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WINSTONCH [101]

3 years ago

15

A light string is wrapped around the edge of the smaller disk, and a 1.50 kgkg block is suspended from the free end of the strin

g. If the block is released from rest at a distance of 2.40 mm above the floor, what is its speed just before it strikes the floor

1 answer:

Mandarinka [93]3 years ago

6 0

Answer:

The speed of the block before it strikes the floor is 0.217 m/s.

Explanation:

Given;

mass of the block, m = 1.5 kg

height above the ground through the block was released, h = 2.4 mm = 2.4 x 10⁻³ m

The speed of the block before it strikes the floor will be maximum.

Let the speed of the block before it strikes the floor = v

Apply the principle of conservation of mechanical energy to determine the speed of the block.

K.E = P.E

¹/₂mv² = mgh

¹/₂v² = gh

v² = 2gh

v = √2gh

v = √(2 x 9.8 x 2.4 x 10⁻³)

v = 0.217 m/s

Therefore, the speed of the block before it strikes the floor is 0.217 m/s.

You might be interested in

A crane lifts a 100 kg concrete block to a vertical height of 25 m. Calculate

Nana76 [90]

Answer:

Solon,

total mass (kg)= 100kg

height (h)= 25m

acceleration due to gravity = 9.8m/s²

so,

work done =m*g*h

= 100*9.8*25

= 24,500 joule

5 0

3 years ago

You pull straight up on the string of a yo-yo with a force 0.35 N, and while your hand is moving up a distance 0.16 m, the yo-yo

jarptica [38.1K]

Answer:

a) 0.138J

b) 3.58m/S

c) (1.52J)(I)

Explanation:

a) to find the increase in the translational kinetic energy you can use the relation

$\Delta E_k=W=W_g-W_p$

where Wp is the work done by the person and Wg is the work done by the gravitational force

By replacing Wp=Fh1 and Wg=mgh2, being h1 the distance of the motion of the hand and h2 the distance of the yo-yo, m is the mass of the yo-yo, then you obtain:

$Wp=(0.35N)(0.16m)=0.056J\\\\Wg=(0.062kg)(9.8\frac{m}{s^2})(0.32m)=0.19J\\\\\Delta E_k=W=0.19J-0.056J=0.138J$

the change in the translational kinetic energy is 0.138J

b) the new speed of the yo-yo is obtained by using the previous result and the formula for the kinetic energy of an object:

$\Delta E_k=\frac{1}{2}mv_f^2-\frac{1}{2}mv_o^2$

where vf is the final speed, vo is the initial speed. By doing vf the subject of the formula and replacing you get:

$v_f=\sqrt{\frac{2}{m}}\sqrt{\Delta E_k+(1/2)mv_o^2}\\\\v_f=\sqrt{\frac{2}{0.062kg}}\sqrt{0.138J+1/2(0.062kg)(2.9m/s)^2}=3.58\frac{m}{s}$

the new speed is 3.58m/s

c) in this case what you can compute is the quotient between the initial rotational energy and the final rotational energy

$\frac{E_{fr}}{E_{fr}}=\frac{1/2I\omega_f^2}{1/2I\omega_o^2}=\frac{\omega_f^2}{\omega_o^2}\\\\\omega_f=\frac{v_f}{r}\\\\\omega_o=\frac{v_o}{r}\\\\\frac{E_{fr}}{E_{fr}}=\frac{v_f^2}{v_o^2}=\frac{(3.58m/s)}{(2.9m/s)^2}=1.52J$

hence, the change in Er is about 1.52J times the initial rotational energy

5 0

3 years ago

Read 2 more answers

A packing crate with mass 80.0 kg is at rest on a horizontal, frictionless surface. At t = 0 a net horizontal force in the +x-di

Nataly [62]

Answer:

Final speed of the crate is 15 m/s

Explanation:

As we know that constant force F = 80 N is applied on the object for t = 12 s

Now we can use definition of force to find the speed after t = 12 s

$F . t = m(v_f - v_i)$

so here we know that object is at rest initially so we have

$80 (12) = 80( v_f - 0)$

$v_f = 12 m/s$

Now for next 6 s the force decreases to ZERO linearly

so we can write the force equation as

$F = 80 - \frac{40}{3} t$

now again by same equation we have

$\int F .dt = m(v_f - v_i)$

$\int (80 - (40/3)t) dt = 80(v_f - 12)$

$80 t - \frac{40t^2}{6} = 80(v_f - 12)$

put t = 6 s

$480 - 240 = 80(v_f - 12)$

$v_f = 12 + 3$

$v_f = 15 m/s$

6 0

3 years ago

A cubical box, 5.00 cm on each side, is immersed in a fluid. The gauge pressure at the top surface of the box is 594 Pa and the

Zolol [24]

Answer:

The density of the fluid is 1100 kg/m³.

Explanation:

Given that,

Height = 5.00 cm

Pressure at top =594 Pa

Pressure at bottom = 1133 Pa

We need to calculate the change in pressure

Using formula of change in pressure

$\Delta P=P_{b}-P_{t}$

Where, $P_{b}$ = Pressure at bottom

$P_{t}$ = Pressure at top

put the value into the formula

$\Delta P=1133-594$

$\Delta P=539\ Pa$

Using formula of pressure for density

$\Delta P = \rho g h$

$\rho =\dfrac{\Delta P}{gh}$

Where, $\rho$ = density

P = pressure

h = height

Put the value in to the formula

$\rho =\dfrac{539}{5.00\times10^{-2}\times9.8}$

$\rho =1100\ kg/m^3$

Hence, The density of the fluid is 1100 kg/m³.

4 0

4 years ago

A traveling electromagnetic wave in a vacuum has an electric field amplitude of 93.3 V/m. Calculate the intensity S of this wave

7nadin3 [17]

Answer:

Intensity = 11.56W/m²

The energy flowing through the given area is 4.55 J

Explanation:

The expression for the intensity of the electromagnetic wave is,

$I = \frac{1}{2} C{ {\varepsilon _0}E_m^2$

Here, $\varepsilon _0$ is the permittivity of the free space,

$E_m$ is the electric field amplitude and

c is the speed of the light.

substitute

⁸m/s for c

8.85×10 −12 C² /N⋅m² for ${\varepsilon _0}$

and 93.3 V/m for ${E_{\rm{m}$

$I = \frac{1}{2} \times (3\times10^8)\times(8.85\times10^-^1^2)(93.3)\\\\I = 11.56W/m^2$

The expression for the energy is,

E = I×A×t

Here, I is the intensity of the electromagnetic wave,

A is the area, and

t is the time.

Substitute

11.56W/m² for I

0.0287m ² for A

13.7s for t

$E = (11.56)\times(0,0287)\times(13.7)\\E = 4.55J$

The energy flowing through the given area is 4.55 J

5 0

4 years ago