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Vladimir79 [104]
3 years ago
7

What is the mass of a rock that weighs 49 newtons?

Physics
1 answer:
ycow [4]3 years ago
8 0
The answer is (1 kilogram) hope it helps and have a great day!
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A rectangular loop with dimensions 4.20 cm by 9.50 cm carries current I. The current in the loop produces a magnetic field at th
tiny-mole [99]

Answer:

I=1.48 A

Explanation:

Given that

B=3.1 x 10⁻5 T

b= 4.2 cm

l= 9.5 cm

The relationship for magnetic field  and current given as

B=\dfrac{2\mu _oI}{\pi}D

Where

D=\dfrac{\sqrt{l^2+b^2}}{lb}

By putting the values

D=\dfrac{\sqrt{l^2+b^2}}{lb}

D=\dfrac{\sqrt{0.042^2+0.095^2}}{0.042\times 0.095}

D=26.03 m⁻¹

B=\dfrac{2\mu _oI}{\pi}D

3.1\times 10^{-5}=\dfrac{2\times 4\times \pi \times 10^{-7} I}{\pi}\times 26.03

I=\dfrac{3.1\times 10^{-5}}{{2\times 4\times 10^{-7} }\times 26.03}

I=1.48 A

8 0
3 years ago
You drop a steel ball bearing, with a radius of 2.40 mm, into a beaker of honey. Note that honey has a viscosity of 6.00 Pa/s an
Stells [14]

Answer:

The “terminal speed” of the ball bearing is 5.609 m/s

Explanation:

Radius of the steel ball R = 2.40 mm

Viscosity of honey η = 6.0 Pa/s

\text { Viscosity has Density } \sigma=1360 \mathrm{kg} / \mathrm{m}^{3}

\text { Steel has a density } \rho=7800 \mathrm{kg} / \mathrm{m}^{3}

\left.\mathrm{g}=9.8 \mathrm{m} / \mathrm{s}^{2} \text { (g is referred to as the acceleration of gravity. Its value is } 9.8 \mathrm{m} / \mathrm{s}^{2} \text { on Earth }\right)

While calculating the terminal speed in liquids where density is high the stokes law is used for viscous force and buoyant force is taken into consideration for effective weight of the object. So the expression for terminal speed (Vt)

V_{t}=\frac{2 \mathrm{R}^{2}(\rho-\sigma) \mathrm{g}}{9 \eta}

Substitute the given values to find "terminal speed"

\mathrm{V}_{\mathrm{t}}=\frac{2 \times 0.0024^{2}(7800-1360) 9.8}{9 \times 6}

\mathrm{V}_{\mathrm{t}}=\frac{0.0048 \times 6440 \times 9.8}{54}

\mathrm{V}_{\mathrm{t}}=\frac{302.9376}{54}

\mathrm{V}_{\mathrm{t}}=5.609 \mathrm{m} / \mathrm{s}

The “terminal speed” of the ball bearing is 5.609 m/s

7 0
3 years ago
Objects with unlike charges what each other
DerKrebs [107]
The answer is attract. Hope it helps! :)
6 0
3 years ago
Read 2 more answers
Two masses (5.3kg and 7.5kg) are fastened together with a small amount of explosive. They are loaded into a spring gun that is t
Vsevolod [243]

Answer:

8.2

Explanation:

8 0
3 years ago
A small ball is attached to the lower end of a 0.800-m-long string, and the other end of the string is tied to a horizontal rod.
mestny [16]

Answer:

a = 17.68 m/s²

Explanation:

given,

length of the string, L = 0.8 m

angle made with vertical, θ = 61°

time to complete 1 rev, t = 1.25 s

radial acceleration = ?

first we have to calculate the radius of the circle

 R = L sin θ

 R = 0.8 x sin 61°

 R = 0.7 m

now, calculating at the angular velocity

\omega =\dfrac{2\pi}{T}

\omega =\dfrac{2\pi}{1.25}

  ω = 5.026 rad/s

now, radial acceleration

 a = r ω²

 a = 0.7 x 5.026²

a = 17.68 m/s²

hence, the radial acceleration of the ball is equal to 17.68 rad/s²

7 0
3 years ago
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