How does the kinetic energy of an object relate to its mass and velocity?

A mass is tied to a string and swung in a horizontal circle with a constant angular speed. show answer No Attempt If this speed

Liono4ka [1.6K]

Answer:

The tension in the string is quadrupled i.e. increased by a factor of 4.

Explanation:

The tension in the string is the centripetal force. This force is given by

$F = \dfrac{mv^2}{r}$

m is the mass, v is the velocity and r is the radius.

It follows that $F \propto v^2$ , provided m and r are constant.

When v is doubled, the new force, $F_1$ , is

$F_1 = \dfrac{m(2v)^2}{r} = \dfrac{4mv^2}{r} = 4\dfrac{mv^2}{r} = 4F$

Hence, the tension in the string is quadrupled.

8 0

3 years ago

You are tasked with calibrating the springs for a pinball machine. Your method of testing the springs is by attaching masses ont

choli [55]

Answer:

$490.5\ \text{N/m}$

Explanation:

m = Mass attached to spring = 14 kg

g = Acceleration due to gravity = $9.81\ \text{m/s}^2$

x = Displacement of spring = $78-50=28\ \text{cm}$

k = Spring constant

The force balance of the system is given by

$kx=mg\\\Rightarrow k=\dfrac{mg}{x}\\\Rightarrow k=\dfrac{14\times 9.81}{0.28}\\\Rightarrow k=490.5\ \text{N/m}$

The spring constant for that spring is $490.5\ \text{N/m}$ .

6 0

2 years ago

HELP!!

raketka [301]

Hello!

Since the two weights are <em>off</em> the table, the block will move towards letter F.

I hope this helps :))

5 0

2 years ago

Read 2 more answers

You have two rocks made of the same material that are at the same

mr Goodwill [35]

Answer:

the rocks have the same amount of thermal energy

4 0

2 years ago

Read 2 more answers

(a) A light-rail commuter train accelerates at a rate of 1.35 m/s2 . How long does it take to reach its top speed of 80.0 km/h,

Mashcka [7]

Answer:

(a) Time t = 16.46 sec

(b) Time t =13.466 sec

(c) Deceleration = $2.677m/sec^2$

Explanation:

(a) As the train starts from rest its initial velocity u = 0 m/sec

Acceleration $a=1.35m/sec^2$

Final speed v = 80 km/hr

$80km/hr=\frac{80\times 1000}{3600sec}=22.22m/sec$

From first equation of motion v =u+at

So $t=\frac{v-u}{a}=\frac{22.22-0}{1.35}=16.46 sec$

(b) Now initial speed u = 22.22 m/sec

As finally train comes to rest so final speed v=0 m/sec

Deceleration $a=1.65m/sec^2$

So $t=\frac{v-u}{a}=\frac{0-22.22}{-1.65}=13.466 sec$

(c) We have given that initial velocity = 80 km/hr = 22.22 m/sec

Final velocity v = 0 m/sec

Time t = 8.30 sec

So acceleration is given by

$a=\frac{v-u}{t}=\frac{0-22.22}{8.3}=-2.6771m/sec^2$

As acceleration is negative so it is a deceleration

7 0

3 years ago