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3 years ago

A 15 cm3 block of gold weighs 2.8 N. It is carefully submerged in a tank of mercury. One cm3 of mercury weighs 0.13 N.

1 answer:

5 0

The density of gold is 2.8 N / 15 = 0.18 units
The density of mercury is 0.13/1 = 0.13 units

Since the density of gold is more than density of mercury, it will sink.
Since the gold will sink, it will displace mercury, but less than its own volume

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A 500 lines per mm diffraction grating is illuminated by light of wavelength 580 nm . what is the maximum diffraction order seen

vampirchik [111]

The maximum diffraction order seen is 3.

<h3>What is the maximum diffraction order seen?</h3>

We know that the maximum angle of diffraction Q_m of the furthest bright fringe < Q = 90 degrees.

Here we need to compute the nth bright fringe for which is approximated to 90 degrees.

The angle of nth bright fringe is given by;

sin(Q_m) = n(λ)N

Approximating Q_m ≈ 90 degrees.

sin (90) = nλN

n = sin (90) / (λN)

n = 1 / ((580 x 10⁻⁶)500)

n = 3.5 orders

Since, we knew that Q_m < 90 degrees, we will choose n = 3 as the maximum number of orders.

Thus, the maximum diffraction order seen is 3.

Learn more about maximum diffraction here: brainly.com/question/14703089

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6 0

1 year ago

Please show work with the answer and thank you

Alika [10]

Answer:

its your homework

Explanation:

you aren't getting any free answers out of me because it sounds like your trying to be smart your welcome

(Not smarter than me)

7 0

2 years ago

Calculate the total energy of 2.0 kg object moving horizontally at 10 m/s 50 meters above the surface

mina [271]

We have:

Total Energy: KE + GPE
KE (Kinetic Energy) = $\frac{1}{2} m*v^2$
GPE (Gravitational Potential Energy) = $m*g*h$

Data:
m (mass) = 2.0 Kg
v (speed) = 10 m/s
h (height) = 50 m
Use: g (gravity) = 10 m/s²

Formula:

Total Energy: KE + GPE
$TE = \frac{1}{2} m*v^2 + m*g*h$

Solving:
$TE = \frac{1}{2} m*v^2 + m*g*h$
$TE = \frac{1}{2} *2.0*10^2 + 2.0*10*50$
$TE = \frac{2.0*100}{2} + 1000$
$TE = \frac{200}{2} + 1000$
$TE = 100 + 1000$
$\boxed{\boxed{TE = 1100\:Joule}}\end{array}}\qquad\quad\checkmark$

5 0

2 years ago

Read 2 more answers

Alarge plate is fabricated from a steel alloy that has a plane strain fracture toughness of 82.4 MPa m1/2. If the plate is expos

Korvikt [17]

Answer:

minimum length of a surface crack is 18.3 mm

Explanation:

Given data

plane strain fracture toughness K = 82.4 MPa m1/2

stress σ = 345 MPa

Y = 1

to find out

the minimum length of a surface crack

solution

we will calculate length by this formula

length = 1/π ( K / σ Y)²

put all value

length = 1/π ( K / σ Y)²

length = 1/π ( 82.4 $10^{3/2}$ / 345× 1)²

length = 18.3 mm

minimum length of a surface crack is 18.3 mm

4 0

2 years ago

A particle is moving along the x-axis. Its position as a function of time is given as x=bt-ct^2a) What must be the units of the

lisabon 2012 [21]

Answer:

We are given x= bt +ct²

A. bxt= m

Because m/s*s= m

So b= m/s and c= m/s²

x= bt-ct²

So at x=0 t=0

x=0 t= 2

We have

bt = ct² so t = b/c at x= 0

So b-2ct= 0

B. To find velocity we use

dx / dt = b - 2 Ct

C. At rest wen V= 0

We have t= b/2c

D. To find acceleration we use

dv / dt = - 2C

3 0

3 years ago