All categories

3 years ago

A bus travels north on some busy city streets for 2.5 km, and a trip

Physics

1 answer:

d1i1m1o1n [39]3 years ago

8 0

Answer:

V = 4.63 m/s

V = 11.31 m/s

Explanation:

Given,

The distance traveled by the bus, towards north, d = 2.5 km

= 2500 m

The time taken by the trip is, t = 9 min

= 540 s

The velocity of the bus,

V = d / t

= 2500 / 540

= 4.63 m/s

At another point, the bus travels at a constant speed of v = 18 m/s

Therefore the velocity becomes

V = (4.63 + 18)/2

= 11.31 m/s

Hence, the velocity of the bus, V = 11.31 m/s

You might be interested in

. Consider the equation =0+0+02/2+03/6+04/24+5/120, where s is a length and t is a time. What are the dimensions and SI units of

Olegator [25]

Answer:

See Explanation

Explanation:

Given

$s=s_0+v_0t+\frac{a_0t^2}{2}+ \frac{j_0t^3}{6}+\frac{S_0t^4}{24}+\frac{ct^5}{120}$

Solving (a): Units and dimension of $s_0$

From the question, we understand that:

$s \to L$ --- length

$t \to T$ --- time

Remove the other terms of the equation, we have:

$s=s_0$

Rewrite as:

$s_0=s$

This implies that $s_0$ has the same unit and dimension as $s$

Hence:

$s_0 \to L$ --- dimension

$s_o \to$ Length (meters, kilometers, etc.)

Solving (b): Units and dimension of $v_0$

Remove the other terms of the equation, we have:

$s=v_0t$

Rewrite as:

$v_0t = s$

Make $v_0$ the subject

$v_0 = \frac{s}{t}$

Replace s and t with their units

$v_0 = \frac{L}{T}$

$v_0 = LT^{-1}$

Hence:

$v_0 \to LT^{-1}$ --- dimension

$v_0 \to$ $m/s$ --- unit

Solving (c): Units and dimension of $a_0$

Remove the other terms of the equation, we have:

$s=\frac{a_0t^2}{2}$

Rewrite as:

$\frac{a_0t^2}{2} = s_0$

Make $a_0$ the subject

$a_0 = \frac{2s_0}{t^2}$

Replace s and t with their units [ignore all constants]

$a_0 = \frac{L}{T^2}\\$

$a_0 = LT^{-2$

Hence:

$a_0 = LT^{-2$ --- dimension

$a_0 \to$ $m/s^2$ --- acceleration

Solving (d): Units and dimension of $j_0$

Remove the other terms of the equation, we have:

$s=\frac{j_0t^3}{6}$

Rewrite as:

$\frac{j_0t^3}{6} = s$

Make $j_0$ the subject

$j_0 = \frac{6s}{t^3}$

Replace s and t with their units [Ignore all constants]

$j_0 = \frac{L}{T^3}$

$j_0 = LT^{-3}$

Hence:

$j_0 = LT^{-3}$ --- dimension

$j_0 \to$ $m/s^3$ --- unit

Solving (e): Units and dimension of $s_0$

Remove the other terms of the equation, we have:

$s=\frac{S_0t^4}{24}$

Rewrite as:

$\frac{S_0t^4}{24} = s$

Make $S_0$ the subject

$S_0 = \frac{24s}{t^4}$

Replace s and t with their units [ignore all constants]

$S_0 = \frac{L}{T^4}$

$S_0 = LT^{-4$

Hence:

$S_0 = LT^{-4$ --- dimension

$S_0 \to$ $m/s^4$ --- unit

Solving (e): Units and dimension of $c$

Ignore other terms of the equation, we have:

$s=\frac{ct^5}{120}$

Rewrite as:

$\frac{ct^5}{120} = s$

Make $c$ the subject

$c = \frac{120s}{t^5}$

Replace s and t with their units [Ignore all constants]

$c = \frac{L}{T^5}$

$c = LT^{-5}$

Hence:

$c \to LT^{-5}$ --- dimension

$c \to m/s^5$ --- units

4 0

3 years ago

A spacecraft is traveling in interplanetary space at a constant velocity. What is the estimated distance traveled by the spacecr

vekshin1

i would say that the correct answer would be cc

3 0

3 years ago

Read 2 more answers

A man attaches a divider to an outdoor faucet so that water flows through a single pipe of radius 9.25 mm into four pipes, each

irinina [24]

Answer:

1.24 m/s

Explanation:

Metric unit conversion:

9.25 mm = 0.00925 m

5 mm = 0.005 m

The volume rate that flow through the single pipe is

$\dot{V} = vA = 1.45 * \pi * 0.00925^2 = 0.00039 m^3/s$

This volume rate should be constant and divided into the 4 narrower pipes, each of them would have a volume rate of

$\dot{V_n} = \dot{V} / 4 = 0.00039 / 4 = 9.74\times10^{-5} m^3/s$

So the flow speed of each of the narrower pipe is:

$v_n = \frac{\dot{V_n}}{A_n} = \frac{\dot{V_n}}{\pi r_n^2}$

$v_n = \frac{9.74\times10^{-5}}{\pi 0.005^2} = 1.24 m/s$

8 0

4 years ago

Which best describes what occurs when an object takes in a wave as the wave hits it?

iren2701 [21]

Answer: B. Absorption

5 0

3 years ago

Read 2 more answers

How to measure the volume of a baseball bat ( need answers ASAP )

vaieri [72.5K]

<em>Measure the amount of water it displaces.</em>

This won't be easy, because the bat floats in water. But I think you can get around that little problem like this:

-- Get some kind of a tank or tub that's big enough to hold the whole bat under water.

-- Get a heavy weight, like a big wrench or a small rock.

-- Fill the tub almost to the tippy top with water.

-- Slip the heavy weight into the tub, slowly. Some water will run over the top and out of the tub. That's OK ... it's exactly what you want. If NO water runs over the top, pour some more in, until it runs out and then stops. You want the tub full to the brimmy rim with the rock at the bottom of it.

-- Take the heavy weight out of the tub.

-- Now set the tub into a bigger tub or a deep pan. The next time it overflows and some water runs out of it, you'll need to catch that water and measure it.

-- Get a short piece of heavy string. Tie the heavy weight to somewhere near the middle of the bat.

-- Slowly slide the bat into the water, with the rock tied to it. The bat needs to go complete underwater.

-- Some more water will run over the top and out of the tub, and INTO the lower tub. Wait until the overflow stops and everything settles down again.

-- Take the bat (tied to the weight) out of the tub. Slowly and carefully, so that your hand or your arm doesn't make any MORE water run over and out.

-- Lift the upper tub out of the lower tub.

-- Take the lower tub, with the overflow water in it. Using a kitchen measuring cup, or a saucepan or a bottle, or anything else with liquid amounts marked on it, measure how much water overflowed into the lower tub.

THAT amount is the volume of the bat.

You may have to do some units conversions. Like if you need the volume of the bat in cm³ and you used measuring vessels marked in fluid ounces. But you can find all those conversion factors with a search on Floogle.

8 0

3 years ago