Answer:
Steven has to row at a speed to reach the same horizontal spot at the other side of the river is, V = 6 m/s
Explanation:
Given data,
The river flowing south at the rate, v = 3 m/s
To reach the other side directly across the river, he aims the raft, Ф = 30°
The speed of his raft across the river is given by the formula,
V = v / Sin Ф
= 3 / Sin 30°
= 6 m/s
Steven has to row at a speed to reach the same horizontal spot at the other side of the river is, V = 6 m/s
Answer:
-5 V
Explanation:
The charged particle (which is positively charged) moves from point A to B, and its kinetic energy increases: it means that the particle is following the direction of the field, so its potential energy is decreasing (because it's been converted into potential energy), therefore it is moving from a point at higher potential (A) to a point at lower potential (B). This means that the value
vb−va
is negative.
We can calculate the potential difference between the two points by using the law of conservation of energy:
![\Delta K+ \Delta U=0\\\Delta K + q\Delta V=0](https://tex.z-dn.net/?f=%5CDelta%20K%2B%20%5CDelta%20U%3D0%5C%5C%5CDelta%20K%20%2B%20q%5CDelta%20V%3D0)
where:
is the change in kinetic energy of the particle
is the charge of the particle
is the potential difference
Re-arranging the equation, we can find the value of the potential difference:
![\Delta V=V_b-V_a = -\frac{\Delta K}{q}=-\frac{1.6\cdot 10^{-18} J}{3.2\cdot 10^{-19} C}=-5 V](https://tex.z-dn.net/?f=%5CDelta%20V%3DV_b-V_a%20%3D%20-%5Cfrac%7B%5CDelta%20K%7D%7Bq%7D%3D-%5Cfrac%7B1.6%5Ccdot%2010%5E%7B-18%7D%20J%7D%7B3.2%5Ccdot%2010%5E%7B-19%7D%20C%7D%3D-5%20V)
That you have thrown a ball with kinetic energy upwards at an increasing velocity rate
Subduction is, "<span>the sideways and downward movement of the edge of a plate of the earth's crust into the mantle beneath another plate." The basalt would most likely be swallowed up into the ground.
Hope this is what you were looking for! :)
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Answer:
Shown by explanation;
Explanation:
The heat of the sample = mass ×specific heat capacity of the sample × temperature change(∆T)
Assumption;I assume the mass of the samples are : 109g and 192g
∆T= 30.1-21=8.9°c.
The heat of the samples are for 109g are:
0.109 × 4186 × 8.9 =4060.84J
For 0.192g are;
∆T= 67-30.1-=36.9°c
0.192 × 4186×36.9=29656.97J