The part of the computer that provides access to the internet is

The gravitional energy of an object is always measured realative to the ?

zaharov [31]

I would say that the answer would be MASS.

4 0

2 years ago

A 5.45-g combustible sample is burned in a calorimeter. the heat generated changes the temperature of 555 g of water from 20.5°c

Y_Kistochka [10]

Given:
m = 555 g, the mass of water in the calorimeter
ΔT = 39.5 - 20.5 = 19 °C, temperature change
c = 4.18 J/(°C-g), specific heat of water

Assume that all generated heat goes into heating the water.
Then the energy released is
Q = mcΔT
= (555 g)*(4.18 J/(°C-g)*(19 °C)
= 44,078.1 J
= 44,100 J (approximately)

Answer: 44,100 J

3 0

2 years ago

A 65.0 kg diver is 4.90 m above the water, falling at speed of 6.40 m/s. Calculate her kinetic energy as she hits the water. (Ne

mojhsa [17]

Answer:

4452.5 J.

Explanation:

The diver have both kinetic and potential energy.

Ek = 1/2mv² ................. Equation 1

Where Ek = Kinetic Energy of the diver, m = mass of the diver, v = velocity of the diver.

Given: m = 65 kg, v = 6.4 m/s.

Substitute into equation 1

Ek = 1/2(65)(6.4²)

Ek = 1331.2 J.

Also,

Ep = mgh ............................ Equation 2

Where Ep = Potential energy of the diver when its above the water, h = height of the diver above the water, g = acceleration due to gravity.

Given: m = 65 kg, h = 4.9 m, g = 9.8 m/s²

Substitute into equation 2.

Ep = 65(4.9)(9.8)

Ep = 3121.3 J.

Note: When she hits the water, the potential energy is converted to kinetic energy.

E = Ek+Ep

Where E = Kinetic energy of the diver when she hits the water.

E = 1331.2+3121.3

E = 4452.5 J.

3 0

3 years ago

Can anyone help me with these questions? TIA!<br> (Don’t actually answer please! :) )

nataly862011 [7]

<h2>Hey There!</h2><h2>_____________________________________</h2><h2>Question 7: </h2>

$\huge\text{Graphs:}$

The graph of

• The I-V for Ohmic Metal wire conductor at constant temperature always shows a straight line between the Current(I) plotted at Y axis and Voltage(V) plotted at X axis. Picture 1

• The I-V graph for Diode shows that first the current is zero but as we increase the potential difference(voltage), it results in the increase in the current. Picture 2

<h2>_____________________________________ </h2><h2>Question 8: </h2>

$\Large\textbf{Diode:}$

A diode is a device that allows current to flow in only one direction.

$\Large\textbf{Forward and Reverse Biasing:}$

Forward Bias, When a diode is forward bias (a voltage in the "forward" direction) then the P-side of the diode is attached to the positive terminal and N-side is fixed to the negative side of the battery which is connected, current flows freely through the device. The forward bias decreases the thickness of potential barrier(The potential barrier barrier in which the charge requires additional force for crossing the region)

Reverse Bias, When a diode is Reverse bias(a voltage in the "backward direction) then the P-side of the diode is connected to the negative terminal and N-side is connected to the positive terminal of the battery which is connected. The reverse bias increases the thickness of the potential barrier resulting in the flow of no current.

$\Large\textbf{Answer to the Question "Resistance"}$

The Forward bias decreases the resistance of the diode whereas the reversed bias increases the resistance of the diode. As in forward biasing the current is easily flowing through the circuit whereas reverse bias does not allow the current to flow through it.

<h2>_____________________________________ </h2><h2>Best Regards, </h2><h2>'Borz' </h2>

8 0

2 years ago

On a frictionless horizontal air table, puck A (with mass 0.254 kg ) is moving toward puck B (with mass 0.367 kg ), which is ini

irinina [24]

Answer:

$v_a=0.8176 m/s$

$\Delta K=0.07969 J - 0.0849 J = -0.00521 J$

Explanation:

According to the law of conservation of linear momentum, the total momentum of both pucks won't be changed regardless of their interaction if no external forces are acting on the system.

Being $m_a$ and $m_b$ the masses of pucks a and b respectively, the initial momentum of the system is