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2 years ago

9

Anna applies a force of 19.5 newtons to push a book placed on a table. If the normal force of the book is 51.7 newtons, what is

the coefficient of kinetic friction?

1 answer:

GarryVolchara [31]2 years ago

6 0

that would be given by

[email protected]

@ representing coefficient of kinetic friction.

thus 19.5/51.7 = 0.377

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Which lanyard provides an impact force of less than 1,800 pounds, as recommended by good practices?

enyata [817]

Answer:

Energy absorbing lanyard as per OSHA

Explanation:

Energy absorbing lanyard if working over 6 feet in height so you don't break your back when you fall.

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2 years ago

A hot air balloon must be designed to support a basket, cords, and one person for a total payload weight of 1300 N plus the addi

RSB [31]

Answer:

r = 4.44 m

Explanation:

For this exercise we use the Archimedes principle, which states that the buoyant force is equal to the weight of the dislodged fluid

B = ρ g V

Now let's use Newton's equilibrium relationship

B - W = 0

B = W

The weight of the system is the weight of the man and his accessories (W₁) plus the material weight of the ball (W)

σ = W / A

W = σ A

The area of a sphere is

A = 4π r²

W = W₁ + σ 4π r²

The volume of a sphere is

V = 4/3 π r³

Let's replace

ρ g 4/3 π r³ = W₁ + σ 4π r²

If we use the ideal gas equation

P V = n RT

P = ρ RT

ρ = P / RT

P / RT g 4/3 π r³ - σ 4 π r² = W₁

r² 4π (P/3RT r - σ) = W₁

Let's replace the values

r² 4π (1.01 10⁵ / (3 8.314 (70 + 273)) r - 0.060) = 13000

r² (11.81 r -0.060) = 13000 / 4pi

r² (11.81 r - 0.060) = 1034.51

As the independent term is very small we can despise it, to find the solution

r = 4.44 m

3 0

2 years ago

A uniform piece of wire, 20 cm long, is bent in a right angle in the center to give it an L-shape. How far from the bend is the

zlopas [31]

Answer:

the center of mass is 7.07 cm apart from the bend

Explanation:

the centre of mass of a wire of length L is L/2 ( assuming uniform density). Then initially the x coordinate of the centre of mass is

x₁ = L/2 = 20 cm /2 = 10 cm

when the wire is bent in a right angle the coordinates of the new centre of mass will be

x₂ = L₂/2

y₂= L₂/2

where L₂ is the length of the horizontal piece and vertical piece . Then L₂=L/2

x₂ = L₂/2 = L/4 = 20 cm/4 = 5 cm

y₂= L₂/2 = L/4 = 20 cm/4 = 5 cm

x₂=y₂=X

locating the bend in the origin (0,0) the distance to the centre of mass is

d = √(x₂²+y₂²) = √(2X²) = √2*X=√2*5cm = 7.07 cm

d = 7.07 cm

5 0

2 years ago

Read 2 more answers

The magnetic field of an electromagnetic wave in a vacuum is Bz =(2.4μT)sin((1.05×107)x−ωt), where x is in m and t is in s. You

tatiyna

Answer:

Explanation:

Given

$B_z=(2.4\mu T)\sin (1.05\times 10^7x-\omega t)$

Em wave is in the form of

$B=B_0\sin (kx-\omega t)$

where $\omega =frequency\ of\ oscillation$

$k=wave\ constant$

$B_0=Maximum\ value\ of\ Magnetic\ Field$

Wave constant for EM wave k is

$k=1.05\times 10^7 m^{-1}$

Wavelength of wave $\lambda =\frac{2\pi }{k}$

$\lambda =\frac{2\pi }{1.05\times 10^7}$

$\lambda =5.98\times 10^{-7} m$

7 0

2 years ago

A wire with a length of 150 m and a radius of 0.15 mm carries a current with a uniform current density of 2.8 x 10^7A/m^2. The c

Mrac [35]

Answer:

The current is 2.0 A.

(A) is correct option.

Explanation:

Given that,

Length = 150 m

Radius = 0.15 mm

Current density $J=2.8\times10^{7}\ A/m^2$

We need to calculate the current

Using formula of current density

$J = \dfrac{I}{A}$

$I=J\timesA$

Where, J = current density

A = area

I = current

Put the value into the formula

$I=2.8\times10^{7}\times\pi\times(0.15\times10^{-3})^2$

$I=1.97=2.0\ A$

Hence, The current is 2.0 A.

7 0

1 year ago