If you decrease the frequency of a vibrating object to one fourth ,its period becomes

An archer puts a 0.30 kg arrow to the bowstring. An average force of 201 N is exerted to draw the string back 1.3 m.a. Assuming

Vlad [161]

Answer:

Explanation:

Given

mass of archer $m=0.3\ kg$

Average force $F_{avg}=201\ N$

extension in arrow $x=1.3\ m$

Work done to stretch the bow with arrow

$W=F\cdot x$

$W=201\times 1.3=261.3\ m$

This work done is converted into kinetic Energy of arrow

$W=\frac{1}{2}mv^2$

where v= velocity of arrow

$261.3=\frac{1}{2}\times 0.3\times v^2$

$v=\sqrt{1742}$

$v=41.73\ m/s$

(b)if arrow is thrown vertically upward then this energy is converted to Potential energy

$W=mgh$

$261.3=0.3\times 9.8\times h$

$h=\frac{261.3}{0.3\times 9.8}$

$h=88.87\ m$

4 0

3 years ago

Richard is driving home to visit his parents. 135{\rm mi} of the trip are on the interstate highway where the speed limit is 65{

nydimaria [60]

Answer:

time spent = 0.2276

Explanation:

given data

distance = 135 mi

usual speed = 65 mph

today speed = 73 mph

solution

we get here time that is express as

time = $\frac{distance}{speed}$ ...................1

usual time = $\frac{135}{65}$ = 2.0769 h

today time = $\frac{135}{73}$ = 1.8493 h

so we get here time spent as

time spent = 2.0769 h - 1.8493 h

time spent = 0.2276

6 0

3 years ago

As gravity continues to crush the inner core, the fusion of carbon and oxygen begins. What temperature is needed for the fusion

Olin [163]

Answer:

600,000,000 degree C

Explanation:

This stage is the last stage and is refereed to as supernova. In the beginning of this stage, gravity pulls the inner core and crush it, due to which fusion of atoms starts. Carbon and Oxygen fuse together and the temperature is about of 600,000,000 degree C.

The most heavier atom that can be formed out of this fusion is the iron. The moment all the atoms becomes of iron, no further fusion is possible hence that body emits radiation of high intensity and collapse causing a big supernova.

8 0

2 years ago

A 2100 g block is pushed by an external force against a spring (with a 22 N/cm spring constant) until the spring is compressed b

Vilka [71]

Answer:

6.5e-4 m

Explanation:

We need to solve this question using law of conservation of energy

Energy at the bottom of the incline= energy at the point where the block will stop

Therefore, Energy at the bottom of the incline consists of the potential energy stored in spring and gravitational potential energy= $\frac{1}{2} kx^{2} +PE1$

Energy at the point where the block will stop consists of only gravitational potential energy= $PE2$

Hence from Energy at the bottom of the incline= energy at the point where the block will stop

⇒ $\frac{1}{2} kx^{2} +PE1=PE2$

⇒ $PE2-PE1=\frac{1}{2} kx^{2}$

Also $PE2-PE2=mgh$

where $m$ is the mass of block

$g$ is acceleration due to gravity=9.8 m/s

$h$ is the difference in height between two positions

⇒ $mgh=\frac{1}{2} kx^{2}$

Given m=2100kg

k=22N/cm=2200N/m

x=11cm=0.11 m

∴ $2100*9.8*h=\frac{1}{2}*2200*0.11^{2}$

⇒ $20580*h=13.31$

⇒ $h=\frac{13.31}{20580}$

⇒h=0.0006467m= $6.5e-4$

7 0

3 years ago

A series of pulses, each of amplitude 0.1 m, is sent down a string that is attached to a post at one end. The pulses are reflect

AleksAgata [21]

Answer:

A_resulting = 0.2 m

Explanation:

Let's analyze the impact of the pulse with the pole, this is a fixed obstacle that does not move therefore by the law of action and reluctant, the force that the pole applies on the rope is of equal magnitude to the force of the rope on the pole (pulse), but opposite directional, so the reflected pulse reverses its direction and sense.

With this information we analyze a point on the string where the incident pulse is and each reflected with an amplitude A = 0.1 m, the resulting is

A_res = 2A

A_resultant = 2 .01

A_resulting = 0.2 m

8 0

2 years ago