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laiz [17]
3 years ago
13

IF THERE ARE ONLY 118 ELEMENTS, HOW DO YOU ACCOUNT FOR THE MANY MILLIONS OF THINGS THAT WE HAVE IN OUR UNIVERSE?

Physics
1 answer:
bogdanovich [222]3 years ago
3 0

Answer:

theres only 118 elements that are discovered. now that they're the only ones out there

Explanation:

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A gas is confined in a cylinder with a piston. What happens when work is done on the gas.
goblinko [34]
It combusts in to a vapor

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3 years ago
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jeka94

الملك - الإتاوات - المحاكم - المجلس الملكي - القانون الروماني

5 0
3 years ago
A person is standing on a level floor. His head, upper
BabaBlast [244]

Answer:

y_{cg} = 1.03 m

Explanation:

Given data:

weigh (head+arms + head) w_1 = 438 N

centre of gravity y_1= 1.28 m

weigh (upper leg) w_2 = 144 N

Center of gravity y_2 = 0.760 m

weigh ( lower leg + feet) = 87 N

centre of gravity = y_3 = 0.250 m

location of center of gravity = \frac{w_1 y_1 + w_2 y_2 + w_3 y_3}{w_1 +W_2 +w_3}

y_{cg} = \frac{438 \times 1.28 + 144\times 0.760 + 87 \times 0.250}{438+144+87}

y_{cg} = 1.03 m

8 0
4 years ago
A parallel-plate capacitor in air has a plate separation of 1.25 cm and a plate area of 25.0 cm2. The plates are charged to a po
Archy [21]

Answer:

(a) Since net charge remains same,after immersion Q is same

(b) I. 14.56pF ii. 3.05V

(c) ΔU = 5.204nJ

Explanation:

a)

C = kεA/d

k=1 for air

ε is 8.85x10-12F/m

A = .0025m2

d = .125m

C = 8.85x10-12x.0025/.125 = 1.77x10-13F = 0.177pF

Q = CV = .177pF * 244V = 43.188pC

Since net charge remains same,after immersion Q is same

b)

C = kεA/d, for distilled water k is approx. 80

Cwater = Cair x k

=0.177pF x 80 = 14.16pF

Q is same and C is changed V=Q/c holds. where Q is still 43.188pC and C is now 14.16pF, so V = 43.188pC/14.16pF = 3.05V

c) Change in energy: ΔU = Uwater - Uair

Uwater = Q2/2C = (43.188)2/2x.177pF = 5.27nJ

Uair = Q2/2C = (43.188)2/2x14.16pF = 0.066nJ

ΔU = 5.204nJ

6 0
3 years ago
A block of mass M1 = 288.0 kg sits on an inclined plane and is connected to a bucket with a massless string over a massless and
AlekseyPX

Answer:

M2 = 278.06 kg

Explanation:

We calculate the weight of M1

W=m*g

Where

m: mass (kg)

g: acceleration due to gravity (m/s²)

W₁=288* 9.8= 2822.4 N

Look at the attached graphic

We calculate the x-y components of the weight :

W₁x= 2822.4*sin41° N =1851.66 N

W₁y= 2822.4 *cos41° N = 2130.09 N

We apply Newton's first law for the balance in M1:

Σ Fy=0

Fn-W₁y=0  ,   Fn: normal force

Fn=W₁y=2130.09N

Friction Force = Ff=μs *Fn = 0.41*2130.09 =873.34 N

Σ Fx=0

T- W₁x- Ff=0

T= 1851.66 + 873.34

T= 1851.66 + 873.34

T=2725 N

We apply Newton's first law for the balance in M2:

Σ Fy=0

T- W₂ =0

W₂ = T = 2725 N

W₂ = M2*g

M2 = W₂/g

M2 = 2725/9.8

M2 = 278.06 kg

6 0
3 years ago
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