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3 years ago

When a metal element bonds with a non mental element they are known as ____ bond

Physics

1 answer:

frozen [14]3 years ago

4 0

Answer:

ionic

Explanation:

ionic bond is when a metal and nonmetal are bonded together

metallic bond is when a two metal are bonded

covalent is nonmetal and nonmetal bond or idk

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According to Ohm’s law, which combination of units is the same as the unit for resistance? volt ÷ ampere ampere × volt volt + am

Maksim231197 [3]

Answer:

volt ÷ ampere

Explanation:

The mathematical form of Ohms law is given by :

V = IR

Where V is voltage

I is current

R is resistance

$R=\dfrac{V}{I}$

The unit of voltage is volt and that of current is ampere

Unit of resistance :

$R=\dfrac{\text{volt}}{\text{ampere}}$

So, volt ÷ ampere is the same as the unit of resistance. Hence, the correct option is (a).

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2 years ago

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3 years ago

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Allisa [31]

Answer:58600

Explanation:

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2 years ago

an object is thrown with an initial horizontal velocity of 10 meters per second and take approximately 9 seconds to reach the gr

pantera1 [17]

The horizontal velocity was constant, so:

$s = vt$

$s = 10\cdot9$

$s = 90$

it traveled 90meters

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3 years ago

A 70.0-kg person throws a 0.0480-kg snowball forward with a ground speed of 33.5 m/s. A second person, with a mass of 55.0 kg, c

saw5 [17]

Answer:

The final velocity of the thrower is $\bf{3.88~m/s}$ and the final velocity of the catcher is $\bf{0.029~m/s}$ .

Explanation:

Given:

The mass of the thrower, $m_{t} = 70~Kg$ .

The mass of the catcher, $m_{c} = 55~Kg$ .

The mass of the ball, $m_{b} = 0.0480~Kg$ .

Initial velocity of the thrower, $v_{it} = 3.90~m/s$

Final velocity of the ball, $v_{fb} = 33.5~m/s$

Initial velocity of the catcher, $v_{ic} = 0~m/s$

Consider that the final velocity of the thrower is $v_{ft}$ . From the conservation of momentum,

$&& m_{t}v_{ft} + m_{b}v_{fb} = (m_{t} + m_{b})v_{it}\\&or,& v_{ft} = \dfrac{(m_{t} + m_{b})v_{it} - m_{b}v_{fb}}{m_{t}}\\&or,& v_{ft} = \dfrac{(70 + 0.0480)(3.90) - (0.0480)(33.5)}{70}\\&or,& v_{ft} = 3.88~m/s$

Consider that the final velocity of the catcher is $v_{fc}$ . From the conservation of momentum,

$&& (m_{c} + m_{b})v_{fc} = m_{b}v_{it}\\&or,& v_{fc} = \dfrac{m_{b}v_{it}}{(m_{c} + m_{b})}\\&or,& v_{fc} = \dfrac{(0.048)(33.5)}{(55.0 + 0.0480)}\\&or,& v_{fc} = 0.029~m/s$

Thus, the final velocity of thrower is $3.88~m/s$ and that for the catcher is $0.029~m/s$ .

8 0

3 years ago