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3 years ago

A simple pendulum of length 2.5 m makes 5.0 complete swings in 16 s. What is the acceleration of gravity at the location?

Physics

1 answer:

Norma-Jean [14]3 years ago

8 0

<h2>The acceleration of gravity at the location is 9.64 m/s²</h2>

Explanation:

Length of pendulum = 2.5 m

Time taken for 5 swings = 16 seconds

Time taken for 1 swing = 3.2 seconds

Period of pendulum = 3.2 seconds.

We have equation for period of simple pendulum as

$T=2\pi \sqrt{\frac{l}{g}}$

Where l is the length of pendulum and g is acceleration due to gravity.

Substituting

$T=2\pi \sqrt{\frac{l}{g}}\\\\3.2=2\pi \sqrt{\frac{2.5}{g}}\\\\g=\frac{4\pi^2 \times 2.5}{3.2^2}\\\\g=9.64m/s^2$

The acceleration of gravity at the location is 9.64 m/s²

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Three identical balls are thrown from the same height off a tall building. They are all thrown with the same speed, but in diffe

koban [17]

Answer:

three balls have the same speed

Explanation:

In a parabolic motion we have that

$v_{x}=v_{0}cos\alpha\\v_{y}=-gt+v_{0}sin\alpha\\v=\sqrt{v_{x}^{2}+v_{y}^{2}}\\$

but the time just before the balls hit the ground is

$t=\frac{2v_{0}sin\alpha}{g}$

Hence we have

ball A

$v=\sqrt{v_{0}^{2}cos^{2}(0)+(-g\frac{2v_{0}sin(0)}{g}+v_{0}sin(0))^{2}}\\v=\sqrt{v_{0}^{2}}=v_{0}$

ball B

$v=\sqrt{v_{0}^{2}cos^{2}(45)+(-g\frac{2v_{0}sin(45)}{g}+v_{0}sin(45))^{2}}\\v=\sqrt{v_{0}^{2}(\frac{\sqrt{2}}{2})^{2}+v_{0}^{2}(\frac{\sqrt{2}}{2})^{2}}\\v=\sqrt{v_{0}^{2}}=v_{0}$

ball C

$v=\sqrt{v_{0}^{2}cos^{2}(-45)+(-g\frac{2v_{0}sin(-45)}{g}+v_{0}sin(-45))^{2}}\\v=\sqrt{v_{0}^{2}(\frac{\sqrt{2}}{2})^{2}+v_{0}^{2}(\frac{\sqrt{2}}{2})^{2}}\\v=\sqrt{v_{0}^{2}}=v_{0}$

Hence, all three balls have the same speed just before hit the groug

vA=vB=vC

I hope this is useful for you

regards

5 0

3 years ago

An elevator cab is pulled directly upward by a single cable. The elevator cab and its single occupant have a mass of 2300 kg. Wh

Murrr4er [49]

Answer:

15.64 KN

Explanation:

mass of the elevator cab with a single occupant= 2300 kg

acceleration relative to the cab $a_{ce}$ = 6.80 m/s^2

acceleration of the coin relative to the cab in the opposite direction of motion of cab so we can consider it as a= -6.80 m/s^2

The acceleration of elevator cab relative to the ground $a_{cg}$

now we can say that

$a_{ce} +a_{eg} =a_{cg}$

=-6.80+ a_{eg}= -9.8

[tex]a_{eg}=-9.8+6.80=-3.8

The forces that act on elevator cab are tension and gravitational, applying newtons second law

T- mg= ma_{eg}

Then the tension in the cable is

T= 2300(-3.8)+2300×9.8= 15640 N= 15.64 KN

therefore tension in the string will be 15.64 KN

3 0

4 years ago

A defibrillator containing a 18.4 μF capacitor is used to shock the heart of a patient by holding it to the patient's chest. Jus

gregori [183]

Answer:

The energy E is 1603.008 J.

Explanation:

Given that,

Capacitor = 18.4 μF

Voltage = 13.2 kV

We need to calculate the energy

Using formula of energy

$E=W=\int{Q}\ dV$ .....(I)

We know that,

$Q=CV$

Put the value of Q in equation (I)

$E=\int{CV}dV$

On integration

$E=\dfrac{CV^2}{2}$

Put the value into the formula

$E=\dfrac{18.4\times10^{-6}\times(13.2\times10^{3})^2}{2}$

$E=1603.008\ J$

Hence, The energy E is 1603.008 J.

3 0

3 years ago

Can some please help me

Brilliant_brown [7]

Answer:

Classification

– Drug (highly regulated)

§ A substance which changes body structure or function.

§ Stimulate hormone secretions

§ Looks like medicine and/or is administered differently than foods

– Dietary Supplement (not highly regulated)

§ Highly refined products

§ No positive nutritional value

Explanation:

https://web.cortland.edu/buckenmeyerp/Lecture14.html

3 0

3 years ago

A racehorse coming out of the gate accelerates from rest to a velocity of 12.0

Natali [406]

Answer:

5.22 m/s²

Explanation:

From the question given above, the following data were obtained:

Initial velocity (u) = 0 m/s

Final velocity (v) = 12 m/s

Time (t) = 2.3 s

Acceleration (a) =.?

Acceleration is defined as the change of velocity with time. Mathematically, it is expressed as:

Acceleration (a) = final velocity (v) – initial velocity (u) / time (t)

a = (v – u) /t

With the above formula, we can obtain the acceleration of the racehorse as follow:

Initial velocity (u) = 0 m/s

Final velocity (v) = 12 m/s

Time (t) = 2.3 s

Acceleration (a) =.?

a = (v – u) /t

a = (12 – 0) / 2.3

a = 12 /2.3

a = 5.22 m/s²

Therefore, the average acceleration of the racehorse is 5.22 m/s²

6 0

3 years ago