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sattari [20]
3 years ago
7

Student 1 and student 2 talk about the speed of the truck

Physics
1 answer:
vodka [1.7K]3 years ago
7 0

Answer:

Explanation:

any more explanation?

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Three point charges are fixed in place in a right triangle, as shown in the figure.
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3 years ago
I need a correct answer plzzzzzzzzzzzzzzzzzzzz
olya-2409 [2.1K]

Answer:

option 1

Explanation:

i just used the SOH CAH TOA, and since the given is tan=opposite/adjacent, that should be the answer

4 0
3 years ago
If a force of 50 Newton’s was applied to an object with a mass of 500 grams, what will the objects acceleration be?
lapo4ka [179]

Answer:  100 m/s^2

F=ma

Explanation:

50N = 50 kg*m/s^2

500g = 0.5 kg

F=ma

a = F/m

a = (50 kg*m/s^2)/(0.5 kg)

a = 100 m/s^2

5 0
2 years ago
Suppose an object’s initial velocity is 10 m/s and its final velocity is 4 m/s. Mass is constant.
Ostrovityanka [42]

The correct answer is:

Work is negative, the environment did work on the object, and the energy of the system decreases.

In fact, the work-energy theorem states that the work done by the system is equal to its variation of kinetic energy:

W=\Delta K=K_f -K_i

In this problem, the variation of kinetic energy \Delta K is negative (because the final velocity is less than the initial velocity), so the work is negative, and this means that the environment did work on the object, and its energy decreased.

3 0
3 years ago
Read 2 more answers
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
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