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Mademuasel [1]
3 years ago
9

Signment

Physics
1 answer:
jeka943 years ago
3 0

Answer:

v = 5.15 m/s

Explanation:

At constant velocity, the cable tension will equal the car weight of 984(9.81) = 9,653 N

As the cable tension is less than this value, the car must be accelerating downward.

7730 = 984(9.81 - a)

a = 1.95 m/s²

kinematic equations s = ut + ½at² and v = u + at

-5.00 = u(4.00) + ½(-1.95)4.00²

u = 2.65 m/s    the car's initial velocity was upward at 2.65 m/s

v = 2.65 + (-1.95)(4.00)

v = -5.15 m/s

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notsponge [240]

Answer:

The water pressure at the bottom of the jar will increase by 1000 Pa.

Explanation:

The Pascal principle states that:

In a fluid, a change in pressure at any point in the fluid is transmitted equally throughout the fluid, as it is occuring everywhere.

If we apply this principle at the case mentioned in the problem, we can say that:

- An initial pressure of 1000 Pa is applied on the top of the fluid (the water)

- According to Pascal's law, this pressure is transmitted with equal intensity (1000 Pa) to every point of the fluid

- So, the water pressure at the bottom of the jar will also increase by the same amount, 1000 Pa

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3 years ago
1. The slope of a velocity-time graph is __________
Naddik [55]

Answer:

d. Displacement

Explanation:

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3 years ago
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Consider a thin rod of mass 3.2 kg, length 1.2 m and uniform density. The rod is pivoted at one end on a frictionless horizontal
notsponge [240]

Answer:

the angular acceleration is 9.7 rad/s^{2}

Explanation:

given information:

mass of thin rod, m = 3.2 kg

the length of the rod, L = 1.2

angle, θ = 38

to find the acceleration of the rod, we can use the torque's formula as below,

τ = Iα

where

τ = torque

I = inertia

σ = acceleration

moment inertia of this rod, I

I = \frac{1}{3} mL^{2}

τ = F d, d = \frac{L}{2}cosθ

τ = m g \frac{L}{2}cosθ

now we can substitute the both equation,

τ = Iα

α = τ/I

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Help! Help!
andreyandreev [35.5K]

Answer

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Explanation:

4 0
3 years ago
Two identical charges,2.0m apart,exert forces of magnitude 4.0 N on each other.What is the value of either charge?
storchak [24]

Answer:

\large \boxed{42\, \mu \text{C}}$

Explanation:

The formula for the force exerted between two charges is

F=k \dfrac{ q_1q_2}{r^2}

where k is the Coulomb constant.

The charges are identical, so we can write the formula as

F=k\dfrac{q^{2}}{r^2}

\begin{array}{rcl}\text{4.0 N}& = & 8.988 \times 10^{9}\text{ N$\cdot$m$^{2}$C$^{-2}$} \times \dfrac{q^{2}}{\text{(2.0 m)}^{2}}\\\\4.0 & = & 2.25 \times 10^{9}\text{ C$^{-2}$} \times q^{2}\\\\q^{2} & = & \dfrac{4.0}{2.25 \times 10^{9}\text{ C$^{-2}$}}\\\\& = & 1.78 \times 10^{-9} \text{ C}^{2}\\q & = & 4.2 \times 10^{-5} \text{ C}\\& = & 42\, \mu \text{C}\\\end{array}\\\text{Each charge has a value of $\large \boxed{\mathbf{42\, \mu }\textbf{C}}$}

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3 years ago
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