When one object pushes or pulls another object, the first object is_______________________

Kobotan [32]

Its an electrochemical cell that derives electrical energy from spontaneous redox reactions taking place within the cell.

8 0

3 years ago

A ray of red light in air is incident at an angle of 30. on a

Vlad [161]

Answer:

20 degrees.

Explanation:

From Snell’s law of refraction:

sinθ1•n1 = sinθ2•n2

where θ1 is the incidence angle, θ2 is the refraction angle, n1 is the refraction index of light in medium1, and n2 is the refraction index for virgin olive oil. The incidence angle of the red light is θ1 = 30 degrees.

The red light is in air as medium1, so n1 (air) = 1.00029

So, to find θ2, the refracted angle:

sinθ1•1.00029 = sinθ2•1.464

sin(30)•1.00029 / 1.464 = sinθ2

0.5•1.00029 / 1.464 = sinθ2

sinθ2 = 0.3416291

θ2 = arcsin(0.3416291)

θ2 = 19.976 degrees

To the nearest degree,

θ2 = 20 degrees.

8 0

3 years ago

You pull straight up on the string of a yo-yo with a force 0.35 N, and while your hand is moving up a distance 0.16 m, the yo-yo

jarptica [38.1K]

Answer:

a) 0.138J

b) 3.58m/S

c) (1.52J)(I)

Explanation:

a) to find the increase in the translational kinetic energy you can use the relation

$\Delta E_k=W=W_g-W_p$

where Wp is the work done by the person and Wg is the work done by the gravitational force

By replacing Wp=Fh1 and Wg=mgh2, being h1 the distance of the motion of the hand and h2 the distance of the yo-yo, m is the mass of the yo-yo, then you obtain:

$Wp=(0.35N)(0.16m)=0.056J\\\\Wg=(0.062kg)(9.8\frac{m}{s^2})(0.32m)=0.19J\\\\\Delta E_k=W=0.19J-0.056J=0.138J$

the change in the translational kinetic energy is 0.138J

b) the new speed of the yo-yo is obtained by using the previous result and the formula for the kinetic energy of an object:

$\Delta E_k=\frac{1}{2}mv_f^2-\frac{1}{2}mv_o^2$

where vf is the final speed, vo is the initial speed. By doing vf the subject of the formula and replacing you get:

$v_f=\sqrt{\frac{2}{m}}\sqrt{\Delta E_k+(1/2)mv_o^2}\\\\v_f=\sqrt{\frac{2}{0.062kg}}\sqrt{0.138J+1/2(0.062kg)(2.9m/s)^2}=3.58\frac{m}{s}$

the new speed is 3.58m/s

c) in this case what you can compute is the quotient between the initial rotational energy and the final rotational energy

$\frac{E_{fr}}{E_{fr}}=\frac{1/2I\omega_f^2}{1/2I\omega_o^2}=\frac{\omega_f^2}{\omega_o^2}\\\\\omega_f=\frac{v_f}{r}\\\\\omega_o=\frac{v_o}{r}\\\\\frac{E_{fr}}{E_{fr}}=\frac{v_f^2}{v_o^2}=\frac{(3.58m/s)}{(2.9m/s)^2}=1.52J$

hence, the change in Er is about 1.52J times the initial rotational energy

5 0

3 years ago

Read 2 more answers

A stone is thrown straight upward and reaches a maximum height of 31.8 m above its

atroni [7]

Answer:

aral ka muna ng mabuti para maintindihan mo

7 0

3 years ago

A capacitor with a capacitance of 50µf when connected to a battery of 400 V. The charge and energy stored on it is? a. 0.05 C an

nekit [7.7K]

Answer:

c. 0.02 C and 4 J

Explanation:

Applying,

Q = CV................ Equation 1

Where Q = Charge, C = Capacitance of the capacitor, V = Voltage.

From the question,

Given: C = 50 μF = 50×10⁻⁶ F, V = 400 V

Substitute these values into equation 1

Q = (50×10⁻⁶)(400)

Q = 0.02 C.

Also Applying

E = CV²/2............. Equation 2

Where E = Energy stored.

Therefore,

E = (50×10⁻⁶ )(400²)/2

E = 4 J

Hence the right option is c. 0.02 C and 4 J

3 0

3 years ago

When one object pushes or pulls another object, the first object is________________________ a force on the second object.