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igor_vitrenko [27]

2 years ago

11

The SI unit for force is the newton. Which of the following is equivalent to a newton? A. kg/m B. kg·m/s2 C. kg·m/s D. kg·m/s

2 answers:

bearhunter [10]2 years ago

8 0

I may be wrong but A

Bad White [126]2 years ago

6 0

<u>Answer:</u> The unit of force, newton is equivalent to $kg.m/s^2$

<u>Explanation:</u>

Force is defined as the push or pull on an object with some mass that causes change in its velocity. It is also defined as the mass multiplied by the acceleration of the object. The S.I unit of force is Newton.

Mathematically,

$F=m\times a$

The S.I unit of mass is kilograms

The S.I unit of acceleration is $m/s^2$ (meters per second square)

So, the unit of force becomes: $kg.m/s^2$

Hence, the unit of force, newton is equivalent to $kg.m/s^2$

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Temka [501]

Wherever a river loses energy

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2 years ago

A 0.0427 kg racquet-ball is moving

Gwar [14]

Answer:

Mass of the box = 0.9433 kg

Explanation:

Mass of racket-ball $(m_1)$ = 0.00427 kg

Velocity of racket-ball before collision $(v_{1i})$ = 22.3 m/s

Velocity of racket-ball after collision with box $(v_{1f})$ = -11.5 m/s

[Since ball is bouncing back, so velocity is taken negative.]

Velocity of the box before collision $v_{2i}$ = 0 m/s

<em>[Since the box is stationary, so velocity is taken zero]</em>

Velocity of box moving forward after collision $v_{2f}$ = 1.53 m/s

To find the mas of the box $m_2$ .

By law of conservation of momentum we have:

Momentum before collision = Momentum after collision

This can be written as:

$p_i=p_f$

$m_1v_{1i}+m_2v_{2i}=m_1v_{1f}+m_2v_{2f}$

We can plugin the given value to find $m_2$

$(0.0427\times 22.3)+(m_2\times 0)=(0.0427\times (-11.5))(m_2\times 1.53)$

$0.9522+0=-0.4911+1.53m_2$

Adding both sides by 0.4911

$0.9522+0.4911=-0.4911+0.4911+1.53m_2$

$1.4433=1.53m_2$

Dividing both sides by 1.53.

$\frac{1.4433}{1.53}=\frac{1.53m_2}{1.53}$

$0.9433=m_2$

∴ $m_2=0.9433$ kg

Mass of the box = 0.9433 kg (Answer)

4 0

2 years ago

schepotkina [342]

Answer:

Explanation:

Length if the bar is 1m=100cm

The tip of the bar serves as fulcrum

A force of 20N (upward) is applied at the tip of the other end. Then, the force is 100cm from the fulcrum

The crate lid is 2cm from the fulcrum, let the force (downward) acting on the crate be F.

Using moment

Sum of the moments of all forces about any point in the plane must be zero.

Let take moment about the fulcrum

100×20-F×2=0

2000-2F=0

2F=2000

Then, F=1000N

The force acting in the crate lid is 1000N

Option D is correct

7 0

2 years ago

a ball is dropped from rest at a height of 89m above the ground. (a)what is it's speed just before it hits the ground? (b) how l

kykrilka [37]

Answer:

(a) 41.75m/s

(b) 4.26s

Explanation:

Let:

Distance, D = 89m

Gravity, $g$ = 9.8 m/ $s^{2}$

Initial Velocity, $u$ = 0m/s

Final Velocity, $v$ = ?

Time Taken, $t$ = ?

With the distance formula, which is

D = $ut$ + $\frac{1}{2} gt^2$

and by substituting what we already know, we have:

89 = $\frac{1}{2}$ ×9.8× $t^{2}$

With the equation above, we can solve for $t$ :

$t=\sqrt{\frac{89(2)}{9.8}} \\t=\sqrt{\frac{178}{9.8} } \\t=\sqrt{18.16} \\t=4.26 seconds$

Now that we have solved $t$ , we can use the following velocity formula to solve for $v$ :

$v = u + at$ , where $a$ is also equals to $g$ , so we have

$v = u + gt$

By substituting $u = 0$ , $g = 9.8$ , and $t = 4.26$ ,

We have:

$v = 0 + 9.8(4.26)\\v = 41.75m/s$

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kumpel [21]

Answer:

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Explanation:

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2 years ago