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4 years ago

675 km equals how much cm ?

1 answer:

7 0

KHDMDCM.
Now go from Kilometer to Centimeter: 5.
Move the decimal 5 places to the right: 67,500,000 centimeters.
Hope this helps :)

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A satellite of mass 5460 kg orbits the Earth and has a period of 6520 s

Nikolay [14]

Answer:

what if I do and b then someone else c I don't have enough time pls

8 0

3 years ago

The gravitational field strength on earth is 10n/kg. find the weight of an object of mass 25kg

vodka [1.7K]

Answer:

250N

Explanation:

weight = Mass(in kg) × Gravitational field strength

25 × 10 = 250N

6 0

3 years ago

Read 2 more answers

Please help as fast you can

Ilya [14]

Answer:

d a

Explanation:

7 0

3 years ago

hovering mosquito is hit by a raindrop that is 45 times as massive and falling at 8.1 m/s , a typical raindrop speed. How fast i

Y_Kistochka [10]

Answer:

7.92 m/s

Explanation:

$m_1$ = Mass of raindrop = $45m_2$

$m_2$ = Mass of mosquito

$u_1$ = Initial Velocity of raindrop = 8.1 m/s

$u_2$ = Initial Velocity of mosquito = 0 m/s

$v$ = Velocity of center of mass

For elastic collision

$m_1u_1 + m_2u_2 =(m_1 + m_2)v\\\Rightarrow v=\frac{m_1u_1 + m_2u_2}{m_1 + m_2}\\\Rightarrow v=\frac{45m_2\times 8.1 + m_2\times 0}{45m_2 + m_2}\\\Rightarrow v=\frac{364.5m_2}{46m_2}\\\Rightarrow v=7.92\ m/s$

Hence, the velocity of the attached mosquito, falling immediately afterward is 7.92 m/s

4 0

3 years ago

One strategy in a snowball fight is to throw a snowball at a high angle over level ground. While your opponent is watching the f

Andreas93 [3]

Answer:

Part a)

$\theta_2 = 15 degree$

Part b)

$\Delta t = 2.88 s$

Explanation:

Part a)

In order to have same range for same initial speed we can say

$R_1 = R_2$

$\frac{v^2 sin2\theta_1}{g} = \frac{v^2 sin2\theta_2}{g}$

so after comparing above we will have

$\theta_1 = 90 - \theta$

so we have

$75 = 90 - \theta_2$

$\theta_2 = 15 degree$

Part b)

Time of flight for the first ball is given as

$T_1 = \frac{2vsin\theta}{g}$

$T_1 = \frac{2(20)sin75}{9.81}$

$T_1 = 3.94 s$

Now for other angle of projection time is given as

$T_2 = \frac{2(20)sin15}{9.81}$

$T_2 = 1.05 s$

So here the time lag between two is given as

$\Delta t = T_1 - T_2$

$\Delta t = 3.94 - 1.05$

$\Delta t = 2.88 s$

5 0

3 years ago