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3 years ago

What is the mass of an object if a force of 34 N produces an acceleration of 4.0 m/s squared

Physics

1 answer:

mrs_skeptik [129]3 years ago

6 0

Answer:

8.5 kg

Explanation:

F = ma

34 N = m(4.0m/s^2)

m = 8.5 kg

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A 100-lb load hangs from three cables of equal length that are anchored at the points (minus4,0,0), (2,2 StartRoot 3 EndRoo

Anarel [89]

Answer:

$\vec{F}_{cable_1} = (-19.245 \ lbf ,0, 33.333 \ lbf)$
$\vec{F}_{cable_2} = (9.622 \ lbf , 13.608 \ lbf ,33.333 \ lbf)$
$\vec{F}_{cable_3} = (9.622 \ lbf , -13.608 \ lbf ,33.333 \ lbf)$

Explanation:

The mass of the load is

$m_{load} = 100 \ lb$

As the mass hangs, the cables must be tight, so, we can obtain the vector parallel to the cable as:

$\vec{r}_{cable} = \vec{r}_{anchored} - \vec{r}_{load}$

where $\vec{r}_{load}$ is the position of the load and $\vec{r}_{anchored}$ is the point where the cable is anchored.

So, for our cables

$\vec{r}_{cable_1} = (-4,0,0) - (0,0,-4\sqrt{3})=(-4,0,4\sqrt{3})$

$\vec{r}_{cable_2} = (2,2\sqrt{3},0) - (0,0,-4\sqrt{3})=(2,2\sqrt{2},4\sqrt{3})$

$\vec{r}_{cable_3} = (2,-2\sqrt{3},0) - (0,0,-4\sqrt{3})=(2,-2\sqrt{2},4\sqrt{3})$

We know that the forces must be in this directions, so we can write

$\vec{F}_i=k_i \vec{r}_{cable_i}$

We also know, as the system is in equilibrium, the sum of the forces must be zero:

$\vec{F}_{cable_1}+\vec{F}_{cable_2}+\vec{F}_{cable_3}+\vec{W}=0$

where $\vec{W}$ is the weight,

$\vec{W} = (0,0,-100 \ lbf)$

So, we get:

$k_1 (-4,0,4\sqrt{3}) + k_2 (2,2\sqrt{2},4\sqrt{3}) + k_3 (2,-2\sqrt{2},4\sqrt{3}) + (0,0,-100 \ lbf) = (0,0,0)$

This gives us the following equations:

$-4 \ k_1 + 2 \ k_2 + 2 \ k_3 = 0$
$2\sqrt{2} \ k_2 -2\sqrt{2} \ k_3 = 0$
$4\sqrt{3} \ k_1 + 4\sqrt{3} \ k_2 + 4\sqrt{3} \ k_3 -100 \ lbf = 0$

From equation [2] is clear that $k_2 = k_3$ , we can see that

$2\sqrt{2} \ k_2 = 2\sqrt{2} \ k_3$

$\frac{2\sqrt{2} \ k_2}{2\sqrt{2} } = \frac{2\sqrt{2} \ k_3}{2\sqrt{2} }$

$k_2 = k_3$

Now, putting this in equation [1]

$-4 \ k_1 + 2 \ k_2 + 2 \ k_3 = -4 \ k_1 + 2 \ k_3 + 2 \ k_3 = -4 \ k_1 + 4 \ k_3 = 0$

$4 \ k_1 = 4 \ k_3$

$\ k_1 = \ k_3$

Taking this result to the equation [3]

$4\sqrt{3} \ k_1 + 4\sqrt{3} \ k_2 + 4\sqrt{3} \ k_3 -100 \ lbf = 0$

$4\sqrt{3} \ k_3 + 4\sqrt{3} \ k_3 + 4\sqrt{3} \ k_3 = 100 \ lbf$

$3 * (4\sqrt{3} \ k_3) = 100 \ lbf$

$k_3 = \frac{100 \ lbf}{ 12 \sqrt{3}}$

$k_1 = k_2 = k_3 = \frac{100 \ lbf}{ 12 \sqrt{3}}$

So, the forces are:

$\vec{F}_{cable_1} = k_1 (-4,0,4\sqrt{3})$

$\vec{F}_{cable_1} = \frac{100 \ lbf}{ 12 \sqrt{3}} (-4,0,4\sqrt{3})$

$\vec{F}_{cable_1} = (-\frac{100 \ lbf}{ 3 \sqrt{3}},0,\frac{100 \ lbf}{3})$

$\vec{F}_{cable_1} = (-19.245 \ lbf ,0, 33.333 \ lbf)$

$\vec{F}_{cable_2} = k_2 (2,2\sqrt{2},4\sqrt{3})$

$\vec{F}_{cable_2} = \frac{100 \ lbf}{ 12 \sqrt{3}} (2,2\sqrt{2},4\sqrt{3})$

$\vec{F}_{cable_2} = (9.622 \ lbf , 13.608 \ lbf ,33.333 \ lbf)$

$\vec{F}_{cable_3} = k_3 (2,-2\sqrt{2},4\sqrt{3})$

$\vec{F}_{cable_3} = \frac{100 \ lbf}{ 12 \sqrt{3}} (2,-2\sqrt{2},4\sqrt{3})$

$\vec{F}_{cable_3} = (9.622 \ lbf , -13.608 \ lbf ,33.333 \ lbf)$

5 0

4 years ago

A new planet has been discovered that has a mass one-sixth that of Earth and a radius that is six times that of Earth. Determine

Soloha48 [4]

Answer:

0.045 m/s²

Explanation:

Let the mass of Earth be 'M' and radius be 'R'.

Given:

Mass of the new planet (m) = one-sixth of Earth's mass = $\frac{M}{6}$

Radius of new planet (r) = 6 times Earth's radius = $6R$

We know that, acceleration due to gravity of a planet of mass 'M' and radius 'R' is given as:

$g=\dfrac{GM}{R^2}$

Now, this is acceleration due to gravity on Earth.

Now, acceleration due to gravity of new planet is given as:

$g_{new}=\dfrac{Gm}{r^2}\\\\g_{new}=\dfrac{G\times\frac{M}{6}}{(6R)^2}\\\\g_{new}=\dfrac{GM}{6\times 36R^2}\\\\g_{new}=\frac{1}{216}(\frac{GM}{R^2})=\frac{1}{216}\times g
$

Now, the value of 'g' on Earth is approximately 9.8 m/s². So,

$g_{new}=\frac{9.8}{216}=0.045\ m/s^2$

Therefore, the free fall acceleration on the surface of this planet is 0.045 m/s².

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4 years ago

Are nerve cells in direct contact with other types of tissues ?<br><br> Please help.

Ber [7]

Answer:

Yes, they are.

Explanation:

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3 years ago

A falling object is said to reach a constant velocity (terminal velocity)..

Lana71 [14]

C

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3 years ago

Three astronauts outside a spaceship and that they decide to play catch. All the astronauts weigh the same on Earth and are equa

guajiro [1.7K]

Answer:

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3 years ago

What is the mass of an object if a force of 34 N produces an acceleration of 4.0 m/s squared​

What is the mass of an object if a force of 34 N produces an acceleration of 4.0 m/s squared