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3 years ago

11

Determine whether each substance will sink or float in corn syrup, which has a density of 1.36 g/cm3. Write “sink” or “float” in

the blanks. Gasoline: Water: Honey: Titanium:

2 answers:

Tatiana [17]3 years ago

3 0

Explanation:

A substance will floats if it is having lower density than the density of the liquid in which it is placed.
A substance will sink if it is having density greater than the density of the liquid in which it is kept.

Density of corn syrup = $1.36 g/cm^3$

1) Density of gasoline = $0.748 g/cm^3$

Density of the gasoline is less than the the density of corn syrup which means it will float in corn syrup.

2) Density of water = $1 g/cm^3$

Density of the water is less than the the density of corn syrup which means it will float in corn syrup.

3) Density of honey = $1.45 g/cm^3$

Density of the gasoline is more than the the density of corn syrup which means it will sink in corn syrup.

4) Density of titanium = $4.506 g/cm^3$

Density of the titanium is more than the the density of corn syrup which means it will sink in corn syrup.

denis23 [38]3 years ago

3 0

Answer:

1. float

2.float

3.sink

4. sink

Explanation:

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1 year ago

Two blocks with masses 1 and 2 are connected by a massless string that passes over a massless pulley as shown. 1 has a mass of 2

Bess [88]

Answer:

The acceleration of $M_2$ is $a = 0.7156 m/s^2$

Explanation:

From the question we are told that

The mass of first block is $M_1 = 2.25 \ kg$

The angle of inclination of first block is $\theta _1 = 43.5^o$

The coefficient of kinetic friction of the first block is $\mu_1 = 0.205$

The mass of the second block is $M_2 = 5.45 \ kg$

The angle of inclination of the second block is $\theta _2 = 32.5^o$

The coefficient of kinetic friction of the second block is $\mu _2 = 0.105$

The acceleration of $M_1 \ and\ M_2$ are same

The force acting on the mass $M_1$ is mathematically represented as

$F_1 = T - M_1gsin \theta_1 - \mu_1 M_1 g cos\theta_1$

=> $M_1 a = T - M_1gsin \theta_1 - \mu_1 M_1 g cos\theta_1$

Where T is the tension on the rope

The force acting on the mass $M_2$ is mathematically represented as

$F_2 = M_2gsin \theta_2 - T -\mu_2 M_2 g cos\theta_2$

$M_2 a = M_2gsin \theta_2 - T -\mu_2 M_2 g cos\theta_2$

At equilibrium

$F_1 = F_2$

So

$T - M_1gsin \theta_1 - \mu_1 M_1 g cos\theta_1 =M_2gsin \theta_2 - T -\mu_2 M_2 g cos\theta_2$

making a the subject of the formula

$a = \frac{M_2 g sin \theta_2 - M_1 g sin \theta_1 - \mu_1 M_1g cos \theta - \mu_2 M_2 g cos \theta_2 }{M_1 +M_2}$

substituting values $a = \frac{(5.45) (9.8) sin (32.5) - (2.25) (9.8) sin (43.5) - (0.205)*(2.25) *9.8cos (43.5) - (0.105)*(5.45) *(9.8) cos(32.5) }{2.25 +5.45}$

=> $a = 0.7156 m/s^2$

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3 years ago