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3 years ago

5

A big box of sausages (30 kg) is lifted from the ground to the top shelf of the freezer. If the box is lifted at a constant spee

d, a distance of 1.75 m, what work is done against gravity?

1 answer:

Lubov Fominskaja [6]3 years ago

5 0

Answer:

Work done to lift the box is 515.03 J

Explanation:

By work energy theorem we know that work done by all forces is equal to change in kinetic energy

So we have

$W_g + W_{ex} = \Delta K$

so we have

$-mgh + W_{ex} = 0$

so we have

$W_{ex} = mgh$

$W_{ex} = 30(9.8)1.75$

$W_{ex} = 515.03 J$

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The figure above shows the net force exerted on an object as a function of the position of the object. The object starts from re

weqwewe [10]

Answer:

0.06 Kg

Explanation:

From the question given above, the following data were obtained:

Initial velocity (u) = 0 m/s

Final velocity (v) = 3.0 m/s

Distance (s) = 0.09 m

Net Force (F) = 3 N

Mass (m) =?

Next, we shall determine the acceleration of the object. This can be obtained as follow:

Initial velocity (u) = 0 m/s

Final velocity (v) = 3.0 m/s

Distance (s) = 0.09 m

Acceleration (a) =?

v² = u² + 2as

3² = 0² + (2 × a × 0.09)

9 = 0 + 0.18a

9 = 0.18a

Divide both side by 0.18

a = 9 / 0.18

a = 50 m/s²

Finally, we shall determine the mass of the object. This can be obtained as follow:

Net Force (F) = 3 N

Acceleration (a) = 50 N

Mass (m) =?

F = ma

3 = m × 50

Divide both side by 50

m = 3 / 50

m = 0.06 Kg

Therefore, the mass of the object is 0.06 Kg

5 0

3 years ago

The interior space of large box is kept at 30 C. The walls of the box are 3 m high and have a ‘sandwich’ construction consisting

White raven [17]

Answer:

$\frac{\dot Q}{A} =20.129\ W.m^{-2}$

$T_1=27.58\ ^{\circ}C$ & $T_2=2.41875\ ^{\circ}C$

Explanation:

Given:

interior temperature of box, $T_i=30^{\circ}C$

height of the walls of box, $h=3\ m$

thickness of each layer of bi-layered plywood, $x_p=1.25\ cm=0.0125\ m$

thermal conductivity of plywood, $k_p=0.104\ W.m^{-1}.K^{-1}$

thickness of sandwiched Styrofoam, $x_s=5\ cm=0.05\ m$

thermal conductivity of Styrofoam, $k_s=0.04\ W.m^{-1}.K^{-1}$

exterior temperature, $T_o=0^{\circ}C$

<u>From the Fourier's law of conduction:</u>

$\dot Q=\frac{dT}{(\frac{x}{kA}) }$

$\dot Q=\frac{dT}{R_{th} }$ ....................................(1)

<u>Now calculating the equivalent thermal resistance for conductivity using electrical analogy:</u>

$R_{th}=R_p+R_s+R_p$

$R_{th}=\frac{x_p}{k_p.A}+\frac{x_s}{k_s.A}+\frac{x_p}{k_p.A}$

$R_{th}=\frac{1}{A} (\frac{x_p}{k_p}+\frac{x_s}{k_s}+\frac{x_p}{k_p})$

$R_{th}=\frac{1}{A} (\frac{0.0125}{0.104}+\frac{0.05}{0.04}+\frac{0.0125}{0.104})$

$R_{th}=\frac{1.4904}{A}$ .....................(2)

Putting the value from (2) into (1):

$\dot Q=\frac{30-0}{\frac{1.4904}{A} }$

$\dot Q=\frac{30\ A}{1.4904}$

$\frac{\dot Q}{A} =20.129\ W.m^{-2}$ is the heat per unit area of the wall.

The heat flux remains constant because the area is constant.

<u>For plywood-Styrofoam interface from inside:</u>

$\frac{\dot Q}{A} =k_p.\frac{T_i-T_1}{x_p}$

$20.129=0.104\times \frac{30-T_1}{0.0125}$

$T_1=27.58\ ^{\circ}C$

&<u>For Styrofoam-plywood interface from inside:</u>

$\frac{\dot Q}{A} =k_s.\frac{T_1-T_2}{x_s}$

$20.129=0.04\times \frac{27.58-T_2}{0.05}$

$T_2=2.41875\ ^{\circ}C$

4 0

3 years ago

Two 10 kg pucks head straight towards each other with velocities of 10 m/s and -20 m/s. They collide and stick together. Calcula

RUDIKE [14]

The final velocity of the two pucks is -5 m/s

Explanation:

We can solve the problem by using the law of conservation of momentum.

In fact, in absence of external force, the total momentum of the two pucks before and after the collision must be conserved - so we can write:

$m_1 u_1 + m_2 u_2 = (m_1 +m_2)v$

where

$m_1 = m_2 = m = 10 kg$ is the mass of each puck

$u_1 = 10 m/s$ is the initial velocity of the 1st puck

$u_2 = -20 m/s$ is the initial velocity of the 2nd puck

v is the final velocity of the two pucks sticking together

Re-arranging the equation and solving for v, we find:

$mu_1 + mu_2 = (m+m)v\\u_1 + u_2 = 2v\\v=\frac{u_1+u_2}{2}=\frac{10-20}{2}=-5 m/s$

Learn more about momentum:

brainly.com/question/7973509

brainly.com/question/6573742

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brainly.com/question/9484203

#LearnwithBrainly

8 0

3 years ago

A single loop of current is immersed in an externally applied uniform magnetic field of 3 Tesla oriented in the positive y direc

vlabodo [156]

Answer:

mu=12Tm^2

Explanation:

the magnetic moment mu of a single loop is given by:

$\mu = I A B$

where I is the current, B is the magnetic field and A is the area of the loop. By replacing we obtain:

$\mu=(0.5A)(4m*2m)(3T)=12Tm^2$

hope this helps!!

6 0

3 years ago

33 POINTS How can I get the temperature? SOUND SPEED 340 m/s = 331 m/s + (0,6xTemperature)

inessss [21]

(340-331)/0.6 = temp

9/0.6

90/6

30/2

15 degrees

6 0

3 years ago