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Ksju [112]
3 years ago
11

A 2.50 kg ball is attached to a 3.00 m bar and swung in a vertical circle. If the ball does not leave the circular loop, what mi

nimum speed must it have at the top of the arc?
Physics
1 answer:
dezoksy [38]3 years ago
6 0

Answer:

5.42 m/s

Explanation:

At minimum speed, the tension in the bar will be 0 when the ball is at the top of the arc, so the only force is gravity pulling down.

Sum of forces towards the center of the circle:

∑F = ma

mg = m v²/r

v = √(gr)

v = √(9.8 m/s² × 3.00 m)

v = 5.42 m/s

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Used an app called Mephyso to do the calculation.

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A 0.25 kg skeet (clay target) is fired at an angle of 30 degrees to the horizon with a speed of 25 m/s. When it reaches the maxi
kozerog [31]

Answer:

6.51 m and 37.13 m

Explanation:

from the question we were given

mass of skeet = 0.25 kg

speed of skeet = 25 m/s

angle = 30 degrees to the horizon

mass of pellet = 15 g

speed of pellet = 2000 m/s

without being hit by the pellet, the x and y components of the skeet velocity are  

Vx = 25 cos 30 = 21.65

Vy = 25 sin 30 = 12.5

now

V = U + (a x t)

where V = final velocity, U = initial velocity , a = acceleration, t = time and s = distance

-25 sin 30 = 25 sin 30 + (-9.8 x t)

-12.5 = 12.5 - 9.8 t

t = 2.55 s

also

V^2 = U^2 + 2as  ( s = vertical distance and V = 0 )

0 = (25 sin 30)^2 + 2 x (-9.8) x Y

19.6 Y = 156.25

Y =7.97 m

the distance traveled without the pellet hitting the skeet can be gotten using distance = speed x time

distance = 21.65 x 2.55 = 55.2 m

applying the conservation of linear momentum

on the x axis : (Ms x Us) + (Mp x Up) = (Ms x Vx) + (Mp x Vx)  ...equation 1

on the y axis :   (Ms x Us) + (Mp x Up) = (Ms x Vy) + (Mp x Vy) ...equation 2

(0.25 x 25 cos 30) + 0 = (0.25 +0.015) Vx

 Vx = 20.42m/s

0 + (0.015 x 200) = (0.25 + 0.015) Vy

 Vy = 11.32 m/s

now V^2 = U^2 + 2 as

 0 = 11.3^2 + (2 x (-9.8) x s)

s = 6.51 m                          

  • to find the extra distance moved after collision we apply

s = ut + 1/2 at^2

-7.98 = 11.32t + 1/2 (-9.8) t^2

4.9 t^2 - 11.32t + 7.98

t =  3.17 s

recall that distance = speed x time

distance = 20.42 x 3.17 = 64.73 m

the distance of the skeet before being hit would be half of the distance it travels without being hit, this is because the skeet was hit at its maximum height = 55.2 /2

= 27.6 m

therefore the extra distance traveled would be the change in distance = 64.73 -27.6 = 37.13 m

5 0
3 years ago
Plz give me the answer of qiestion ofno 31
77julia77 [94]
The resistance between A and B is 10 ohms.
5 0
3 years ago
A square is 1.0 m on a side. Point charges of +4.0 μC are placed in two diagonally opposite corners. In the other two corners ar
finlep [7]

Answer:

<em>B) 1.0 × 10^5 V</em>

Explanation:

<u>Electric Potential Due To Point Charges </u>

The electric potential produced from a point charge Q at a distance r from the charge is

\displaystyle V=k\frac{Q}{r}

The total electric potential for a system of point charges is equal to the sum of their individual potentials. This is a scalar sum, so direction is not relevant.

We must compute the total electric potential in the center of the square. We need to know the distance from all the corners to the center. The diagonal of the square is

d=\sqrt2 a

where a is the length of the side.

The distance from any corner to the center is half the diagonal, thus

\displaystyle r=\frac{d}{2}=\frac{a}{\sqrt{2}}

\displaystyle r=\frac{1}{\sqrt{2}}=0.707\ m

The total potential is  

V_t=V_1+V_2+V_3+V_4

Where V1 and V2 are produced by the +4\mu C charges and V3 and V4 are produced by the two opposite charges of \pm 3\mu\ C. Since all the distances are equal, and the charges producing V3 and V4 are opposite, V3 and V4 cancel each other. We only need to compute V1 or V2, since they are equal, but they won't cancel.

\displaystyle V_1=V_2=k\frac{Q}{r}=9\times 10^9 \frac{4\times 10^{-6}}{0.707}

V_1=V_2=50912\ V

The total potential is

V_t=50912\ V+50912\ V=1\times 10^5\ V

\boxed{V_t=1\times 10^5\ V}

6 0
3 years ago
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