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Alex777 [14]
2 years ago
12

Pls help if u do it 100 points

Physics
1 answer:
Bumek [7]2 years ago
3 0

Answer:

hhuuufuhrhuuuryheu7uhhehfu

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While finding the spring constant, if X1 = 12 cm, X2 = 15 cm, and hanging mass = 22 grams, the value of spring constant K would
Pavel [41]

Answer:

If x₁=12 cm then k=1.7985 N/m

If x₂=15 cm then k=1.4388 N/m

Explanation:

Hanging mass= 22 g=0.022 kg

Acceleration due to gravity g=9.81 m/s²

If x₁=displacement= 12 cm=0.12 m

k= spring constant

F=ma\\\Rightarrow F=0.022\times 9.81\\\Rightarrow F=0.21582\ N

\text {For spring}\\F=kx\\\Rightarrow 0.21582=k\times 0.012\\\Rightarrow k=1.7985\ N/m\\

∴k = 1.7985 N/m

If x₂=15 cm=0.15 m

Force of the hanging mass is same however the spring constant will change

\text {For spring}\\F=kx\\\Rightarrow 0.21582=k\times 0.015\\\Rightarrow k=1.4388\ N/m\\

∴k = 1.4388 N/m

As the mass is not changing the spring constant has to change. That means that here there are two spring one with k=1.7985 N/m and the other with k= 1.4388 N/m

4 0
3 years ago
a flag of mass 2.5 kg is supported by a single rope. A strong horizontal wind exerts a force of 12 N on the flag. Calculate the
tatuchka [14]
The free-body diagram of the forces acting on the flag is in the picture in attachment.

We have: the weight, downward, with magnitude
W=mg = (2.5 kg)(9.81 m/s^2)=24.5 N
the force of the wind F, acting horizontally, with intensity
F=12 N
and the tension T of the rope. To write the conditions of equilibrium, we must decompose T on both x- and y-axis (x-axis is taken horizontally whil y-axis is taken vertically):
T \cos \alpha -F=0
T \sin \alpha -W=
By dividing the second equation by the first one, we get
\tan \alpha =  \frac{W}{F}= \frac{24.5 N}{12 N}=2.04
From which we find
\alpha = 63.8 ^{\circ}
which is the angle of the rope with respect to the horizontal.

By replacing this value into the first equation, we can also find the tension of the rope:
T= \frac{F}{\cos \alpha}= \frac{12 N}{\cos 63.8^{\circ}}=27.2 N




7 0
3 years ago
You lift a 10-N physics book up in the air a distance of 1.0 m, at a constant velocity
Anarel [89]

The book is lifted upward, but gravity points down, so the work done by gravity must be negative (so you can eliminate options 1 and 3).

The force exerted on the book by gravity has magnitude

<em>F</em> = <em>mg</em> = (10 N) (9.80 m/s^2) = 9.8 N ≈ 10 N

You raise the book 1.0 m in the opposite direction, so the work done is

<em>W</em> = (10 N) (-1.0 m) = -10 J

5 0
3 years ago
What should a free-body diagram look like for a skydiver who has opened his parachute and is now slowing down as he falls?
mote1985 [20]
The last choice. Two arrows and the arrow up is shorter than the arrow down. Since the guy is falling and he’s opened his chute, he’s slowing down but he’s still falling meaning the force of gravity is stronger than the air resistance.
8 0
3 years ago
Read 2 more answers
Sometimes the tar of a newly paved road gets sticky on a very hot summer day. What causes this?
Ksivusya [100]
I’m pretty sure it is caused by the heat of the sun warming it up back into its original state of tar
6 0
3 years ago
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