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Paladinen [302]
3 years ago
6

Which term below describes a measurement of how hard an object pushes against a surface?

Physics
2 answers:
Scilla [17]3 years ago
8 0

I hope the answer is D. Pressure

expeople1 [14]3 years ago
3 0

Answer:

like the other guy, the answer is pressure

Explanation:

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The sun _____.
enyata [817]

The sun <u><em>appears</em></u> brighter than any other star.

(It isn't really, but it looks that way because it's much much much much much much closer to us than any other star.)

7 0
3 years ago
A bullet starting from rest accelerates uniformly at a rate of 1,250 meters per square second. What is the bullet's speed after
Tatiana [17]

Answer:

125,000

Explanation:

8 0
3 years ago
Assume that a pendulum used to drive a grandfather clock has a length L0=1.00m and a mass M at temperature T=20.00°C. It can be
Sedaia [141]

Answer:

The period will change a 0,036 % relative to its initial state

Explanation:

When the rod expands by heat its moment of inertia increases, but since there was no applied rotational force to the pendulum , the angular momentum remains constant. In other words:

ζ= Δ(Iω)/Δt, where ζ is the applied torque, I is moment of inertia, ω is angular velocity and t is time.

since there was no torque ( no rotational force applied)

ζ=0 → Δ(Iω)=0 → I₂ω₂ -I₁ω₁ = 0 → I₁ω₁ = I₂ω₂

thus

I₂/I₁ =ω₁/ω₂ , (2) represents final state and (1) initial state

we know also that ω=2π/T , where T is the period of the pendulum

I₂/I₁ =ω₁/ω₂ = (2π/T₁)/(2π/T₂)= T₂/T₁

Therefore to calculate the change in the period we have to calculate the moments of inertia. Looking at tables, can be found that the moment of inertia of a rod that rotates around an end is

I = 1/3 ML²

Therefore since the mass M is the same before and after the expansion

I₁ = 1/3 ML₁² , I₂ = 1/3 ML₂²  → I₂/I₁ = (1/3 ML₂²)/(1/3 ML₁²)= L₂²/L₁²= (L₂/L₁)²

since

L₂= L₁ (1+αΔT) , L₂/L₁=1+αΔT  , where ΔT is the change in temperature

now putting all together

T₂/T₁=I₂/I₁=(L₂/L₁)² = (1+αΔT) ²

finally

%change in period =(T₂-T₁)/T₁ = T₂/T₁ - 1 = (1+αΔT) ² -1

%change in period =(1+αΔT) ² -1 =[ 1+18×10⁻⁶ °C⁻¹ *10 °C]² -1 = 3,6 ×10⁻⁴ = 3,6 ×10⁻² %  = 0,036 %

4 0
3 years ago
A machine does 1500 joules of work in 30 second What is the power of the machine
lesya [120]
This is simple as power in watts is equal to joules per second so we can do 1500 joules divided by 30 seconds which equals 50 watts
8 0
3 years ago
During a single-replacement reaction, a more-active metal will replace a less-active metal in a compound. Which single-replaceme
son4ous [18]

Answer:

Mg will replace Ag in a compound

Explanation:

A single replacement reaction is driven by the position of ions on the activity series.

As a rule of thumb, the position of metal ions on the activity series determines their reactivity.

Metal ions that are above another are more reactive and they will displace those that are lower.

Generally, activity increases as we go up the group.

Mg ions are higher than Ag ions on the series so, Mg will displace Ag from a solution.

8 0
3 years ago
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