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yanalaym [24]
3 years ago
11

An object with mass m is spinning in uniform circular motion in a circle perpendicular to the ground with radius r at a circular

velocity of 2 m/s. At the height of its spin, the forces acting on it, f, can be calculated by summing the force due to change in direction and the force due to gravity. The force due to gravity is the mass of the object multiplied by 9.8 m/s2s2. The force due to change in direction is the product of the square of the velocity of the object, the mass of the object, and the reciprocal of the radius. What is m in terms of r and f?
A) f.r4+9.8rf.r4+9.8r
B) f4r+9.8f4r+9.8
C) f.r4r+9.8f.r4r+9.8
D) 4r+9.8f4r+9.8f
E) (f.r)−49.8r
Physics
1 answer:
LekaFEV [45]3 years ago
6 0

Answer:

m = f*r / (9.8*r + 4)

Explanation:

Given

f = (m*g) + (m*v²/r)

⇒  f = m*(g + (v²/r)) = (m/r)*(g*r + v²)

⇒  m = f*r / (g*r + v²)

⇒  m = f*r / (9.8*r + 2²)

⇒  m = f*r / (9.8*r + 4)

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sukhopar [10]

Answer:

The answer is B, contact!

Explanation:

The two objects are contacting (or interacting) each other.

6 0
3 years ago
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If for every action force an equal and opposite reaction force exists, how can anything ever be accelerated?
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If there are two equal and opposite forces on the SAME THING, then the thing doesn't accelerate. You're right about that. But the action and reaction forces act on two different things. The bullet and the rifle. The ball and the bat. The airplane and the air. etc. So BOTH can accelerate.
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4 years ago
Cart A, with a mass of 0.20 kg, travels on a horizontal air trackat 3.0m/s and hits cart B, which has a mass of 0.40 kg and is i
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Answer:

\large \boxed{\text{C. 2.3 m/s}}

Explanation:

Data:

m_{\text{A} } = \text{0.20 kg};\,v_{\text{Ai}} = \text{3.0 m/s}\\m_{\text{B} } = \text{0.40 kg};\,v_{\text{Bi}} = \text{2.0 m/s}\\

Calculation:

This is a perfectly inelastic collision.  The two carts stick together after the collision and move with a common final velocity.

The conservation of momentum equation is

\begin{array}{rcl}m_{\text{A}}v_{\text{Ai}} +m_{\text{B}} v_{\text{Bi}}&=&(m_{\text{A}}  + m_{\text{B}})v_{\text{f}}\\0.20\times 3.0 + 0.40\times 2.0 & = & (0.20 + 0.40)v_{\text{f}}\\0.60 + 0.80 & = & 0.60v_{\text{f}}\\1.40 & = & 0.60v_{\text{f}}\\v_{\text{f}}&=& \dfrac{1.40}{0.60}\\\\& = & \textbf{2.3 m/s}\\\end{array}\\\text{The centre of mass has a velocity of $\large \boxed{\textbf{2.3 m/s}}$}

3 0
4 years ago
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sergij07 [2.7K]

Answer:

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Explanation:

We can calculate the temperature change of each substance by using the equation:

\Delta T=\frac{Q}{mC_s}

where

Q = 200.0 J is the heat provided to the substance

m is the mass of the substance

C_s is the specific heat of the substance

Let's apply the formula for each substance:

A) m = 50.0 g, Cs = 0.903 J/g°C

\Delta T=\frac{200}{(50)(0.903)}=4.4^{\circ}C

B) m = 50.0 g, Cs = 0.385 J/g°C

\Delta T=\frac{200}{(50)(0.385)}=10.4^{\circ}C

C) m = 25.0 g, Cs = 0.79 J/g°C

\Delta T=\frac{200}{(25)(0.79)}=10.1^{\circ}C

D) m = 25.0 g, Cs = 0.128 J/g°C

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E) m = 25.0 g, Cs = 0.235 J/g°C

\Delta T=\frac{200}{(25)(0.235)}=34.0^{\circ}C

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