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2 years ago

HELP ASAP ILL GIVE BRAINLIST

Physics

2 answers:

velikii [3]2 years ago

7 0

Are you Brainliest cuz i sure ain’t

OlgaM077 [116]2 years ago

4 0

Answer:

Each of the joints represents a degree of freedom in the manipulator system and allows translation and rotary motion :) Hope this helps

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4 what is the difference between an array's size declarator and a subscript?

Ber [7]

What is the difference between asize declarator and a subscript? The size declarator is ... When writing a function that accepts a two-dimensional array as an argument, which size declarator must you provide in the parameter for thearray? The second size ...

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3 years ago

Some gamma ray bursts are hypothesized to come from mergers of neutron stars or black holes. if this hypothesis is correct, what

Umnica [9.8K]

We should see (and have now detected with LIGO) gravitational waves

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3 years ago

Water is pumped steadily out of a flooded basement at a speed of 5.4 m/s through a uniform hose of radius 0.83 cm. The hose pass

Gala2k [10]

To solve this problem it is necessary to apply the concepts related to the flow as a function of the volume in a certain time, as well as the potential and kinetic energy that act on the pump and the fluid.

The work done would be defined as

$\Delta W = \Delta PE + \Delta KE$

Where,

PE = Potential Energy

KE = Kinetic Energy

$\Delta W = (\Delta m)gh+\frac{1}{2}(\Delta m)v^2$

Where,

m = Mass

g = Gravitational energy

h = Height

v = Velocity

Considering power as the change of energy as a function of time we will then have to

$P = \frac{\Delta W}{\Delta t}$

$P = \frac{\Delta m}{\Delta t}(gh+\frac{1}{2}v^2)$

The rate of mass flow is,

$\frac{\Delta m}{\Delta t} = \rho_w Av$

Where,

$\rho_w$ = Density of water

A = Area of the hose $\rightarrow A=\pi r^2$

The given radius is 0.83cm or $0.83 * 10^{-2}$ m, so the Area would be

$A = \pi (0.83*10^{-2})^2$

$A = 0.0002164m^2$

We have then that,

$\frac{\Delta m}{\Delta t} = \rho_w Av$

$\frac{\Delta m}{\Delta t} = (1000)(0.0002164)(5.4)$

$\frac{\Delta m}{\Delta t} = 1.16856kg/s$

Final the power of the pump would be,

$P = \frac{\Delta m}{\Delta t}(gh+\frac{1}{2}v^2)$

$P = (1.16856)((9.8)(3.5)+\frac{1}{2}5.4^2)$

$P = 57.1192W$

Therefore the power of the pump is 57.11W

6 0

3 years ago

How much heat is needed to change 1.25 kg of steak at 100°C to water at 100°C?

cricket20 [7]

The heat required to change 1.25 kg of steak is 2825 kJ /kg.

Explanation:

Given, mass m = 1.25 kg, Temperature t = 100 degree celsius

To calculate the heat required,

Q = m $\times$ L

where m represents the mass in kg,

L represents the heat of vaporization.

When a material in the liquid state is given energy, it changes its phase from liquid to vapor and the energy absorbed in this process is called heat of the vaporization. The heat of vaporization of the water is about 2260 kJ/kg.

Q = 1.25 $\times$ 2260

Q = 2825 kJ /kg.

7 0

3 years ago

A stone is thrown in the upwards direction at the velocity of 4 m/s. It attains a certain height and then it falls back. During

Sati [7]

Answer: 0.8 m

Explanation:

Velocity of throw = 4m/s

Maximum Height attained(h) =?

Downward acceleration experienced = 10m/s^2

Using the relation:

v^2 = u^2 + 2aS

v = final Velocity = 0 (at maximum height)

u = Initial Velocity = 4

a = g downward acceleration = - 10

0 = 4^2 + 2(-10)(S)

0 = 16 - 20S

20S = 16

S = 16 / 20

S = 0.8m

Maximum Height attained = 0.8m

3 0

3 years ago