A ladder is balanced against a wall without moving. What<br> must be true about this ladder?

A 1.2kg stone is tied to a string and swung in a vertical circle with a radius of 0.75m. The string can withstand a tension of 4

san4es73 [151]

Hello!

We can begin by summing the forces acting on the stone when it is at the bottom of its trajectory.

Refer to the free-body diagram in the image below for clarification.

We have the force of tension (produced by the string) and the force of gravity acting in opposite directions, so:
$\Sigma F = T - F_g$

The net force is equivalent to the centripetal force experienced by the stone. Recall the equation for centripetal force for uniform circular motion:
$F_c = \frac{mv^2}{r}$

m = mass of object (1.2 kg)
v = velocity of object (? m/s)
r = radius of circle (0.75 m)

The centripetal force is the resultant of the forces of tension and gravity, and points upward (same direction as the tension force) since the tension force is greater.

Therefore:
$\frac{mv^2}{r} = T - Mg$

We can solve the equation for 'v':

$mv^2 = r(T - Mg) \\\\v^2 = \frac{r(T - Mg)}{m}\\\\v = \sqrt{\frac{r(T - Mg)}{m}}$

Plug in values and solve.

$v = \sqrt{\frac{(0.75)(40 - 1.2(9.8))}{1.2}} = \boxed{4.201 \frac{m}{s}}$

3 0

2 years ago

Your friend wants your advice about his new transformation fitness plan. He has found a plan that sounds promising to him, but h

cestrela7 [59]

What is the question? I think that you answered it yourself...

3 0

3 years ago

a car moving on a road passes by kilometer 218 at 10:15 am and milestone 236 at 10:30 am. determine an average scalar speed in k

Leno4ka [110]

Answer:

v?

Explanation:

v = ∆s / ∆t

v =?

∆s = 236 - 218 = 18km

=t = 10: 30 - 10:15 = 15 minutes to hour 15/60 = 0.25h

v = 18 / 0.25

v = 72km / h

speed of 72km / h

5 0

3 years ago

A cube (V = 12 cm³) is lowered in water at 0.2 m.s-¹, using a string. Density of the cube is 4170 kg.m-³. What is the tension in

Bingel [31]

hi hai na usne bheji to make sure to Manas he has to

6 0

2 years ago

What is the electric field strength just outside the surface of a conducting sphere carrying surface charge density 1.4 μC/m2μC/

Nataly [62]

Answer:

$E=158.19\frac{kN}{m}$

Explanation:

Gauss's theorem states that the flux of the electric field through a closed surface is equal to the the charge enclosed by the surface divided by the vacuum permittivity:

$\int\limits{\vec{E}\cdot \vec{dS}} \,=\frac{q}{\epsilon_0}$

The direction of the electric field ( $\vec{E}$ ) just outside of a conductor is parallel to its surface ( $\vec{S}$ ):

$\int\limits{\vec{E}\cdot \vec{dS}} \,=\int\limits{EdScos(0^\circ)} \,=E\int\limits{dS} \,=ES$

Recall that the surface charge density is defined as:

$\sigma=\frac{q}{S}$

Now, we get the electric field strength:

$ES=\frac{q}{\epsilon_0}\\E=\frac{q}{S\epsilon_0}\\E=\frac{\sigma}{\epsilon_0}\\E=\frac{1.4*10^{-6}\frac{C}{m^2}}{8.85*10^{-12\frac{C^2}{N\cdot m^2}}}\\\\E=158192.09\frac{N}{C}=158.19\frac{kN}{C}$

3 0

4 years ago

A ladder is balanced against a wall without moving. What must be true about this ladder?