All categories

3 years ago

I traveled 6.00 km to school in 0.250 hours . What was my average speed

Physics

1 answer:

SOVA2 [1]3 years ago

5 0

Answer: 6.67 m/s

Explanation:

Recall that average speed is the total distance covered by a moving body divided by the total time taken. The SI unit is metres per second.

i.e Average speed = distance/time taken

Given that:

Distance covered = 6.00 km

Convert kilometers to meters

If 1 km = 1000 m

6.00 km = (6.00 x 1000) = 6000m

Time taken to school = 0.250 hours

Convert time in hours to seconds

If 1 hour = 60 minutes & 1 minute = 60 seconds

Then, 0.250 hours = (0.250 x 60 x 60)

= 900 seconds

Average speed = ?

Average speed = 6000m/900s

= 6.67 m/s

Thus, your average speed to school is 6.67 m/s

You might be interested in

The aqueduct passes under Johnson Road in Lancaster through a siphon. The maximum capacity of the aqueduct is 350 m3/s. The heig

Mariulka [41]

Answer:

D ≈ 8.45 m

L ≈ 100.02 m

Explanation:

Given

Q = 350 m³/s (volumetric water flow rate passing through the stretch of channel, maximum capacity of the aqueduct)

y₁ - y₂ = h = 2.00 m (the height difference from the upper to the lower channels)

x = 100.00 m (distance between the upper and the lower channels)

We assume that:

the upper and the lower channels are at the same pressure (the atmospheric pressure).
the velocity of water in the upper channel is zero (v₁ = 0 m/s).
y₁ = 2.00 m (height of the upper channel)
y₂ = 0.00 m (height of the lower channel)
g = 9.81 m/s²
ρ = 1000 Kg/m³ (density of water)

We apply Bernoulli's equation as follows between the point 1 (the upper channel) and the point 2 (the lower channel):

P₁ + (ρ*v₁²/2) + ρ*g*y₁ = P₂ + (ρ*v₂²/2) + ρ*g*y₂

Plugging the known values into the equation and simplifying we get

Patm + (1000 Kg/m³*(0 m/s)²/2) + (1000 Kg/m³)*(9.81 m/s²)*(2 m) = Patm + (1000 Kg/m³*v₂²/2) + (1000 Kg/m³)*(9.81 m/s²)*(0 m)

⇒ v₂ = 6.264 m/s

then we apply the formula

Q = v*A ⇒ A = Q/v ⇒ A = Q/v₂

⇒ A = (350 m³/s)/(6.264 m/s)

⇒ A = 55.873 m²

then, we get the diameter of the pipe as follows

A = π*D²/4 ⇒ D = 2*√(A/π)

⇒ D = 2*√(55.873 m²/π)

⇒ D = 8.434 m ≈ 8.45 m

Now, the length of the pipe can be obtained as follows

L² = x² + h²

⇒ L² = (100.00 m)² + (2.00 m)²

⇒ L ≈ 100.02 m

6 0

3 years ago

Read 2 more answers

Two bicycle tires are set rolling with the same initial speed of 3.30 m/s along a long, straight road, and the distance each tra

vredina [299]

Answer:

At low pressure- $\mu_{k}=0.02315$

At high pressure- $\mu_{k}=0.00445$

Explanation:

Initial speed, $V_{i}=3.3 m/s$

Final speed, $V_{f}=3.3/2= 1.65 m/s$

Net horizontal force due to rolling friction $F_{net}=\mu_{k}$ mg where m is mass, g is acceleration due to gravity, $\mu_{k}$ is coefficient of rolling friction

From kinematic relation, $V_{f}^{2}- V_{i}^{2}=2ad$

For each tire,

$V_{f}^{2}- V_{i}^{2}=2\mu_{k}gd$

Making $\mu_{k}$ the subject

$\mu_{k}=\frac {V_{f}^{2}- V_{i}^{2}}{2gd}$

Under low pressure of 40 Psi, d=18 m

$\mu_{k}=\frac {1.65^{2}- 3.3^{2}}{2*9.8*18}=-0.02315$

Therefore, $\mu_{k}=0.02315$

At a pressure of 105 Psi, d=93.7

$\mu_{k}=\frac {1.65^{2}- 3.3^{2}}{2*9.8*93.7}=-0.00445$

Therefore, $\mu_{k}=0.00445$

4 0

3 years ago

An automobile engine delivers 55.0 hp. How much time will it take for the engine to do 6.22 × 105 J of work? One horsepower is e

Gennadij [26K]

Answer:

15.2 s

Explanation:

Convert hp to W:

55.0 hp × 746 W/hp = 41,030 W

Power = energy / time

41030 W = 6.22×10⁵ J / t

t = 15.2 s

8 0

3 years ago

A circular parallel plate capacitor is constructed with a radius of 0.52 mm, a plate separation of 0.013 mm, and filled with an

olya-2409 [2.1K]

Answer:

289282

Explanation:

r = Radius of plate = 0.52 mm

d = Plate separation = 0.013 mm

A = Area = $\pi r^2$

V = Potential applied = 2 mV

k = Dielectric constant = 40

$\epsilon_0$ = Electric constant = $8.854\times 10^{-12}\ \text{F/m}$

Capacitance is given by

$C=\dfrac{k\epsilon_0A}{d}$

Charge is given by

$Q=CV\\\Rightarrow Q=\dfrac{k\epsilon_0AV}{d}\\\Rightarrow Q=\dfrac{40\times 8.854\times 10^{-12}\times\pi \times (0.52\times 10^{-3})^2\times 2\times 10^{-3}}{0.013\times 10^{-3}}\\\Rightarrow Q=4.6285\times 10^{-14}\ \text{C}$

Number of electron is given by

$n=\dfrac{Q}{e}\\\Rightarrow n=\dfrac{4.6285\times 10^{-14}}{1.6\times10^{-19}}\\\Rightarrow n=289281.25\ \text{electrons}$

The number of charge carriers that will accumulate on this capacitor is approximately 289282.

6 0

2 years ago

How do i do this???/

Gekata [30.6K]

Start at the big tree, and take the instructions one at a time: 1). Walk 15 west. 2). From there, turn around and walk 2 east. 3). From there, walk 7 more east. Where are you from the big tree now ? Aren't you 6 west of it ? I mean, let's see some common sense here !

5 0

3 years ago

Read 2 more answers