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3 years ago

Pe=___

Physics

1 answer:

Nesterboy [21]3 years ago

8 0

Answer:

H = start height (v = 0)

h = present height

v = present speed

assuming no friction

total energy = PE + KE

mgH = mgh + .5mv^2

if PE = KE then

mgH = mgh + mgh

h = H/2

potential energy = kinetic energy when object is at half its start height.

Explanation:

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The equipotential surfaces surrounding a point charge are concentric spheres with the charge at the center. If the electric pote

tester [92]

Answer:

1.06 m

Explanation:

Since the charge is at the centre of two concentric spheres, we use the formula for electric potential due to a point charge. V = kq/r. Let r₁ be the radius of the sphere with potential, V₁ = 200 V and r₂ be the radius of the sphere with potential, V₂ = 82.0 V. From V = kq/r, r = kq/V. So that r₁ = kq/V₁ and r₂ = kq/V₂. The magnitude of the difference r₁ - r₂ is the distance between the two surfaces. q the charge equals 1.63 × 10⁻⁸ C

r₂ - r₁ = kq/V₂ - kq/V₁ = kq(1/V₂ - 1/V₁) = 1.63 × 10⁻⁸ × 9 × 10⁹ (1/82 -1/200) m = 1.63 × 10⁻⁸ × 9 × 10⁹ (0.0122 - 0.005) = 1.63 × 10⁻⁸ × 9 × 10⁹(0.0072) m = 1.06 m

The distance between them is 1.06 m

8 0

3 years ago

APEX What is one advantage of using primary sources when doing research on an

MissTica

Answer:

Explanation:

They are first hand sources so they are more reliable and detailed...

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3 years ago

Imagina que compras una placa rectangular de metal de 2mm de alto, 10mm de ancho x 50mm de largo, y una masa de 0.02kg. El vende

shusha [124]

Answer:

Densidad de la placa = 20 g/cm³.

La placa no es de oro.

Explanation:

Para encontrar la densidad de la placa rectangular primero debemos hallar su volumen:

$V = 2 mm*10 mm*50 mm = 1000 mm^{3}*\frac{1 cm^{3}}{(10 mm)^{3}} = 1 cm^{3}$

Ahora, encontremos al densidad de la placa:

$d = \frac{m}{V} = \frac{20 g}{1 cm^{3}} = 20 g/cm^{3}$

Dado que la densidad del oro es 19.32 g/cm³ y que la densidad de la placa rectangular calculada es 20 g/cm³, podemos decir que dicha placa no es de oro.

Espero que te sea de utilidad!

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3 years ago

sing a rope that will snap if the tension in it exceeds 361 N, you need to lower a bundle of old roofing material weighing 455 N

yaroslaw [1]

Answer:

a)-2m/s^2

b)27.2m/s

Explanation:

Hello! The first step to solve this problem is to find the mass of the block remembering that the definition of weight force is mass by gravity (g=9.8m / s ^ 2)

W=455N=weight

W=mg

W=455N=weight

$m=\frac{W}{g} =\frac{455}{9.81}=46.38kg$

The second step is to draw the free body diagram of the body (see attached image) and use Newton's second law that states that the sum of the forces is equal to mass by acceleration

$a=\frac{T-W}{m} =\frac{361-455}{46.38kg} =-2m/s^2$

for point b we use the equations of motion with constant acceleration to find the velocity

$Vf=\sqrt{X(2)(a)+Vo^2}$

Where

Vf = final speed

Vo = Initial speed =0

A = acceleration =2m/s

X = displacement =6.8m

Solving

$Vf=\sqrt{(6.8)(2)(2)+0^2}=27.2m/s$

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3 years ago

What is the strength of the electric field between two parallel conducting plates separated by 1.00 cm and having a potential di

yan [13]

What is the strength of the electric field between two parallel conducting plates separated by 1.00 cm and having a potential difference (voltage) between them of 1.50×10^4v ?

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2 years ago