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zlopas
2 days ago
6

A silver block of silver block of density 10.5 g/cm3 has a volume of 30 cm3. Which of the following is the correct mass of the b

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Physics
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Suppose a rectangular piece of aluminum has a length D, and its square cross section has the dimensions W XW, where D (W x W) to
kicyunya [3294]

Answer:

R₂ / R₁ = D / L

Explanation:

The resistance for a metal can be calculated by

        R = ρ L / A

Where ρ indicates the resistivity of aluminum, L is the resistance's length, and A indicates the cross-sectional area

We use this formula for both configurations

For small face measurements (W x W)

The length is

         L = W

Area  

         A = W W = W²

         R₁ = ρ W / W² = ρ / W

For larger face measurements (D x L)

       Length L = D= 2W

       Area     A = W L

     R₂ = ρ D / WL = ρ 2W / W L = 2 ρ/L

From this, we find the relation to be

    R₂ / R₁ = 2W²/L

6 0
3 months ago
A fire hose held near the ground shoots water at a speed of 6.5 m/s. At what angle(s) should the nozzle point in order that the
inna [3103]
The velocity of water can be decomposed into its vertical and horizontal components:
v_x = 6.5 cos \theta \\ v_y = 6.5 sin \theta

The vertical component will exhibit a parabolic trajectory due to gravity, while the horizontal component will be linear:
y(t) = -4.9t^2 + (6.5sin \theta) t \\ \\ x(t) = (6.5 cos \theta) t
To determine when the water reaches the ground 2.5m away, set y= 0 and x = 2.5
-4.9t^2 + (6.5sin \theta) t=0 \\ \\ t = \frac{6.5}{4.9} sin \theta \\ \\(6.5 cos \theta)(\frac{6.5}{4.9} sin \theta) = 2.5 \\ \\ sin \theta cos \theta = 0.29 \\ \\ sin 2\theta = 0.58 \\ \\ 2\theta = 35.4, 144.6 \\ \\ \theta = 17.7,72.3
8 0
3 months ago
A roundabout in a fairground requires an input power of 2.5 kW when operating at a constant angular velocity of 0.47 rad s–1 . (
ValentinkaMS [3465]
Since the roundabout operates at a constant angular velocity, the input power equals the frictional power. Given that the frictional power is 2.5 kW, we can express this as frictional torque multiplied by angular velocity: frictional torque x 0.47 = 2.5 kW. Therefore, solving for frictional torque gives us 2.5 / 0.47 kN.m, which amounts to approximately 5.32 kN.m, leading to a rounded value of 5 kN.m. When the power supply is interrupted, the roundabout experiences deceleration due to the influence of the frictional torque.
5 0
2 months ago
What is the equation describing the motion of a mass on the end of a spring which is stretched 8.8 cm from equilibrium and then
Softa [3030]
The equation for simple harmonic motion (SHM) is as follows.
6 0
3 months ago
Read 2 more answers
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