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3 years ago

A 5- kg object experiences forces as shown in the diagram. Which statement best describes the motion of the object

Physics

1 answer:

rosijanka [135]3 years ago

8 0

Answer: D. The object would have an acceleration of 6 m/s/s to the right.

Explanation:

I got it right on USA test prep

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3 years ago

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Formula for weight on moon compared to weight on earth

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3 years ago

You may have noticed runaway truck lanes while driving in the mountains. These gravel-filled lanes are designed to stop trucks t

kati45 [8]

Answer:

The coefficient of kinetic friction $\mu_k = 0.724$

Explanation:

From the question we are told that

The length of the lane is $l = 36.0 \ m$

The speed of the truck is $v = 22.6\ m/s$

Generally from the work-energy theorem we have that

$\Delta KE = N * \mu_k * l$

Here N is the normal force acting on the truck which is mathematically represented as

$\Delta KE$ is the change in kinetic energy which is mathematically represented as

$\Delta KE = \frac{1}{2} * m * v^2$

=> $\Delta KE = 0.5 * m * 22.6^2$

=> $\Delta KE = 255.38m$

$255.38m = m * 9.8 * \mu_k * 36.0$

=> $255.38 = 352.8 * \mu_k$

=> $\mu_k = 0.724$

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3 years ago

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

Andre45 [30]

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,

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