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2 years ago

6

Can someone help me w this ?

1 answer:

lukranit [14]2 years ago

7 0

Answer:

i can't sorry

Explanation:

I didn't really pay attention in that class

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Verizon [17]

Q 24: Newton's first law of motion states that anything that stands still will stay still unless unbalanced force is acted upon that object. Newton's second law of motion states that the velocity of an object changes when subjected to external Force. Newton's third law of motion says that for every action there is an equal and opposite reaction. when bumper cars crash into each other one car goes one way and the other goes the other way.

Q 25: if the bumper cars doubled their Mass the motion would be halfed but if the net force is doubled the motion will double.

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3 years ago

What happens when an electron absorbs energy?

Klio2033 [76]

Answer:

C

Explanation:

Its because electron does its thing with energy to

its orbital

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3 years ago

A 6 N and a 10 N force act on an object. The moment arm of the 6 N force is 0.2 m. If the 10 N force produces five times the tor

Levart [38]

Answer:

The moment arm is 0.6 m

Explanation:

Given that,

First force $F_{1}=6\ N$

Second force $F_{2}=10\ N$

Distance r = 0.2 m

We need to calculate the moment arm

Using formula of torque

$\tau=Force\times lever\ arm$

So, Here,

$\tau_{2}=5 \tau_{1}$

We know that,

The torque is the product of the force and distance.

Put the value of torque in the equation

$F_{2}\times d_{2}=5\times F_{1}\times r_{1}$

$r_{2}=\dfrac{5\times F_{1}\times r_{1}}{F_{2}}$

Where, $F_{1}$ =First force

$F_{1}$ =First force

$F_{2}$ =Second force

$r_{1}$ = distance

Put the value into the formula

$r_{2}=\dfrac{5\times6\times0.2}{10}$

$r_{2}=0.6\ m$

Hence, The moment arm is 0.6 m

6 0

3 years ago

A 4.0-kg object is supported by an aluminum wire of length 2.0 m and diameter 2.0 mm. How much will the wire stretch?

forsale [732]

Answer:

The extension of the wire is 0.362 mm.

Explanation:

Given;

mass of the object, m = 4.0 kg

length of the aluminum wire, L = 2.0 m

diameter of the wire, d = 2.0 mm

radius of the wire, r = d/2 = 1.0 mm = 0.001 m

The area of the wire is given by;

A = πr²

A = π(0.001)² = 3.142 x 10⁻⁶ m²

The downward force of the object on the wire is given by;

F = mg

F = 4 x 9.8 = 39.2 N

The Young's modulus of aluminum is given by;

$Y = \frac{stress}{strain}\\\\Y = \frac{F/A}{e/L}\\\\Y = \frac{FL}{Ae} \\\\e = \frac{FL}{AY}$

Where;

Young's modulus of elasticity of aluminum = 69 x 10⁹ N/m²

$e = \frac{FL}{AY} \\\\e = \frac{(39.2)(2)}{(3.142*10^{-6})(69*10^9)} \\\\e = 0.000362 \ m\\\\e = 0.362 \ mm$

Therefore, the extension of the wire is 0.362 mm.

8 0

3 years ago

Why are overuse injuries particularly frustrating set-backs?

olganol [36]

Because they are caused by your exercise

7 0

3 years ago

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