 )

( m₂ - m₁ ) = 220 x ( 8.29 - 7.82 )

= 103.4 kg .

8 0

4 years ago

A 3.35 kg object initially moving in the positive x direction with a velocity of 4.90 m s collides with and sticks to a 1.88 kg

ahrayia [7]

Answer:

The final components of velocity of the composite object is 3.33 m/s.

Explanation:

Given;

mass of the first object, m₁ = 3.35 kg

initial velocity of the first object, u₁ = 4.90 m/s in positive x-direction

mass of the second object, m₂ = 1.88 kg

initial velocity of the second object, u₂ = 3.12 m/s in negative y-direction

initial momentum of the first object, P₁ = 3.35 x 4.9 = 16.415 kgm/s

initial momentum of the second object, P₂ = 1.88 x 3.12 = 5.8656 kgm/s

The resultant velocity of the two objects is given by;

R² = 16.415² + 5.8656²

R² = 303.858

R = √303.858

R = 17.432 kgm/s

Apply the principle of conservation of linear momentum for inelastic collision;

total initial momentum before = total final momentum after collision

P₁(x) + P₂(y) = Pf

R = Pf

R = v(m₁ + m₂)

17.432 = v(m₁ + m₂)

where;

v is the final components of velocity of the composite object

$v = \frac{17.432}{m_1 + m_2} \\\\v = \frac{17.432}{3.35+1.88} \\\\v = 3.33 \ m/s$

Therefore, the final components of velocity of the composite object is 3.33 m/s.

8 0

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

What causes the formation of ocean currents?

aliina [53]

Ocean currents can be generated by wind, density differences in watermasses caused by temperature and salinity variations, gravity, and events such as earthquakes

5 0

3 years ago

A 3 kg object is attached to a 1000 N/m spring. The spring is compressed 0.10 m and then the spring launches the object horizont

Viktor [21]

Answer:

1.826m/s

Explanation:

E=1/2*k*(∆L)^2

E=1/2*mV^2

6 0

3 years ago