If a 50 N force pulls on a 10 kg object,how much acceleration will occur?

Chemistry

2 answers:

Zinaida [17]3 years ago

6 0

Answer:

The value of acceleration occurred is $5 m/s^2$ .

Explanation:

Force is a push or pull on an object with mass which causes in change in its velocity Or it is the product of its mass and acceleration by which that object is moving.

Force = $mass\times acceleration$

F = ma

We have:

F = 50 N, a = ? , m = 10 kg

Putting in the values we get,

$50 N=10 kg\times a$

$a=\frac{50 N}{10 kg}=5 m/s^2$

The value of acceleration occurred is $5 m/s^2$ .

meriva3 years ago

4 0

Assuming that the force of friction does not equal to the force at which you are pulling at, hence Fnet is not zero, to solve for acceleration need to use the following equation:

Fnet = ma
a = Fnet/m
a = 50 N/10 kg
a = 5 N/kg or a = 5 m/s^2.

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A solution made by dissolving 33 mg of insulin in 6.5 mL of water has an osmotic pressure of 15.5 mmHg at 25°C. Calculate the mo

Liula [17]

Answer: The molar mass of the insulin is 6087.2 g/mol

Explanation:

To calculate the concentration of solute, we use the equation for osmotic pressure, which is:

$\pi=iMRT$

Or,

$\pi=i\times \frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}\times RT$

where,

$\pi$ = osmotic pressure of the solution = 15.5 mmHg

i = Van't hoff factor = 1 (for non-electrolytes)

Mass of solute (insulin) = 33 mg = 0.033 g (Conversion factor: 1 g = 1000 mg)

Volume of solution = 6.5 mL

R = Gas constant = $62.364\text{ L.mmHg }mol^{-1}K^{-1}$

T = temperature of the solution = $25^oC=[273+25]=298K$

Putting values in above equation, we get:

$15.5mmHg=1\times \frac{0.033\times 1000}{\text{Molar mass of insulin}\times 6.5}\times 62.364\text{ L.mmHg }mol^{-1}K^{-1}\times 298K\\\\\text{molar mass of insulin}=\frac{1\times 0.033\times 1000\times 62.364\times 298}{15.5\times 6.5}=6087.2g/mol$