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4 years ago

Nothing can travel faster than _____.

Physics

1 answer:

Tema [17]4 years ago

3 0

D, which is around the speed of light

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Which mass is the same as 0.031 g?

LenaWriter [7]

0.031 g is equal to 31 grams

6 0

3 years ago

Read 2 more answers

Calculate the speed of the ball, vo in m/s, just after the launch. A bowling ball of mass m = 1.5 kg is launched from a spring c

klemol [59]

Answer:

$v_0=17.3m/s$

Explanation:

In this problem we have three important moments; the instant in which the ball is released (1), the instant in which the ball starts to fly freely (2) and the instant in which has its maximum height (3). From the conservation of mechanical energy, the total energy in each moment has to be the same. In (1), it is only elastic potential energy; in (2) and (3) are both gravitational potential energy and kinetic energy. Writing this and substituting by known values, we obtain:

$E_1=E_2=E_3\\\\U_e_1=U_g_2+K_2=U_g_3+K_3\\\\\frac{1}{2}kd^2=mg(d\sin\theta)+\frac{1}{2}mv_0^2=mgh+\frac{1}{2}m(v_0\cos\theta)^2$

Since we only care about the velocity $v_0$ , we can keep only the second and third parts of the equation and solve:

$mgd\sin\theta+\frac{1}{2}mv_0^2=mgh+\frac{1}{2}mv_0^2\cos^2\theta\\\\\frac{1}{2}mv_0^2(1-\cos^2\theta)=mg(h-d\sin\theta)\\\\v_0=\sqrt{\frac{2g(h-d\sin\theta)}{1-\cos^2\theta}}\\\\v_0=\sqrt{\frac{2(9.8m/s^2)(4.4m-(0.21m)\sin32\°)}{1-\cos^232\°}}\\\\v_0=17.3m/s$

So, the speed of the ball just after the launch is 17.3m/s.

4 0

3 years ago

Suppose you performed the experiment in atmosphere of Argon at 25 deg. C, (viscosity of argon is 2.26X10^-5 N.s/m^2 at that temp

yuradex [85]

Answer:

2*10^9electrons

Explanation:

Remember that the net force will be zero at terminal voltege so

Mg = 6πrng

At 35v

We have

qvr = 6πrng

q= 6 x 3.142* nx 2.6*10^-5/35

q,= 3.2x 10^ - 10C

So using n= q/e

= 3.2x 10^ - 10C/1.6*10-19

= 2*10^9electrons

7 0

3 years ago

9. How much work is done when a 15kg box is lifted to a height of 2 meters?

torisob [31]

Answer: W = 294 J

Explanation: Solution:

Work is expressed as the product of force and the distance of the object.

W = Fd where F = mg

W= Fd

= mg d

= 15 kg ( 9.8 m/s²) ( 2m )

= 294 J

5 0

3 years ago

A saturated solution is formed by keeping 48 gram of a salt in 200 gram of water at 35°C. calculate the solubility of the salt.

viva [34]

The solubility of the salt. : 24 g/100 g solvent(water)

<h3>Further explanation</h3>

Given

48 gram of a salt in 200 gram of water at 35°C

Required

The solubility of the salt.

Solution

Solubility : the amount of a substance that can be dissolved in a solvent

Can be formulated(for 100 g solution)⇒grams solute per 100 g of solvent(100 ml water)

$\tt solubility=\dfrac{mass~solute}{mass~solvent}\times 100$

Input the value :

$\tt solubilty=\dfrac{48~}{200~g}\times 100=24~g$

4 0

3 years ago