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3 years ago

Match the units in the right hand column with the terms in the right hand column

Physics

2 answers:

Sergio [31]3 years ago

6 0

Answer:

Explanation:

Hope this helps

Check if right by the way

marissa [1.9K]3 years ago

6 0

Answer:

Explanation:

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A 7 kg suitcase and a 12 kg suitcase are dropped from the same second story window at the same time. If there is no wind resista

Natali5045456 [20]

Explanation:

the correct answer is C.

if there is no wind resistance; in vaccum both will hit the ground at the same time.

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3 years ago

What is the Sl unit for momentum? O kg• m O kg O kg• m O kg 2 m

Andrej [43]

Explanation:

<h2>The S. I. unit of momentum is Kg. m/sec</h2>

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8 0

3 years ago

on the surface of theearth ,the weight of a boy is 400N but on a mountainpeak his weight is 360N,Calculatethe value of ''g' on t

andreyandreev [35.5K]

The value of g at sea level is 9.81 ms^-2.

The boy's mass is constant wherever he is in the universe but his weight will depend on the strength gravity where he is.

By proportion its value on the mountain peak is (360 /400) * 9.81

= 0.9 * 9.81 = 8.83 ms^-2 to nearest hundredth, (answer).

7 0

3 years ago

Two particles, with charges of 20.0 nC and -20.0 nC, are fixed at points with coordinates <0, 4.00 cm> and <0, -4.00 cm

Dmitry [639]

Answer:

Explanation:

Potential energy of the system of charges

= 9 x 10⁹ x [ q₁q₂ / r₁₂ + q₂q₃ / r₂₃ + q₁q₃ / r₁₃ ]

here r₁₂ , r₂₃ , r₁₃ are distance between 1 st and 2 nd charge , 2 nd and 3 rd charge and fist and third charge.

r₁₂ = 8 cm , r₂₃ = 4 cm , r₁₃ = 4 cm.

q₁ = 20 x 10⁻⁹ C , q₂ = - 20 x 10⁻⁹ C , q₃ = 10 x 10⁻⁹ C

Potential energy = 9 x 10⁹ x [ - 400 x 10⁻¹⁸ / .08 + -200x10⁻¹⁸ / .04 + 200 x 10¹⁸ / .04 ]

= 9 x 10⁹ x - 400 x 10⁻¹⁸ / .08

= 45 x 10⁻⁶ J .

Potential at the point of fourth charge due to three charges of 20 nC , - 20 nC and 10 nC at the centre

9 x 10⁹ [ 20 x 10⁻⁹ / .05 + - 20 x 10⁻⁹ / .05 + 10 x 10⁻⁹ / .03 ]

= 9 x 10⁹ x 10 x 10⁻⁹ / .03

= 3000 V .

potential energy of fourth particle = charge x potential

= 3000 x 40 x 10⁻⁹ = 12 x 10⁻⁵ J .

kinetic energy at infinity = 12 x 10⁻⁵ J

1/2 m v² = 12 x 10⁻⁵ J

.5 x 2 x 10⁻¹³ x v² = 12 x 10⁻⁵

v² = 12 x 10⁸

v = 3.46 x 10⁴ m/s

= 9 x 10⁹

5 0

3 years ago

When a 25-kg crate is pushed across a frictionless horizontal floor with a force of 200 N, directed 20 below the horizontal, th

Fofino [41]

Answer:

Option E is correct 310N

Explanation:

Given that the force used to push the crate is F = 200N

The force directed 20° below the horizontal

Mass of crate is m = 25kg

Weight of the crate can be determine using

W = mg

g is gravitational constant =9.8m/s²

W = 25×9.8

W = 245 N

Check attachment. For free body diagram and better understanding

Using newton second law along the vertical axis since we want to find the normal force

ΣFy = m•ay

ay = 0, since the body is not moving in the vertical or y direction

N—W—F•Sin20 = 0

N = W+F•Sin20

N = 245+ 200Sin20

N = 245 + 68.4

N = 313.4 N

The normal force is approximately 310 N to the nearest ten

3 0

3 years ago

Match the units in the right hand column with the terms in the right hand column​

Match the units in the right hand column with the terms in the right hand column