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Vinvika [58]
3 years ago
15

Instrument used to measure the length of a football field​

Physics
1 answer:
r-ruslan [8.4K]3 years ago
6 0

Answer:

If you need to measure much longer lengths - for example the length of a football pitch - then you could use a trundle wheel. You use it by pushing the wheel along the ground. It clicks every time it measures one metre.

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What are some of the forces that are found in skateboarding?
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Normal force, friction force, gravitational force
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3 years ago
Two identical 0.400 kg masses are pressed against opposite ends of a light spring of force constant 1.75 N/cm, compressing the s
nlexa [21]

Answer:

0.853 m/s

Explanation:

Total energy stored in the spring = Total kinetic energy of the masses.

1/2ke² = 1/2m'v².................... Equation 1

Where k = spring constant of the spring, e = extension, m' = total mass, v = speed of the masses.

make v the subject of the equation,

v = e[√(k/m')].................... Equation 2

Given: e = 39 cm = 0.39 m, m' = 0.4+0.4 = 0.8 kg, k = 1.75 N/cm = 175 N/m.

Substitute into equation 2

v = 0.39[√(1.75/0.8)

v = 0.39[2.1875]

v = 0.853 m/s

Hence the speed of each mass = 0.853 m/s

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Why does it take the water in a swimming pool some time to heat up during a hot day?
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3 years ago
If a 160W light bulb consumes 1000J of electrical energy in a given time, then in the same time interval, how much energy will a
NNADVOKAT [17]
160w/4 = 40w
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7 0
2 years ago
Read 2 more answers
A wire of resistivity ρ must be replaced in a circuit by a wire of the same material but four times as long. If, however, the to
Elanso [62]

Answer:

E. two times the original diameter

Explanation:

Resistance of a wire is:

R = ρ L/A

where ρ is the resistivity of the material, L is the length, and A is the cross-sectional area.

For a round wire with diameter d:

R = ρ L / (¼ π d²)

The two wires must have the same resistance, so:

ρ₁ L₁ / (¼ π d₁²) = ρ₂ L₂ / (¼ π d₂²)

The wires are made of the same material, so ρ₁ = ρ₂:

L₁ / (¼ π d₁²) = L₂ / (¼ π d₂²)

The new length is four times the old, so 4 L₁ = L₂:

L₁ / (¼ π d₁²) = 4 L₁ / (¼ π d₂²)

1 / (¼ π d₁²) = 4 / (¼ π d₂²)

Solving:

1 / (d₁²) = 4 / (d₂²)

(d₂²) / (d₁²) = 4

(d₂ / d₁)² = 4

d₂ / d₁ = 2

So the new wire must have a diameter twice as large as the old wire.

3 0
3 years ago
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