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hodyreva [135]
3 years ago
6

Identify which of the following quantities can be described fully by its magnitude.

Physics
1 answer:
goldenfox [79]3 years ago
4 0

Answer:

Energy

Explanation:

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A current of 1.41 A in a long, straight wire produces a magnetic field of 5.61 uT at a certain distance from the wire. Find
pshichka [43]

Answer:

0.050 m

Explanation:

The strength of the magnetic field produced by a current-carrying wire is given by

B=\frac{\mu_0 I}{2\pi r}

where

\mu_0=4\pi \cdot 10^{-7} H/m is the vacuum permeability

I is the current in the wire

r is the distance from the wire

And the magnetic field around the wire forms concentric circles, and it is tangential to the circles.

In this problem, we have:

I=1.41 A (current in the wire)

B=5.61\mu T=5.61\cdot 10^{-6} T (strength of magnetic field)

Solving  for r, we find the distance  from the wire:

r=\frac{\mu_0 I}{2\pi B}=\frac{(4\pi \cdot 10^{-7})(1.41)}{2\pi (5.61\cdot 10^{-6})}=0.050 m

4 0
4 years ago
Your friend, who is age 15 and has a resting heart rate of 70, wishes to start a jogging program. What would you suggest as a pr
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I think the right answer for this question is : 138 because the boy is just 15 year old but he has resting heart rate of 70
6 0
4 years ago
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A car starts from rest and accelerates at a constant rate for a distance of 130m in 7.0 s. At what rate did the car accelerate?
FrozenT [24]

Answer:

18.57

Explanation:

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3 years ago
In which item is energy stored in the form of gravitational potential energy
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Stretched bow string
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3 years ago
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Water is circulating through a closed system of pipes in a two floor apartment. On the first floor, the water has a gauge pressu
anastassius [24]

Answer:

The value of gauge pressure at outlet = -38557.224 pascal

Explanation:

Apply Bernoulli' s Equation

\frac{P_{1}}{9810} + \frac{V_{1} ^{2}}{19.62} + h_{1} = \frac{P_{2}}{9810} + \frac{V_{2} ^{2}}{19.62} + h_{2} --------------(1)

Where

P_{1} =  Gauge pressure at inlet = 3.70105 pascal

V_{1} = velocity at inlet =  2.4 \frac{m}{sec}

P_{2} = Gauge pressure at outlet = we have to calculate

V_{2} = velocity at outlet = 3.5 \frac{m}{sec}

h_{2} - h_{1} = 3.6 m

Put all the values in equation (1) we get,

⇒ \frac{3.70105}{9810} + \frac{2.4 ^{2}}{19.62} = \frac{P_{2}}{9810} + \frac{3.5 ^{2}}{19.62} + 3.6

⇒ 0.294 = \frac{P_{2}}{9810} + 0.6244 + 3.6

⇒ \frac{P_{2}}{9810} = 0.294 - 0.6244 - 3.6

⇒ \frac{P_{2}}{9810} = - 3.9304

⇒ P_{2} = - 38557.224 pascal

This is the value of gauge pressure at outlet.

3 0
4 years ago
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