Answer:
D. All of the above
Explanation:
Think about it.
Can you see how big or small an object is? Unless in very special cases, the ordinary person can.
Can you see the color of an object? Unless you are color blind (missing color recepticles in your eyes), then yes you can.
Can you see the motion of an object? Unless the object is not moving, or you cannot see, then yes.
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The question is incomplete. Here is the entire question.
A jetboat is drifting with a speed of 5.0m/s when the driver turns on the motor. The motor runs for 6.0s causing a constant leftward acceleration of magnitude 4.0m/s². What is the displacement of the boat over the 6.0 seconds time interval?
Answer: Δx = - 42m
Explanation: The jetboat is moving with an acceleration during the time interval, so it is a <u>linear</u> <u>motion</u> <u>with</u> <u>constant</u> <u>acceleration</u>.
For this "type" of motion, displacement (Δx) can be determined by:
is the initial velocity
a is acceleration and can be positive or negative, according to the referential.
For Referential, let's assume rightward is positive.
Calculating displacement:
= - 42
Displacement of the boat for t=6.0s interval is = - 42m, i.e., 42 m to the left.
Answer:
Explained
Explanation:
Although Atom are electrically neutral. But atom atom is combination of nucleus and electrons. The nucleus of the atom is composed neutron and positively charged protons. On the outside of nucleus at some distance are the electrons which are negatively charged. So, there is difference in position of the two differently charge species. So, this way a electrically neutral atom can exert a electrostatic force on other electrically neutral atom
Answer:
31.6 m/s
Explanation:
Mass is conserved, so the mass flow at the outlet of the pump equals the mass flow at the nozzle.
m₁ = m₂
ρQ₁ = ρQ₂
Q₁ = Q₂
v₁A₁ = v₂A₂
v₁ πd₁²/4 = v₂ πd₂²/4
v₁ d₁² = v₂ d₂²
Now use Bernoulli equation:
P₁ + ½ ρ v₁² + ρgh₁ = P₂ + ½ ρ v₂² + ρgh₂
Since h₁ = 0 and P₂ = 0:
P₁ + ½ ρ v₁² = ½ ρ v₂² + ρgh₂
Writing v₁ in terms of v₂:
P₁ + ½ ρ (v₂ d₂²/d₁²)² = ½ ρ v₂² + ρgh₂
P₁ + ½ ρ (d₂/d₁)⁴ v₂² = ½ ρ v₂² + ρgh₂
P₁ − ρgh₂ = ½ ρ (1 − (d₂/d₁)⁴) v₂²
Plugging in values:
579,160 Pa − (1000 kg/m³)(9.8 m/s²)(15 m) = ½ (1000 kg/m³) (1 − (1.99 in / 3.28 in)⁴) v₂²
v₂ = 31.6 m/s
