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2 years ago

Will mark brainlist

Physics

1 answer:

Ksenya-84 [330]2 years ago

4 0

Answer:

5.00 x 10^9 j

Explanation:

mass = 100 kg

speed / velocity = 10 km/s x 1000 = 10,000m

K.E =

$\frac{m{v}^{2} }{2}$

10,000x 100/ 2

= 5000000000

= 5.0 x 10^9j

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A student says that since the atomic theory is just a theory, it should not be considered useful. Which statement best argues ag

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Theories are usually backed up with a lot of evidence. If the evidence is well studied then it is useful information.

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3 years ago

A jet airliner moving initially at 889 mph

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Answer:

1500 mph

Explanation:

Take east to be +x and north to be +y.

The x component of the velocity is:

vₓ = 889 cos 0° + 830 cos 59°

vₓ = 1316.5 mph

The y component of the velocity is:

vᵧ = 889 sin 0° + 830 sin 59°

vᵧ = 711.4 mph

The speed is found with Pythagorean theorem:

v² = vₓ² + vᵧ²

v² = (1316.5 mph)² + (711.4 mph)²

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Rounded to two significant figures, the jet's speed relative to the ground is 1500 mph.

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3 years ago

NASA launches a probe with a mass of 15,000 kg to another planet more massive than Earth. Which statement is true about the prob

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Answer: The weight of the probe will increase.

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2 years ago

What force is most responsible for the path of an object during projectile motion?

Levart [38]

The gravitational force.

In fact, the motion of the projectile is composed by two independent motions:
- on the horizontal direction, it is a uniform motion (with constant speed)
- on the vertical direction, it is a uniformly accelerated motion, where the vertical acceleration g is given by the gravity exerted by the Earth on the projectile.

For this reason, the composition of the two motions results in a parabolic trajectory.

7 0

3 years ago

A conical container of radius 6 ft and height 24 ft is filled to a height of 19 ft of a liquid weighing 64.4 lb divided by ft cu

kobusy [5.1K]

Answer:

Part (i) work required to pump the contents to the rim is 281,913.733 lb.ft

Part (ii) work required to pump the liquid to a level of 5ft above the cone's rim is 426,484.878 lb.ft

Explanation:

The center of mass of a uniform solid right circular cone of height h lies on the axis of symmetry at a distance of h/4 from the base and 3h/4 from the top.

Center mass of the liquid Z = (24-19)ft + 19/4 = 5ft + 4.75ft = 9.75 ft

Mass of liquid in the cone = volume × density (ρ) = ¹/₃.π.r².h.ρ

where;

r is the radius of the liquid surface = [6*(19/24)]ft = 4.75ft

ρ is the density of liquid = 64.4 lb/ft³

h is the height of the liquid = 19 ft

Mass of liquid in the cone = ¹/₃ × π × (4.75)² × 19 × 64.4 = 28,914.229 lbs

Part (i) work required to pump the contents to the rim

Work required = 28,914.229 lbs × 9.75 ft = 281,913.733 lb.ft

Part (ii) work required to pump the liquid to a level of 5 ft above the cone's rim

Extra work required = 28,914.229 lb × 5ft = 144571.145 lb.ft

Total work required = (281,913.733 + 144571.145) lb.ft

= 426,484.878 lb.ft

5 0

3 years ago