The earth pulls the moon towards it because of the earth has more mass and therefore moe

A 0.700-kg ball is on the end of a rope that is 2.30 m in length. The ball and rope are attached to a pole and the entire appara

otez555 [7]

Answer:

The tangential speed of the ball is 11.213 m/s

Explanation:

The radius is equal:

$r=2.3*sin70=2.161m$ (ball rotates in a circle)

If the system is in equilibrium, the tension is:

$Tcos70=mg\\Tsin70=\frac{mv^{2} }{r}$

Replacing:

$\frac{mg}{cos70} sin70=\frac{mv^{2} }{r} \\Clearing-v:\\v=\sqrt{rgtan70}$

Replacing:

$v=\sqrt{2.161x^{2}*9.8*tan70 } =11.213m/s$

7 0

3 years ago

A small block with a mass of 0.0600 kg is attached to a cord passing through a hole in a frictionless, horizontal surface (Figur

pogonyaev

Answer with Explanation:

We are given that mass of block=0.0600 kg

Initial speed of block=0.63 m/s

Distance of block from the hole when the block is revolved=0.47 m

Final speed=3.29 m/s

Distance of block from the hole when the block is revolved= $9\times 10^{-2}m$

a.We have to find the tension in the cord in the original situation when the block has speed = $v_0=0.63 m/s$

$T=\frac{mv^2}{r}$

Because tension is equal to centripetal force

Substitute the values

$T=\frac{0.06\times (0.63)^2}{0.47}=0.05 N$

b. $v=3.29 m/s$

$T=\frac{mv^2}{r}=\frac{0.06\times (3.29)^2}{0.09}=7.2 N$

c.Work don=Final K.E-Initial K.E

$W=\frac{1}{2}m(v^2-v^2_0)$

$W=\frac{1}{2}(0.06)((3.29)^2-(0.63)^2)$

$W=0.31 J$

4 0

3 years ago

Read 2 more answers

A cylinder has a height of 5.3 cm and a diameter of 3.0 cm. what is its volume?

anzhelika [568]

The cylinder has a volume of 37.46 cubic cm

3 0

3 years ago

What is the gravitational potential energy of a 0.550-kg projectile flying with 335 m/s, 72 meters above the ground?

Fofino [41]

Answer:

GPE = 388.08 Joules.

Explanation:

Given the following data;

Mass = 0.550kg

Speed = 335 m/s

Height = 72 meters

We know that acceleration due to gravity, g is equal to 9.8 m/s²

To find the gravitational potential energy;

Gravitational potential energy (GPE) is an energy possessed by an object or body due to its position above the earth.

Mathematically, gravitational potential energy is given by the formula;

$G.P.E = mgh$

Where;

G.P.E represents potential energy measured in Joules.

m represents the mass of an object.

g represents acceleration due to gravity measured in meters per seconds square.

h represents the height measured in meters.

Substituting into the formula, we have;

$G.P.E = 0.550 * 9.8 * 72$

GPE = 388.08 Joules.

4 0

3 years ago

A negative charge of 20 x 10-6C and another charge of 15 x 10-6C are separated by as distance of 0.7 m.

denpristay [2]

Answer:

Approximately $5.5\; \rm N$ , assuming that the volume of these two charged objects is negligible.

Explanation:

Assume that the dimensions of these two charged objects is much smaller than the distance between them. Hence, Coulomb's Law would give a good estimate of the electrostatic force between these two objects regardless of their exact shapes.

Let $q_1$ and $q_2$ denote the magnitude of two point charges (where the volume of both charged object is negligible.) In this question, $q_1 = 20 \times 10^{-6}\; \rm C$ and $q_2 = 15 \times 10^{-6}\; \rm C$ .

Let $r$ denote the distance between these two point charges. In this question, $r = 0.7\; \rm m$ .

Let $k$ denote the Coulomb constant. In standard units, $k \approx 8.98755\times 10^{9}\; \rm kg \cdot m^{3}\cdot s^{-2}\cdot C^{-2}$ .

By Coulomb's Law, the magnitude of electrostatic force (electric force) between these two point charges would be:

$\begin{aligned}F &= \frac{k \cdot q_1 \cdot q_2}{r^{2}}\end{aligned}$ .

Substitute in the values and evaluate:

$\begin{aligned}F &= \frac{k \cdot q_1 \cdot q_2}{r^{2}}\\ &\approx 8.98755 \times 10^{9}\; \rm kg \cdot m^{3}\cdot s^{-2}\cdot C^{-2} \\ &\quad \times 20\times 10^{-6}\; \rm C\\ &\quad \times 15\times 10^{-6}\; \rm C \\ &\quad \times \frac{1}{{(0.7\; \rm m)}^{2}}\\ &\approx 5.5\; \rm N \end{aligned}$ .

8 0

3 years ago