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2 years ago

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A 1 kg ball is hung at the end of a rod 1 m long. If the system balances at a point on the rod one third of the distance from th

e end holding the mass, what is the mass of the rod?

1 answer:

Anarel [89]2 years ago

8 0

Answer:

2 kg

Explanation:

Assuming the rod's mass is uniformly distributed, the center of mass is at half the length.

Sum of the moments at the balance point:

-(Mg)(L/3) + (mg)(L/2 − L/3) = 0

(Mg)(L/3) = (mg)(L/2 − L/3)

(Mg)(L/3) = (mg)(L/6)

2M = m

M = 1 kg, so m = 2 kg.

The mass of the rod is 2 kg.

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What is a concentration?

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Answer/Explanation:

In chemistry, concentration is the abundance of a constituent divided by the total volume of a mixture. Several types of mathematical description can be distinguished: mass concentration, molar concentration, number concentration, and volume concentration.

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A fixed system of charges exerts a force of magnitude 22 N on a 4.0 C charge. The 4.0 C charge is replaced with a 8.0 C charge.

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Answer:

44 N

Explanation:

The electrostatic forces between two charges is given by:

$F=k\frac{q_1 q_2}{r^2}$

where

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q1 and q2 are the two charges

r is their separation

We notice that the force is directly proportional to the charges.

In this problem, initially we have a force of

F = 22 N

on a q2 = 4.0 C, exerted by a charge q1.

If the charge is doubled,

q2 = 8.0 C

This means that the force will also double, so it will be

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The football team is selling mulch as a fundraiser. Which two options are

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Suppose a tidal basin is 6 m above the ocean at low tide and that the area of the basin is 5×107 m2. Estimate the gravitational

Neko [114]

Answer:

Potential energy will be $176.58\times 10^{11}j$

Explanation:

We have given the height of the basin is h = 6 m

Area of the basin $A=5\times 10^7m^2$

Volume $V=area \times height=5\times 10^7\times 6=30\times 10^7m^3$

Density $\rho =1000kg/m^3$

We know that mass is given by $m=\rho V=1000\times 30\times 10^7=3\times 10^{11}kg$

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2 years ago

Two carts, one twice as heavy as the other, are at rest on a horizontal track. A person pushes each cart for 8 s. Ignoring frict

GalinKa [24]

Answer:

The correct option is: B that is 1/2 K

Explanation:

Given:

Two carts of different masses, same force were applied for same duration of time.

Mass of the lighter cart = $m$

Mass of the heavier cart = $2m$

We have to find the relationship between their kinetic energy:

Let the KE of cart having mass m be "K".

and KE of cart having mass m be "K1".

As it is given regarding Force and time so we have to bring in picture the concept of momentum Δp and find a relation with KE.

Numerical analysis.

⇒ $KE = \frac{mv^2}{2}$

⇒ $KE = \frac{mv^2}{2}\times \frac{m}{m}$

⇒ $KE = \frac{m^2v^2}{2}\times \frac{1}{m}$

⇒ $KE = \frac{(mv)^2}{2}\times \frac{1}{m}$

⇒ $KE = \frac{(\triangle p)^2}{2}\times \frac{1}{m}$

⇒ $KE = \frac{(\triangle p)^2}{2m}=\frac{(F\times t)^2}{2m}$

Now,

Kinetic energies and their ratios in terms of momentum or impulse.

KE (K) of mass m.

⇒ $K=\frac{(F\times t)^2}{2m}$ ...equation (i)

KE (K1) of mass 2m.

⇒ $K_1=\frac{(F\times t)^2}{2\times 2m}$

⇒ $K_1=\frac{(F\times t)^2}{4m}$ ...equation (ii)

Lets divide K1 and K to find the relationship between the two carts's KE.

⇒ $\frac{K_1}{K} =\frac{(F\times t)^2}{4m} \times \frac{2m}{(F\times t)^2}$

⇒ $\frac{K_1}{K} =\frac{2m}{4m}$

⇒ $\frac{K_1}{K} =\frac{2}{4}$

⇒ $\frac{K_1}{K} =\frac{1}{2}$

⇒ $K_1=\frac{K}{2}$

⇒ $K_1=\frac{1}{2}K$

The kinetic energy of the heavy cart after the push compared to the kinetic energy of the light cart is 1/2 K.

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