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3 years ago

What must happen to an object in order to accelerate it?

Physics

1 answer:

White raven [17]3 years ago

6 0

Answer:

For an object to accelerate, appropriate force must be applied to the object to cause it to change it’s velocity if it’s already in motion. However, to cause it to overcome static friction if it is at rest, and cause it to change it’s velocity

Explanation:

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How are element and compound are alike and different?

Dovator [93]

Elements is the simplest form of a pure substance.They cannot be broken down into any thing else ,by either physical or chemical means.In contrasts compounds are pure substances like elements ,they are made up of two or more elements.Just like elements they can be broken down into simpler substances by chemical means,but unlike elements they can only be broken down by chemical means .not both physical and chemical means like elements can.
(I hope that this was helpful.)

5 0

3 years ago

Why hydraulic system know as forcee multiplier

Norma-Jean [14]

A hydraulic jack can be force multiplier because it can take a small force

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3 years ago

Read 2 more answers

A long cylindrical capacitor is made of a central wire of radius a = 2.50 mm surrounded by a conducting shell of radius b = 7.50

asambeis [7]

Answer:

The capacitance per unit length is $5.06\times10^{-11}\ F/m$

(b) is correct option.

Explanation:

Given that,

Radius a= 2.50 mm

Radius b=7.50 mm

Dielectric constant = 3.68

Potential difference = 120 V

We need to calculate charge per length for the capacitance

Using formula of charge per length

$\lambda=\dfrac{4\pi\epsilon_{0}\Delta V}{2 ln(\dfrac{r_{2}}{r_{1}})}$

Put the value into the formula

$\lambda=\dfrac{120}{9\times10^{9}\times2 ln(\dfrac{7.50\times10^{-3}}{2.50\times10^{-3}})}$

$\lambda=6.068\times10^{-9}\ C/m$

We know that,

$\lambda=\dfrac{Q}{L}$

We need to calculate the capacitance per unit length

Using formula of capacitance per unit length

$C=\dfrac{\dfrac{Q}{L}}{\Delta V}$

$C=\dfrac{6.068\times10^{-9}}{120}$

$C=5.06\times10^{-11}\ F/m$

Hence, The capacitance per unit length is $5.06\times10^{-11}\ F/m$

7 0

4 years ago

Water enters a student's house 10.0 m above the ground through a pipe with a cross section area of 1.00 x 10-4m2 at ground. Insi

dezoksy [38]

Answer:

(a). V₁ = 10m/s (velocity inside the house), V₂ = 5m/s (velocity at ground level)

(b). P₂ = 236500 Pa

Explanation:

This is quite straight-forward so let us begin by defining the terms given.

Given that;

The cross-section area inside the student's house A₁ = 0.50 0.50 x 10-4m2.

Let us make the velocity of water inside the house be V₁

such that the Volume of water entering the per second is = A₁V₁

Therefore, in 90sec:

45 L = 90 A₁V₁

V₁ = 45 * 10⁻³m³ / 90*0.5*10⁻⁴

V₁ = 10m/s (velocity of water inside the house)

From the continuity equation we have that;

A₁V₁ = A₂V₂

0.5*10⁻⁴ * 10 = 1*10⁻⁴ V₂

V₂ = 5m/s (velocity at ground level)

(b). We are told to calculate the water pressure in the pipeline at the ground level.

Using Bernoulli's equation;

P₁ + pgh₁ + 1/2PV₁² (inside) = P₂ + pgh₂ + 1/2PV₂² (ground level)

1.01*10⁵ + 1000*9.8*10 + 1/2*1000*(10)² = P₂ + 0 + 1/2*1000*(5)²

P₂ (pressure) = 1.01*10⁵Pa

Therefore we have;

101000 + 98000 + 50000 = P₂ + 12500

P₂ = 236500 Pa

cheers I hope this helped !!

3 0

3 years ago

The diagram shows a skydiver at different points of her jump. At what point would her

8_murik_8 [283]

I’m pretty sure it’s B! :)

6 0

3 years ago