2 years ago

Charged objects have a _________ charge.A) net, B) moving, C) positive, D) negative

1 answer:

5 0

C. positive, this is right because when something it is charged kinda how when you charge your phone it has energy now so that means it is positive but when its dead it is negative

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What is the net force

nadya68 [22]

Answer:

20 N in West direction.

Explanation:

opposite forces cancel each other. so 20 N in north and 20N in south cancel each other. In west and east direction...

70N in west-50N in east= 20N in west

3 0

2 years ago

What is the relationship of an object's acceleration, its mass and the force applied to it?

Anon25 [30]

Answer:

F = ma

Explanation:

Newtons second law

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2 years ago

1. Identify this mystery element: I am very reactive, I desperately want to get rid of my electron. When I make a compound with

galina1969 [7]

1.) NaCl (Sodium Chlorine)
2.) Ne (Neon) lights.

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2 years ago

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A 0.111 kg hockey puck moving at 55 m/s is caught by a 80. kg goalie at rest. with what speed does the goalie slide on the frict

madreJ [45]

Answer:

0.076 m/s

Explanation:

Momentum is conserved:

m v = (m + M) V

(0.111 kg) (55 m/s) = (0.111 kg + 80. kg) V

V = 0.076 m/s

After catching the puck, the goalie slides at 0.076 m/s.

8 0

2 years ago

50 grams of ice cubes at -15°C are used to chill a water at 30°C with mass mH20 = 200 g. Assume that the water is kept in a foam

Arada [10]

Answer : The final temperature is, $25.0^oC$

Explanation :

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

$q_1=-q_2$

$m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)$

where,

$c_1$ = specific heat of ice = $2.09J/g^oC$

$c_2$ = specific heat of water = $4.18J/g^oC$

$m_1$ = mass of ice = 50 g

$m_2$ = mass of water = 200 g

$T_f$ = final temperature = ?

$T_1$ = initial temperature of ice = $-15^oC$

$T_2$ = initial temperature of water = $30^oC$

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

$50g\times 2.09J/g^oC\times (T_f-(-15))^oC=-200g\times 4.184J/g^oC\times (T_f-30)^oC$

$T_f=25.0^oC$

Therefore, the final temperature is, $25.0^oC$

5 0

2 years ago