Gravity has traditionally been defined as a force of attraction between things that have what

The skateboarder has a mass of 100 kg. When traveling downward from E to D, he reaches a velocity of 11 m/s. Calculate his kinet

patriot [66]

6050 J is the kinetic energy at D

<u>Explanation:</u>

In physics, the object's kinetic energy (K.E) defined as the energy it possesses during movement. It can be defined as the required work to accelerate a certain body weight in order to rest at a certain speed. When the body receives this energy as it speeds up (accelerates), it retains this energy unless speed varies. The equation is given as,

$K . E=\frac{1}{2} \times m \times v^{2}$

Where,

m - mass of an object

v - velocity of the object

Here,

Given data:

m = 100 kg

v = 11 m/s

By substituting the given values in the above equation, we get

$K . E=\frac{1}{2} \times 100 \times(11)^{2}=\frac{1}{2} \times 100 \times 121=\frac{12100}{2}=6050\ \mathrm{J}$

6 0

3 years ago

A thin rod of length 1.4 m and mass 140 g is suspended freely from one end. It is pulled to one side and then allowed to swing l

m_a_m_a [10]

Answer:

a The kinetic energy is $KE = 0.0543 J$

b The height of the center of mass above that position is $h = 1.372 \ m$

Explanation:

From the question we are told that

The length of the rod is $L = 1.4m$

The mass of the rod $m = 140 = \frac{140}{1000} = 0.140 \ kg$

The angular speed at the lowest point is $w = 1.09 \ rad/s$

Generally moment of inertia of the rod about an axis that passes through its one end is

$I = \frac{mL^2}{3}$

Substituting values

$I = \frac{(0.140) (1.4)^2}{3}$

$I = 0.0915 \ kg \cdot m^2$

Generally the kinetic energy rod is mathematically represented as

$KE = \frac{1}{2} Iw^2$

$KE = \frac{1}{2} (0.0915) (1.09)^2$

$KE = 0.0543 J$

From the law of conservation of energy

The kinetic energy of the rod during motion = The potential energy of the rod at the highest point

Therefore

$KE = PE = mgh$

$0.0543 = mgh$

$h = \frac{0.0543}{9.8 * 0.140}$

$h = 1.372 \ m$

5 0

2 years ago

Read 2 more answers

HELP ME PLEASE! according ohm's law if the resistance remains constant what would happen if you decrease the amount of voltage b

SpyIntel [72]

B.
technically it would depend if the resistors were in series or parallel but B is the answer.

7 0

3 years ago

How much energy is stored in a 2.80-cm-diameter, 14.0-cm-long solenoid that has 200 turns of wire and carries a current of 0.800

ozzi

Answer:

The energy stored in the solenoid is 7.078 x 10⁻⁵ J

Explanation:

Given;

diameter of the solenoid, d = 2.80 cm

radius of the solenoid, r = d/2 = 1.4 cm

length of the solenoid, L = 14 cm = 0.14 m

number of turns, N = 200 turns

current in the solenoid, I = 0.8 A

The cross sectional area of the solenoid is given as;

$A = \pi r^2\\\\A = \pi (0.014)^2\\\\A = 6.16*10^{-4} \ m^2$

The inductance of the solenoid is given by;

$L = \frac{\mu_0 N^2A}{l} \\\\L = \frac{(4\pi*10^{-7})(200^2)(6.16*10^{-4})}{0.14}\\\\L = 2.212*10^{-4} \ H$

The energy stored in the solenoid is given by;

E = ¹/₂LI²

E = ¹/₂(2.212 x 10⁻⁴)(0.8)²

E = 7.078 x 10⁻⁵ J

Therefore, the energy stored in the solenoid is 7.078 x 10⁻⁵ J

8 0

3 years ago

A spring has a equilibrium length of 10.0 cm. When a force of 40.0 N is applied to the spring, the spring has a length of 14.0 c

mote1985 [20]

Answer:

The value of the spring constant of this spring is 1000 N/m

Explanation:

Given;

equilibrium length of the spring, L = 10.0 cm

new length of the spring, L₀ = 14 cm

applied force on the spring, F = 40 N

extension of the spring due to applied force, e = L₀ - L = 14 cm - 10 cm = 4 cm

From Hook's law

Force applied to a spring is directly proportional to the extension produced, provided the elastic limit is not exceeded.

F ∝ e

F = ke

where;

k is the spring constant

k = F / e

k = 40 / 0.04

k = 1000 N/m

Therefore, the value of the spring constant of this spring is 1000 N/m

7 0

3 years ago