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satela [25.4K]
2 years ago
15

You drive 6.0 km at 50 km/h and then another 6.0 km at 90 km/h. Your average speed over the 12 km drive will be ____

Physics
1 answer:
Elis [28]2 years ago
5 0

Answer:

The average speed is less than 70 km/h.

(B) is correct option.

Explanation:

Given that,

distance = 6.0 km

Speed = 50 km/h

Speed = 90 km/h

We need to calculate the time in 6.0 km distance

Using formula of time

t = \dfrac{d}{v}

Put the value in to the formula

t=\dfrac{6.0}{50}

t=0.12\ hr

We need to calculate the time in another distance

Using formula of time

t = \dfrac{d}{v}

Put the value in to the formula

t=\dfrac{6.0}{90}

t=0.067\ hr

We need to calculate the average speed

Using formula of average speed

v=\dfrac{D}{T}

Where, D = total distance

T = total time

Put the value into the formula

v=\dfrac{12}{0.12+0.067}

v=64.17\ km/hr

Hence, The average speed is less than 70 km/h.

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3 years ago
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Sam drives her scooter 7 kilometres north. She stops for lunch and then drives
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Answer:

1. Distance travelled = 12 km.

2. Displacement = 8.6 km

Explanation:

From the question given above, the following data were obtained:

Distance 1 (d₁) = 7 km

Distance 2 (d₂) = 5 km

Total distance =?

Displacement =?

1. Determination of the distance travelled.

Distance 1 (d₁) = 7 km

Distance 2 (d₂) = 5 km

Total distance (dₜ) =?

dₜ = d₁ + d₂

dₜ = 7 + 5

dₜ = 12 km

2. Determination of the displacement.

In the attached photo, R is the displacement.

We can obtain the value of R by using the pythagoras theory as illustrated below:

R² = 7² + 5²

R² = 49 + 25

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Take the square root of both side

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The SAF operates the M113 Ultra APC.
HACTEHA [7]

The volume of the object must be no larger than 11.15 m^3.

Explanation:

In order for an object to be able to float in water, its density must be equal or smaller than the water density.

The density of water is:

\rho = 1000 kg/m^3

This means that the density of the object must be no larger than this value.

We also know that the density of an object is given by

\rho = \frac{m}{V}

where

m is the mass of the object

V is its volume

For the object in this problem, the mass is

m=1.115\cdot 10^4 kg

Therefore, we can re-arrange the equation to find its volume:

V=\frac{m}{\rho}=\frac{1.115\cdot 10^4}{1000}=11.15 m^3

So, the volume of the object must be no larger than 11.15 m^3.

Learn more about density:

brainly.com/question/5055270

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