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2 years ago

A robot that can move a hand left and right and rotate the hand 360º is said to have

Physics

1 answer:

Luda [366]2 years ago

3 0

You've described two (2) axes of motion.
The third one would have been up-and-down.

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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

How much of the universe is believed to consist of ordinary atoms that can be detected?

Margaret [11]

The universe believed to be has no limit in space and the elements, molecules and different kind of gases are found in this area. So the percentage of that is believed that and ordinary atom can be detected is C. 3% i hope you understand my answer.

4 0

3 years ago

Estimate the wavelength corresponding to maximum emission from each of the following surfaces: the sun, a tungsten filament at 2

igomit [66]

Answer

Applying Wein's displacement

$\lamda_{max}\ T = 2898 \mu_mK$

1) for sun T = 5800 K

$\lambda_{max} = \dfrac{2898}{5800}$

$\lambda_{max} = 0.5 \mu_m$

2) for tungsten T = 2500 K

$\lambda_{max} = \dfrac{2898}{2500}$

$\lambda_{max} = 1.16 \mu_m$

3) for heated metal T = 1500 K

$\lambda_{max} = \dfrac{2898}{1500}$

$\lambda_{max} = 1.93 \mu_m$

4) for human skin T = 305 K

$\lambda_{max} = \dfrac{2898}{305}$

$\lambda_{max} = 9.50 \mu_m$

5) for cryogenically cooled metal T = 60 K

$\lambda_{max} = \dfrac{2898}{60}$

$\lambda_{max} = 48.3 \mu_m$

range of different spectrum

UV ----0.01-0.4

visible----0.4-0.7

infrared------0.7-100

for sun T = 5800

λ 0.01 0.4 0.7 100

λT 58 2320 4060 5.8 x 10⁵

F 0 0.125 0.491 1

fractions

for UV = 0.125

for visible = 0.441-0.125 = 0.366

for infrared = 1 -0.491 = 0.509

8 0

3 years ago

What is the strategy you use to solve word problems in physics?

Inga [223]

Answer:

The strategy we would like you to learn has five major steps: Focus the Problem, Physics Description, Plan a Solution, Execute the Plan, and Evaluate the Solution. Let's take a detailed look at each of these steps and then do an sample problem following the strategy.

3 0

2 years ago

In a certain region of space, a uniform electric field has a magnitude of 4.30 x 104 n/c and points in the positive x direction.

denis23 [38]

The magnetic force exerted by a field E to a charge q is given by F=Eq. In this case, F=4.30*10^4*(6.80mu C). 1mu C=10^-6C, so F=4.30*6.80=10^-2=0.29N. The direction is in the x direction, the direction that the field is applied because the charge is positive.

5 0

3 years ago