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3 years ago

How did the theory of relativity change the law of conservation of energy?

1 answer:

3 0

I believe that the first one is correct because if you have a ball that Weighs 5 pounds and you push it down a ramp it will hit the end with more force than a ball that weighs 2 pounds

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(a) when rebuilding her car's engine, a physics major must exert 300 n of force to insert a dry steel piston into a steel cylind

Vilka [71]

There are some missing data in the text of the problem. I've found them online:
a) coefficient of friction dry steel piston - steel cilinder: 0.3
b) coefficient of friction with oil in between the surfaces: 0.03

Solution:
a) The force F applied by the person (300 N) must be at least equal to the frictional force, given by:
$F_f = \mu N$
where $\mu$ is the coefficient of friction, while N is the normal force. So we have:
$F=\mu N$
since we know that F=300 N and $\mu=0.3$ , we can find N, the magnitude of the normal force:
$N= \frac{F}{\mu}= \frac{300 N}{0.3}=1000 N$

b) The problem is identical to that of the first part; however, this time the coefficienct of friction is $\mu=0.03$ due to the presence of the oil. Therefore, we have:
$N= \frac{F}{\mu}= \frac{300 N}{0.03}=10000 N$

8 0

2 years ago

A 40 kg girl and an 8.4 kg sled are on the surface of a frozen lake, 15 m apart. By means of a rope, the girl exerts a 5.2 N for

stealth61 [152]

Answer:

(a) $a_s=0.62\frac{m}{s^2}$

(b) $a_s=0.13\frac{m}{s^2}$

Explanation:

(a) According to Newton's second law, the acceleration of a body is directly proportional to the force exerted on it and inversely proportional to it's mass.

$a_s=\frac{F}{m_s}\\a_s=\frac{5.2N}{8.4kg}\\a_s=0.62\frac{m}{s^2}$

(b) According to Newton's third law, the force that the sled exerts on the girl is equal in magnitude but opposite in the direction of the force that the girl exerts on the sled:

$a_g=\frac{F}{m_g}\\a_g=\frac{5.2N}{40kg}\\a_g=0.13\frac{m}{s^2}$

$x_f=x_0+v_0t \pm \frac{at^2}{2}$

For the girl, we have $x_0=0$ and $v_0=0$ . So:

$x_f_g=\frac{a_gt^2}{2}(1)$

For the sled, we have $v_0=0$ . So:

$x_f_s=x_0_s-\frac{a_st^2}{2}(2)$

When they meet, the final positions are the same. So, equaling (1) and (2) and solving for t:

$x_0_s-\frac{a_st^2}{2}=\frac{a_st^2}{2}\\t^2(a_g+a_s)=2x_0_s\\t=\sqrt{\frac{2x_s_0}{a_g+a_s}}\\t=\sqrt{\frac{2(15m)}{0.13\frac{m}{s^2}+0.62\frac{m}{s^2}}}\\t=6.32s$

Now, we solve (1) for $x_f_g$

$x_f_g=\frac{0.13\frac{m}{s^2}(6.32s)^2}{2}\\x_f_g=2.6m\\x_f=2.6m$

5 0

2 years ago

Two stones resembling diamonds are suspected of being fakes. To determine if the stones might be real, the mass and volume of ea

dimaraw [331]

Answer:

stone A is diamond.

Explanation:

given,

Volume of the two stone = 0.15 cm³

Mass of stone A = 0.52 g

Mass of stone B = 0.42 g

Density of the diamond = 3.5 g/cm³

So, to find which stone is gold we have to calculate the density of both the stone.

We know,

$density$ $\density = \dfrac{mass}{volume}$

density of stone A

$\rho_A = \dfrac{0.52}{0.15}$

$\rho_A = 3.467\ g/cm^3$

density of stone B.

$\rho_B = \dfrac{0.42}{0.15}$

$\rho_B = 2.8\ g/cm^3$

Hence, the density of the stone A is the equal to Diamond then stone A is diamond.

6 0

2 years ago

If it requires 8.0 J of work to stretch a particular spring by 2.0 cm from its equilibrium length, how much more work will be re

jenyasd209 [6]

Answer:

The amount of work done required to stretch spring by additional 4 cm is 64 J.

Explanation:

The energy used for stretching spring is given by the relation :

$E = \frac{1}{2}kx^{2}$ .......(1)

Here k is spring constant and x is the displacement of spring from its equilibrium position.

For stretch spring by 2.0 cm or 0.02 m, we need 8.0 J of energy. Hence, substitute the suitable values in equation (1).

$8 = \frac{1}{2}\timesk\times k \times(0.02)^{2}$

k = 4 x 10⁴ N/m

Energy needed to stretch a spring by 6.0 cm can be determine by the equation (1).

Substitute 0.06 m for x and 4 x 10⁴ N/m for k in equation (1).

$E = \frac{1}{2}\times4\times10^{4}\times (0.06)^{2}$

E = 72 J

But we already have 8.0 J. So, the extra energy needed to stretch spring by additional 4 cm is :

E = ( 72 - 8 ) J = 64 J

7 0

2 years ago

Shorter the vibrating part more will be the pitch. How?

Lina20 [59]

Answer:When the length of a string is changed, it will vibrate with a different frequency.Shorter strings have higher frequency and therefore higher pitch.

6 0

2 years ago