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3 years ago

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You need to know the height of a tower, but darkness obscures the ceiling. You note that a pendulum extending from the ceiling a

lmost touches the floor and that its period is 26 s. The acceleration of gravity is 9.81 m/s ^ 2 How tall is the tower ? Answer in units of m .

1 answer:

Drupady [299]3 years ago

3 0

U gotta add like all of them bro

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Initially a car accelerates at 2 m/s2 for x seconds. The car then travels at a velocity of -6 m/s for x seconds. If the car disp

Luda [366]

Answer:

The time travel is

t=8 s

Explanation:

$a= 2 \frac{m}{s^{2} } \\v=-6 \frac{m}{s} \\x=16m$

$x_{f}=x_{o}+v_{o}*t+\frac{1}{2} *a*t^{2}$

$x_{f}=0-6*t+\frac{1}{2} *2*t^{2}$

$t^{2}-6*t-16=0\\ using :\\\frac{-b+/-\sqrt{b^{2}-4*c*a } }{2} \\\frac{-(-6)+/-\sqrt{(-6)^{2}-4*(-16)*(1) } }{2}=\frac{3}{2} +/- \frac{10}{2} \\t_{1} = 2s \\t_{2} = 8s$

Check

$t_{2}=8s$

$x_{f}=x_{o}+v_{o}*t+\frac{1}{2} *a*t^{2}$

$x_{f}=0-6*+\frac{1}{2} *2*8^{2}$

$x_{f}=-48+64\\x_{f}=16$

5 0

3 years ago

A grandfather clock depends on the period of a pendulum to keep correct time.(ii) Suppose a grandfather clock is calibrated corr

ANEK [815]

The grandfather clock will now run slow (Option A).

<h3>What is Time Period of an oscillation?</h3>

The time period of an oscillation refers to the time taken by an object to complete one oscillation.
It is the inverse of frequency of oscillation; denoted by "T".

Now,

$\sqrt{\frac{L}{g}}$ , where L is the length and g is the gravitational constant, is the formula for a pendulum's period.
The period will increase as one climbs a very tall mountain because g will slightly decrease.
Due to this and the previous issue, the clock runs slowly and it seems that one second is longer than it actually is.

Hence, the grandfather clock will now run slow (Option A).

To learn more about the time period of an oscillation, refer to the link: brainly.com/question/26449711

#SPJ4

6 0

1 year ago

A 47-kg packing crate is pulled with constant speed across a rough floor with a rope that is at an angle of 37 degrees above the

levacccp [35]

Tension in the rope due to applied force will be given as

$F = 142 N$

angle of applied force with horizontal is 37 degree

displacement along the floor = 6.1 m

so here we can use the formula of work done

$W = F d cos\theta$

now we can plug in all values above

$W = 142 * 6.1 * cos37$

$W = 691.8 J$

So here work done to pull is given by 691.8 J

8 0

3 years ago

50 points: can someone help me with this question please: A classroom is 5.345m long, 4.128m wide, and 3.859m high. What is the

lesya [120]

this is basically the same as volume, no?

So, 5.345*4.128*3.859=85.145

5 0

3 years ago

Read 2 more answers

Summerize how friction and inertia act on an object sliding on a flat surface

natta225 [31]

Inertia is when a object in motion will stay in motion or in a standing still state unless acted upon by a unbalancing force.
Friction is when a object slows down because it is rubbing against another object.

If a object is sliding across a surface, theoretically, it would not stop but because it is on a flat surface it would experience friction, this will disperse some of the kinetic energy that it has thus slowing the object down eventually, after some time, to a stop.

Hope this helps! :)

5 0

3 years ago