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2 years ago

How much heat does it take to raise the temperature of 7.0 kg of water from

Physics

2 answers:

aksik [14]2 years ago

6 0

Answer:

non of the above

Explanation:

Quantity of heat = mass× specific heat× change in temperature

m= 7kg c= 4.18 temp= 46-25=21°

.......H= 7×4.18×21= 614.46kJ

Leviafan [203]2 years ago

3 0

Answer:610 KJ

Explanation:A P E X answers

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Ajoba and Prav drive to work. Ajoba drives 45 miles in 2.5 hours. Prav drives 74 km in 1 hour 15 min. Work out the difference be

Serggg [28]

Explanation:

We have,

Ajoba and Prav drive to work. Ajoba drives 45 miles in 2.5 hours. Prav drives 74 km in 1 hour 15 min.

1 mile = 1.6 km

45 miles = 72.42 km

74 miles = 119.0 km

1 hour 15 min means 1.25 hours

Average speed of Ajoba is :

$v_1=\dfrac{72.42 }{2.5}=28.96\ km/h$

Average speed of Prav,

$v_2=\dfrac{119}{1.25}=95.2\ km/h$

Difference in average speed of Ajoba and Prav is :

$v=v_2-v_1\\\\v=v_2-v_1\\\\v=95.2-28.96\\\\v=66.24\ km/h$

So, the difference in average speed of Ajoba and Prav is 66.24 km/h.

7 0

2 years ago

If the final speed of an object as is strikes the ground is 77 m/s and it was in the air for 6.5 seconds. What was the initial d

azamat

Answer:

The answer to your question is: 13.2 m/s

Explanation:

final speed (fs) = 77 m/s

t = 6.5 s

gravity (g) = 9.81 m/s2

initial speed (is) = ?

Formula

fs = is + gt from this equation we clear "is" = fs - gt

Substitution is = 77 - (9,81)(6.5)

Process is = 77 - 63.8

is = 13.2 m/s

8 0

2 years ago

(See picture) may I have help!!?

algol13

Picture is blurry…. try re uploading it

6 0

1 year ago

A voltage V is applied to the primary coil of a step-up transformer with a 3:1 ratio of turns between its primary and secondary

Snezhnost [94]

Explanation:

Let $N_p\ and\ N_s$ are the number of turns in primary and secondary coil of the transformer such that,

$\dfrac{N_p}{N_s}=\dfrac{1}{3}$

A resistor R connected to the secondary dissipates a power $P_s=100\ W$

For a transformer, $\dfrac{N_s}{N_p}=\dfrac{V_s}{V_p}$

$V_s=(\dfrac{N_s}{N_p})V_p$

$V_s=3V_p$ ...............(1)

The power dissipated through the secondary coil is :

$P_s=\dfrac{V_s^2}{R}$

$100\ W=\dfrac{V_s^2}{R}$

$V_p^2=\dfrac{100R}{9}$ .............(2)

Let $N_p'\ and\ N_s'$ are the new number of turns in primary and secondary coil of the transformer such that,

$\dfrac{N_p'}{N_s'}=\dfrac{1}{24}$

New voltage is :

$V_s'=(\dfrac{N_s'}{N_p'})V_p'$

$V_s'=24V_p$ ...............(3)

So, new power dissipated is $P_s'$

$P_s'=\dfrac{V_s'^2}{R}$

$P_s'=\dfrac{(24V_p)^2}{R}$

$P_s'=24^2\times \dfrac{(V_p)^2}{R}$

$P_s'=24^2\times \dfrac{(\dfrac{100R}{9})}{R}$

$P_s'=6400\ Watts$

So, the new power dissipated by the same resistor is 6400 watts. Hence, this is the required solution.

3 0

2 years ago

Which of the following is true concerning nuclear fusion?

Zielflug [23.3K]

Choice-'a' is a slippery, misleading, ambiguous statement,
but it's less wrong than any of the other choices on this list.

6 0

2 years ago