What's the relationship between potential energy and a position of a rock

Assuming no air resistance, all projectiles have:

Kitty [74]

Answer: Option C

C) accelerated vertical motion and constant horizontal motion.

Explanation:

If there is no air resistance then during the projectile movement the only force that causes an acceleration is the gravitational force.

We know that this force produces an acceleration of 9.8 m / s ^ 2 in the projectile.

As the gravitational force attracts the object towards the earth, then the acceleration that this force produces is always in the vertical direction. In the horizontal direction the object is not accelerated (because there is no air resistance).

Therefore the correct answer is option C.

"accelerated vertical motion and constant horizontal motion".

3 0

3 years ago

A 1.80-m -long uniform bar that weighs 531 N is suspended in a horizontal position by two vertical wires that are attached to th

Readme [11.4K]

Answer:

(a) 498.4 Hz

(b) 442 Hz

Solution:

As per the question:

Length of the wire, L = 1.80 m

Weight of the bar, W = 531 N

The position of the copper wire from the left to the right hand end, x = 0.40 m

Length of each wire, l = 0.600 m

Radius of the circular cross-section, R = 0.250 mm = $.250\times 10^{- 3}\ m$

Now,

Applying the equilibrium condition at the left end for torque:

$T_{Al}.0 + T_{C}(L - x) = W\frac{L}{2}$

$T_{C}(1.80 - 0.40) = 531\times \frac{1.80}{2}$

$T_{C} = 341.357\ Nm$

The weight of the wire balances the tension in both the wires collectively:

$W = T_{Al} + T_{C}$

$531 = T_{Al} + 341.357$

$T_{Al} = 189.643\ Nm$

Now,

The fundamental frequency is given by:

$f = \frac{1}{2L}\sqrt{\frac{T}{\mu}}$

where

$\mu = A\rho = \pi R^{2}\rho$

(a) For the fundamental frequency of Aluminium:

$f = \frac{1}{2L}\sqrt{\frac{T_{Al}}{\mu}}$

$f = \frac{1}{2L}\sqrt{\frac{T_{Al}}{\pi R^{2}\rho_{Al}}}$

where

$\rho_{l} = 2.70\times 10^{3}\ kg/m^{3}$

$f = \frac{1}{2\times 0.600}\sqrt{\frac{189.643}{\pi 0.250\times 10^{- 3}^{2}\times 2.70\times 10^{3}}} = 498.4\ Hz$

(b) For the fundamental frequency of Copper:

$f = \frac{1}{2L}\sqrt{\frac{T_{C}}{\mu}}$

$f = \frac{1}{2L}\sqrt{\frac{T_{C}}{\pi R^{2}\rho_{C}}}$

where

$\rho_{C} = 8.90\times 10^{3}\ kg/m^{3}$

$f = \frac{1}{2\times 0.600}\sqrt{\frac{341.357}{\pi 0.250\times 10^{- 3}^{2}\times 2.70\times 10^{3}}} = 442\ Hz$

7 0

4 years ago

One part of a freely swinging magnet always points

Sladkaya [172]

One part of a freely swinging magnet always points to the Earth's magnetic pole in the Northern Hemisphere.

8 0

4 years ago

Read 2 more answers

Will a pair of parallel current-carrying wires exert forces on each other? 1. yes; the wires are electrically charged. 2. no; th

Naddika [18.5K]

Number 3.

When you have two parallel wire and the charges are moving, they create a magnetic field around the wire (right-hand rule, thumb points in the direction of current).

If you want the math, here ya go:

$\frac{F}{L} = \frac{4 \pi *10^{-7}II' }{2 \pi r}$

F: magnetic force on the wire, L: length of the wire I: current in one wire I': current in other wire r: distance of the wire.

As we can see, two wires that has current through it will generate a force on each other.

7 0

4 years ago

Air friction damps a tuning fork so the amplitude decreases by 1/10 per second. Resonances] 10 Hz. what % chnge in frequency res

aleksandr82 [10.1K]

Answer:

The percentage change in frequency is 10%.

Explanation:

Given that,

Amplitude decreases by 1/10 per second.

Resonance = 10 Hz

We need to calculate the change in frequency

Using formula of change in frequency