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Ghella [55]
3 years ago
13

A motorboat accelerates uniformly from a ve- locity of 7.0 m/s to the west to a velocity of 2.4 m/s to the west. If its accelera

tion was 2.7 m/s2 to the east, how far did it travel during the acceleration? Answer in units of m.
Physics
1 answer:
Lena [83]3 years ago
8 0

Answer:

Distance, d = 8 meters

Explanation:

Given that,

Initial velocity of the motorboat, u = 7 m/s (west)

Final velocity of the motorboat, v = 2.4 m/s (west)

Acceleration of the motorboat, a=-2.7\ m/s^2 (east)

We need to find the distance covered by it. It can be calculated as :

v^2-u^2=2as

(2.4)^2-(7)^2=2\times (-2.7)s

s = 8 meters

So, the distance covered by the motor boat is 8 meters. Hence, this is the required solution.

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A car traveling west in a straight line on a highway decreases its speed from 30 meters per second to 23 meters per second in 2
garik1379 [7]
U1= 30m/s
u2= 23m/s
t=2s
a=(u2-u1)/t
a=-7/2=-3.5 m/s^2
6 0
3 years ago
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A closed system’s internal energy changes by 178 J as a result of being heated with 658 J of energy. The energy used to do work
nika2105 [10]

Answer:

480J

Explanation:

Using the formula:

Delta U = Q - W

Q:Heat (J)

Delta U: Changes in internal Energy (J)

W:Work (J)

We can plug in the give numbers, Q and W.

Delta U = 658J - 178J = 480J

6 0
3 years ago
An electric furnace runs 13 hours a day to heat a house during January (31 days). The heating element has a resistance of 7.2 an
liq [111]

Answer:

cost of running the furnace during January is $5619.62

Explanation:

given data

runs a day = 13 hours

January days = 31 days

resistance = 7.2 ohm

current = 16.7 A

cost of electricity = $0.10/kWh

to find out

cost of running the furnace during January

solution

first we get her power consumed by furnace that is

Power consumed = \frac{I^2}{R}  ........1

put here value we get

Power consumed = \frac{16.7^2}{7.2}

Power consumed = 38.7347 W

and

Power consumed by furnace in one hour is

Power consumed by furnace in one hour is = Power consumed × 3600

Power consumed by furnace in one hour is = 38.7347 × 3600  

Power consumed by furnace in one hour is 139.445kWh

and

Power consumed by furnace in the month of January is

Power consumed by furnace in the month of January = 139.445kWh × 13 hours × 31 days

Power consumed by furnace in the month of January = 56196.335 kWh

so

cost of running the furnace during January is = $0.10/kWh × 56196.335 kWh

cost of running the furnace during January is $5619.62

4 0
3 years ago
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What are the characteristics of high energy waves?
Temka [501]
Amplitude and frequency
4 0
3 years ago
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Hey help me plzzzzz i will mark brainliest​
vodomira [7]

Answer:

The answer to your question is given below.

Explanation:

Mechanical advantage (MA) = Load (L)/Effort (E)

MA = L/E

Velocity ratio (VR) = Distance moved by load (l) / Distance moved by effort (e)

VR = l/e

Efficiency = work done by machine (Wd) /work put into the machine (Wp) x 100

Efficiency = Wd/Wp x100

Recall:

Work = Force x distance

Therefore,

Work done by machine (wd) = load (L) x distance (l)

Wd = L x l

Work put into the machine (Wp) = effort (E) x distance (e)

Wp = E x e

Note: the load and effort are measured in Newton (N), while the distance is measured in metre (m)

Efficiency = Wd/Wp x100

Efficiency = (L x l) / (E x e) x 100

Rearrange

Efficiency = L/E ÷ l/e x 100

But:

MA = L/E

VR = l/e

Therefore,

Efficiency = L/E ÷ l/e x 100

Efficiency = MA ÷ VR x 100

Efficiency = MA / VR x 100

7 0
2 years ago
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