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diamong [38]
1 year ago
7

Drying of Cassava (Tapioca) Root. Tapioca flour is used in many countries for bread and similar products. The flour is made by d

rying coarse granules of the cassava root containing 66 wt % moisture to 5% moisture and then grinding to produce a flour. How many kg of granules must be dried and how much water removed to produce 5000 kg/h of flour?
Physics
1 answer:
svet-max [94.6K]1 year ago
5 0
Yea it would be 500 minus 10 is 490
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The three point charges +4.0 μC, -5.0 μC, and -9.0 μC are placed on the x-axis at the points x = 0 cm, x = 40 cm, and x = 120 cm
ale4655 [162]

Answer:

 

Explanation:

4μC will attract -9μC towards the centre and -5μC will repel it away from the centre.  Both these forces are opposite to each other.

Force due to 4μC on -9μC towards the centre

= k x Q₁ Q₂/R² = 9 X 10⁹ X 4 X 10⁻⁶ X 9 X 10⁻⁶ / (1.2)² = 225 X 10⁻³ N/C

Force due to -5μC on -9μC  away from the centre

= 9 x 10⁹ x 5 x 10⁻⁶x 9 x 10⁻⁶/( 0.8)² = 632.8 x 10⁻³ .N/C

Ner field =407.8 N/C.

6 0
2 years ago
Read 2 more answers
Two disks with the same rotational inertia i are spinning about the same frictionless shaft, with the same angular speed ω, but
valentina_108 [34]

Answer:

3. none of these

Explanation:

The rotational kinetic energy of an object is given by:

K=\frac{1}{2}I \omega^2

where

I is the moment of inertia

\omega is the angular speed

In this problem, we have two objects rotating, so the total rotational kinetic energy will be the sum of the rotational energies of each object.

For disk 1:

K_1 = \frac{1}{2}I (\omega)^2 = \frac{1}{2}I\omega^2

For disk 2:

K_2 = \frac{1}{2}I(-\omega)^2 = \frac{1}{2}I\omega^2

so the total energy is

K=K_1 + K_2 = \frac{1}{2}I\omega^2 + \frac{1}{2}I\omega^2 = I\omega^2

So, none of the options is correct.

5 0
1 year ago
A large crate is suspended from the end of a vertical rope. Is the tension in the rope greater when the crate is at rest or when
choli [55]

Answer:

Part a)

the tension force is equal to the weight of the crate

Part b)

tension force is more than the weight of the crate while accelerating upwards

tension force is less than the weight of crate if it is accelerating downwards

Explanation:

Part a)

When large crate is suspended at rest or moving with uniform speed then it is given as

F_t - mg = ma

here since speed is constant or it is at rest

so we will have

a = 0

F_t = mg

so the tension force is equal to the weight of the crate

Part b)

Now let say the crate is accelerating upwards

now we can say

F_t - mg = ma

F_t = mg + ma

so tension force is more than the weight of the crate

Now if the crate is accelerating downwards

F_t - mg = -ma

F_t = mg - ma

so tension force is less than the weight of crate if it is accelerating downwards

4 0
2 years ago
The resistivity of a semiconductor can be modified by adding different amounts of impurities. A rod of semiconducting material o
zavuch27 [327]

Answer:

pp

Explanation:

7 0
1 year ago
Two objects are placed in thermal contact and are allowed to come to equilibrium in isolation. the heat capacity of object a is
Harman [31]
Given:
Ca = 3Cb                      (1)
where
Ca =  heat capacity of object A
Cb =  heat capacity f object B

Also,
Ta = 2Tb                     (2)
where
Ta = initial temperature of object A
Tb = initial temperature of object B.

Let
Tf =  final equilibrium temperature of both objects,
Ma = mass of object A,
Mb = mass of object B.

Assuming that all heat exchange occurs exclusively between the two objects, then energy balance requires that
Ma*Ca*(Ta - Tf) = Mb*Cb*(Tf - Tb)           (3)

Substitute (1) and (2) into (3).
Ma*(3Cb)*(2Tb - Tf) = Mb*Cb*(Tf - Tb)
3(Ma/Mb)*(2Tb - Tf) = Tf - Tb

Define k = Ma/Mb, the ratio f the masses.
Then
3k(2Tb - Tf) = Tf - Tb
Tf(1+3k) = Tb(1+6k)
Tf = [(1+6k)/(1+3k)]*Tb

Answer:
T_{f} =( \frac{1+6k}{1+3k} )T_{b}= \frac{1}{2}( \frac{1+6k}{1+3k})T_{a}
where
k= \frac{M_{a}}{M_{b}} 
7 0
2 years ago
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