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3 years ago

11

A 6000 kg roller coaster goes around a loop of radius 30 m at 6 m/s. What is the centripetal acceleration?

1 answer:

Margarita [4]3 years ago

7 0

Answer:

The answer to the question is 7200

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As the temperature increases the number of effective collisions between reacting particles in a chemical reaction

Reil [10]

Answer: option A) initially increases, then decreases.

Justification:

The increase of the rate of effective collisions among particles as the temperature increases is explained by the collision theory in virtue of the increase of the kinetic energy.

This is, as the temperature increase so the kinetic energy increase and the higher the kinetic energy the greater the number of collisions and the greater the chances that this energy overcome the activation energy (the energy needed to start the reaction).

Now, as the reaction progress the number of reactants particles naturally decrease (some of them have been converted into product) so this lower number of particles means lower concentration which means lower collisions and, thereafter, a decrease in the reaction rate.

8 0

3 years ago

A mass is tied to a string and swung in a horizontal circle with a constant angular speed. show answer No Attempt If this speed

Liono4ka [1.6K]

Answer:

The tension in the string is quadrupled i.e. increased by a factor of 4.

Explanation:

The tension in the string is the centripetal force. This force is given by

$F = \dfrac{mv^2}{r}$

m is the mass, v is the velocity and r is the radius.

It follows that $F \propto v^2$ , provided m and r are constant.

When v is doubled, the new force, $F_1$ , is

$F_1 = \dfrac{m(2v)^2}{r} = \dfrac{4mv^2}{r} = 4\dfrac{mv^2}{r} = 4F$

Hence, the tension in the string is quadrupled.

8 0

3 years ago

7nadin3 [17]

Answer:

2.5 m/s²

Explanation:

Using the formula, v = u + at ( v = Final velocity; u = Initial velocity; t = Time; a = Acceleration)

25 = 0 + 10a

a = 25/10 = 2.5 m/s²

8 0

3 years ago

A thin uniform rod (length = 1.2 m, mass = 2.0 kg) is pivoted about a horizontal, frictionless pin through one end of the rod. (

Anika [276]

Answer:

$a=9.8 rad/s^{2}$

Explanation:

Torque, $\tau$ is given by

$\tau=Fr$ where F is force and r is perpendicular distance

$R=0.5Lcos\theta$ where $\theta$ is the angle of inclination

Torque, $\tau$ can also be found by

$\tau=Ia$ where I is moment of inertia and a is angular acceleration

Therefore, Fr=Ia and F=mg where m is mass and g is acceleration due to gravity

Making a the subject, $a=\frac {Fr}{I}=\frac {mgr}{I}$ and already I is given as

$I=\frac {mL^{2}}{3} and r is 0.5Lcos\theta$ hence

$a=\frac {0.5mgLcos\theta}{1/3 mL^{2}}$

$a=\frac {3gcos\theta}{2L}$

Taking g as 9.81, $\theta$ is given as 37 and L is 1.2

$a=\frac {3*9.81cos37}{2*1.2}=9.7932679419$

$a=9.8 rad/s^{2}$

4 0

3 years ago

A tennis ball is dropped from a roof. If it takes 38.9 seconds to reach the ground, how fast is the ball moving just before it h

Ilya [14]

Answer:

The velocity of the ball before it hits the ground is 381.2 m/s

Explanation:

Given;

time taken to reach the ground, t = 38.9 s

The height of fall is given by;

h = ¹/₂gt²

h = ¹/₂(9.8)(38.9)²

h = 7414.73 m

The velocity of the ball before it hits the ground is given as;

v² = u² + 2gh

where;

u is the initial velocity of the on the root = 0

v is the final velocity of the ball before it hits the ground

v² = 2gh

v = √2gh

v = √(2 x 9.8 x 7414.73 )

v = 381.2 m/s

Therefore, the velocity of the ball before it hits the ground is 381.2 m/s

5 0

2 years ago