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Anarel [89]
3 years ago
6

The center of gravity of a basketball is located _______.

Physics
1 answer:
zhenek [66]3 years ago
5 0

Answer:

at its centre

Explanation:

The centre of gravity is a point inside the body where entire weight os the body is said to be concentrated. For geometrical symmetric bodies it lies at the geometric centre of the body.

The base ball is a spherical body, so its centre of gravity lies at its centre.

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According to a certain estimate, the depth N(t), in centimeters, of the water in a certain tank at t hours past 2:00 in the morn
never [62]

Answer:

b) 7.00

Explanation:

N( t ) = -20( t - 5 )²

dN/ dt = -20 x 2 ( t - 5 )

For maximum N ( depth )

dN/dt = 0

- 40 ( t - 5 ) = 0

t = 5

So at 2 + 5 = 7 .00 am depth of water reaches its maximum.

6 0
3 years ago
A small cube of metal measures 19.0 mm on a side and weighs 79.6 g. What is the density of the metal in g/cm3?
Nady [450]

Answer:

density of cube =11.605 g/cm³

Explanation:

density of a substance is the mass per unit volume of that substance.

the density of a substance = \frac{mass}{volume}

volume of a cube = l³,

l = 19.0mm , lets convert mm to cm

1mm = 0.1cm, thus, 19mm =19*0.1 =1.9cm

length of cube =1.9cm

volume of cube = 1.9³

density of cube = \frac{79.6}{1.9^{3} }

density of cube =11.605 g/cm³

8 0
3 years ago
Air enters a turbine operating at steady state at 8 bar, 1600 K and expands to 0.8 bar. The turbine is well insulated, and kinet
kobusy [5.1K]

Answer:

the maximum theoretical work that could be developed by the turbine is 775.140kJ/kg

Explanation:

To solve this problem it is necessary to apply the concepts related to the adiabatic process that relate the temperature and pressure variables

Mathematically this can be determined as

\frac{T_2}{T_1} = (\frac{P_2}{P_1})^{(\frac{\gamma-1}{\gamma})}

Where

Temperature at inlet of turbine

Temperature at exit of turbine

Pressure at exit of turbine

Pressure at exit of turbine

The steady flow Energy equation for an open system is given as follows:

m_i = m_0 = mm(h_i+\frac{V_i^2}{2}+gZ_i)+Q = m(h_0+\frac{V_0^2}{2}+gZ_0)+W

Where,

m = mass

m(i) = mass at inlet

m(o)= Mass at outlet

h(i)= Enthalpy at inlet

h(o)= Enthalpy at outlet

W = Work done

Q = Heat transferred

v(i) = Velocity at inlet

v(o)= Velocity at outlet

Z(i)= Height at inlet

Z(o)= Height at outlet

For the insulated system with neglecting kinetic and potential energy effects

h_i = h_0 + WW = h_i -h_0

Using the relation T-P we can find the final temperature:

\frac{T_2}{T_1} = (\frac{P_2}{P_1})^{(\frac{\gamma-1}{\gamma})}\\

\frac{T_2}{1600K} = (\frac{0.8bar}{8nar})^{(\frac{1.4-1}{1.4})}\\ = 828.716K

From this point we can find the work done using the value of the specific heat of the air that is 1,005kJ / kgK

W = h_i -h_0W = C_p (T_1-T_2)W = 1.005(1600 - 828.716)W = 775.140kJ/Kg

the maximum theoretical work that could be developed by the turbine is 775.140kJ/kg

4 0
3 years ago
How should the student change the circuit to give negative values for current and
dmitriy555 [2]

Answer:

Flip the cell.

Explanation:

This reverses direction of energy transfer.

Alternatively, flip ammeters and voltmeters to give negative readings.

what do penguins eat for lunch?

Ice-burgers!

3 0
2 years ago
To tugboats pull a disabled supertanker. Each tug exerts a constant force of 1.50•10^6 N, one at an angle 19.0° west of north, a
irina1246 [14]

Answer:

Work done by a tug boat, W = 1.735 x 10⁸ J

Explanation:

Given,

The of each tugboat, F = 1.5 x 10⁶ N

The angle of each tugboat forms with the resultant force, θ = 19°

The displacement of the supertanker, s = 710 m

The individual tugboat will be responsible for the displacement, d = 710/2

                                                                                                               = 355 m

The displacement component in each tugboat direction = 355 · sin θ meter

Therefore, the work done by each tugboat is

                                           W = F x S    joules

Substituting the values in the above equation

                                            W = 1.5 x 10⁶  x  355 · sin θ

                                                = 1.735 x 10⁸ J

Hence, the work done by each tugboat is, W = 1.735 x 10⁸ J                        

6 0
3 years ago
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