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Verdich [7]
3 years ago
11

A steady 45 N horizontal force is applied to a 15 kg object a table. The object slides against a friction force of 30 N. Calcula

te the acceleration of the object in m/s.
Physics
1 answer:
lisov135 [29]3 years ago
6 0

The net force on an object subject to friction is equal to the sum of the applied force and the frictional force.

Mathematically,

F_{N} = ma = F_{applied} - f_{fr}

Here, m is mass of object and a is its acceleration. We take frictional force negative because it opposes the motion of object.

Given, m=15\ kg , F_{applied} =45\ N and f_{fr} = 30\ N

Substituting these values in above formula, we get

15\ kg\times a = 45\ N -30\ N=15\ N \\\\a=\frac{15\ N}{15\ kg} =1\ m/s^2.

Thus, the acceleration of an object is 1\ m/s^2.


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For adults, the RDA of the amino acid lysine is 12 mg per kg of body weight. How many grams per day should a 78kg adult receive?
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RDA stands for Recommended Daily Allowance. To determine the amount needed of a certain adult per day, we simply multiply the mass of the adult to the value of RDA. For this case, we do as follows:

Daily requirement = 12 mg/kg ( 78 kg ) = 936 mg of lysine = 936000 g
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a machine with a mechanical advantage of 2.5 requires an input force of 120 newtons. What output force is applied by this machin
creativ13 [48]
If your machine has a mechanical advantage of 2.5, then WHATEVER force you apply to the input, the force at the output will be 2.5 times as great.

If you apply 1 newton to the machine's input, the output force is

                 (2.5 x 1 newton)  =  2.5 newtons.

If you apply 120 newtons to the machine's input, the output force is

                 (2.5 x 120 newtons)  =  300 newtons.

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3 years ago
Which of these is an uneven rhythm movement? Question 5 options:
Musya8 [376]

Answer:

Sliding would be an uneven rhythm because Galloping and skipping has a constant flow of the same movement and same noise.

Explanation:

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An insulated beaker with negligible mass contains 0.250 kg of water at 75.0C. How many kilograms of ice at -20.0C must be droppe
kkurt [141]

Answer:

The amount of kilograms of ice at -20.0°C that must be dropped into the water to make the final temperature of the system 40.0°C = 0.0674 kg

Explanation:

Heat gained by ice in taking the total temperature to 40°C = Heat lost by the water

Total Heat gained by ice = Heat used by ice to move from -20°C to 0°C + Heat used to melt at 0°C + Heat used to reach 40°C from 0°C

To do this, we require the specific heat capacity of ice, latent heat of ice and the specific heat capacity of water. All will be obtained from literature.

Specific heat capacity of ice = Cᵢ = 2108 J/kg.°C

Latent heat of ice = L = 334000 J/kg

Specific heat capacity of water = C = 4186 J/kg.°C

Heat gained by ice in taking the total temperature to 40°C = mCᵢ ΔT + mL + mC ΔT = m(2108)(0 - (-20)) + m(334000) + m(4186)(40 - 0) = 42160m + 334000m + 167440m = 543600 m

Heat lost by water = mC ΔT = 0.25 (4186)(75 - 40) = 36627.5 J

543600 m = 36627.5

m = 0.0674 kg = 67.4 g of ice.

3 0
3 years ago
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