Answer:
<h2><em>
6000 counts per second</em></h2>
Explanation:
If a sample emits 2000 counts per second when the detector is 1 meter from the sample, then;
2000 counts per second = 1 meter ... 1
In order to know the number of counts per second that would be observed when the detector is 3 meters from the sample, we will have;
x count per second = 3 meter ... 2
Solving the two expressions simultaneously for x we will have;
2000 counts per second = 1 meter
x counts per second = 3 meter
Cross multiply to get x
2000 * 3 = 1* x
6000 = x
<em></em>
<em>This shows that 6000 counts per second would be observed when the detector is 3 meters from the sample</em>
Answer:
0.75%
Explanation:
Measured value of melting point of potassium thiocyanate = 174.5 °C
Actual value of melting point of potassium thiocyanate = 173.2 °C
<em>Error in the reading = |Experimental value - Theoretical value|</em>
<em>= |174.5 - 173.2|</em>
<em>= |1.3|</em>
<em>Percentage error = (Error / Theoretical value) × 100</em>
<em>= (1.3 / 173.2)×100</em>
<em>= 0.75 %</em>
∴ Percentage error in the reading is 0.75%
Answer:
The answer is "The object's speed relative to S can be greater than or less than its speed relative to S', depending on the actual values."
Explanation:
The S' frame and the object are moving in a positive direction. The object is moving with respect to the S frame so the S frame the rest frame
take the velocity of the object with respect to the rest frame as v and the velocity of the S' frame with respect S frame as v2
relative velocity of the object to the S' frame would be
Vrel = v2- v
This means the Vrel of the object with respect to the S' frame is less than the Vrel of the object with respect to the S frame
However is the S' velocity is greater than that of the object then the Vrel of the object with respect to the S' frame is greater than the Vrel of the object with respect to the S frame.
This would mean the second option is the answer, the relative speed of the object depends on the actual values.
We assign the variables: T as tension and x the angle of the string
The <span>centripetal acceleration is expressed as v²/r=4.87²/0.9 and (0.163x4.87²)/0.9 = </span><span>T+0.163gcosx, giving T=(0.163x4.87²)/0.9 – 0.163x9.8cosx.
</span>
<span>(1)At the bottom of the circle x=π and T=(0.163x4.87²)/0.9 – .163*9.8cosπ=5.893N. </span>
<span>(2)Here x=π/2 and T=(0.163x4.87²)/0.9 – 0.163x9.8cosπ/2=4.295N. </span>
<span>(3)Here x=0 and T=(0.163x4.87²)/0.9 – 0.163x9.8cos0=2.698N. </span>
<span>(4)We have T=(0.163v²)/0.9 – 0.163x9.8cosx.
</span><span>This minimum v is obtained when T=0 </span><span>and v verifies (0.163xv²)/0.9 – 0.163x9.8=0, resulting to v=2.970 m/s.</span>
Answer:
Energy is essentially work done by an object or on object.
From,
W = Fd
It's directly proportional to mass.
from,
K. E = 1/2mv²
Energy is directly proportional to mass.
P. E = mgh
Energy is directly proportional to mass.
H = mc∆T
Energy is directly proportional to mass.
Thus increasing mass will increase the energy also imparted on another object since all the above eqns show that relationship.
And for 2 moving bodies
K.Ei = K.Ef(energy conservation)
m1u²1 + m2u²2 = m1v²1 + m2v²2
The relationship is the same that the greater mass the greater the impact.
