The diagram below show the alignment of the sun moon and earth

7.1 Project Guidelines 2021

Ede4ka [16]

I’m not sure if this will help but I found: https://prezi.com/l0fa6du3b9kp/going-off-the-grid-assignment/?fallback=1 and

3 0

2 years ago

True or false. when objects collide , some momentum is lost

ycow [4]

Answer:

It is neither false nor true. When they collide some of one of the objects goes to the other object.

Explanation:

5 0

2 years ago

Read 2 more answers

An example of constant velocity

pashok25 [27]

Some examples of constant velocity (or at least almost- constant velocity) motion include (among many others): • A car traveling at constant speed without changing direction. A hockey puck sliding across ice. A space probe that is drifting through interstellar space.

4 0

2 years ago

A stone is thrown with an initial speed of 11.5 m/s at an angle of 50.0 above the horizontal from the top of a 30.0-m-tall build

rewona [7]

Answer:

The magnitude of the horizontal displacement of the rock is 7.39 m/s.

Explanation:

Given that,

Initial speed = 11.5 m/s

Angle = 50.0

Height = 30.0 m

We need to calculate the horizontal displacement of the rock

Using formula of horizontal component

$v_{x}=u\cos\theta$

Put the value into the formula

$v_{x}=11.5\times\cos50$

$v_{x}=7.39\ m/s$

Hence, The magnitude of the horizontal displacement of the rock is 7.39 m/s.

8 0

3 years ago

Daring Darless wishes to cross the Grand Canyon of the Snake River by being shot from a cannon. She wishes to be launched at 65

guajiro [1.7K]

Answer:

She must be launched with minimum speed of <u>57.67 m/s</u> to clear the 520 m gap.

Step-by-step explanation:

Given:

The angle of projection of the projectile is, $\theta =65$ °

Range of the projectile is, $R=520$ m.

Acceleration due to gravity, $g=9.8\ m/s^2$

The minimum speed to cross the gap is the initial speed of the projectile and can be determined using the formula for range of projectile.

The range of projectile is given as:

$R=\frac{v_{0}^2\sin2\theta}{g}$

Plug in all the given values and solve for minimum speed, $v_0$ .

$520=\frac{v_{0}^2\sin(2(65))}{9.8}\\520\times 9.8=v_{0}^2\sin(130)\\5096=1.532v_{0}^2\\v_0^2=\frac{5096}{1.532}\\v_0^2=3326.371\\v_0=\sqrt{3326.371}=57.67\textrm{ m/s}$

Therefore, she must be launched with minimum speed of 57.67 m/s to clear the 520 m gap.

3 0

3 years ago