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Anna35 [415]
3 years ago
15

In our usual coordinate system( +x to the right, +y up (away from the center of the Earth), +z out of the page toward you), what

is the vector gravitational force on a 31 kg object sitting on the ground? (From the momentum principle you can conclude that a force of the same magnitude is exerted upward by the ground on the object if the momentum isn't changing. In Chapter 4 we'll see more about the origin of this force.)
Physics
1 answer:
skad [1K]3 years ago
5 0

Answer:

The vector gravitational force is  W = 303.8\  N( -y)

Explanation:

From the question we are told that

      The mass of the object is  m  = 31 \ kg

      The acceleration due to gravity experienced by the object is g =  9.8 m/s^2

Generally the vector gravitational force on the object is mathematically represented as

                       W = 31 *  9.8

                       W = 303.8\  N (-y)

The  -y indicates that the direction of the force is towards the center of the earth

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Answer:

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Explanation:

Hope this helps :)

pls mark brainliest :P

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Describe the differences among ultraviolet waves, visible light waves, and infrared waves. how are these waves alike?
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Read 2 more answers
A laboratory technician drops a 72.0 g sample of unknown solid material, at a temperature of 80.0°C, into a calorimeter. The cal
Natalija [7]

Answer : The specific heat of unknown sample is, 8748.78J/kg^oC

Explanation :

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

q_1=-[q_2+q_3]

m_1\times c_1\times (T_f-T_1)=-[m_2\times c_2\times (T_f-T_2)+m_3\times c_3\times (T_f-T_2)]

where,

c_1 = specific heat of unknown sample = ?

c_2 = specific heat of water = 4186J/kg^oC

c_3 = specific heat of copper = 390J/kg^oC

m_1 = mass of unknown sample = 72.0 g  = 0.072 kg

m_2 = mass of water = 203 g  = 0.203 kg

m_2 = mass of copper = 187 g  = 0.187 kg

T_f = final temperature of calorimeter = 39.4^oC

T_1 = initial temperature of unknown sample = 80.0^oC

T_2 = initial temperature of water and copper = 11.0^oC

Now put all the given values in the above formula, we get

0.072kg\times c_1\times (39.4-80.0)^oC=-[(0.203kg\times 4186J/kg^oC\times (39.4-11.0)^oC)+(0.187kg\times 390J/kg^oC\times (39.4-11.0)^oC)]

c_1=8748.78J/kg^oC

Therefore, the specific heat of unknown sample is, 8748.78J/kg^oC

7 0
3 years ago
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