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marusya05 [52]
3 years ago
13

When a hammer exerts a force on a nail, how does the amount of force

Physics
1 answer:
taurus [48]3 years ago
7 0

Answer:

10 pzs

Explanation:

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A slender, uniform metal rod of mass M and length lis pivoted without friction about an axis through its midpoint andperpendicul
lys-0071 [83]

Answer:

Explanation:

Moment of inertia of the metal rod pivoted in the middle

= M l² / 12

If the spring is compressed by small distance x twisting the rod by angle θ

restoring force by spring

= k x

moment of torque  about axis

= k x l /2

= k θ( l /2 )²     ( x / .5 l = θ )

=

moment of torque = moment of inertia of  rod  x angular acceleration

k θ( l /2 )²   = M l² / 12 d²θ/dt²

d²θ/dt² = 3 k/M  θ

acceleration =  ω² θ

ω² = 3 k/M

ω = √ 3 k / M

8 0
3 years ago
Read 2 more answers
Quick! how did Daniel hale Williams contribute to science?
My name is Ann [436]
He preformed the first ever successful open heart surgery
3 0
3 years ago
Which factors could be potential sources of error in the experiment? Check all that apply.
mote1985 [20]

Answer:

1. energy lost in the lever due to friction

3. visual estimation of height of the beanbag

5. position of the fulcrum for the lever affecting transfer of energy

Explanation:

Edge 2021

8 0
3 years ago
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4. The blades on a fan have a frequency of 15 Hz.
vichka [17]

Answer:

a) 4500 cycles b) 0.0667s c) 6.67s

Explanation:

a) 15 Hz= 15 cycles/ s

   5 mins= 300s

   15 cycles/s * 300s= 4500 cycles

b) Period= 1/ frequency

   Period= 1/ 15 cycles/s

   Period= 0.0667s

c) Period * number of revolutions= time

  0.0667 * 100= 6.67s

6 0
3 years ago
car traveling on a flat (unbanked), circular track accelerates uniformly from rest with a tangential acceleration of a. The car
Luba_88 [7]

Answer:

0.572

Explanation:

First examine the force of friction at the slipping point where Ff = µsFN = µsmg.

the mass of the car is unknown,

The only force on the car that is not completely in the vertical direction is friction, so let us consider the sums of forces in the tangential and centerward directions.

First the tangential direction

∑Ft =Fft =mat

And then in the centerward direction ∑Fc =Ffc =mac =mv²t/r

Going back to our constant acceleration equations we see that v²t = v²ti +2at∆x = 2at πr/2

So going backwards and plugging in Ffc =m2atπr/ 2r =πmat

Ff = √(F2ft +F2fc)= matp √(1+π²)

µs = Ff /mg = at /g √(1+π²)=

1.70m/s/2 9.80 m/s² x√(1+π²)= 0.572

7 0
3 years ago
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