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2 years ago

An astronaut drops a feather from 1.2 m above

Physics

1 answer:

vfiekz [6]2 years ago

7 0

This is a free fall and in free fall we use this formula:

d = (1 ÷ 2) × g × t², where d is the distance, g is the gravitational acceleration and t is the time.

In our case,

We are already given the moon's gravitational acceleration and we are going to substitute it with g. Let's leave the unknown alone, which is t.

t = √(2d ÷ g)

If we perform the formula, t is found to be √(2d ÷ g) = √(2 × 1.2 ÷ 1.62) ≅ 1.217 s

I am sorry for my bad English and if there is anything that you do not understand please let me know.

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A piece of luggage is being loaded onto an airplane by way of an inclined conveyor belt. The bag, which has a mass of 15.0 kg, t

LenKa [72]

Answer:

a) W = - 318.26 J, b) W = 0 , c) W = 318.275 J , d) W = 318.275 J , e) W = 0

Explanation:

The work is defined by

W = F .ds = F ds cos θ

Bold indicate vectors

We create a reference system where the x-axis is parallel to the ramp and the axis and perpendicular, in the attached we see a scheme of the forces

Let's use trigonometry to break down weight

sin θ = Wₓ / W

Wₓ = W sin 60

cos θ = Wy / W

Wy = W cos 60

X axis

How the body is going at constant speed

fr - Wₓ = 0

fr = mg sin 60

fr = 15 9.8 sin 60

fr = 127.31 N

Y Axis

N - Wy = 0

N = mg cos 60

N = 15 9.8 cos 60

N = 73.5 N

Let's calculate the different jobs

a) The work of the force of gravity is

W = mg L cos θ

Where the angles are between the weight and the displacement is

θ = 60 + 90 = 150

W = 15 9.8 2.50 cos 150

W = - 318.26 J

b) The work of the normal force

From Newton's equations

N = Wy = W cos 60

N = mg cos 60

W = N L cos 90

W = 0

c) The work of the friction force

W = fr L cos 0

W = 127.31 2.50

W = 318.275 J

d) as the body is going at constant speed the force of the tape is equal to the force of friction

W = F L cos 0

W = 127.31 2.50

W = 318.275 J

e) the net force

F ’= fr - Wx = 0

W = F ’L cos 0

W = 0

4 0

2 years ago

An electrical conductor is an element with __________ electrons in its outer orbit.

Setler [38]

An electric conductor is an element with free electrons in its outer orbit

5 0

2 years ago

Small frogs that are good jumpers are capable of remarkable accelerations. One species reaches a takeoff speed of3.7 m/s in60 ms

MAVERICK [17]

We know that acceleration is change in velocity by time taken for that change.

In this case velocity change is 3.7 m/s

Time taken for this change = 60 ms = $6 *10^{-3} seconds$

So acceleration of frog = $\frac{3.7}{60*10^{-3}}$

= 61.66 m/ $s^2$

So acceleration of frog is 61.66 m/ $s^2$

o it is evident that frog is capable of remarkable accelerations.

8 0

2 years ago

Read 2 more answers

A large crate sits on the floor of a warehouse. Paul and Bob apply constant horizontal forces to the crate. The force applied by

Delicious77 [7]

Answer:

W = -510.98J

Explanation:

Force = 43N, 61° SW

Displacement = 12m, 22° NE

Work done is given as:

W = F*d*cosA

where A = angle between force and displacement.

Angle between force and displacement, A = 61 + 90 + 22 = 172°

W = 43 * 12 * cos172

W = -510.98J

The negative sign shows that the work done is in the opposite direction of the force applied to it.

6 0

2 years ago

In a supermarket, you place a 22.3-N (around 5 lb) bag of oranges on a scale, and the scale starts to oscillate at 2.7 Hz. What

allsm [11]

Answer:

Force constant, k = 653.3 N/m

Explanation:

It is given that,

Weight of the bag of oranges on a scale, W = 22.3 N

Let m is the mass of the bag of oranges,

$m=\dfrac{W}{g}$

$m=\dfrac{22.3}{9.8}$

m = 2.27 kg

Frequency of the oscillation of the scale, f = 2.7 Hz

We need to find the force constant (spring constant) of the spring of the scale. We know that the formula of the frequency of oscillation of the spring is given by :

$f=\dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}}$

$k=4\pi^2 f^2m$

$k=4\pi^2 \times (2.7)^2\times 2.27$

k = 653.3 N/m

So, the force constant of the spring of the scale is 653.3 N/m. Hence, this is the required solution.

7 0

3 years ago