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Levart [38]
4 years ago
14

On a planet far, far away, an astronaut picks up a rock. The rock has a mass of 4.80 kg , and on this particular planet its weig

ht is 41.5 N . You may want to review (Pages 121 - 129) . Part A If the astronaut exerts an upward force of 48.1 N on the rock, what is its acceleration
Physics
1 answer:
Oksanka [162]4 years ago
3 0

Answer:

   a = 1,375 m / s²

Explanation:

For this exercise we use Newton's second law where the force applied by the astronaut is F = 48.1, the body weight is W = 41.5 N and the body mass is m = 4.80 kg

              F - W = m a

              a = (F -W) / m

let's calculate

              a = (48.1 - 41.5) / 4.80

              a = 1,375 m / s²

this acceleration is directed upwards

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If there is a huge boulder on your lawn and you want to determine its density, but it is
Olin [163]

There are different options here but all of them work by approximating and assuming.

i) that the boulder is above ground.

ii) that the bottom surface of the boulder is known.

iii) the shape of the boulder is taken into account.

The most accurate way is measuring it by displacement method but the boulder is immovable hence the volume can be calculated by measuring the boulder or a waterproof box to be built around the boulder and calculate the volume occupied by boulder.

All the above methods are estimating methods.

*Another way to find the density is through specific gravity.

S.G = <u>Density</u><u> </u><u>of</u><u> </u><u>object</u>

Density of water

If the material that makes the boulder is known that is if it's stone or a mineral then the specific gravity can be found.

If the boulder is purely rock then S.G lies between 3 - 3.5 and the density of water is known thus the density of the boulder can be found without moving the boulder.

This is what I think after correction and allthe best!

3 0
2 years ago
Read 2 more answers
PLEASE HELP!!! GIVING BRAINLIEST!! ill also answer questions that you have posted if you answer this correctly!!!! (40pts)
melisa1 [442]
D.)downward toward the center of earth
7 0
3 years ago
The sense of equilibrium responds to movements of the head. It is divided into two types--static and dynamic equilibrium. Descri
Sonja [21]

Answer:

nodding of head ,yes- static equilibrium

nodding of head, no- dynamic equilibrium.

Explanation:

static equilibrium monitors head position when body is not moving .

dynamic equilibrium monitors the angular or rotational movements of the head when body moves.

5 0
3 years ago
Read 2 more answers
Is it better to wire a house using a series circuit or a parallel circuit?
Svetach [21]

Answer: its better to use parallel because, in parallel connection there will be more advantages than a series connection. and also the electronic devices are wired in series so thats why you should use parralel in house wiring

so its c.

because the parallel is wired through the whole house so if one of the circuits fail youŕe not screwed

Explanation:

5 0
3 years ago
A 36.0 kg box initially at rest is pushed 5.00 m along a rough, horizontal floor with a constant applied horizontal force of 130
konstantin123 [22]

Answer:

(a) W = 650J

(b) Wf = 529.2J

(c) W = 0J

(d) W = 0J

(e) ΔK = 120.8J

(f) v2 = 2.58 m/s

Explanation:

(a) In order to find the work done by the applied force you use the following formula:

W=Fd      (1)

F: applied force = 130N

d: distance = 5.0m

W=(130N)(5.0m)=650J

The work done by the applied force is 650J

(b) The increase in the internal energy of the box-floor system is given by the work done of the friction force, which is calculated as follow:

W_f=F_fd=\mu Mgd       (2)

μ: coefficient of friction = 0.300

M: mass of the box = 36.0kg

g: gravitational constant = 9.8 m/s^2

W_f=(0.300)(36.0kg)(9.8m/s^2)(5.0m)=529.2J

The increase in the internal energy is 529.2J

(c) The normal force does not make work on the box because the normal force is perpendicular to the motion of the box.

W = 0J

(d) The same for the work done by the normal force. The work done by the gravitational force is zero because the motion of the box is perpendicular o the direction of the gravitational force.

(e) The change in the kinetic energy is given by the net work on the box. You use the following formula:

\Delta K=W_T         (3)

You calculate the total work:

W_T=Fd-F_fd=(F-F_f)d     (4)

F: applied force = 130N

Ff: friction force

d: distance = 5.00m

The friction force is:

F_f=(0.300)(36.0kg)(9.8m/s^2)=105.84N

Next, you replace the values of all parameters in the equation (4):

W_T=(130N-105.84N)(5.00m)=120.80J

\Delta K=120.80J

The change in the kinetic energy of the box is 120.8J

(e) The final speed of the box is calculated by using the equation (3):

W_T=\frac{1}{2}M(v_2^2-v_1^2)       (5)

v2: final speed of the box

v1: initial speed of the box = 0 m/s

You solve the equation (5) for v2:

v_2 = \sqrt{\frac{2W_T}{M}}=\sqrt{\frac{2(120.8J)}{36.0kg}}=2.58\frac{m}{s}

The final speed of the box is 2.58m/s

5 0
3 years ago
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