Answer:
Explanation:
Some common conductors are copper, aluminum, gold, and silver. Some common insulators are glass, air, plastic, rubber, and wood.
Answer:
Explanation:
Let the volume below water be v . Then
buoyant force = v d g where d is density of water , g is acceleration due to gravity
= v x 1000 x g
weight of wood piece = volume x density of wood x g
= .6 x 600 x g
for equilibrium while floating
buoyant force = weight
= v x 1000 x g = .6 x 600 x g
v = .36 m²
volume above water or volume exposed = .6 - .36
= .24 m²
When immersed completely ,
buoyant force = .6 x 1000 x 9.8
= 5880 N
weight of wood
= .6 x 600 x g
= 3528 N
buoyant force is more than the weight . In order to equalise them for floating with full volume in water
weight required = 5880 - 3528
= 2352 N.
Measured reading is 1.65 kg
actual reading was 1.72 kg
error in the reading is 1.72 - 1.65 = 0.07 kg
Now percentage error is given as



<em>so this is -4.1% error as the measured reading is less than the actual value</em>
Answer:
The sample rate determines the maximum receive bandwidth of a direct digital conversion SDR. An analog signal must be sampled at twice the rate of the highest frequency component of the signal by an analog-to-digital converter so that the signal can be accurately reproduced.
Explanation:
To develop this problem it is necessary to apply the concepts related to Work and energy conservation.
By definition we know that the work done by a particle is subject to the force and distance traveled. That is to say,

Where,
F= Force
d = Distance
On the other hand we know that the potential energy of a body is given based on height and weight, that is

The total work done would be given by the conservation and sum of these energies, that is to say

PART A) Applying the work formula,

PART B) Applying the height equation and considering that there is an angle in the distance of 25 degrees and the component we are interested in is the vertical, then



The net work would then be given by



Therefore the net work done by these 2 forces on the cannonball while it is in the cannon barrel is 3355J