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Evgen [1.6K]
3 years ago
9

EX 6-1 A ball is twirled on a 0.870 - m-long string with a constant speed of 3.36 m / s . Calculate the acceleration of the ball

. Be sure to specify the direction of the acceleration.
Physics
1 answer:
Elina [12.6K]3 years ago
4 0

Answer:

a=12.97\ m/s^2

Explanation:

Given that,

The length of a string, l = 0.87 m

Speed of the ball, v = 3.36 m/s

We need to find the acceleration of the ball. The acceleration acting on the ball is centripetal acceleration. It is given by :

a=\dfrac{v^2}{r}\\\\a=\dfrac{(3.36)^2}{0.87}\\\\=12.97\ m/s^2

So, the acceleration of the ball is 12.97\ m/s^2.

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You are trying to find out how high you have to pitch a water balloon in order for it to burst when it hits the ground. You disc
FrozenT [24]

Answer:

The balloon hit the ground with velocity -15.34 m/s

Explanation:

<em>Lets explain how to solve the problem</em>

You found that the best height to pitch a water balloon in order for it to

burst when it hits the ground is 12 meters.

We consider that the 12 meters is the maximum height, so the velocity

at this height is zero.

To find the velocity when the balloon hits the ground lets use the rule

<em>v² = u² + 2gh</em>, where v is the final velocity, u is the initial velocity, g is

the acceleration of gravity and h is the height.

u = 0 , h = 12 m , g = 9.8 m/s²

<em>Substitute these values in the equation above</em>

v² = 0 + 2(9.8)(12)

v² = 235.2

<em>Take square root for both sides</em>

v = ± \sqrt{235.2}

The velocity is downward, then it's a negative value

Then v = -15.34 m/s

<em>The balloon hit the ground with velocity -15.34 m/s</em>

6 0
3 years ago
Plss help
Genrish500 [490]

Answer:

b

Explanation:

imagine urself on an elevator dont you feel lighter

3 0
3 years ago
1.A body of certain mass is kept at a height of 15m from the ground, taking the value of g as 10m/s find out the mass of the bod
ehidna [41]

Answer:

1.#potential energy = PE, m = mass in kg, g = force of gravity, h= vertical height above the ground.  ** means to the power of ie exponent. * means multiply.

PE = mgh

300 = m(10)(15)

m = 300/(10)(15)

m= 2kg

2. KE = 1/2 mv**2

     = 1/2(50)(50)**2

     = 2500 joules

Explanation

Is as in solution

4 0
3 years ago
A motorist traveling with a constant speed of 15 m/s (about 34 mi/h) passes a school-crossing corner, where the speed limit is 0
nikklg [1K]

Answer:

(a) 10 s

(b) 30 m/s

(c) 150 m

Explanation:

The motorist's position at time t is:

x = 15t

The officer's position at time t is:

x = ½ (3) t² = 1.5 t²

(a) When they have the same position, the time is:

15t = 1.5 t²

t = 0 or 10 s

(b) The officer's speed is:

v = 3t

v = 30 m/s

(c) The position is:

x = 15t = 150 m

6 0
3 years ago
a system of four particles moves along one dimension. the center of mass of the system is at rest, and the particles do not inte
Zinaida [17]

The velocity of the particle 4 at time, t = 2.83 s, is -14.1 m/s.

The given parameters;

m1 = 1.45 kg, v1(t) = (6.09m/s) + (0.299m/s^2) × t

m2 = 2.81 kg, v2(t) = (7.83m/s) + (0.357m/s^2) × t

m3 = 3.89 kg, v3(t) = (8.09m/s) + (0.405m/s^2) × t

m4 = 5.03kg

The velocity of the center mass of the particles is calculated as;

McmVcm = m1v1 + m2v2 +m3v3+m4v4

Vcm= m1v1 + m2v2 +m3v3 +m4v4/ Mcm

0 = m1v1 + m2v2 +m3v3 +m4v4/ Mcm

m1v1 + m2v2 +m3v3+m4v4 = 0

m4v4 = -(m1v1 + m2v2 +m3v3)

v4 =-(m1v1 + m2v2 +m3v3)/ m4

The velocity of particle 1 at time, t = 2.83 s;

vi = 6.09 + 0.299× 2.83

v1 = 6.94 m/s

The velocity of particle 3 at time, t = 2.83 s;

v2 = 7.83 + 0.357 × 2.83

v2 = 8.84 m/s

The velocity of particle 3 at time, t = 2.83 s;

v3 = 8.09 + 0.405 × 2.83

v3 = 9.24 m/s

The velocity of particle 3 at time, t = 2.83 s;

v4 = - (m1v1 + m2v2 + m3v3)/m4

v4 = -(1.45×6.94 + 2.81×8.84 + 3.89×9.24)/5.03

v4 = -14.4 m/s

Thus, the velocity of the particle 4 at time, t = 2.83 s, is -14.1 m/s.

Learn more about Velocity here:

brainly.com/question/18084516

#SPJ4

5 0
1 year ago
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