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2 years ago

Which of the following is always a part of speed, velocity, and acceleration?

1 answer:

3 0

Speed = distance/time
Velocity = displacement/time
Acceleration = change in velocity/time

Time is a part of all three quantities.

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A Submerged Ball Part A A ball of mass mb and volume V is lowered on a string into a fluid of density Pi (Figure 1) Assume that

frozen [14]

Answer:

T = mg - $\rho_f$ ×V×g

Explanation:

Here we have the mass of the ball is given as $m_b$

The density of the fluid = $\rho_f$

The volume of the ball = V

Acceleration due to gravity = g

According to Archimedes's principle, the upthrust = Weight of fluid displaced

Tension in string = T = weight of ball - upthrust

T = mg - $\rho_f$ ×V×g

5 0

3 years ago

Vectors a and b have scalar product â6.00, and their vector product has magnitude +9.00. what is the angle between these two vec

salantis [7]

Answer:

Value of angle between vector a and b is $56.30^{\circ}$ .

Explanation:

Vectors a and b have scalar product 6.00

Let $\theta$ be the angle between a and b.

$\vec{a}.\vec{b} = 6$

ab cos $\theta$ = 6 ...(1)

Vectors a and b have magnitude of vector product 9.00

$\vec{a} \times\vec{b} = 9$

ab sin $\theta$ = 9 ...(2)

Dividing equation (2) by (1) we get

$\frac{ab sin \theta}{ab cos \theta} = \frac{9}{6}$

tan $\theta$ = 1.5

$\theta = tan ^{-1} (1.5)$

$\theta$ = $56.30^{\circ}$

Thus, value of angle between vector a and b is $56.30^{\circ}$ .

3 0

3 years ago

3. What are the two types of mixtures?

ioda

Answer:

heterogeneous and homogeneous

Explanation:

Hetero you can separate whilst homo is combined and cant be separated

7 0

2 years ago

What is the rarefaction of a longitudinal

Softa [21]

The answer Is B! Good luck

8 0

2 years ago

A heat engine has a maximum possible efficiency of 0.780. If it operates between a deep lake with a constant temperature of-24.8

Volgvan

Answer : The temperature of the hot reservoir (in Kelvins) is 1128.18 K

Explanation :

Efficiency of carnot heat engine : It is the ratio of work done by the system to the system to the amount of heat transferred to the system at the higher temperature.

Formula used for efficiency of the heat engine.

$\eta =1-\frac{T_c}{T_h}$

where,

$\eta$ = efficiency = 0.780

$T_h$ = Temperature of hot reservoir = ?

$T_c$ = Temperature of cold reservoir = $-24.8^oC=273+(-24.8)=248.2K$

Now put all the given values in the above expression, we get:

$\eta =1-\frac{T_c}{T_h}$

$0.780=1-\frac{248.2K}{T_h}$

$T_h=1128.18K$

Therefore, the temperature of the hot reservoir (in Kelvins) is 1128.18 K

8 0

2 years ago