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3 years ago

How is your guys day going?

Physics

2 answers:

vichka [17]3 years ago

7 0

Answer:

Awesome and u made it even better when you gave me these free points

Explanation:

faltersainse [42]3 years ago

4 0

Good i’m tired how about you

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A land rover drops from rest a vertical distance of 1000 m to the surface of planet in 15 seconds what is the magnitude of the a

umka21 [38]

Answer:

66.6 or it could be 66.66 one of those

Explanation:

6 0

3 years ago

A comet is in an elliptical orbit around the Sun. Its closest approach to the Sun is a distance of 4.7 1010 m (inside the orbit

Lubov Fominskaja [6]

Answer:

58515.9 m/s

Explanation:

We are given that

$d_1=4.7\times 10^{10} m$

$v_i=9.5\times 10^4 m/s$

$d_2=6\times 10^{12} m$

We have to find the speed (vf).

Work done by surrounding particles=W=0 Therefore, initial energy is equal to final energy.

$K_i+U_i=K_f+U_f$

$\frac{1}{2}mv^2_i-\frac{GmM}{d_1}=\frac{1}{2}mv^2_f-\frac{GmM}{d_2}$

$\frac{1}{2}v^2_i-\frac{GM}{d_1}+\frac{GM}{d_2}=\frac{1}{2}v^2_f$

$v^2_f=2(\frac{1}{2}v^2_i-\frac{GM}{d_1}+\frac{GM}{d_2})$

$v_f=\sqrt{2(\frac{1}{2}v^2_i-\frac{GM}{d_1}+\frac{GM}{d_2})}$

Using the formula

$v_f=\sqrt{v^2_i+2GM(\frac{1}{d_2}-\frac{1}{d_1})}$

$v_f=\sqrt{(9.5\times 10^4)^2+2\times 6.7\times 10^{-11}\times 1.98\times 10^{30}(\frac{1}{6\times 10^{12}}-\frac{1}{4.7\times 10^{10})}$

Where mass of sun= $M=1.98\times 10^{30} kg$

$G=6.7\times 10^{-11}$

$v_f=58515.9 m/s$

4 0

3 years ago

Now assume that Alice and Bob are twins, and Alice left Earth and Bob stayed behind fixing his spaceship. If Alice spent some ti

alexgriva [62]

我們的確認為我可以幫你們解決問題，我們要不要買不起房屋貸款土地

8 0

3 years ago

What type of music classes would you lime to see next year?

Alexus [3.1K]

Answer:

If I were you, I would recommend a trumpet its the easiest instrument to play butits my opinion its your decision.

Explanation:

8 0

3 years ago

As a train enters the station it slows down from 40 m/s to a 10 m/s in 5 seconds

lora16 [44]

Answer:

See the answers below.

Explanation:

To solve this problem we must use the following equation of kinematics.

$v_{f}=v_{o}-a*t$

where:

Vf = final velocity = 10 [m/s]

Vo = initial velocity = 40 [m/s]

t = time = 5 [s]

a = acceleration [m/s²]

Now replacing:

$10=40-a*5\\40-10=a*5\\30=5*a\\a=6[m/s^{2}]$

Note: The negative sign in the above equation means that the velecity is decreasing.

To solve this second part we must use the following equation of kinematics.

$v_{f}^{2} =v_{o}^{2} -2*a*x\\$

where:

x = distance [m]

$(10)^{2} =(40)^{2} -2*6*x\\100=1600-12*x\\12*x=1600-100\\12*x=1500\\x=125[m]$

7 0

2 years ago

How is your guys day going?​

How is your guys day going?