Which technology has allowed for space exploration

Physics

2 answers:

Margarita [4]2 years ago

6 0

It should be satellites but I’m not sure

Olegator [25]2 years ago

6 0

Answer:

I think it's the

The hubble telescope

You might be interested in

Vector ????⃗ has a magnitude of 16.6 and is at an angle of 50.5∘ counterclockwise from the +x‑axis. Vector ????⃗ has a magnitude

natka813 [3]

Answer:

For vector u, x component = 10.558 and y component =12.808

unit vector = 0.636 i+ 0.7716 j

For vector v, x component = 23.6316 and y component = -6.464

unit vector = 0.9645 i-0.2638 j

Explanation:

Let the vector u has magnitude 16.6

u makes an angle of 50.5° from x axis

So $u_x=ucos\Theta =16.6\times cos50.5=10.558$

Vertical component $u_y=usin\Theta =16.6\times sin50.5=12.808$

So vector u will be u = 10.558 i+12.808 j

Unit vector $u=\frac{10.558i+12.808j}{\sqrt{10.558^2+12.808^2}}=0.636i+0.7716j$

Now in second case let vector v has a magnitude of 24.5

Making an angle with -15.3° from x axis

So horizontal component $v_x=vcos\Theta =24.5\times cos(-15.3)=23.6316$

Vertical component $v_y=vsin\Theta =24.5\times sin(-15.3)=-6.464$

So vector v will be 23.6316 i - 6.464 j

Unit vector of v $=\frac{23.6316i-6.464}{\sqrt{23.6316^2+6.464^2}}=0.9645i-0.2638j$

8 0

2 years ago

Explain how the fountain pen is filled up with ink

Leno4ka [110]

Answer: A piston-filling fountain pen has a piston — just like in a car — inside the barrel. This piston goes down to expel air or ink and then back up, pulling ink into the barrel. The typical process is very simple, assuming the pen is clean and dry: Push the piston down, expelling any air in the barrel

8 0

2 years ago

Friction provides the force needed for a car to travel around a flat, circular race track. What is the maximum speed at which a

blagie [28]

Answer:

The maximum speed at which the car can safety travel around the track is 18.6m/s.

Explanation:

Since the car is in circular motion, there has to be a centripetal force $F_c$ . In this case, the only force that applies for that is the static frictional force $f_s$ between the tires and the track. Then, we can write that: