Identify the false statement: Select one:

A person weighing 55 kg walks by applying 160 N of force on the ground, while pushing a 10-kg object. If the person accelerates

Sav [3153]

Answer:

I'm uncertain

Explanation:

since I didn't provide a correct answer, continue with my inquiries and you can use 'I'm uncertain' for 100 points.

6 0

3 months ago

In order for the ball to be able to make a complete circle around the peg, there must be sufficient speed at the top of its arc

kicyunya [3294]

Answer:

Explanation:

Let T represent the tension in the swing.

At the peak $mg-T=\frac{mv^2}{r}$

where v denotes the velocity needed to maintain the circular motion.

r equals the distance from the rotation point to the center of the ball, which is L+\frac{d}{2} (with d being the ball's diameter).

The threshold velocity can be expressed as $mg-0=\frac{mv^2}{r}$

To determine the velocity at the bottom, we can use energy conservation principles at both the top and bottom positions.

At the top $E_T=mg\times 2L+\frac{mv^2}{2}$

Energy at the bottom $E_b=\frac{mv_0^2}{2}$

By comparing the two states using conservation of energy, we find $v_0^2=4gr+gr$

$v_0^2=5gr$

$v_0=\sqrt{5gr}$

$v_0=\sqrt{5g\left ( \frac{d}{2}+L\right )}$

6 0

2 months ago

A uniformly charged spherical droplet of mercury has electric potential Vbig throughout the droplet. The droplet then breaks int

kicyunya [3294]

Answer:

$\frac{V_{big}}{V_{small}} = n^{2/3}$

Explanation:

Let the charge on the large droplet be denoted as Q.

When the radius of the droplet is R, the electric potential for the larger droplet can be expressed as:

$V_{big} = \frac{KQ}{R}$

If it splits into n identical droplets, let the charge of each be "q" and their radius be "r".

Applying volume conservation gives us:

$\frac{4}{3}\pi R^3 = n(\frac{4}{3}\pi r^3)$

$r = \frac{R}{n^{1/3}}$

Now, the potential for the smaller droplets is given as:

$V_{small} = \frac{kq}{r}$

$V_{small} = \frac{K(Q/n)}{\frac{R}{n^{1/3}}}$

$V_{small} = \frac{1}{n^{2/3}}\frac{KQ}{R}$

$\frac{V_{big}}{V_{small}} = n^{2/3}$

7 0

2 months ago

A (1.25+A) kg bowling ball is hung on a (2.50+B) m long rope. It is then pulled back until the rope makes an angle of (12.0+C)o

Ostrovityanka [3204]

Answer:

F = 0.535 N

Explanation:

We will apply energy concepts, considering both the peak and the bottom of the path.

Top

Em₀ = U = mg y

Bottom

$Em_{f}$ = K = ½ m v²

Emo = $Em_{f}$

mg y = ½ m v²

v = √ (2gy)

y = L - L cos θ

v = √ (2g L (1 - cos θ))

Next, we will employ Newton's second law at the lowest point where the acceleration is centripetal.

F = ma

a = v² / r

For the turning radius, the cable length is r = L.

F = m 2g (1 - cos θ)

Now, let's find the result.

F = 2 1.25 9.8 (1 - cos 12)

F = 0.535 N

7 0

2 months ago