Which of the following is NOT usually published with a scientific report?

A space probe lands on a newly discovered planet. A small canister is released from the probe and falls a distance of 3 m in 0.5

ivanzaharov [21]

Answer:

24m/s²

Explanation:

Given

Distance S = 3m

Time of fall = 0.5sec

Required

Acceleration due to gravity

Using the equation of motion

S = ut+1/2gt²

Substitute the given values

3 = 0+1/2g(0.5)²

3 = 1/2(0.25)g

3 = 0.125g

g = 3/0.125

g = 24

Hence the value for the acceleration of gravity on this new planet is 24m/s²

3 0

2 years ago

Will give Brianliest answer. help.

dimaraw [331]

Answer:6.2

Explanation:

3 0

2 years ago

I really need help with this to be able to pass this last semester this is about (Circular Motion ) on physics

ankoles [38]

Question 1
To find centripetal acceleration, use the formula : centripetal acceleration = v^2/r
so answer would be (3.71)^2/42.85=0.32 (2d.p.)
Question 2
Force =ma
a= (9.98)^2/31.77=3.1350
Force= 3.1350 * 56.63 = 177.54 (2 d.p.)

4 0

3 years ago

Read 2 more answers

A roller coaster car is traveling at a constant 3 m/s when it reaches a downward slope. On the slope, the car accelerates at a c

kirill [66]

Answer:

Velocity of the car at the bottom of the slope: approximately $20.3\; \rm m \cdot s^{-2}$ .

It would take approximately $3.9\; \rm s$ for the car to travel from the top of the slope to the bottom.

Explanation:

The time of the travel needs to be found. Hence, make use of the SUVAT equation that does not include time.

Let $v$ denote the final velocity of the car.
Let $u$ denote the initial velocity of the car.
Let $a$ denote the acceleration of the car.
Let $x$ denote the distance that this car travelled.

$v^2 - u^2 = 2\, a\cdot x$ .

Given:

$u = 3\; \rm m \cdot s^{-1}$ .
$a = 4.5\; \rm m \cdot s^{-2}$ .
$x = 45\; \rm m$ .

Rearrange the equation $v^2 - u^2 = 2\, a\cdot x$ and solve for $v$ :

$\begin{aligned}v &= \sqrt{2\, a \cdot x + u^2} \\ &= \sqrt{2 \times 4.5\; \rm m \cdot s^{-2} \times 45\; \rm m + \left(3\; \rm m \cdot s^{-1}\right)^{2}} \\ &\approx 20.3\; \rm m \cdot s^{-1}\end{aligned}$ .

Calculate the time required for reaching this speed from $u = 3\; \rm m \cdot s^{-1}$ at $a = 4.5\; \rm m \cdot s^{-2}$ :

$\begin{aligned}t &= \frac{v - u}{a} \\ &\approx \frac{20.3\; \rm m \cdot s^{-1} - 3\; \rm m \cdot s^{-1}}{4.5\; \rm m \cdot s^{-2}} \approx 3.9\; \rm m \cdot s^{-1}\end{aligned}$ .

3 0

3 years ago

Your friend's car is stuck in the snow. You push very hard and become very 1 tired but the car won't move. Did you do work?

crimeas [40]

Answer:

Yes you worked

Explanation:

You may have not seen a good outcome but you did put effort in.

7 0

3 years ago