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Aliun [14]
4 years ago
14

Daniel takes his two dogs, Pauli the Pointer and Newton the Newfoundland, out to a field and lets them loose to exercise. Both d

ogs sprint away in different directions while Daniel stands still. From Daniel's point of view, Newton runs due North at 3.40 m/s, but from Pauli's point of view, Newton appears to be moving at 1.10 m/s due East. What must Pauli's velocity relative to Daniel be for this to be true? Express your answer in terms of the ???? ‑ and ???? ‑components if North is the +???? ‑direction and East is the +???? ‑direction.
Physics
1 answer:
DedPeter [7]4 years ago
7 0

Answer:

Explanation:

3.4 m/s due North, -1.1 m/s due East

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An α-particle has a charge of +2e and a mass of 6.64 × 10-27 kg. It is accelerated from rest through a potential difference that
zzz [600]

Answer with Explanation:

We are given that

Charge on alpha particle=q=2 e=2\times 1.6\times 10^{-19} C

1 e=1.6\times 10^{-19} C

Mass of alpha particle=m=6.64\times 10^{-27} kg

Potential difference,V=1.97\times 10^6 V

Magnetic field,B=3.49 T

a.Speed of alpha particle=v=\sqrt{\frac{2 qV}{m}}

By using the formula

v=\sqrt{\frac{2\times 2\times 1.6\times 10^{-19}\times 1.97\times 10^6}{6.64\times 10^{-27}}

v=1.38\times 10^7 m/s

b.Magnetic force,F=qvB=2\times 1.6\times 10^{-19}\times 1.38\times 10^7\times 3.49=1.5\times 10^{-11} N

F=1.5\times 10^{-11} N

c.Radius of circular path, r=\frac{mv^2}{F}

r= \frac{6.64\times 10^{-27}\times (1.38\times 10^7)^2}{1.5\times 10^{-11}}

r=0.084 m

5 0
3 years ago
Two football players with mass 75kg and 100kg run directly toward each other with speeds of 6 m/s and 8 m/s respectively, If the
Rina8888 [55]

Answer:

2 m/s

Explanation:

From the law of conservation of momentum,

Total momentum before collision = total momentum after collision

mu+m'u' = V(m+m') .................................Equation 1

Where m = mass of the first player, u = initial speed of the first player, m' = mass of the second player, u' = initial speed of the second player, V = combined speed of both players.

Making V the subject of the equation,

V = (mu+m'u')/(m+m')................ Equation 2

Note: Taking the direction of the first player as positive.

Given: m = 75 kg, m' = 100 kg, u = 6 m/s, u' = -8 m/s (opposite the first player),

Substituting into equation 2

V = [(75×6)+(100×(--8))]/(75+100)

V = (450-800)/175

V = 350/175

V = - 2 m/s.

Note: The negative signs tells that the combined speed is in the direction of the second player.

Hence the combined speed of the two players = 2 m/s

8 0
4 years ago
The speed of water at the tap of lower storey is more than that in the upper storey​
nadezda [96]

Answer:

Pressure of liquid in container is given by, P= height × density × acceleration of gravity.

At the lower storey, the height of the liquid from the open end is great, since height is directly proportional to pressure, the pressure exerted by liquid is maximum hence increase in velocity of flow.

Unlike the upper storey where the height of water is less hence the pressure exerted by the liquid is minimum which decreases the velocity / speed of liquid flow

4 0
3 years ago
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