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Tanzania [10]
3 years ago
12

If you use the same force to push a motorcycle as you would push a bike which one would have more acceleration and why explain u

sing Newton's second law
Physics
2 answers:
zaharov [31]3 years ago
6 0

Answer:

The bike would have more acceleration

Explanation:

Accourding to newtons first law a force is equal to its mass multiplied by its acceleration (f=ma) therefore an object with a higher mass compared to an object with a lower mass would experience less acceleration.

Eg.

F=50N

Motorbike M=200kg

F=ma

50=200 x a

50/200=a

0.25m/s/s =a

Bike M=35kg

F=ma

50=35 x a

50/35= a

1.43m/s/s=a

Elenna [48]3 years ago
4 0

Answer:

the bike

Explanation:

it has less mass than the motorcycle so so it would have more acceleration

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Negligible weights is a change so minor or insignificant to be deemed to have no effect on weight or balance.

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What causes a solar eclipse? Select all that apply.
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I think it is either D) or E)

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A supertanker filled with oil has a total mass of 6.1 x108 kg. If the dimensions of the ship are those of a rectangular box 300
IrinaVladis [17]

Answer:

The bottom of the sea is 25 m below sea level.

Explanation:

Given data

Mass = 6.1 × 10^{8} \ kg

\rho_{sea} = 1020\  \frac{kg}{m^{3} }

We know that Buoyant force on the tank is equal to gravity force of the tank.

F_B = F_g

(\rho_{Fluid}) (g) (V_{disp}) = m g

(\rho_{Fluid})  (V_{disp}) = m

1020 × V_{disp} = 6.1 × 10^{8}

V_{disp} = 598039.21 m^{3}

We know that

V_{disp} = W × L × H

598039.21 = 300 × 80 × H

H = 25 m

Therefore the bottom of the sea is 25 m below sea level.

7 0
3 years ago
Consider two points in an electric field. The potential at point 1, V1, is 33 V. The potential at point 2, V2, is 175 V. An elec
Mnenie [13.5K]

Answer:

ΔU  = e(V₂ - V₁) and its value ΔU = -2.275 × 10⁻²¹ J

Explanation:

Since the electric potential at point 1 is V₁ = 33 V and the electric potential at point 2 is V₂ = 175 V, when the electron is accelerated from point 1 to point 2, there is a change in electric potential ΔV which is given by ΔV = V₂ - V₁.

Substituting the values of the variables into the equation, we have

ΔV = V₂ - V₁.

ΔV = 175 V - 33 V.

ΔV = 142 V

The change in electric potential energy ΔU = eΔV = e(V₂ - V₁) where e = electron charge = -1.602 × 10⁻¹⁹ C and ΔV = electric potential change from point 1 to point 2 = 142 V.

So, substituting the values of the variables into the equation, we have

ΔU = eΔV

ΔU = eΔV

ΔU = -1.602 × 10⁻¹⁹ C × 142 V

ΔU = -227.484 × 10⁻¹⁹ J

ΔU = -2.27484 × 10⁻²¹ J

ΔU ≅ -2.275 × 10⁻²¹ J

So, the required equation for the electric potential energy change is

ΔU  = e(V₂ - V₁) and its value ΔU = -2.275 × 10⁻²¹ J

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