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MAXImum [283]
4 years ago
6

What is the density of a 15g block of metal with a volume of 2cm3?

Physics
2 answers:
UkoKoshka [18]4 years ago
5 0

Answer: Density = 7.5 g / cm³

Explanation: Density is the ratio between mass and volume.

D= m / V

= 15 g / 2 cm³

= 7.5 g / cm³

JulsSmile [24]4 years ago
4 0

Answer:

density =mass/volume

Explanation:

density=15/2

density=7.5 g/cm³

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Assume that the loop is initially positioned at θ=30∘θ=30∘ and the current flowing into the loop is 0.500 AA . If the magnitude
labwork [276]

Answer:\tau=1.03\times 10^{-4}\ N-m

Torque,

Explanation:

Given that,

The loop is positioned at an angle of 30 degrees.

Current in the loop, I = 0.5 A

The magnitude of the magnetic field is 0.300 T, B = 0.3 T

We need to find the net torque about the vertical axis of the current loop due to the interaction of the current with the magnetic field. We know that the torque is given by :

\tau=NIAB\ \sin\theta

Let us assume that, A=0.0008\ m^2

\theta is the angle between normal and the magnetic field, \theta=90^{\circ}-30^{\circ}=60^{\circ}

Torque is given by :

\tau=1\times 0.5\ A\times 0.0008\ m^2\times 0.3\ T\ \sin(60)\\\\\tau=1.03\times 10^{-4}\ N-m

So, the net torque about the vertical axis is 1.03\times 10^{-4}\ N-m. Hence, this is the required solution.

4 0
3 years ago
An airplane in a holding pattern flies at constant altitude along a circular path of radius 3.38 km. If the airplane rounds half
Andrej [43]

Answer:

a) 6076 m

b) 43.33 m/s

c) 68 m/s

Explanation:

(a) If the airplane rounds half the circle in 156s, its displacement is the circle diameter in 156s, or twice the circle's radius

s = 2r = 2* 3.38km = 6.76 km or 6760 m

(b) The average velocity would be displacement over unit of time

v = s/t = 6760 / 156 = 43.33 m/s

(c) The length of the chord it's swept in 156s is half of the circle perimeter

c = πr = π3.38 = 10.62 km or 10620 m

The airplane average speed is its chord length over a unit of time

c / t = c / 156 = 68 m/s

4 0
3 years ago
Acceleration increases over time once a force is applied to the object. Determine the acceleration at 3.5 sec? A) 3 m/s2 B) 6 m/
Novay_Z [31]

Answer: Acceleration does not increase over time once a force is applied to the object. It depends on the force and the object's mass. If those don't change, then the acceleration is constant.

Explanation:. F = M • a

We don't know either of those numbers, so we can't answer the question.

4 0
3 years ago
Read 2 more answers
Consider a box sitting in the back of a pickup. The pickup accelerates to the right, and because the bed of the pickup is sticky
andreyandreev [35.5K]
The force would be coming from the right causing the box the lean/ slide to left, if it wasnt sticky.
8 0
4 years ago
An aluminum wire is held between two clamps under zero tension at room temperature. Reducing the temperature, which results in a
Blababa [14]

Answer:

\frac{\Delta L}{L} =5.37\times 10^{-4}

Explanation:

Given:

  • cross sectional area of the wire, A=5.75\times 10^{-6}\ m^2
  • density of aluminium wire, \rho=2.7\times 10^3\ kg.m^{-3}
  • young's modulus of the material, E=7\times 10^{10}\ N.m^{-2}
  • wave speed, v=118\ m.s^{-1}

<u>We have mathematical expression for strain as:</u>

\frac{\Delta L}{L} =\frac{\sigma}{E} ...............................(1)

and since, \sigma =\frac{T}{A}

where, T = tension force in the wire

equation (1) becomes:

\frac{\Delta L}{L} =\frac{T}{A.E} ............................(2)

<u>Also velocity ofwave in tensed wire:</u>

v=\sqrt{\frac{T}{\mu} } ...................................(3)

where: \mu= linear mass density of the wire

\therefore \mu=\rho \times A

Now, equation (3) becomes

v=\sqrt{\frac{T}{\rho \times A} }

T=v^2.\rho \times A ............................(4)

Using eq. (2) & (4) for tension T

v^2.\rho \times A=A.E\times \frac{\Delta L}{L}

\frac{\Delta L}{L} =\frac{v^2.\rho}{E}

putting the respective values

\frac{\Delta L}{L} =\frac{118^2\times 2.7\times 10^3}{7\times 10^{10}}

\frac{\Delta L}{L} =5.37\times 10^{-4}

6 0
3 years ago
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