TP2 2) An example of a 3rd gen lever is:

Tarzan swings back and forth on a long vine with a period of 7.27 s. how long is the vine?(unit=m)

lana [24]

Answer:

Tarzan, who weighs 688N, swings from a cliff at the end of a convenient vine that is 18m long. From the top of the cliff to the bottom of the swing he descends by 3.2m.

Explanation:

3 0

3 years ago

The energy released from condensation in thunderstorms can be very large. Calculate the energy (in J) released into the atmosphe

makkiz [27]

Answer:

7.19 * 10^14J

Explanation:

Given that

Density of water Pwater= 1000kg/m3

R=2.1km = 2.1*10^3m

H= 2.3cm. = 2.3*10^-2m

Lv water= 2256 * 10^3J/kg

First, mass of water need to be calculated, using an imaginary cylinder

Density= Mass /Volume

Mass= Density* Volume

Volume of a cylinder= πR2h

Therefore mass= PπR2H

= 1000 * π * (2.1 *10^3)^2 * (2.3 * 10^-2)

= 3.18 *10^8

Heat Released Qv = mLV

= 3.18*10^8 * 2236*10^3

= 7.19 * 10^14J

7 0

3 years ago

Read 2 more answers

A projectile is launched horizontally from a 20-m tall edifice with a vox of 25 m/s. How long will it take for the projectile to

NISA [10]

Answer:

a) First let's analyze the vertical problem:

When the projectile is on the air, the only vertical force acting on it is the gravitational force, then the acceleration of the projectile is the gravitational acceleration, and we can write this as:

a(t) = -9.8m/s^2

To get the vertical velocity we need to integrate over time to get:

v(t) = (-9.8m/s^2)*t + v0

where v0 is the initial vertical velocity because the object is thrown horizontally, we do not have any initial vertical velocity, then v0 = 0m/s

v(t) = (-9.8m/s^2)*t

To get the vertical position equation we need to integrate over time again, to get:

p(t) = (1/2)*(-9.8m/s^2)*t^2 + p0

where p0 is the initial position, in this case is the height of the edifice, 20m

then:

p(t) = (-4.9m/s^2)*t^2+ 20m

The projectile will hit the ground when p(t) = 0m, then we need to solve:

(-4.9m/s^2)*t^2+ 20m = 0m

20m = (4.9m/s^2)*t^2

√(20m/ (4.9m/s^2)) = t = 2.02 seconds

The correct option is a.

b) The range will be the total horizontal distance traveled by the projectile, as we do not have any horizontal force, we know that the horizontal velocity is 25 m/s constant.

Now we can use the relationship:

distance = speed*time

We know that the projectile travels for 2.02 seconds, then the total distance that it travels is:

distance = 2.02s*25m/s = 50.5m

Here the correct option is a.

c) Again, the horizontal velocity never changes, is 25m/s constantly, then here the correct option is option b. 25m/s

d) Here we need to evaluate the velocity equation in t = 2.02 seconds, this is the velocity of the projectile when it hits the ground.

v(2.02s) = (-9.8m/s^2)*2.02s = -19.796 m/s

The velocity is negative because it goes down, and it matches with option d, so I suppose that the correct option here is option d (because the sign depends on how you think the problem)

4 0

2 years ago

A flutist assembles her flute in a room where the speed of sound is 342m/s . When she plays the note A, it is in perfect tune wi

USPshnik [31]

Answer:

a.3Hz

b.0.0034m

Explanation:

First, we know the flute is an open pipe, because open pipe as both end open and a close organ pipe as only one end close.

The formula relating the length and he frequency is giving as

$f=\frac{nv}{2l}\\$ .

a.we first determine the length of the flute at the fundamental frequency i.e when n=1 and when the speed is in the 342m/s

Hence from

$f=\frac{nv}{2l}\\\\l=\frac{342}{2*440}\\ l=0.389m\\$ .

since the value of the length will remain constant, we now use the value to determine the frequency when the air becomes hotter and the speed becomes 345m/s.

$f=\frac{nv}{2l} \\f=\frac{345}{2*0.389}\\f=443.4Hz$

Hence the require beat is

$B=/f_{1}-f_{2}/\\B=/440-443/\\B=3Hz$ .

b. since the length is dependent also on the speed and frequency, we determine the new length when she plays with a fundamental frequency when the speed of sound is 345m/s

using the formula

$L_{new}=\frac{v}{2f}\\\\L_{new}=\frac{345}{2*440}\\\\L_{new}=0.39204$

Now to determine the extension,

$L_{extend}=L_{new}-L_{old}\\L_{extend}=0.39204- 0.38864\\L_{extend}=0.0034m\\$

4 0

3 years ago

A block attached to a spring with an unknown spring constant oscillates with a period of 2.0 s. What is the period if

Zigmanuir [339]

Answer:

a) If the mass is doubled, then the period is increased by $\sqrt{2}$ . Hence, the period of the system is 2.828 seconds.

b) If the mass is halved, then the period is reduced by $\frac{\sqrt{2}}{2}$ . Hence, the period of the system is 1.414 seconds.

c) The period of the system does not depend on amplitude. Hence, the period of the system is 2 seconds.

d) If the spring constant is doubled, then the period is reduced by $\frac{\sqrt{2}}{2}$ . Hence, the period of the system is 1.414 seconds.

Explanation:

The statement is incomplete. We proceed to present the complete statement: A block attached to a spring with unknown spring constant oscillates with a period of 2.00 s. What is the period if a. The mass is doubled? b. The mass is halved? c. The amplitude is doubled? d. The spring constant is doubled?

We have a block-spring system, whose angular frequency ( $\omega$ ) is defined by the following formula:

$\omega = \sqrt{\frac{k}{m} }$ (1)

Where:

$k$ - Spring constant, measured in newtons per meter.

$m$ - Mass, measured in kilograms.

And the period ( $T$ ), measured in seconds, is determined by the following expression:

$T = \frac{2\pi}{\omega}$ (2)

By applying (1) in (2), we get the following formula:

$T = 2\pi\cdot \sqrt{\frac{m}{k} }$

a) If the mass is doubled, then the period is increased by $\sqrt{2}$ . Hence, the period of the system is 2.828 seconds.

b) If the mass is halved, then the period is reduced by $\frac{\sqrt{2}}{2}$ . Hence, the period of the system is 1.414 seconds.

c) The period of the system does not depend on amplitude. Hence, the period of the system is 2 seconds.

d) If the spring constant is doubled, then the period is reduced by $\frac{\sqrt{2}}{2}$ . Hence, the period of the system is 1.414 seconds.

8 0

3 years ago