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stealth61 [152]
3 years ago
9

The function​ s(t) represents the position of an object at time t moving along a line. Suppose s( 1 )=123 and s( 3 )=173. Find t

he average velocity of the object over the interval of time [1,3]?
Physics
1 answer:
Wittaler [7]3 years ago
3 0

Answer:

The average velocity is v_{average}=25.

Explanation:

The average velocity is calculated in the following way:

If s(t) represents the position of an object at time t,

and s(t_{1})=a ; s(t_{2})=b, the average velocity is defined in that interval as:

v_{average}= \frac{final.position-initial.position}{elapsed.time}=\frac{b-a}{t_{2}-t_{1}}

Taking the data from the question:

v_{average}=\frac{173-123}{3-1}=\frac{50}{2}=25.

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Part complete How long must a simple pendulum be if it is to make exactly one swing per five seconds?
shtirl [24]

Answer:

L=6.21m

Explanation:

For the simple pendulum problem we need to remember that:

\frac{d^{2}\theta}{dt^{2}}+\frac{g}{L}sin(\theta)=0,

where \theta is the angular position, t is time, g is the gravity, and L is the length of the pendulum. We also need to remember that there is a relationship between the angular frequency and the length of the pendulum:

\omega^{2}=\frac{g}{L},

where \omega is the angular frequency.

There is also an equation that relates the oscillation period and the angular frequeny:

\omega=\frac{2\pi}{T},

where T is the oscillation period. Now, we can easily solve for L:

(\frac{2\pi}{T})^{2}=\frac{g}{L}\\\\L=g(\frac{T}{2\pi})^{2}\\\\L=9.8(\frac{5}{2\pi})^{2}\\\\L=6.21m

3 0
3 years ago
A girl throws a rock horizontally at 10 m/s from the top of a building, 22 m above street level. Assuming free fall conditions a
MAXImum [283]

Answer:21.18 m

Explanation:

Given

initial speed u=10 m/s

height of building h=22 m

time taken to complete 22 m

h=ut+\frac{1}{2}at^2

initial vertical velocity =0

22=\frac{1}{2}gt^2

t=\sqrt{\frac{22\times 2}{g}}

t=2.11 s

Horizontal Distance moved

R=u_x\times t

R=10\times 2.11

R=21.18 m

6 0
2 years ago
When distances were carefully measured from the Earth to globular clusters above and below the Milky Way plane (where our view o
uranmaximum [27]

Answer:

A) Spherically symmetric about a point in the constellation Sagittarius and concentrated in that direction

Explanation:

The globular clusters are present mainly in the direction of Sagittarius with the center of the system of globular cluster being measured as a spherical cluster cloud such that the center of the Milky Way can be taken as being in the Sagittarius constellation

4 0
3 years ago
The motion of an object undergoing constant acceleration can be modeled by the kinematic equations. One such equation is xf=xi+v
Arturiano [62]

Answer:

a = 1.72 m/s²

Explanation:

The given kinematic equation is the 2nd equation of motion. The equation is as follows:

xf = xi + (Vi)(t) + (1/2)(a)t²

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xf = the final position =  5000 m

xi = the initial position = 1000 m

Vi = the initial velocity = 15 m/s

t = the time taken = 60 s

a = acceleration = ?

Therefore,

5000 m = 1000 m + (15 m/s)(60 s) + (1/2)(a)(60 s)²

5000 m = 1000 m + 900 m + a(1800 s²)

5000 m = 1900 m + a(1800 s²)

5000 m - 1900 m = a(1800 s²)

a(1800 s²) = 3100 m

a = 3100 m/1800 s²

<u>a = 1.72 m/s²</u>

5 0
3 years ago
A uniform rod of length 50cm and mass 0.2kg is placed on a fulcrum at a distance of 40cm from the left end of the rod. At what d
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I’m pretty sure it’s b
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