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2 years ago

An object has a gravitational force of 111N acting on it. If it had 4 times more mass, what would the force on it be?

Physics

1 answer:

Nastasia [14]2 years ago

4 0

Answer:

What is the gravity equation? — It is equal to 6.674×10-11 N·m²/kg². Did you notice that this equation is similar to the formula in Coulomb's law?

Explanation:

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School homework about multiplying fractions worth 30 brainly points

RoseWind [281]

Answer:

Explanation:

1. $\frac{1}{3}$ x $\frac{6}{7}$ = $\frac{6}{21}$

= $\frac{2}{7}$

b. $\frac{6}{8}$ + $\frac{4}{9}$ = $\frac{24}{72}$

= $\frac{1}{3}$

c. $\frac{10}{15}$ x $\frac{3}{4}$ = $\frac{30}{60}$

= $\frac{1}{2}$

d. $\frac{7}{10}$ of $\frac{5}{10}$ = $\frac{7}{10}$ x $\frac{5}{10}$

= $\frac{35}{100}$

= $\frac{7}{20}$

e. $\frac{3}{8}$ of $\frac{4}{6}$ = $\frac{3}{8}$ x $\frac{4}{6}$

= $\frac{12}{48}$

= $\frac{1}{4}$

f. $\frac{7}{12}$ of $\frac{9}{14}$ = $\frac{7}{12}$ x $\frac{9}{14}$

= $\frac{63}{168}$

= $\frac{3}{8}$

2. $\frac{22}{6}$ x $\frac{3}{11}$ = $\frac{66}{66}$

= 1

b. $\frac{15}{6}$ x $\frac{4}{5}$ = $\frac{60}{30}$

= 2

c. $\frac{25}{8}$ x $\frac{4}{10}$ = $\frac{100}{80}$

= $\frac{5}{4}$

d. $\frac{33}{12}$ x $\frac{4}{15}$ = $\frac{132}{180}$

= $\frac{11}{15}$

4 0

3 years ago

A scientist wishes to generate a chemical reaction in his laboratory. The temperature values in his laboratory

8090 [49]

At the same temperature . . .

 Fahrenheit reading = (1.8 times Celsius reading) + 32 .

F = (1.8 x 232) + 32

F = 417.6 + 32

F = 449.6°

8 0

3 years ago

Read 2 more answers

A stone is dropped into a river from a bridge at a height h above the water. Another stone is thrown vertically down at a time t

Mumz [18]

Answer:

$v_{y_0} = \frac{\frac{g}{2}t(t - 2\sqrt{\frac{2h}{g}})}{\sqrt{\frac{2h}{g}} - t}$

Explanation:

We will apply the equations of kinematics to both stones separately.

First stone:

Let us denote the time spent after the second stone is thrown as 'T'.

$y - y_0 = v_{y_0}(t+T) + \frac{1}{2}a(t+T)^2\\0 - h = 0 + \frac{1}{2}(-g)(t+T)^2\\(t+T)^2 = \frac{2h}{g}\\T = \sqrt{\frac{2h}{g}}-t$

Second stone:

$y - y_0 = v_{y_0}T + \frac{1}{2}aT^2\\0 - h = v_{y_0}T -\frac{1}{2}gT^2\\-h = v_{y_0}(\sqrt{\frac{2h}{g}} - t) - \frac{g}{2}(\sqrt{\frac{2h}{g}} - t)^2\\-h = v_{y_0}(\sqrt{\frac{2h}{g}} - t) - \frac{g}{2}(\frac{2h}{g} + t^2 - 2t\sqrt{\frac{2h}{g}})\\-h = v_{y_0}\sqrt{\frac{2h}{g}} - v_{y_0}t - h -\frac{g}{2}t^2 + gt\sqrt{\frac{2h}{g}}\\v_{y_0}(\sqrt{\frac{2h}{g}} - t) = \frac{g}{2}t^2 - gt\sqrt{\frac{2h}{g}}\\v_{y_0} = \frac{\frac{g}{2}t(t - 2\sqrt{\frac{2h}{g}})}{\sqrt{\frac{2h}{g}} - t}$

6 0

3 years ago

Read 2 more answers

The weight of a block on the inclined plane is 500 N and the angle of incline is 30 degrees. What is the magnitude of the force

yanalaym [24]

Answer:

250 N

433 N

Explanation:

N = Normal force by the surface of the inclined plane

W = Weight of the block = 500 N

f = static frictional force acting on the block

Parallel to incline, force equation is given as

f = W Sin30

f = (500) Sin30

f = 250 N

Perpendicular to incline force equation is given

N = W Cos30

N = (500) Cos30

N = 433 N

3 0

3 years ago

Please help with this!!!!!

34kurt

(1.00 atm) (0.1156 L) = (n) (0.08206 L atm / mol K) (273 K) I hoped that helped

4 0

3 years ago