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kupik [55]
3 years ago
5

A 1430 kg car speeds up from 7.50 m/s to 11.0 m/s in 9.30 s. Ignoring friction, how much power did that require?(unit=W)PLEASE H

ELP
Physics
1 answer:
AlexFokin [52]3 years ago
7 0

Answer:

4982.8 W

Explanation:

Given the mass of the car = 1430 kg

initial speed = 7.5m/sec

final speed = 11.0 m/sec

time = 9.30sec

Assuming constant accelration a We know that

v=u+at

11=7.5+9.3a

a=0.376 m/sec^2

We know that F=ma

F= 1430*0.376 = 538.2N

We know that v^2-u^2=2as

11^2-7.5^2=2*0.376*s

s= 86.10m

We know that work(W)=F*s

W= 538.2*86.10 = 46339.02J = 46.34KJ

We know that power= \frac{work}{time} = \frac{46.34}{9.3}=4.982 KW

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0.733J/g°C

Explanation:

Using the formula

Q=mcΔθ

Q=38.5J, c=? , m=17.5g , Δθ=3°C

c= Q/(mΔθ)

c=\frac{38.5}{17.5*3}

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tekilochka [14]

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3 years ago
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Please help on this one!!
posledela
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someone help pls. Two students, Mia and Peter, leave school to meet at the local coffee shop. Peter decides to jog to the coffee
cluponka [151]

Answer:

1) The distance further it takes Peter to arrive at the Coffee shop than Mia is 1.24 km

2) Mia's average speed is 6.00 km/hour

Peter's average speed is 8.48 km/hour

4) Mia's average velocity = Peter's average velocity = 6.00 km/hour

Explanation:

The given information from the diagram are;

The distance Peter jogs from school to the flower shop = 2.00 km

The distance Peter jogs from the Flower shop to the Coffee shop = 2.24 km.

The distance Mia walks from school directly to the Coffee shop = 3.00 km

The time it takes both Peter and Mia to arrive at the coffee shop = 30 minutes = 0.5 hour

1) The total distance Peter travels from school to the Coffee shop = 2.00 km + 2.24 km = 4.24 km

The distance Mia travels from school to the Coffee shop = 3.00 km

The distance further it takes Peter to arrive at the Coffee shop than Mia = 4.24 km - 3.00 km = 1.24 km

The distance further it takes Peter to arrive at the Coffee shop than Mia = 1.24 km

2) Average \ speed = \dfrac{Total \ distance \ traveled}{Total \ time \ taken \  in \ the \ journey}

Therefore, \ Mia's \ average \ speed = \dfrac{3.00 \ km}{0.5 \ hour}= 6.00 \ km/hour

Mia's average speed = 6.00 km/hour

Peter's \ average \ speed = \dfrac{4.24 \ km}{0.5 \ hour}= 8.48 \ km/hour

Peter's average speed = 8.48 km/hour

4) Average \ velocicty = \dfrac{Displacement }{Time  \ taken}

The displacement from the School to the Coffee shop is 3.00 km for both Mia and Peter

The time it takes both Peter and Mia to arrive at the Coffee shop from the school is 30 minutes = 0.5 hour

Therefore, \ Mia's \ average \ velocity = \dfrac{3.00 \ km}{0.5 \ hour}= 6.00 \ km/hour

Mia's average velocity = 6.00 km/hour

Peter's \ average \ velocity = \dfrac{3.00 \ km}{0.5 \ hour}= 6.00 \ km/hour

Therefore, Peter's average velocity is also = 6.00 km/hour

6 0
3 years ago
Iron + Hydrochloric acid --> ?
marusya05 [52]
<h3>Iron - Fe</h3>

<h3>Hydrochloric Acid- HCl</h3>

<h2><u>Solution</u></h2>

\bold{Fe  +2HCl \rightarrow FeCl _{}{ \tiny2}  + H{ \tiny{2}}}

\therefore \bold{\fbox{{Balanced}}}

Iron + Hydrochloric Acid \rightarrow Ferrous Chloride + Hydrogen

<h2>Hope This Helps You ❤️</h2>
8 0
3 years ago
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