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3 years ago

What is the mass of 2830 Newton's on Earth?

Physics

1 answer:

irina [24]3 years ago

5 0

Answer:

288.8kg

Explanation:

Given parameters:

Weight = 2830N

Unknown:

mass of this weight = ?

Solution:

Weight is the vertical component of force that acts on a body.

Weight = mass x acceleration due to gravity

So;

Mass = $\frac{2830}{9.8}$ = 288.8kg

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A 36-cm-circumference loop of wire is placed between the poles of an electromagnet. When a field of 0.76 T is switched on in a t

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Answer:

R = 25.3ohms

Explanation:

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3 years ago

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In a playground, there is a small merry-go-round of radius 1.20 m and mass 160 kg. Its radius of gyration is 91.0 cm. (Radius of

aksik [14]

Answer:

a) 145.6kgm^2

b) 158.4kg-m^2/s

c) 0.76rads/s

Explanation:

Complete qestion: a) the rotational inertia of the merry-go-round about its axis of rotation

(b) the magnitude of the angular momentum of the child, while running, about the axis of rotation of the merry-go-round and

a) From I = MK^2

I = (160Kg)(0.91m)^2

I = 145.6kgm^2

b) The magnitude of the angular momentum is given by:

L= r × p The raduis and momentum are perpendicular.

L = r × mc

L = (1.20m)(44.0kg)(3.0m/s)

L = 158.4kg-m^2/s

c) The total moment of inertia comprises of the merry- go - round and the child. the angular speed is given by:

L = Iw

158.4kgm^2/s = [145kgm^2 + ( 44.0kg)(1.20)^2]

w = 158.6/208.96

w = 0.76rad/s

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3 years ago

At the train station, you notice a large horizontal spring at the end of the track where the train comes in. This is a safety de

e-lub [12.9K]

Assume a maximum stopping acceleration of g/2 where g is acceleration due to gravity.

Answer:

2.99 m/s

Explanation:

Stopping distance, s = 3 ft = 0.914 m

final velocity, v = 0

a = g/2 = 4.9 m/s²

Use third equation of motion:

$v^2-u^2 = 2as$

substitute the values to find the speed of train:

$0 -u^2 = 2\times -4.9 \times 0.914 \\u^2=8.96 \\u=2.99 m/s$

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3 years ago

Assume that at sea-level the air pressure is 1.0 atm and the air density is 1.3 kg/m3.

scoundrel [369]

Answer

Pressure, P = 1 atm

air density, ρ = 1.3 kg/m³

a) height of the atmosphere when the density is constant

Pressure at sea level = 1 atm = 101300 Pa

we know

P = ρ g h

$h = \dfrac{P}{\rho\ g}$

$h = \dfrac{101300}{1.3\times 9.8}$

h = 7951.33 m

height of the atmosphere will be equal to 7951.33 m

b) when air density decreased linearly to zero.

at x = 0 air density = 0

at x= h ρ_l = ρ_sl

assuming density is zero at x - distance

$\rho_x = \dfrac{\rho_{sl}}{h}\times x$

now, Pressure at depth x

$dP = \rho_x g dx$

$dP = \dfrac{\rho_{sl}}{h}\times x g dx$

integrating both side

$P = g\dfrac{\rho_{sl}}{h}\times \int_0^h x dx$

$P =\dfrac{\rho_{sl}\times g h}{2}$

now,

$h=\dfrac{2P}{\rho_{sl}\times g}$

$h=\dfrac{2\times 101300}{1.3\times 9.8}$

h = 15902.67 m

height of the atmosphere is equal to 15902.67 m.

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3 years ago

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If it was removed it would not be a predator

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