All categories

3 years ago

A spring deforms 5 cm when we attach a 2 kg object:

Physics

1 answer:

kramer3 years ago

7 0

Answer:

[a]

F=ke

E= 5/100=0.05m, m=2kg, F=mg[2x9.8]=19.6N

Spring constant[K] =F/e

K=19.6/0.05

K=392N/m

You might be interested in

Imagine an infinite earth with a hole dripped through it. You fall in and accelerate at g~10m/s/s. How long until you reach the

Soloha48 [4]

An object with non-zero mass (even negligible mass is non-zero) will never reach the speed of light. Due to relativistic effects, each "unit" of acceleration becomes less effective at increasing your velocity (relative to some other object, of course) as your relative velocity approaches the speed of light.

And even if there was a way, If you would accelerate to the 99,99% of the speed light in just 1 second, you would experience a G-force of aprox. 30,600,000 g's which is enough to kill you in a few seconds

6 0

3 years ago

In practice, if a voltmeter was connected across any combination of the terminals, the potential difference would be less than w

VladimirAG [237]

Answer:

This difference is kept to a minimum because the resistance in transformers is a few tens of ohms and the resistance of modern voltmeters is of the order of MΩ.

Explanation:

A voltmeter is built by a galvanometer and a resistance in series, this set is connected in parallel to the resistance where the voltage is to be measured, therefore the voltage is divided between the voltmeter and the element to be measured, consequently the measured voltage It is less than the calculated one, since for them the resistance of the voltmeter is assumed infinite.

This difference is kept to a minimum because the resistance in transformers is a few tens of ohms and the resistance of modern voltmeters is of the order of MΩ.

8 0

3 years ago

for an ideal monoatomic gas, the internal energy U os due to the kinetic energy and U=3/2RT per mole.show that cv=3/2R per mole

sladkih [1.3K]

Answer:

i. Cv =3R/2

ii. Cp = 5R/2

Explanation:

i. Cv = Molar heat capacity at constant volume

Since the internal energy of the ideal monoatomic gas is U = 3/2RT and Cv = dU/dT

Differentiating U with respect to T, we have

= d(3/2RT)/dT

= 3R/2

ii. Cp - Molar heat capacity at constant pressure

Cp = Cv + R

substituting Cv into the equation, we have

Cp = 3R/2 + R

taking L.C.M

Cp = (3R + 2R)/2

Cp = 5R/2

3 0

3 years ago

A spherical balloon has a radius of 6.55 m and is filled with helium. The density of helium is 0.179 kg/m3, and the density of a

Ksju [112]

Answer:

$M_1 = 317.7 kg$

Explanation:

Mass of the helium gas filled inside the volume of balloon is given as

$m = \rho V$

$m = 0.179(\frac{4}{3}\pi R^3)$

$m = 0.179(\frac{4}{3}\pi 6.55^3)$

$m = 210.7 kg$

now total mass of balloon + helium inside balloon is given as

$M = 210.7 + 990$

$M = 1200.7 kg$

now we know that total weight of balloon + cargo = buoyancy force on the balloon

so we will have

$(M + M_1)g = \rho_{air} V g$

$(1200.7 + M_1) = (\frac{4}{3}\pi 6.55^3) (1.29)$

$1200.7 + M_1 = 1518.4$

$M_1 = 317.7 kg$

3 0

3 years ago

a spring gun initially compressed 2cm fires a 0.01kg dart straight up into the air. if the dart reaches a height it 5.5m determi

Vikki [24]

Answer:

2697.75N/m

Explanation:

Step one

This problem bothers on energy stored in a spring.

Step two

Given data

Compression x= 2cm

To meter = 2/100= 0.02m

Mass m= 0.01kg

Height h= 5.5m

K=?

Let us assume g= 9.81m/s²

Step three

According to the principle of conservation of energy

We know that the the energy stored in a spring is

E= 1/2kx²

1/2kx²= mgh

Making k subject of formula we have

kx²= 2mgh

k= 2mgh/x²

k= (2*0.01*9.81*5.5)/0.02²

k= 1.0791/0.0004

k= 2697.75N/m

Hence the spring constant k is 2697.75N/m

7 0

3 years ago