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3 years ago

7

WHOEVER GET IT RIGHT GETS 50 POINTS

1 answer:

AVprozaik [17]3 years ago

3 0

Answer:

ANSWER BELOW I

I

V

Remember that w=mg where w is weight in Newtons, m is mass in kilograms, and g is gravity in

m/s2

. For example, for Earth, 445 N = 45.4 × 9.8

m/s2

:Notice that the x-axis values will be gravity in

m/s2

, which is already given in the table, and the y-axis values will be the weight in Newtons. Remember to round your weights to a whole number, and to enter the points starting with the lowest gravity (moon, then Mars, then Venus, then Earth).

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Which of these accurately describes the products of a reaction?

stiv31 [10]

The one that accurately describes the products of a reaction is : B. new substances that are present at the end of a reaction
For example the process of photosynthesis transform CO2 and other nutrients into O2 and H2O

hope this helps

3 0

2 years ago

Read 2 more answers

A light pointer is stuck to the rubber sheet so that it pivots about a point P near the

deff fn [24]

I’m not going to church tomorrow or Friday I don’t want to go go back up

8 0

1 year ago

If a cart of 10 kg mass has a force of 5 newtons exerted on it, what is its acceleration? m/s2

Elza [17]

Answer:

= 0.5 m/s²

Explanation:

According to Newton's second law of motion, the resultant force is directly proportion to the rate of change of linear momentum.

Therefore;<em> F = ma , where F is the Force, m is the mass and a is the acceleration.</em>

<em>Thus; a = F/m</em>

<em>but; F = 5 N, and m = 10 kg</em>

<em> a = 5 /10</em>

<u>= 0.5 m/s²</u>

4 0

2 years ago

Read 2 more answers

A train whistle is heard at 300 Hz as the train approaches town. The train cuts its speed in half as it nears the station, and t

spin [16.1K]

Answer:

The speed of the train before and after slowing down is 22.12 m/s and 11.06 m/s, respectively.

Explanation:

We can calculate the speed of the train using the Doppler equation:

$f = f_{0}\frac{v + v_{o}}{v - v_{s}}$

Where:

f₀: is the emitted frequency

f: is the frequency heard by the observer

v: is the speed of the sound = 343 m/s

$v_{o}$ : is the speed of the observer = 0 (it is heard in the town)

$v_{s}$ : is the speed of the source =?

The frequency of the train before slowing down is given by:

$f_{b} = f_{0}\frac{v}{v - v_{s_{b}}}$ (1)

Now, the frequency of the train after slowing down is:

$f_{a} = f_{0}\frac{v}{v - v_{s_{a}}}$ (2)

Dividing equation (1) by (2) we have:

$\frac{f_{b}}{f_{a}} = \frac{f_{0}\frac{v}{v - v_{s_{b}}}}{f_{0}\frac{v}{v - v_{s_{a}}}}$

$\frac{f_{b}}{f_{a}} = \frac{v - v_{s_{a}}}{v - v_{s_{b}}}$ (3)

Also, we know that the speed of the train when it is slowing down is half the initial speed so:

$v_{s_{b}} = 2v_{s_{a}}$ (4)

Now, by entering equation (4) into (3) we have:

$\frac{f_{b}}{f_{a}} = \frac{v - v_{s_{a}}}{v - 2v_{s_{a}}}$

$\frac{300 Hz}{290 Hz} = \frac{343 m/s - v_{s_{a}}}{343 m/s - 2v_{s_{a}}}$

By solving the above equation for $v_{s_{a}}$ we can find the speed of the train after slowing down:

$v_{s_{a}} = 11.06 m/s$

Finally, the speed of the train before slowing down is:

$v_{s_{b}} = 11.06 m/s*2 = 22.12 m/s$

Therefore, the speed of the train before and after slowing down is 22.12 m/s and 11.06 m/s, respectively.

I hope it helps you!

7 0

2 years ago

A 1kw electric heater is switched on for ten minutes. how much heat does it produce

Ainat [17]

H= P × t

1kW= 1000w

10 min = 600s

H= 1000×600=600,000J

=> 143.40kcal

4 0

2 years ago