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Advocard [28]
3 years ago
11

Steel is an alloy of _____. One of the main reasons steel is used more widely than iron is because it’s_____ than iron. thanks f

or helping
Physics
1 answer:
yarga [219]3 years ago
8 0

1. Iron I think and 2. stronger

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A 2.5 kg tribble is placed in a bucket and whirled in a 1.4 m radius vertical circle at a constant tangential speed. If the forc
Over [174]

Given that,

Mass of a tribble, m = 2.5 kg

Radius, r = 1.4 m

The force on the tribble from the bucket does not exceed 10 times its weight.

To find,

The maximum tangential speed.

Solution,

The force acting on the tribble is equal to the centripetal force.

F = 10mg

The formula for the centripetal force is given by :

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

v is maximum tangential speed

v=\sqrt{\dfrac{Fr}{m}} \\\\v=\sqrt{\dfrac{mgr}{m}} \\\\v=\sqrt{{10gr}} \\\\v=\sqrt{10\times 9.8\times 1.4} \\\\v=11.7\ m/s

So, the maximum tangential speed is 11.7 m/s.

8 0
3 years ago
The plates of a spherical capacitor have radii 6.25 cm and 15.0 crn. The space between the two spheres is filled with a material
aleksklad [387]

Answer:

Capacitance is 0.572×10⁻¹⁰ Farad

Explanation:

Radius = R₁ = 6.25 cm = 6.25×10⁻² m

Radius = R₂ = 15 cm = 15×10⁻² m

Dielectric constant = k = 4.8

Electric constant = ε₀ = 8.854×10⁻¹² F/m

ε/ε₀=k

ε=kε₀

Capacitance\ (C)=\frac{4\pi k\epsilon_0 R_1\times R_2}{R_2-R_1}\\\Rightarrow C=\frac{4\pi 4.8\times 8.854\times 10^{-12}\times 15\times 10^{-2}\times 6.25\times 10^{-2}}{15\times 10^{-2}-6.25\times 10^{-2}}\\\Rightarrow C=0.572\times 10^{-10}\ Farad

∴ Capacitance is 0.572×10⁻¹⁰ Farad

3 0
3 years ago
Which of the following types of rocks could contain organic matter
Furkat [3]
Pet rocks contain organic matter
7 0
3 years ago
Does potential energy increase,kinetic energy decrease when a book is placed on a shelf
DerKrebs [107]
Yes potential increases while kinetic decreases
3 0
3 years ago
Consider the two moving boxcars in Example 5. Car 1 has a mass of m1 = 65000 kg and a velocity of v01 = +0.80 m/s. Car 2 has a m
Amiraneli [1.4K]

Answer:

1.034m/s

Explanation:

We define the two moments to develop the problem. The first before the collision will be determined by the center of velocity mass, while the second by the momentum preservation. Our values are given by,

m_1 = 65000kg\\v_1 = 0.8m/s\\m_2 = 92000kg\\v_2 = 1.2m/s

<em>Part A)</em> We apply the center of mass for velocity in this case, the equation is given by,

V_{cm} = \frac{m_1v_1+m_2v_2}{m_1+m_2}

Substituting,

V_{cm} = \frac{(65000*0.8)+(92000*1.2)}{92000+65000}

V_{cm} = 1.034m/s

Part B)

For the Part B we need to apply conserving momentum equation, this formula is given by,

m_1v_1+m_2v_2 = (m_1+m_2)v_f

Where here v_f is the velocity after the collision.

v_f = \frac{m_1v_1+m_2v_2}{m_1+m_2}

v_f = \frac{(65000*0.8)+(92000*1.2)}{92000+65000}

v_f = 1.034m/s

8 0
3 years ago
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