An 880n box is pushed across a level floor for a distance of 5.0m with a force of 440n. how much work was done on the box

Physics

2 answers:

Simora [160]4 years ago

8 0

W=Fd, F=440N, d=5m. 5x440=2,200J of work

Lunna [17]4 years ago

5 0

Answer: W = 2,200 joules

Explanation: Work is defined as a motion caused by a force, and can be calculated as:

W = F*d

where F is the force applied and d is the distance moved, here F = 440n and d = 5.0m

(the mass of the box does not matter in this case, where there are no friction and the movement is only in the perpendicular axis with respect at the direction in wich the weight pushes)

So the total work is:

w = 5.0m*440N = 2,200j

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How many milliliters of water at 25.0°C with a density of 0.997 g/mL must be mixed with 162 mL of coffee at 94.6°C so that the r

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Answer:

The volume of water is 139 mL.

Explanation:

Due to the Law of conservation of energy, the heat lost by coffee is equal to the heat gained by the water, that is, the sum of heats is equal to zero.

$Q_{coffee} + Q_{water} = 0\\Q_{coffee} = - Q_{water}$

The heat (Q) can be calculated using the following expression:

$Q=c \times m \times \Delta T$

where,

c is the specific heat of each substance

m is the mass of each substance

ΔT is the difference in temperature for each substance

The mass of coffee is:

$162mL.\frac{0.997g}{mL} = 162g$

Then,

$Q_{coffee} = - Q_{water}\\c_{c} \times m_{c} \times \Delta T_{c} = -c_{w} \times m_{w} \times \Delta T_{w}\\m_{c} \times \Delta T_{c} = -m_{w} \times \Delta T_{w}\\m_{w} = \frac{m_{c} \times \Delta T_{c}}{-\Delta T_{w}} \\m_{w}=\frac{162g \times (62.5 \°C - 94.6 \°C ) }{-(62.5 \°C - 25.0 \°C)} \\m_{w} = 139 g$