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Korolek [52]
3 years ago
14

What is the five things to do in P.E

Physics
2 answers:
Nimfa-mama [501]3 years ago
6 0

Answer:

Fucd6

Explanation:

Cuufufcuvjgjvug7fuguguguguf7

sergejj [24]3 years ago
6 0

Answer:

1: Stretches

2: Mile run

3: Sit ups

4:Push ups

5:Planks

Explanation:

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Why do bones weaken as a person gets older
Sonbull [250]

Answer:

The bones aren't as strong as a younger person, because an older person, ages due to time and it also depend on what they do in their past.

Does this help? ^-^"

6 0
2 years ago
A grapefruit falls from a tree and hits the ground 0.76 s later.
Tomtit [17]

Answer:

2.8 m 7.4 m/s

Explanation:

write all the values then use the equations of motion to find the distance and speed. please see attached photo

7 0
3 years ago
I NEED HELP ON THIS QUESTION!
Leto [7]

The second option is the correct one. m/s^2

7 0
3 years ago
Every few hundred years most of the planets line up on the same side of the Sun.(Figure 1)Calculate the total force on the Earth
mylen [45]

Answer: 3.7 \times 10^{-4} N

Explanation:

The gravitational pull between two object is given by:

F = G\frac{Mm}{r^2}

Where M and m are the masses of the object, r is the distance between the masses and G = 6.67× 10⁻¹¹ m³kg⁻¹ s⁻² is the gravitational constant.

We have to calculate the net force on Earth due to Venus, Jupiter and Saturn when they are in one line. It means when they are the closest distance.

F_{net] = G\frac{M_eM_v}{r_v^2}+G\frac{M_eM_j}{r_j^2}+G\frac{M_eM_s}{r_s^2}

Mass of Earth, Me = 5.98 × 10²⁴ kg

Mass of Venus, Mv = 0.815 Me

Mass of Jupiter, Mj = 318 Me

Mass of Saturn, Ms = 95.1 Me

closest distance between Earth and Venus, rv = 38 × 10⁶ km = 0.25 AU

closest distance between Jupiter and Earth, rj = 588 × 10⁶ km = 3.93 AU

closest distance between Earth and Saturn, rs = 1.2 × 10⁹ km = 8.0 AU

where 1 AU = 1.5 × 10¹¹ m

Inserting the values:

F_{net} = G\frac{M_e\times 0.815 M_e}{(0.25AU)^2}+G\frac{M_e\times 318 M_e}{(3.93AU)^2}+G\frac{M_e\times 95.1 M_e}{(8.0AU)^2}\\ \Rightarrow F_{net} = \frac{(GM_e^2)}{(1AU)^2}(\frac{0.815}{0.25^2}+\frac{318}{3.93^2}+\frac{95.1}{8.0^2})=\frac{6.67\times 10^{-11} \times (5.98\times 10^{24})^2}{(1.5\times 10^{11})^2}(35.1) = 3.7 \times 10^{-4} N

4 0
3 years ago
Read 2 more answers
A car is traveling at 8 m/s accelerates at 3.1 m/s^2 to reach a final top speed of 56 m/s. How much time will pass before the ca
AlekseyPX

Please find attached photograph for your answer.

Hope it helps.

Do comment if you have any query.

3 0
3 years ago
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