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tekilochka [14]
3 years ago
12

A train travels 60 miles per hour for 3 hours and then 45 miles per hour for 2 hours. What is the train’s average speed in miles

per hour during the 5‑hour period?
Physics
2 answers:
eimsori [14]3 years ago
5 0

Answer:

Explanation:

Average speed is given by

Total distance/ total time taken

So we need to know the distance in the first travel

Then

First part

Speed=60miles/hour

Time 3hours

Distance = speed × time

Distance = 60×3

Distance = 180 miles

Second part

Speed 45mile/hour

And time is 2hours

Distance = speed × time

Distance = 45× 2

Distance = 90miles

So total distance travelled for both cases is 90+180=270miles

And total time is 5hours

Then,

Average speed = total distance/total time

A.S=270/5

A.S= 54miles/hour

dedylja [7]3 years ago
3 0

Answer:

ah

Explanation:

ahhhhhhhhhhhhh

You might be interested in
A simple pendulum consists of a brass sphere of mass m = 1 kg suspended on a string of length L ≈ 1 meter. The pendulum oscillat
hram777 [196]

Answer:

2. Using a shorter string of length L ′ ≈ 0.25 meters

5. Using a shorter string of length L ′ ≈ 0.5 meters

Explanation:

The period of a pendulum is given by

T=2\pi \sqrt{\frac{L}{g}}

where

L is the length of the pendulum

g is the acceleration due to gravity

We see from the formula that the period of the pendulum depends only on its length, not on its mass or its amplitude of ocillation. Therefore, the only alterations that can change the period of the pendulum are the ones where its length is changed.

Moreover, we notice that the period is proportional to the square of the length: this means that in order to decrease the period of the pendulum (the problem asks us which alterations will reduce the period of the pendulum from 2 s to 1 s), the length of the pendulum should also be reduced.

Therefore, the only alterations that will reduce the period of the pendulum are:

2. Using a shorter string of length L ′ ≈ 0.25 meters

5. Using a shorter string of length L ′ ≈ 0.5 meters

6 0
3 years ago
If the wave represents a sound wave, explain how increasing amplitude will affect the loudness of the sound? If we decrease the
Viktor [21]

Answer:

Explanation:

Think of a sound wave like a wave on the ocean, or lake... It's not really water moving, as much as it's energy moving through the water. Ever see something floating on the water, and notice that it doesn't come in with the wave, but rides over the top and back down into the trough between them? Sound waves are very similar to that. If you looked at a subwoofer speaker being driven at say... 50 cycles a second, you'd actually be able to see the speaker cone moving back and forth. The more power you feed into the speaker, the more it moves back and forth, not more quickly, as that would be a higher frequency, but further in and further out, still at 50 cycles per second. Every time it pushed out, it's compressing the air in front of it... the compressed air moves away from the speaker's cone, but not as a breeze or wind, but as a wave through the air, similar to a wave on the ocean

More power, more amplitude, bigger "wave", louder ( to the human ear) sound.

If you had a big speaker ( subwoofer ) and ran a low frequency signal with enough power in it, you could hold a piece of paper in front of it, and see the piece of paper move in and out at exactly the same frequency as the speaker cone. The farther away from the speaker you got, the less it'd move as the energy of the sound wave dispersed through the room.

Sound is a wave

We hear because our eardrums resonates with this wave I.e. our ear drums will vibrate with the same frequency and amplitude. which is converted to an electrical signal and processed by our brain.

By increasing the amplitude our eardrums also vibrate with a higher amplitude which we experience as a louder sound.

Of course when this amplitude is too high the resulting resonance tears our eardrums so that they can't resonate with the sound wave I.e. we become deaf

6 0
3 years ago
Computers A and B implement the same ISA. Computer A has a clock cycle time of 200 ps and an effective CPI of 1.5 for some progr
lesya692 [45]

Answer:

Computer A is 1.41 times faster than the Computer B

Explanation:

Assume that number of instruction in the program is 1

Clock time  of computer A is CT_{A} =200 ps

Clock time  of computer B is CT_{B} =250 ps

Effective CPI of computer A is CPI_{A} =1.5

Effective CPI of computer B isCPI_{B} =1.7

CPU time of A is

CPU_{time}=instructions \times CPA_{A} \times CT_{A}\\CPU_{time}=1 \times 1.5 \times 200=300 sec

CPU time of B is

CPU_{time}=instructions \times CPA_{B} \times CT_{B}\\CPU_{time}=1 \times 1.7 \times 250=425 sec

Hence Computer A is Faster by \frac{425}{300} =1.41

Computer A is 1.41 times faster than the Computer B

4 0
3 years ago
The driver of a pickup truck accelerates from rest to a speed of 37 mi/hr over a horizontal distance of 215 ft with constant acc
Alona [7]

Answer:

Maximum shearing force developed in each of the two pegs during acceleration is 1830 lbf

Explanation:

First we will find the acceleration of pickup truck.

As, the acceleration is uniform, therefore we can use Newton's third equation of motion:

2as = V_{f}^{2}-V_{i}^{2}

First convert speed into ft/sec

1 mile/hr = 1.47 ft/sec

therefore,

37 mile/hr = 37 x 1.47 ft/sec

37 mile/hr =  54.39 ft/sec

with initial speed 0 ft/sec (starting from rest), using in equation of motion:

a = [(54.39 ft/sec)² - (0 ft/sec)²]/2(215 ft)

a = 6.88 ft/sec²

Now, the total shear force will be given by Newton's second law of motion:

F = ma

F = (460 lbm +72 lbm)(6.88 ft/sec²)

F = 3660 lbf

Now for the max shear force in each of the two pegs we divide total fore by 2:

Force in each peg = F/2 = (3660 lbf)/2

<u>Force in each peg = 1830 lbf</u>

4 0
4 years ago
The three-toed sloth is the slowest moving land mammal. On the ground, the sloth moves at an average speed of 0.037 m/s, conside
Georgia [21]

Answer:

tortoise distance w.r.t sloth is 28.08 m further

Explanation:

given data

average speed v1 = 0.037 m/s

walking speed v2 = 0.076 m/s

time t = 12 min = 720 seconds

to find out

how much tortoise have gone wrt sloth

solution

we find here first tortoise walk that is

distance d1 = v2 × t

distance d1 =  0.076  × 720

distance d1 = 54.72 m

and sloth walk distance

distance d2 = v1 × t

distance d2 = 0.037 × 720

distance d2 = 26.64 m

and so

tortoise distance w.r.t sloth = d1 - d2

tortoise distance w.r.t sloth = 54.72 - 26.64 = 28.08

tortoise distance w.r.t sloth is 28.08 m further

7 0
3 years ago
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