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3 years ago

HELP PLEASE

Physics

2 answers:

Wewaii [24]3 years ago

6 0

Answer:

The solute, salt, must dissolve in the solvent, water

Explanation:

So you would use solvent

tigry1 [53]3 years ago

5 0

The component that’s dissolved is called the solvent

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Why won’t anyone help me please anybody help me I really need help .

irga5000 [103]

Answer:

1➡️ this is the method of decomposition

2➡️ H2 and O2

3➡️ b

sorry if I am wrong

8 0

2 years ago

If an object force of 50 N is used to move an object a distance of 20 m, what distance must the object be moved if the input for

steposvetlana [31]

Answer:

$\ d_{out} = 100 \ m.$

Explanation:

Given data:

$F_{in} = 50 \ \rm N$

$F_{out} = 10 \ \rm N$

$d_{in} = 20 \ m$

Let the distance traveled by the object in the second case be $d_{out}.$

In the given problem, work done by the forces are same in both the cases.

Thus,

$W_{in} = W_{out}$

$F_{in}.d_{in} = F_{out}.d_{out}$

$\Rightarrow \ d_{out} = \frac{F_{in}.d_{in}}{F_{out}}$

$\ d_{out} = \frac{50 \times 20}{10}$

$\ d_{out} = 100 \ m.$

5 0

3 years ago

All of the following are types of symbiotic relationships except for

nevsk [136]

The answer is D. the way I remember it is they all end with -alism.

3 0

3 years ago

Light bulb 1 operates with a filament temperature of 2700 K whereas light bulb 2 has a filament temperature of 2100 K. Both fila

Lemur [1.5K]

Answer:

0.3659

Explanation:

The power (p) is given as:

P = AeσT⁴

where,

A =Area

e = transmittivity

σ = Stefan-boltzmann constant

T = Temperature

since both the bulbs radiate same power

P₁ = P₂

Where, 1 denotes the bulb 1

2 denotes the bulb 2

thus,

A₁e₁σT₁⁴ = A₂e₂σT₂⁴

Now e₁=e₂

⇒A₁T₁⁴ = A₂T₂⁴

$\frac{A_1}{A_2} =\frac{T_{2}^{4}}{T_{1}^{4}}$

substituting the values in the above question we get

$\frac{A_1}{A_2} =\frac{2100_{2}^{4}}{2700_{1}^{4}}$

$\frac{A_1}{A_2} }$ =0.3659

6 0

3 years ago

A tissue sample at 275 K is submerged in 2 kg of liquid nitrogen at 70 K for cryopreservation. The final temperature of the nitr

kirill115 [55]

Answer:

The heat capacity of a sample is 37.7 J/K.

Explanation:

Given that,

Submerged temperature of tissue sample = 275 K

Mass of liquid nitrogen= 2 kg

Temperature = 70 K

Final temperature = 75 K

We need to calculate the heat

Using formula of heat

$Q=mc(T_{f}-T_{i})$

Put the value into the formula

$Q=2\times1.039\times10^{3}\times(75-70)$

$Q=10390\ J$

We need to calculate the heat capacity of a sample

Using formula of heat capacity

$\Delta S=\dfrac{Q}{T}$

Put the value into the formula

$\Delta S=\dfrac{10390}{275}$

$\Delta S=37.7\ J/K$

Hence, The heat capacity of a sample is 37.7 J/K.

7 0

3 years ago

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