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Sav [38]
3 years ago
13

A 20-newton weight is attached to a spring.

Physics
1 answer:
Makovka662 [10]3 years ago
8 0

Answer:

40 N/m

Explanation:

The diagram attached is used to answer the question

We know from Hooke's law that extension is directly proportional to the applied force hence

F=kx where x is extension, F is applied force and k is the spring constant. Making k the subject of the formula then

k=\frac {F}{k}

From the attached diagram extension is given by subtracting unstretched spring from stretched spring hence extension, x=1-0.5=0.5m

Substituting 20 N for F and 0.5 m for x then

k=\frac {20}{0.5}=40 N/m

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Vinvika [58]

Answer:

             

Explanation:

is  A projectile is any object on which the only force acting is gravity and air resistance (drag).

Examples of projectiles are:

baseballs and softballs in the air after being hit by the bat

golf balls hit by a club

objects dropped from aircraft, such as people (skydivers), bombs, crates of food being dropped to refugees

objects launched by cannons, such as cannonballs, shells, and circus performers

Once the baseball, softball, golf ball, skydiver, bomb, crate, cannonball, shell, or clown are no longer touching the bat, club, aircraft, or cannon, and are in the air with only gravity and slight air resistance acting on it, then it is a projectile.

Here is an online projectile motion applets to play with, just for fun.

Unless otherwise stated in a particular problem or discussion, we will be ignoring the effects of air resistance.

The key to understanding the motion of projectiles is that the horizontal motion and the vertical motion of the projectile are independent of each other. So we can write separate equations for the displacement of the projectile in the horizontal (x) and vertical (y) directions.

                         

The only common variable between these two equations is t, the time. Because in projectile problems there is usually no acceleration (i.e. we ignore air resistance) in the horizontal direction, we can write

           

The velocity components follow the same equations we used for one-dimensional motion.

                             

Because there is usually no acceleration in the x direction, the x-velocity is constant.

3 0
3 years ago
A yo-yo is made of two solid cylindrical disks, each of mass 0.055 kg and diameter 0.070 m , joined by a (concentric) thin solid
DedPeter [7]

Answer: IM 95%sure that the answer is B jus took the test got the answer right

Explanation:

6 0
3 years ago
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For a damped simple harmonic oscillator, the block has a mass of 1.2 kg and the spring constant is 9.8 N/m. The damping force is
ArbitrLikvidat [17]

Answer:

a) t=24s

b) number of oscillations= 11

Explanation:

In case of a damped simple harmonic oscillator the equation of motion is

m(d²x/dt²)+b(dx/dt)+kx=0

Therefore on solving the above differential equation we get,

x(t)=A₀e^{\frac{-bt}{2m}}cos(w't+\phi)=A(t)cos(w't+\phi)

where A(t)=A₀e^{\frac{-bt}{2m}}

 A₀ is the amplitude at t=0 and

w' is the angular frequency of damped SHM, which is given by,

w'=\sqrt{\frac{k}{m}-\frac{b^{2}}{4m^{2}} }

Now coming to the problem,

Given: m=1.2 kg

           k=9.8 N/m

           b=210 g/s= 0.21 kg/s

           A₀=13 cm

a) A(t)=A₀/8

⇒A₀e^{\frac{-bt}{2m}} =A₀/8

⇒e^{\frac{bt}{2m}}=8

applying logarithm on both sides

⇒\frac{bt}{2m}=ln(8)

⇒t=\frac{2m*ln(8)}{b}

substituting the values

t=\frac{2*1.2*ln(8)}{0.21}=24s(approx)

b) w'=\sqrt{\frac{k}{m}-\frac{b^{2}}{4m^{2}} }

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T'=\frac{2\pi}{w'}, where T' is time period of damped SHM

⇒T'=\frac{2\pi}{2.86}=2.2s

let n be number of oscillations made

then, nT'=t

⇒n=\frac{24}{2.2}=11(approx)

8 0
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How long does it take a car travelling at 60 m.p.h to cover 5 miles?
k0ka [10]

Answer:

Time = 5 minutes

6 0
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What is the acceleration of a 10 kg mass pushed by a 5 N force
hichkok12 [17]

Answer:

The formula is a = F m so in this case a = 5 10 = 0.5 m s 2

Explanation:

3 0
2 years ago
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