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2 years ago

A spring stretches 0.220 m when a 0.400 kg-mass is hung from it. What is its spring constant? (Mass is not a force )

2 answers:

8 0

We want to know the amount of force that stretches the spring 0.22 m.
That force is the WEIGHT of the mass hung from it.
The weight of the mass is (mass) times (gravity).
To do that calculation, we need to know the value of gravity, but
gravity has different values on every planet. I shall assume that
this whole springy question is taking place on Earth, so that the
value of gravity is 9.8 m/s² .

The weight of the mass is (0.4 kg) x (9.8 m/s²) = 3.92 Newtons.

The spring constant is

(force/length of the stretch)

= (3.92 Newtons) / (0.22 meters)

= (3.92 / 0.22) Newtons/meter

= 17.82 N/m .

romanna [79]2 years ago

8 0

Correct Answer:

17.82 N/m

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Alex goes cruising on his dirt bike. He rides 700m north, 300m east, 400 m north, 600m west, 1200m south, 300m east and finally

Bess [88]

Find Displacement and Distance

displacement ...

north is 700+400+100 =1200m n

south=1200m

1200-1200=0

east is 300+300=600m

west is 600m

600-600=0

back at dtart. displ zero

distance is 700+ 300m + 400 m + 600m + 1200m + 300m + 100m = 3600m

3 0

2 years ago

A record is dropped vertically onto a freely rotating (undriven) turntable. Frictional forces act to bring the record and turnta

BartSMP [9]

Answer:

35%

Explanation:

Assuming no external torques present during the collision between the record and the turntable, total angular momentum must be conserved.

For a rotating body with some angular velocity and moment of inertia, the angular momentum can be expressed as follows;

L = I* ω

So, as initial angular momentum and final angular momentum must be the same, we have:

Li = Lf

⇒ I₁ * ω₁ = I₂ * ω₂ (1)

where I₁ is the rotational inertia of the turntable, and I₂, is the combined rotational inertia of the turntable and the record:

I₂ = I₁ + 0.54 I₁ = 1.54 I₁

We can solve (1) for the new common angular speed, as follows:

ω₂ = ω₁ / 1.54 (2)

The initial rotational kinetic energy is given by definition for the following equation:

Kroti = 1/2 * I₁ * ω₁² (3)

The final rotational kinetic energy takes into account the new rotational inertia and the common final angular speed:

Krotf = 1/2* I₂ * ω₂² = 1/2* 1.54 I₁* (ω₁/1.54)² (4)

Dividing both sides in (3) and (4), we get:

Krotf/Kroti = 1/1.54 = 0.65

This means that the final rotational kinetic energy, has reduced to 0.65 of the initial value, or that has lost 35% of the initial kinetic energy.

8 0

2 years ago

if you apply a Force of F1 to area A1 on one side of a hydraulic jack, and the second side of the jack has an area that is twice

Firlakuza [10]

Answer:

$2F_{1}$

Explanation:

F₁ = Force on one side of the jack

A₁ = Area of cross-section of one side of the jack

F₂ = Force on second side of the jack

A₂ = Area of cross-section of second side of the jack = 2 A₁

Using pascal's law

$\frac{F_{1}}{A_{1}}= \frac{F_{_{2}}}{A_{_{2}}}$

$\frac{F_{1}}{A_{1}}= \frac{F_{_{2}}}{2A_{_{1}}}$

$F_{1}= \frac{F_{2}}{2}\\$

$F_{2}= 2F_{1}$

7 0

2 years ago

For the first 10 seconds a squirrel runs 3 m/s to look for an acorn. The next 5 seconds he eats an acorn that he finds. Afterwar

Gala2k [10]

Distance covered by the squirrel to look for an acorn :

d = ( 3 m/s ) × 10 s = 30 m.

Time taken to eat an Acron is 5 seconds.

Time taken to cover distance of 30 m with 2 m/s speed is :

$T=\dfrac{30}{2}\ s= 15 \ s$

Therefore, total time take to get back to where he started is ( 10+5+15 ) = 30 s.

Hence, this is the required solution.

7 0

2 years ago

A plastic rod is rubbed against a wool shirt, thereby acquiring a charge of −4.9 µc. how many electrons are transferred from the

creativ13 [48]

To find the number of electrons transferred, we should divide the total charge acquired by the rod $Q=-4.9 \mu C=-4.9 \cdot 10^{-6}C$ by the charge of a single electron ( $e=-1.6 \cdot 10^{-19}C$ ), and we find:
$N= \frac{Q}{e}= \frac{-4.9 \cdot 10^{-6}C}{-1.6 \cdot 10^{-19}C} =3.1 \cdot 10^{13}$

5 0

2 years ago