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2 years ago

7

Suppose a cart is being pushed by a certain net force. If the net force is doubled, by how much does the acceleration change? Ex

plain.

1 answer:

Alekssandra [29.7K]2 years ago

5 0

Answer:

The acceleration doubles.

Explanation:

couldnt find an explenation sorry if its wrong -w-

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The _______ energy from the _______ arrives in the form of visible and invisible light. A. Earth's, Sun B. Moon's, Earth C. Sun'

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3 years ago

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2 years ago

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3 years ago

A solution of 0.0027 M K2CrO4 was diluted from 3.00 mL to 100. mL. What is the molarity of the new solution?

Stells [14]

Answer:

[K₂CrO₄] → 8.1×10⁻⁵ M

Explanation:

First of all, you may know that if you dilute, molarity must decrease.

In the first solution we need to calculate the mmoles:

M = mmol/mL

mL . M = mmol

0.0027 mmol/mL . 3mL = 0.0081 mmoles

These mmoles of potassium chromate are in 3 mL but, it stays in 100 mL too.

New molarity is:

0.0081 mmoles / 100mL = 8.1×10⁻⁵ M

4 0

3 years ago

A 32.5 g iron rod, initially at 22.4 ∘C, is submerged into an unknown mass of water at 63.0 ∘C, in an insulated container. The f

a_sh-v [17]

Answer:

$m_{H_2O}=39.0g$

Explanation:

Hello,

In this case, is possible to infer that the thermal equilibrium is governed by the following relationship:

$\Delta H_{iron}=-\Delta H_{H_2O}\\m_{iron}Cp_{iron}(T_{eq}-T_{iron})=-m_{H_2O}Cp_{H_2O}(T_{eq}-T_{H_2O})$

Thus, both iron's and water's heat capacities are: 0.444 and 4.18 J/g°C respectively, so one solves for the mass of water as shown below:

$m_{H_2O}=\frac{m_{iron}Cp_{iron}(T_{eq}-T_{iron})}{-Cp_{H_2O}(T_{eq}-T_{H_2O}} \\\\m_{H_2O}=\frac{32.5g*0.444\frac{J}{g^0C}*(59.7-22.4)^0C}{-4.18\frac{J}{g^0C}*(59.7-63.0)^0C} \\\\m_{H_2O}=39.0g$

Best regards.

8 0

3 years ago