Ask question

Ask question

All categories

3 years ago

11

A block slides down an inclined plane from rest. Initially the block is at 4.5m above the ground. Find the speed of the block wh

en it is 1.5m above the ground. 1) 7.7m/s 2) 9.4m/s 3) 5.4m/s 4) 3.2m/s

1 answer:

WINSTONCH [101]3 years ago

7 0

Since, no external force is acting , so the system is in equilibrium .

Initial total energy = Final total energy

$mg(4.5) = mg(1.5) + \dfrac{mv^2}{2}\\\\\dfrac{v^2}{2}=3\times g \\\\v^2=3\times 9.8\times 2\\\\v = \sqrt{58.8}\ m/s\\\\v = 7.67 \ m/s$ ( Here , g = acceleration due to gravity = 9.8 m/s² )

Therefore, option 1) is correct.

Hence, this is the required solution.

You might be interested in

The electric field strength is 1.70 × 104 N/C inside a parallel-plate capacitor with a 0.800 m spacing. An electron is released

Vesna [10]

Answer:

Here, "v" is the velocity of electron and "V" is the potential.

4 0

3 years ago

What type of energy transfer is made possible by a nine-volt battery in a calculator?

Varvara68 [4.7K]

The right answer for the question that is being asked and shown above is that: "d. electrical to electromagnetic." the type of energy transfer is made possible by a nine-volt battery in a calculator is that of d. electrical to electromagnetic<span>
</span>

5 0

3 years ago

A que profundidad esta nadando una persona dentro de una alberca si la presión absoluta sobre ésta es de 156kPa?

lara31 [8.8K]

Answer:

La persona está nadando en la alberca a una profundida de 5.575 metros.

Explanation:

La presión absoluta ( $P_{tot}$ ) experimentada por la persona es la suma de la presión atmosférica ( $P_{atm}$ ) y la presión hidrostática de la columna de agua de la alberca ( $P_{h}$ ), medidas en kilopascales. Es decir,

$P_{tot} = P_{atm}+P_{h}$ (1)

$P_{tot} = P_{atm} + \frac{\rho\cdot g \cdot z}{1000}$ (2)

Donde:

$\rho$ - Densidad del fluido de la alberca, medida en kilogramos por metro cúbico.

$g$ - Aceleración gravitacional, medida en metros por segundo al cuadrado.

$z$ - Profundidad de la persona en la alberca, medida en metros.

Si sabemos que $P_{atm} = 101.325\,kPa$ , $\rho = 1000\,\frac{kg}{m^{3}}$ , $g = 9.807\,\frac{m}{s^{2}}$ y $P_{tot} = 156\,kPa$ , entonces la profundidad de la persona en la alberca es:

$156 = 101.325 +\frac{(1000)\cdot (9.807)\cdot z}{1000}$

$54.675 = 9.807\cdot z$

$z = 5.575\,m$

La persona está nadando en la alberca a una profundida de 5.575 metros.

5 0

3 years ago

I'll give brainliest can someone help me with 2.4.1-2.4.3?with explanation.

KIM [24]

Answer:

220÷4=550 per hour

2.4.3 550×4=2200

hope it's right

8 0

3 years ago

A size-5 soccer ball of diameter 22.6 cm and mass 426 g rolls up a hill without slipping, reaching a maximum height of 5.00 m ab

Drupady [299]

Answer

According the conservation of energy

$\dfrac{1}{2}mv_i^2+\dfrac{1}{2}I\omega_i^2+0 = mgh + 0$

I for ball = $\dfrac{2}{3}mr^2$

$\dfrac{1}{2}mv_i^2+\dfrac{1}{2} \dfrac{2}{3}mr^2\omega_i^2= mgh$

$\omega_i = \dfrac{v_i}{r}$

$v_i^2+\dfrac{2}{3}r^2(\dfrac{v_i}{r})^2 = 2gh$

$v_i^2+\dfrac{2}{3}v_i^2 = 2gh$

$v_i^2+[1+\dfrac{2}{3}]=2gh$

$v_i^2\dfrac{5}{3}=2gh$

$v_i=\sqrt{\dfrac{6gh}{5}}=\sqrt{\dfrac{6\times 9.8\times 5}{5}}$

$v_i = 7.67\ m/s$

a) $\omega_i = \dfrac{v_i}{R}$

$\omega_i = \dfrac{7.67}{0.113}$

$\omega_i =67.87\ rad/s$

b) $K_{rot} = \dfrac{1}{2}\dfrac{2}{3}mR^2\omega_i^2$

$K_{rot} = \dfrac{1}{3}\times 0.426\times 0.112^2\times 67.87^2$

$K_{rot} = 8.205\ J$

4 0

3 years ago