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Orlov [11]
3 years ago
6

A material that breaks down quickly in the environment is called??

Physics
2 answers:
irinina [24]3 years ago
5 0
A material that breaks down quickly in the environment is called Biodegradable.
Artist 52 [7]3 years ago
3 0
Biodegradable material
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Ted wants to hang a wall clock on the wall by using a string. If the mass of the wall clock is 0. 250 kilograms, what should be
antoniya [11.8K]
Sum of all forces = mass * acceleration

Ft= tension force
Fw= force of gravity (Fw= mass* acceleration of gravity which is 9.8 this only applies to force of gravity)

Ft- Fw = 0 (there is no acceleration)
Ft = Fw
Ft= m*g
Ft= 0.250kg*9.8m/s
Ft= 2.45N

7 0
2 years ago
What is the “p” process for elements heavier than iron
tia_tia [17]
Supernova nucleosynthesis is also thought to be responsible for the creation of rarerelements heavier than iron<span> and nickel, in the last few seconds of a type II supernova event.</span>
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3 years ago
What are the two varieties of friction?
murzikaleks [220]
There are two main types of friction, static friction and kinetic friction. Static friction operates between
6 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
A roller coaster has a "hump" and a "loop" for riders to enjoy (see picture). The top of the hump has a radius of curvature of 1
nikklg [1K]

Answer:

Part a)

F_n = 306 N

Part B)

v = 12.1 m/s

Explanation:

Part A)

At the top of the hump the force on the rider is

1) Normal force

2) weight

so here we know that

mg - F_n = \frac{mv^2}{R}

F_n = mg - \frac{mv^2}{R}

F_n = (100)(9.81) - \frac{100(9^2)}{12}

F_n = 306 N

Part B)

At the top of the loop we will have

F_n + mg = \frac{mv^2}{R}

in order to remain in contact the normal force must be just greater than zero

so we will have

mg = \frac{mv^2}{R}

v = \sqrt{Rg}

v = \sqrt{15\times 9.81}

v = 12.1 m/s

5 0
3 years ago
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