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bearhunter [10]

2 years ago

15

A construction worker pushes a crate horizontally on a frictionless floor with a net force of 10\, \text 10N, start text, N, end

text for 4.0\,\text m4. 0m4, point, 0, start text, m, end text. How much kinetic energy does the crate gain?

1 answer:

Strike441 [17]2 years ago

7 0

Answer:

$k_f$ = 40J

Explanation:

The work and energy theorem says that:

$W_f =k_f-k_i$

where $W_f$ is the work of the force, $k_f$ the final kinetic energy and $k_i$ the initial kinetic energy.

Addittionally, the work of the force is calculate as force multiply by distance and if the crate inittialy is at rest, the initial kinetic energy is zero, so:

$Fd = k_f$

where F is the force and d the distance. Then, replacing values, we get:

$(10N)(4m) = k_f$

$40J = k_f$

it means that the system gain 40J of kinetic energy.

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A 35 g steel ball is held by a ceiling-mounted electromagnet 4.0 m above the floor. A compressed-air cannon sits on the floor, 4

HACTEHA [7]

Answer:

7.9 m/s

Explanation:

When both balls collide, they have spent the same time for their motions.

Motion of steel ball

This is purely under gravity. It is vertical.

Initial velocity, <em>u </em>= 0 m/s

Distance, <em>s</em> = 4.0 m - 1.2 m = 2.8 m

Acceleration, <em>a</em> = g

Using the equation of motion

$s = ut+\frac{1}{2}at^2$

$2.8 \text{ m} = 0+\dfrac{gt^2}{2}$

$t = \sqrt{\dfrac{5.6}{g}}$

Motion of plastic ball

This has two components: a vertical and a horizontal.

The vertical motion is under gravity.

Considering the vertical motion,

Initial velocity, <em>u </em>= ?

Distance, <em>s</em> = 1.2 m

Acceleration, <em>a</em> = -<em>g </em> (It is going up)

Using the equation of motion

$s = ut+\frac{1}{2}at^2$

$1.2\text{ m} = ut-\frac{1}{2}gt^2$

Substituting the value of <em>t</em> from the previous equation,

$1.2\text{ m} = u\sqrt{\dfrac{5.6}{g}}-\dfrac{1}{2}\times g\times\dfrac{5.6}{g}$

$u\sqrt{\dfrac{5.6}{g}} = 4.0$

Taking <em>g</em> = 9.8 m/s²,

$u = \dfrac{4.0}{0.756} = 5.29 \text{ m/s}$

This is the vertical component of the initial velocity

Considering the horizontal motion which is not accelerated,

horizontal component of the initial velocity is horizontal distance ÷ time.

$u_h = \dfrac{4.4\text{ m}}{0.756\text{ s}} = 5.82\text{ m/s}$

The initial velocity is

$v_i = \sqrt{u^2+u_h^2} = \sqrt{(5.29\text{ m/s})^2+(5.82\text{ m/s})^2} = 7.9 \text{ m/s}$

4 0

2 years ago

Consider an alcohol and a mercury thermometer that read exactly 0 oC at the ice point and 100 oC at the steam point. The distanc

Alexeev081 [22]

Answer:

No, both the thermometers will give the different reading.

Explanation:

Given,

Both thermometer has same ice point = $T_i\ =\ 0^o C$
Both thermometer has same steam point = $T_s\ =\ 100^o C$

Distance between the ice point and steam point in both the thermometer is same of 100 division,

All the data given in both the thermometers are same, but the material in the thermometer is different due to this the reading at 60^o C will differ in both the thermometer. Because the reading on both the thermometer is depended upon the thermal expansion of the material inside it, but both the materials are different. Due to this the rise of fluid in the thermometer, i,e,. the volume of the fluid material in the thermometer will depend upon the thermal expansion. Hence both the material alcohol and mercury have the different thermal expansion, therefore the rise of the fluid in the thermometer also differ in both the thermometer.

7 0

2 years ago

You have a 2m long wire which you will make into a thin coil with N loops to generate a magnetic field of 3mT when the current i

Anni [7]

Answer:

<em>radius of the loop = 7.9 mm</em>

<em>number of turns N ≅ 399 turns</em>

Explanation:

length of wire L= 2 m

field strength B = 3 mT = 0.003 T

current I = 12 A

recall that field strength B = μnI

where n is the turn per unit length

vacuum permeability μ = $4\pi *10^{-7} T-m/A$ = 1.256 x 10^-6 T-m/A

imputing values, we have

0.003 = 1.256 x 10^−6 x n x 12

0.003 = 1.507 x 10^-5 x n

n = 199.07 turns per unit length

for a length of 2 m,

number of loop N = 2 x 199.07 = 398.14 ≅ <em>399 turns</em>

since there are approximately 399 turns formed by the 2 m length of wire, it means that each loop is formed by 2/399 = 0.005 m of the wire.

this length is also equal to the circumference of each loop

the circumference of each loop = $2\pi r$

0.005 = 2 x 3.142 x r

r = 0.005/6.284 = $7.9*10^{-4} m$ = 0.0079 m =<em> 7.9 mm</em>

8 0

2 years ago

A moving roller coaster speeds up with constant acceleration for 2.3\,\text{s}2.3s2, point, 3, start text, s, end text until it

Ad libitum [116K]

Answer:

Δx=(v+v0/2)t

Explanation:

We can figure out which kinematic formula to use by choosing the formula that includes the known variables, plus the target unknown.

In this problem, the target unknown is the initial velocity v_0v

0

v, start subscript, 0, end subscript of the roller coaster.

7 0

2 years ago

1. Si tengo medio kilo de fruta y te doy un cuarto y tú me das tres cuartos de kilo, ¿cuánto tengo? 2. Si en una carrera te qued

Kipish [7]

Answer:

1. Tienes 1 kg de fruta.

2. Queda por recorrer 1/4 km.

3. Ambos pesan lo mismo.

Explanation:

1. Tienes 1/2 kg y cuando te doy 1/4 te queda:

$m = \frac{1}{2} - \frac{1}{4} = \frac{1}{4}$

Ahora cuando te doy 3/4 kg te queda en total:

$m_{T} = \frac{1}{4} + \frac{3}{4} = 1 kg$

Por lo tanto, tienes 1 kg de fruta al final.

2. Si falta por recorrer la mitad de la mitad, tenemos:

$d = \frac{1/2}{2} = \frac{1}{4}$

Entonces, queda por recorrer 1/4 km.

3. El peso (P) del hierro es:

$P = m*g$

$P = (1 + 1/2)kg*9.81 m/s^{2} = 14.72 N$

Y el peso de la paja es:

$P = 3/2 kg*9.81 m/s^{2} = 14.72 N$

Por lo tanto, ambos pesan lo mismo.

Espero que te sea de utilidad!

6 0

2 years ago