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nlexa [21]
2 years ago
5

The formula s = 16t2 gives the distance an object falls due to gravity, where s is the distance in feet and t is the time in sec

onds. how many feet will the object fall in 6 seconds?
Physics
1 answer:
Georgia [21]2 years ago
8 0
To determine the distance in units of feet of the object falling at a given time, we simply use the relation given above and substitute the given value of time to the function. We do as follows:

s = 16t^2
s = 16 ( 6 )^2
s = 576 feet

The distance of the object from the ground at 6 seconds is 576 ft.
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Pie
Please post in English so i or someone else can help you.
7 0
2 years ago
A mass m slides down a frictionless ramp and approaches a frictionless loop with radius R. There is a section of the track with
Lana71 [14]

Answer:

   h = 2 R (1 +μ)

Explanation:

This exercise must be solved in parts, first let us know how fast you must reach the curl to stay in the

let's use the mechanical energy conservation agreement

starting point. Lower, just at the curl

       Em₀ = K = ½ m v₁²

final point. Highest point of the curl

        Em_{f} = U = m g y

Find the height y = 2R

      Em₀ = Em_{f}

      ½ m v₁² = m g 2R

       v₁ = √ 4 gR

Any speed greater than this the body remains in the loop.

In the second part we look for the speed that must have when arriving at the part with friction, we use Newton's second law

X axis

    -fr = m a                      (1)

Y Axis  

      N - W = 0

      N = mg

the friction force has the formula

     fr = μ  N

     fr = μ m g

    we substitute 1

    - μ mg = m a

     a = - μ g

having the acceleration, we can use the kinematic relations

    v² = v₀² - 2 a x

    v₀² = v² + 2 a x

the length of this zone is x = 2R

    let's calculate

     v₀ = √ (4 gR + 2 μ g 2R)

     v₀ = √4gR( 1 + μ)

this is the speed so you must reach the area with fricticon

finally have the third part we use energy conservation

starting point. Highest on the ramp without rubbing

     Em₀ = U = m g h

final point. Just before reaching the area with rubbing

     Em_{f} = K = ½ m v₀²

      Em₀ = Em_{f}

     mgh = ½ m 4gR(1 + μ)

       h = ½ 4R (1+ μ)

       h = 2 R (1 +μ)

7 0
2 years ago
A 1000-kg car is slowly picking up speed as it goes around a horizontal curve whose radius is 100 m. The coefficient of static f
Snezhnost [94]

Answer:

18.5 m/s

Explanation:

On a horizontal curve, the frictional force provides the centripetal force that keeps the car in circular motion:

\mu mg = m\frac{v^2}{r}

where

\mu is the coefficient of static friction between the tires and the road

m is the mass of the car

g is the gravitational acceleration

v is the speed of the car

r is the radius of the curve

Re-arranging the equation,

v=\sqrt{\mu gr}

And by substituting the data of the problem, we find the speed at which the car begins to skid:

v=\sqrt{(0.350)(9.8 m/s^2)(100 m)}=18.5 m/s

7 0
2 years ago
Read 2 more answers
A certain alarm clock ticks four times each second, with each tick representing half a period. The balance wheel consists of a t
Semenov [28]

Answer:

a. I=2.77x10^{-8} kg*m^2

b. K=4.37 x10^{-6} N*m

Explanation:

The inertia can be find using

a.

I = m*r^2

m = 0.95 g * \frac{1 kg}{1000g}=9.5x10^{-4} kg

r=0.54 cm * \frac{1m}{100cm} =5.4x10^{-3}m

I = 9.5x10^{-4}kg*(5.4x10^{-3}m)^2

I=2.77x10^{-8} kg*m^2

now to find the torsion constant can use knowing the period of the balance

b.

T=0.5 s

T=2\pi *\sqrt{\frac{I}{K}}

Solve to K'

K = \frac{4\pi^2* I}{T^2}=\frac{4\pi^2*2.7702 kg*m^2}{(0.5s)^2}

K=4.37 x10^{-6} N*m

3 0
2 years ago
Every spring has an equilibrium position. Which statements describe a spring at its equilibrium position? Check all that
alexgriva [62]

Answer:

The elastic potential energy is zero.

The net force acting on the spring is zero.

Explanation:

The equilibrium position of a spring is the position that the spring has when its neither compressed nor stretched - it is also called natural length of the spring.

Let's now analyze the different statements:

The spring constant is zero.  --> false. The spring constant is never zero.

The elastic potential energy is at a maximum  --> false. The elastic potential energy of a spring is given by

E=\frac{1}{2}kx^2

where k is the spring constant and x the displacement. Therefore, the elastic potential energy is maximum when x, the displacement, is maximum.

The elastic potential energy is zero.  --> true. As we saw from the equation above, the elastic potential energy is zero when the displacement is zero (at the equilibrium position).

The displacement of the spring is at a maxi num  --> false, for what we said above

The net force acting on the spring is zero. --> true, as the spring is neither compressed nor stretched

8 0
2 years ago
Read 2 more answers
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