A car goes from point A to point B, five miles away and then returns to point A. The car is going 15 mph.

Outside our solar system the closest star to earth is Proxima century life from the start takes 2200000 minutes to reach earth.

s2008m [1.1K]

Answer:

2200000 = 2.2E6 min for light from Proxima to reach earth

8.3 min from light sun to reach earth

2.2E6/8.3 = 2.56E5 times for light from Proxima

Proxima is about 256,000 times farther away than the sun

Since the sun is about 93,000,000 = 9.3E7 miles from earth

Proxima is then 9.3E7 * 2.56E5 = 2.4E13 miles away

Note - the speed of light is

3.00E8 m/s * 60 s/min / 1000 m/km = 1.8E7 km/min as given

5 0

2 years ago

A parachutist falls 50.0 m without friction. When the parachute opens, he slows down at a rate of 67 m/s*2. If he reaches the gr

KIM [24]

Answer:

3.49 seconds

3.75 seconds

-43200 ft/s²

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

$s=ut+\frac{1}{2}at^2\\\Rightarrow 50=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{50\times 2}{9.81}}\\\Rightarrow t=3.19\ s$

Time the parachutist falls without friction is 3.19 seconds

$v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 9.81\times 50+0^2}\\\Rightarrow v=31.32\ m/s$

Speed of the parachutist when he opens the parachute 31.32 m/s. Now, this will be considered as the initial velocity

$v=u+at\\\Rightarrow 11=31.32+9.81t\\\Rightarrow t=\frac{11-31.32}{-67}=0.3\ s$

So, time the parachutist stayed in the air was 3.19+0.3 = 3.49 seconds

$s=ut+\frac{1}{2}at^2\\\Rightarrow \frac{s}{2}=0t+\frac{1}{2}\times a\times t^2\\\Rightarrow \frac{s}{2}=\frac{1}{2}at^2$

$s=ut+\frac{1}{2}at^2\\\Rightarrow \frac{s}{2}=u1.1+\frac{1}{2}\times a\times 1.1^2$

Now the initial velocity of the last half height will be the final velocity of the first half height.

$v=u+at\\\Rightarrow v=at$

Since the height are equal

$\frac{1}{2}at^2=u1.1+\frac{1}{2}\times a\times 1.1^2\\\Rightarrow \frac{1}{2}at^2=at1.1+\frac{1}{2}\times a\times 1.1^2\\\Rightarrow 0.5t^2-1.1t-0.605=0\\\Rightarrow 500t^2-1100t-605=0$

$t=\frac{11\left(1+\sqrt{2}\right)}{10},\:t=\frac{11\left(1-\sqrt{2}\right)}{10}\\\Rightarrow t=2.65, -0.45$

Time taken to fall the first half is 2.65 seconds

Total time taken to fall is 2.65+1.1 = 3.75 seconds.

When an object is thrown with a velocity upwards then the velocity of the object at the point to where it was thrown becomes equal to the initial velocity.

$v^2-u^2=2as\\\Rightarrow a=\frac{v^2-u^2}{2s}\\\Rightarrow a=\frac{0^2-240^2}{2\times \frac{8}{12}}\\\Rightarrow a=-43200\ ft/s^2$

Magnitude of acceleration is -43200 ft/s²

5 0

3 years ago

An antigen is a protein made by your body to respond to a specific foreign molecule.

Tom [10]

hope this helps it's F

8 0

3 years ago

How many seconds will it take for a the International Space Station to travel 450 km at a rate of 100 m/s?

SVEN [57.7K]

Time = (distance) / (speed)

Time = (450 km) / (100 m/s)

Time = (450,000 m) / (100 m/s)

Time = <em>4500 seconds </em>(that's 75 minutes)

Note:

This is about HALF the speed of the passenger jet you fly in when you go to visit Grandma for Christmas.

If the International Space Station flew at this speed, it would immediately go ker-PLUNK into the ocean.

The speed of the International Space Station in its orbit is more like 3,100 m/s, not 100 m/s.

8 0

3 years ago

The Burj Khalifa is the tallest building in the world at 828 m. How much work would a man with a weight of 700 N do if he climbe

Hoochie [10]

Answer:

579600J

Explanation:

Given parameters:

Height of the building = 828m

Weight of the man = 700N

Unknown:

Work done by the man = ?

Solution:

The work done by the man is the same as the potential energy expended.

Work done:

Work done = Weight x height = 700 x 828

Work done = 579600J

6 0

2 years ago