All categories

3 years ago

5 points !!!is potential chemical energy always transformed into kinetic

1 answer:

5 0

Potential energy is energy stored in an object due to its position or arrangement. Kinetic energy is energy of an object due to its movement - its motion. All types of energy can be transformed into other types of energy.

You might be interested in

Suppose the coefficient of static friction between a quarter and the back wall of a rocket car is 0.383. At what minimum rate wo

djverab [1.8K]

Answer:

25.59 m/s²

Explanation:

Using the formula for the force of static friction:

$f_s = \mu_s N$ --- (1)

where;

$f_s =$ static friction force

$\mu_s =$ coefficient of static friction

N = normal force

Also, recall that:

F = mass × acceleration

Similarly, N = mg

here, due to min. acceleration of the car;

$N = ma_{min}$

From equation (1)

$f_s = \mu_s ma_{min}$

However, there is a need to balance the frictional force by using the force due to the car's acceleration between the quarter and the wall of the rocket.

Thus,

$F = f_s$

$mg = \mu_s ma_{min}$

$a_{min} = \dfrac{mg }{ \mu_s m}$

$a_{min} = \dfrac{g }{ \mu_s }$

where;

$\mu_s = 0.383$ and g = 9.8 m/s²

$a_{min} = \dfrac{9.8 \ m/s^2 }{0.383 }$

$\mathbf{a_{min}= 25.59 \ m/s^2}$

3 0

3 years ago

A 460 g , 6.0-cm-diameter can is filled with uniform, dense food. It rolls across the floor at 1.1 m/s . Part A What is the can'

Reika [66]

Answer:

the can's kinetic energy is 0.42 J

Explanation:

given information:

Mass, m = 460 g = 0.46 kg

diameter, d = 6 cm, so r = d/2 = 6/2 = 3 cm = 0.03 m

velocity, v = 1.1 m/s

the kinetic energy of the can is the total of kinetic energy of the translation and rotational.

KE = $\frac{1}{2}$ I ω^2 + $\frac{1}{2} mv^{2}$

where

I = $\frac{1}{2} mr^{2}$ and ω = $\frac{v}{r}$

thus,

KE = $\frac{1}{2}$ $\frac{1}{2} mr^{2}$ ( $\frac{v}{r}$ )^2 + $\frac{1}{2} mv^{2}$

= $\frac{1}{2}$ $\frac{1}{2} mr^{2}$ $\frac{v^{2} }{r^{2}}$ + $\frac{1}{2} mv^{2}$

= $\frac{1}{4} mv^{2}$ + $\frac{1}{2} mv^{2}$

= $\frac{3}{4} mv^{2}$

= $\frac{3}{4} (0.46) (1.1)^{2}$

= 0.42 J

8 0

3 years ago

Two astronauts on opposite ends of a spaceship are comparing lunches. One has an apple, the other has an orange. They decide to

lbvjy [14]

Answer:

Explanation:

We shall consider direction towards left as positive Let the required velocity be v and let v makes an angle φ

Applying law of conservation of momentum along direction of original motion

m₁ v₁ - m₂ v₂ = m₂v₃ - m₁ v₄

0.132 x 1.25 - .143 x 1.14 = 1.03 cos43 x .143 - v cos θ

v cos θ = .8

Applying law of conservation of momentum along direction perpendicular to direction of original motion

1.03 sin 43 x .143 = .132 x v sinθ

v sinθ = .76

squaring and adding

v² = .76 ² + .8²

v = 1.1 m /s

Tan θ = .76 / .8

θ = 44°

6 0

3 years ago

Briefly describe the characteristics of each soil horizon from the top layer to the bottom layer

inna [77]

<span>There is six horizen. 1. O Horizon - The top, organic layer of soil, 2. A Horizon - The layer called topsoil; 3. E Horizon - This layer is beneath the A Horizon and above the B Horizon. It is made up mostly of sand. 4. B Horizon - Also called the subsoil - this layer is beneath the E Horizon and above the C Horizon. 5. C Horizon - it's called regolith: the layer beneath the B Horizon and above the R Horizon. 6 R Horizon - this is last and the unweathered rock layer that is beneath all the other layers.</span>

7 0

4 years ago

What is the car's speed at the bottom of the dip?The passengers in a roller coaster car feel 50% heavier thantheir true weight a

Rashid [163]

Answer:

v = 14 m/s

Explanation:

given,

radius of dip = 40 m

The passengers in a roller coaster car feel 50% heavier than their true weight.

Apparent weight

$A = W + \dfrac{W}{2}$

$A =\dfrac{3W}{2}$

$A =\dfrac{3mg}{2}$

When the car is at the bottom, the weight will be acting downwards and the centripetal force will also be acting downward where as Normal force which is apparent weight will be acting in upward direction.

now,

$N = m g + \dfrac{mv^2}{r}$

$\dfrac{3mg}{2} = m g + \dfrac{mv^2}{r}$

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

$v = \sqrt{\dfrac{rg}{2}}$

$v = \sqrt{\dfrac{40\times 9.8}{2}}$

v = 14 m/s

8 0

3 years ago