All categories

3 years ago

Can a falling object reach terminal velocity in outer space?

2 answers:

6 0

Answer: lth air resistance acting on an object that has been dropped, the object will eventually reach a terminal velocity, which is around 53 m/s (195 km/h or 122 mph) for a human skydiver (on the moon, the gravitational acceleration is much less on earth, approximately 1.6 m/s2.)

Explanation:

Triss [41]3 years ago

4 0

Answer:

There is no gravity in outer space so how will any object fall in the first place

Also the principle of terminal velocity is limited to free fall where the acceleration is equal to the gravity of the planet

You might be interested in

The Apollo Lunar Module was used to make the transition from the spacecraft to the Moon's surface and back. Consider a similar m

ivolga24 [154]

Answer:

v=6.05 m/s

Explanation:

Given that,

Th initial velocity of the lander, u = 1.2 m/s

The lander is at a height of 1.8 m, d = 1.8 m

We need to find the velocity of the lander at impact. It is a concept based on the conservation of mechanical energy. So,

$\dfrac{1}{2}mv^2-\dfrac{1}{2}mu^2=W\\\\\dfrac{1}{2}mv^2-\dfrac{1}{2}mu^2=F\times d\\\\\dfrac{1}{2}mv^2-\dfrac{1}{2}mu^2=mgd\\\\\dfrac{1}{2}m(v^2-u^2)=mgd\\\\v^2-u^2=2gd$

v is the velocity of the lander at the impact

g is the acceleration due to gravity on the surface of Mars, which is 0.4 times that on the surface of the Earth, g = 0.4 × 9.8 = 3.92 m/s²

So,

$v=\sqrt{u^2+2gd} \\\\v=\sqrt{(1.2)^2+2\times 9.8\times 1.8} \\\\v=6.05\ m/s$

So, the velocity of the lander at the impact is 6.05 m/s.

6 0

3 years ago

The diagram shows a box in the periodic table. 15 P Phosphorus 31.0 What is the atomic number of the element shown? A. 46.0 B. 1

natali 33 [55]

Answer:

b 15

Explanation:

the atomic number is the number of the element

4 0

3 years ago

A stationary police car emits a sound of frequency 1240 HzHz that bounces off of a car on the highway and returns with a frequen

Tju [1.3M]

Answer

given,

frequency from Police car= 1240 Hz

frequency of sound after return = 1275 Hz

Calculating the speed of the car = ?

Using Doppler's effect formula

Frequency received by the other car

$f_1 = \dfrac{f_0(u + v)}{u}$ ..........(1)

u is the speed of sound = 340 m/s

v is the speed of the car

Frequency of the police car received

$f_2= \dfrac{f_1(u)}{u-v}$

now, inserting the value of equation (1)

$f_2= f_0\dfrac{u+v}{u-v}$

$1275=1240\times \dfrac{340+v}{340-v}$

1.02822(340 - v) = 340 + v

2.02822 v = 340 x 0.028822

2.02822 v = 9.799

v = 4.83 m/s

hence, the speed of the car is equal to v = 4.83 m/s

5 0

3 years ago

What do the arrows at point three indicate

Maru [420]

Maybe show a picture ? I don’t get the question .

5 0

3 years ago

Consider the uniform electric field E = (8.0ĵ + 2.0 ) ✕ 103 N/C. What is its electric flux (in N · m2/C) through a circular area

cluponka [151]

Answer:

5.09 x 10⁵ Nm²/C

Explanation:

The electric flux φ through a planar area is defined as the electric field Ε times the component of the area Α perpendicular to the field. i.e

φ = E A

From the question;

E = (8.0j + 2.0k) ✕ 10³ N/C

r = radius of the circular area = 9.0m

A = area of a circle = π r² [Take π = 3.142]

A = 3.142 x 9² = 254.502m²

Now, since the area lies in the x-y plane, only the z-component of the electric field is responsible for the electric flux through the circular area.

Therefore;

φ = (2.0) x 10³ x 254.502

φ = 5.09 x 10⁵ Nm²/C

The electric flux is 5.09 x 10⁵ Nm²/C

4 0

3 years ago