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gtnhenbr [62]
3 years ago
8

What is the weight of a 10-kilogram object on Earth?

Physics
1 answer:
Rasek [7]3 years ago
8 0
98 N weight on object
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The air you breathe is made of 21% oxygen and 78% nitrogen. The oxygen in air is the _____ and the nitrogen is the _____. soluti
siniylev [52]
The oxygen is the Solute, since this is smaller in quantity.

The Nitrogen is the Solvent, since this is more in quantity.

The air is the Solution, this is the mixture itself.
7 0
3 years ago
A toy rocket, launched from the ground, rises vertically with an acceleration of 28 m/s 2 for 9.7 s until its motor stops. Disre
vredina [299]

Answer:

5080.86m

Explanation:

We will divide the problem in parts 1 and 2, and write the equation of accelerated motion with those numbers, taking the upwards direction as positive. For the first part, we have:

y_1=y_{01}+v_{01}t+\frac{a_1t^2}{2}

v_1=v_{01}+a_1t

We must consider that it's launched from the ground (y_{01}=0m) and from rest (v_{01}=0m/s), with an upwards acceleration a_{1}=28m/s^2 that lasts a time t=9.7s.

We calculate then the height achieved in part 1:

y_1=(0m)+(0m/s)t+\frac{(28m/s^2)(9.7s)^2}{2}=1317.26m

And the velocity achieved in part 1:

v_1=(0m/s)+(28m/s^2)(9.7s)=271.6m/s

We do the same for part 2, but now we must consider that the initial height is the one achieved in part 1 (y_{02}=1317.26m) and its initial velocity is the one achieved in part 1 (v_{02}=271.6m/s), now in free fall, which means with a downwards acceleration a_{2}=-9,8m/s^2. For the data we have it's faster to use the formula v_f^2=v_0^2+2ad, where d will be the displacement, or difference between maximum height and starting height of part 2, and the final velocity at maximum height we know must be 0m/s, so we have:

v_{02}^2+2a_2(y_2-y_{02})=v_2^2=0m/s

Then, to get y_2, we do:

2a_2(y_2-y_{02})=-v_{02}^2

y_2-y_{02}=-\frac{v_{02}^2}{2a_2}

y_2=y_{02}-\frac{v_{02}^2}{2a_2}

And we substitute the values:

y_2=y_{02}-\frac{v_{02}^2}{2a_2}=(1317.26m)-\frac{(271.6m/s)^2}{2(-9.8m/s^2)}=5080.86m

3 0
3 years ago
A phonograph record 0.15 m in its radius rotates 18 times per 90 seconds what is the frequency?
ioda

Answer:

The frequency of the phonograph record is 0.2 Hz

Explanation:

The frequency of an object moving in uniform circular motion is the number of completed cycles the object makes in a specified time period

The given parameters of the phonograph record are;

The radius of the record = 0.15 m

The number of times the phonograph record rotates, n = 18 times

The time it takes the phonograph record to rotate the 18 times, t = 90 seconds

The frequency of the phonograph record, f = (The number of times the phonograph record rotates) ÷ (The time it takes the phonograph record to rotate the 18 times)

∴ The frequency of the phonograph record, f = n/t = 18/(90 s) = 0.2 Hz

The frequency of the phonograph record = 0.2 Hz.

6 0
2 years ago
3. A 142 g baseball is thrown at a speed of 42.9 m/s. What is the kinetic energy of the baseball at this moment?
krok68 [10]
Kinetic energy=1/2mv^2
=1/2(142*10^-3)(42.9)^2=130.6=131J
8 0
2 years ago
A paper is placed in between two books can be quickly pulled out without movinf the books. whuch statement explains this phenome
Montano1993 [528]
This is easily explained saying that the frictional force between the books and the paper isn't big enough to produce a displacement in the books. The displacement in the books doesn't happen because the frictional force between the books and the surface they are standing on is bigger than the paper's one.
7 0
2 years ago
Read 2 more answers
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