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3 years ago

15

If you walk eight blocks north and then three blocks south from your home what is your position compared to your home? what dist

ance did you walk?

1 answer:

Solnce55 [7]3 years ago

6 0

Imagine you walk 8 blocks, then you turn back and walk 3 blocks
your position from your home is 8 - 3 = 5

your distance is 8 + 3 = 11

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anice is watching her granddaughter drive a Barbie Jeep with a 6 V battery and an electric motor with 5 ohm of resistance. How m

Scrat [10]

Answer:

Power required will be 7.2 watt

Explanation:

We have given battery of Barble jeep is 6 volt

So potential difference V = 6 volt

Resistance of electric motor R = 5 ohm '

We have to find the power motor using to drive the jeep

Power is given by $P=\frac{V^2}{R}$ , here V is voltage and R is resistance

So $P=\frac{V^2}{R}=\frac{6^2}{5}=7.2watt$

5 0

3 years ago

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An alien spacecraft flies directly over a football stadium at a speed of 0.50c. if the proper length of the field is 100 yards,

Ivahew [28]

Go on way and you will get your answer

6 0

3 years ago

What is the weight of a 42 kg object if the object was on the moon?

sveticcg [70]

(1.6 m/s²)(42 Kg)= 80 N

4 0

3 years ago

Calculate the true mass (in vacuum) of a piece of aluminum whose apparent mass is 4.5000 kgkg when weighed in air. The density o

spin [16.1K]

Answer:

The true weight of the aluminium is $m_{alu} =$ 4.5021 kg

Explanation:

Given data

$m_{app}$ = 4.5 kg

$\rho_{air}$ = 1.29 $\frac{kg}{m^{3} }$

$\rho_{al}$ = 2.7× $10^{3} \frac{kg}{m^{3} }$

The true mass of the aluminium is given by

$m_{alu} = \frac{\rho_{alu}m_{app}}{\rho_{alu} -\rho_{air} }$

Put all the values in above equation we get

$m_{alu} = \frac{(2700)(4.5)}{2700-1.29}$

$m_{alu} =$ 4.5021 kg

Therefore the true weight of the aluminium is $m_{alu} =$ 4.5021 kg

6 0

3 years ago

Coulomb measured the deflection of sphere A when spheres A and B had equal charges and were a distance d apart. He then made the

sdas [7]

Answer:

The new distance is d = 0.447 d₀

Explanation:

The electric out is given by Coulomb's Law

F = k q₁ q₂ / r²

This electric force is in balance with tension.

We reduce the charge of sphere B to 1/5 of its initial value ( $q_{B}$ =q₂ = q₂ / 5) than new distance (d = n d₀)

dat

q₁ = $q_{A}$

q₂ = $q_{B}$

r = d₀

In order for the deviation to maintain the electric force it should not change, so we apply the Coulomb equation for the two points

F = k q₁ q₂ / d₀²

F = k q₁ (q₂ / 5) / (n d₀)²

.k q₁ q₂ / d₀² = q₁ q₂ / (5 n² d₀²)

5 n² = 1

n = √ 1/5

n = 0.447

The new distance is

d = 0.447 d₀

6 0

3 years ago