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3 years ago

6

How much energy must be absorbed by water with a mass of 0.5 kg in order to raise the temperature from 30°C to 65°C? Note: Water

has a specific heat of 4,190 J/kg °C

1 answer:

Setler79 [48]3 years ago

8 0

Answer:

Q = 73325 J or 73.325 KJ

Explanation:

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Can you please answer qeastion

notsponge [240]

Answer:

the mantel

Explanation:

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3 years ago

Engineers at the Space Centre must determine the net force needed for a rockets engine to achieve an acceleration of 70 m/s2. As

Troyanec [42]

Answer:

3,150,000N

Explanation:

According to Newton's second law;

F = mass * acceleration

Given

Mass = 45000kg

acceleration = 70m/s^2

Substitute

F = 45000 * 70

F = 3,150,000N

Hence the force required to be produced by the rocket engines is 3,150,000N

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3 years ago

HELP CHECK ANSWERS ONLY 3 QUESTIONS:

maksim [4K]

Answer:

answer 1 is A, Answer 2 is A, Answer 3 is c

8 0

3 years ago

If the force of a golf club on a golf ball is 200 N forward, what will the force of the ball on the club be? A. 200 N forward B.

Korolek [52]

The ball should put 200 N of force towards the golfer.

Newton's Third Law is every action has an equal and opposite reaction.

It's the ball exerting 200 N of force towards the club as well, but the opposite reaction is that it flies away.

8 0

4 years ago

Read 2 more answers

A student buys a plastic dart gun and tries to find the maximum horizontal range. The student shoots the gun straight up and it

ryzh [129]

Answer:

The value is $R_{max} = 33.54 \ m$

Explanation:

From the question we are told that

The total time of flight is $t = 3.7 \ s$

Generally from kinematic equation

$v = u - g * \frac{t}{2}$

So v is the velocity at maximum height and the value is v = 0 m/s

So

$0 = u - 9.8 * \frac{ 3.7}{2}$

=> $u = 18.13 \ m/s$

Here u is the initial velocity of the dart as it leaves that gun

Gnerally the horizontal range of the dart is mathematically represented as

$R = \frac{u ^2 sin 2\theta }{g}$

For maximum horizontal range the value of $\theta = 45^o$

So

$R_{max} = \frac{ 18.13 ^2 sin 2(45) }{9.8}$

=> $R_{max} = 33.54 \ m$

6 0

3 years ago