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ivolga24 [154]
3 years ago
12

A piece of indium with a mass of 16.6 g is submerged in 46.3 cm3 of water in a graduated cylinder. The water level increases to

48.6 cm3. What is the density of indium to the proper number of significant figures
Physics
1 answer:
blagie [28]3 years ago
8 0

Answer:

the density of indium is  7.2 g/cm^3

Explanation:

The computation of the density of indium is shown below:

Given that

Mass = 16.6 g

Volume = 48.6 c,^3 - 46.3cm^3 = 2.3 cm^3

Based on the above information

As we know that

Density = mass  ÷ volume

So,

= 16.6g ÷ 2.3 cm^3

= 7.2 g/cm^3

hence, the density of indium is  7.2 g/cm^3

We simply applied the above formula so that the correct value could come

And, the same is to be considered  

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If a bag of pretzels says it contains 10 servings, and each serving is 150 calories, how many calories would you consume if you
Sindrei [870]

Answer:

5400

Explanation:

i calculated it

Big  brain

8 0
3 years ago
Suppose an individual is lying on his stomach with sheets of paper stacked on his back. If each sheet of paper has a mass of 0.0
AURORKA [14]

Answer:

N = 177843 sheets

Explanation:

We are given;

Mass;m = 0.0035 kg

Pressure; p = 101325 pa = 101325 N/m²

L = 0.279m

W = 0.216m

The weight of N sheets is N(mg)

Where;

m is the mass of one sheet

N is number of sheets

g is the acceleration due to gravity.

The pressure equals weight divided by the area on which the weight presses:

Thus,

p= F/A = Nmg/(L•W)

Therefore, making N the subject;

N = pLW/(mg)

N = 101325 x 0.279 x 0.216/ (0.0035 x 9.81)

N = 177843

5 0
3 years ago
What is the net force exerted by these two charges on a third charge q3 = 49.5 nC placed between q1 and q2 at x3 = -1.170 m ? Yo
insens350 [35]

Full Question

Consider two point charges located on the x axis: one charge, Q1 = -12.0 nC , is located at x1 = -1.705m ; the second charge, Q2 = 36.5 nC, is at the origin (x=0.0000).

What is the net force exerted by these two charges on a third charge q3 = 49.5 nC placed between q1 and q2 at x3 = -1.170 m ? Your answer may be positive or negative, depending on the direction of the force.

Answer:

6.79E6 N

Explanation:

Given

Q1 is negative and to the left of Q3 the force will be to the left

Q2 is positive and to the right of Q3 the the force will also be to the left

Net Force is calculated as:

Using Coulomb's law

Coulomb's law: F = kqQ / r²

the constant k = 8.99 x 10^9 N m2 / C2

F = -kQ1*Q3/(r1)² -kQ2*Q3/(r2)²)

F = -kQ3(Q1/(r1)² + Q2/(r2)²)

Where

Q1 = -12nC = -12 * 10^-9C

Q2 = 36.5nC = 36.5 * 10^-9C

Q3 = 49.5nC = 49.5 * 10^-9C

x1 = -1.705m

x2 = x = 0

x3 = -1.170m

r1 = x3 - x1

r1 = -1.170 - -1.705

r1 = -1.170 + 1.705

r1 = 0.535

r1² = 0.286225

r2 = x3

r2 = -1.170

r2² = -1.170²

r2² = 1.3689

So,

F = (-8.99 * 10^9)(49.5 *10^-9) [-12 * 10^-9/0.286225 + 36.5 * 10^-9/1.3689]

F = -445.005 (−4.192505895711E−8 + 2.6663744612462E−8)

F = -445.005 * −1.5261314344648E−8

F = -(8.99 * 10^9) * (49.5 * 10^-9) * [ (-12 * 10^-9) /(-1.770 - -1.705)² + (36.5 * 10^-9)/(-1.170)²]

F = -445.005( −0.000002813572941777538)

F = 0.00000679136118994008324

F = 6.79E6 N

3 0
3 years ago
Consider a metal single crystal oriented such that the normal to the slip plane and the slip direction are at angles of 60 and 3
Scilla [17]

Explanation:

Given that,

Angle by the normal to the slip α= 60°

Angle by the slip direction with the tensile axis β= 35°

Shear stress = 6.2 MPa

Applied stress = 12 MPa

We need to calculate the shear stress applied at the slip plane

Using formula of shear stress

\tau=\sigma\cos\alpha\cos\beta

Put the value into the formula

\tau=12\cos60\times\cos35

\tau=4.91\ MPa

Since, the shear stress applied at the slip plane is less than the critical resolved shear stress

So, The crystal will not yield.

Now, We need to calculate the applied stress necessary for the crystal to yield

Using formula of stress

\sigma=\dfrac{\tau_{c}}{\cos\alpha\cos\beta}

Put the value into the formula

\sigma=\dfrac{6.2}{\cos60\cos35}

\sigma=15.13\ MPa

Hence, This is the required solution.

3 0
3 years ago
hydraulic lift is to be used to lift a 2100-kg weight by putting a weight of 25 kg on a piston with a diameter of 10 cm. Determi
trasher [3.6K]

Answer:

840 cm

Explanation:

Note: A hydraulic press operate based on pascal's principle.

From pascal's principle

W₁/d₁ = W₂/d₂...................... Equation 1

Where W₁ and W₂ are the first and second weight, and d₁ and d₂ are the  first and second diameter of the piston.

make d₁ the subject of the equation

d₁ = W₁×d₂/W₂................ Equation 2

Given: W₁ = 2100 kg, W₂ = 25 kg, d₂ = 10 cm = 0.1 m.

Substitute these values into equation 2

d₁ = 2100(0.1)/25

d₁ = 8.4 m

d₁ = 840 cm

3 0
3 years ago
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