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DerKrebs [107]
2 years ago
5

Eating breakfast in the morning increases the ability to learn in school.

Physics
1 answer:
bonufazy [111]2 years ago
4 0

Answer:

Independent Variable - eating reakfast

Dependent Variable - ability to learn

Constant Variable - going to school

Explanation:

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2. Two long parallel wires each carry a current of 5.0 A directed to the east. The two wires are separated by 8.0 cm. What is th
Gre4nikov [31]

Answer:

The magnitude of magnetic field at given point = 5.33 × 10^{-5} T

Explanation:

Given :

Current passing through both wires = 5.0 A

Separation between both wires = 8.0 cm

We have to find magnetic field at a point which is 5 cm from any of wires.

From biot savert law,

We know the magnetic field due to long parallel wires.

⇒ B = \frac{\mu_{0}i }{2\pi R}

Where B = magnetic field due to long wires, \mu_{0} = 4\pi \times10^{-7}, R = perpendicular distance from wire to given point

From any one wire R_{1}  = 5 cm, R_{2}  = 3 cm

so we write,

∴ B = B_{1} + B_{2}

 B = \frac{\mu_{0} i}{2\pi R_{1} } +  \frac{\mu_{0} i}{2\pi R_{2} }

 B =\frac{ 4\pi \times10^{-7} \times5}{2\pi } [\frac{1}{0.03} + \frac{1}{0.05} ]

 B = 5.33\times10^{-5}  T

Therefore, the magnitude of magnetic field at given point = 5.33\times10^{-5} T

3 0
3 years ago
What's is the kinetic energy of a .235-kg baseball thrown at 50.0m/s
san4es73 [151]

Answer:

34.51

Explanation:

k=1/2mv² is the kenetic energy equation to fill is in

k=[1/2(0.235)×50]²

6 0
2 years ago
A car traveling at a speed of 50.0 m/s encounters an emergency and comes to a complete stop. How much time will it take for the
ludmilkaskok [199]

Answer:

t=1.25s

Explanation:

The formula is

a= (V1-V0) /t

t = (V1-V0)/a

V1=50m/s

V0= 0 m/s

a= - 4m/s2

t= (0-50)/-4 = 1.25s

6 0
2 years ago
A mover applies a net force of 28 N to a sofa that has a mass of 70 kg.
Alinara [238K]

Answer: .4 m/s^2= acceleration

Explanation:

f = m*a

We can rearrange this equation to solve for acceleration. Therefore,

a=f/m

a= 28N/70kg

a= 0.4 m/s^2

8 0
3 years ago
When a system does work, what happens to its internal energy
tester [92]
The first law, which deals with changes in the internal energy, thus becomes 0 = Q - W, so Q = W.

If the system does work, the energy comes from heat flowing into the system from the reservoir; if work is done on the system, heat flows out of the system to the reservoir
8 0
3 years ago
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