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4 years ago

Help pleasee i give brainliest

Physics

1 answer:

MA_775_DIABLO [31]4 years ago

5 0

Answer:

A is a solid. C is a gas. In solid an liquid the particals are touching. In C, the particals have less affect on each other because they are so far apart.

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What will happen in the next moment for the girl on the swing?

guajiro [1.7K]

Answer:

Increase, same, same

Explanation:

8 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

(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

Hurricane katrina, which hit the gulf coast of louisiana and mississippi on august 29, 2005, had the second lowest ever recorded

Simora [160]

The unit 'mb' means millibar which is equivalent to 1/1000 of 1 bar. To convert the units from bar to atmospheres (atm) and to inches Hg (inHg), we need to know the conversion factors.

a.) 1 atm = 1.01325 bar

0.92 mb(1 bar/1000 mbar)(1 atm/1.01325 bar) =<em> 9.08×10⁻⁴ atm</em>

b.) 1 bar = 29.53 inHg

0.92 mb(1 bar/1000 mbar)(29.53 inHg/1 bar) =<em> 0.027 inHg</em>

3 0

3 years ago

An alternating-current (AC) source supplies a sinusoidally varying voltage that can be described with the function v of t is equ

Marrrta [24]

Answer:

ω, the angular frequency of the source equals 377 rad/s

Explanation:

From the question, V(t) = V cosωt.

Now, ω = the angular frequency of the sinusoidal wave is given by

ω = 2πf where f = the frequency of the source = 60 Hz

So, the angular frequency of the source ,ω = 2π × the frequency of the source.

So, ω = 2πf

ω = 2π × 60 Hz

ω = 120π rad/s

ω = 376.99 rad/s

ω ≅ 377 rad/s

So, ω, the angular frequency of the source equals 377 rad/s

3 0

3 years ago

At the top of the path, the horozontal velocity is___.

Setler [38]

Answer:

the answer is A have a nice day hope I helped

4 0

3 years ago