All categories

2 years ago

If my weight on Earth is 140lbs, what is my mass?

Physics

1 answer:

SashulF [63]2 years ago

5 0

Answer:

63.57 kg

Explanation:

weight = 140 lbs

Let the mass is m.

1 lbs = 4.45 N

The weight of an object is defined as the force with which our earth attracts the body towards its centre.

Weight is the product of mass of the body and the acceleration due to gravity of that planet.

W = m x g

On earth surface g = 9.8 m/s^2

Now convert lbs in newton

So, 140 lbs = 140 x 4.45 = 623 N

So, m x 9.8 = 623

m = 63.57 kg

Thus, the mass is 63.57 kg.

You might be interested in

find the time taken, if the speed of a train increased from 72 km/hr to 90 km/hr for 234 km. leave your answer in seconds

Airida [17]

Answer:

Time taken = 10400 s

Explanation:

Given:

Initial speed of the train, $u=72\textrm{ km/h}=72\times \frac{5}{18}=20\textrm{ m/s}$

Final speed of the train, $v=90\textrm{ km/h}=90\times \frac{5}{18}=25\textrm{ m/s}$

Displacement of the train, $S=234\textrm{ km}=234\times 1000=234000\textrm{ m}$

Using Newton's equation of motion,

$v - u = at\\a=\frac{v-u}{t}$

Now, using Newton's equation of motion for displacement,

$v^{2}-u^{2}=2aS$

Now, plug in the value of $a=\frac{v-u}{t}$ in the above equation. This gives,

$v^{2}-u^{2}=2\times \frac{v-u}{t}\times S\\(v+u)(v-u)=\frac{2(v-u)S}{t}\\t=\frac{2(v-u)S}{(v+u)(v-u)}\\t=\frac{2S}{v+u}$

Now, plug in 234000 m for $S$ , 25 m/s for $v$ and 20 m/s for $u$ . Solve for $t$ .

$t=\frac{2S}{v+u}\\t=\frac{2\times 234000}{25+20}\\t=\frac{468000}{45}=10400\textrm{ s}$

Therefore, the time taken by the train is 10400 s.

3 0

3 years ago

Look at the divisions between 2 and 4. What level of precision does this stopwatch have based on the divisions marked on its fac

klemol [59]

Answer:

10ths of a second

Explanation:

Because I just did it

8 0

2 years ago

Three equal charge 1.8*10^-8 each are located at the corner of an equilateral triangle ABC side 10cm.calculate the electric pote

Arlecino [84]

Answer:

If all these three charges are positive with a magnitude of $1.8 \times 10^{-8}\; \rm C$ each, the electric potential at the midpoint of segment $\rm AB$ would be approximately $8.3 \times 10^{3}\; \rm V$ .

Explanation:

Convert the unit of the length of each side of this triangle to meters: $10\; \rm cm = 0.10\; \rm m$ .

Distance between the midpoint of $\rm AB$ and each of the three charges:

$d({\rm A}) = 0.050\; \rm m$ .
$d({\rm B}) = 0.050\; \rm m$ .
$d({\rm C}) = \sqrt{3} \times (0.050\; \rm m)$ .

Let $k$ denote Coulomb's constant ( $k \approx 8.99 \times 10^{9}\; \rm N \cdot m^{2} \cdot C^{-2}$ .)

Electric potential due to the charge at $\rm A$ : $\displaystyle \frac{k\, q}{d({\rm A})}$ .

Electric potential due to the charge at $\rm B$ : $\displaystyle \frac{k\, q}{d({\rm B})}$ .

Electric potential due to the charge at $\rm A$ : $\displaystyle \frac{k\, q}{d({\rm C})}$ .

While forces are vectors, electric potentials are scalars. When more than one electric fields are superposed over one another, the resultant electric potential at some point would be the scalar sum of the electric potential at that position due to each of these fields.

Hence, the electric field at the midpoint of $\rm AB$ due to all these three charges would be:

$\begin{aligned}& \frac{k\, q}{d({\rm A})} + \frac{k\, q}{d({\rm B})} + \frac{k\, q}{d({\rm C})} \\ &= k\, \left(\frac{q}{d({\rm A})} + \frac{q}{d({\rm B})} + \frac{q}{d({\rm C})}\right) \\ &\approx 8.99 \times 10^{9}\; \rm N \cdot m^{2} \cdot C^{-2} \\ & \quad \quad \times \left(\frac{1.8 \times 10^{-8} \; \rm C}{0.050\; \rm m} + \frac{1.8 \times 10^{-8} \; \rm C}{0.050\; \rm m} + \frac{1.8 \times 10^{-8} \; \rm C}{\sqrt{3} \times (0.050\; \rm m)}\right) \\ &\approx 8.3 \times 10^{3}\; \rm V\end{aligned}$ .

4 0

2 years ago

in physics lab, a cube slides down a frictionless incline as shown in the figure below, and elastically strikes another cube at

Tema [17]

<span>In the physics lab, a cube slides down a frictionless incline as shown in the figure below, check the image for the complete solution:

</span>

3 0

2 years ago

What are the 3 Newton's law of motions?

irinina [24]

Answer:

laws of motion relate an object’s motion to the forces acting on it. In the first law, an object will not change its motion unless a force acts on it. In the second law, the force on an object is equal to its mass times its acceleration. In the third law, when two objects interact, they apply forces to each other of equal magnitude and opposite direction.

4 0

2 years ago