Does anyone know this???

Please help me on this quickly

skelet666 [1.2K]

Answer:

b

Explanation:

basic knowledge on properties of different forms of matter

3 0

3 years ago

Taking the resistivity of platinoid as 3.3 x 10-7 m, find the resistance of 7.0 m of platinoid wire of average diameter 0.14 cm.

mr_godi [17]

Answer:

$1.5 \Omega$

Explanation:

The resistance of a wire is given by the equation:

$R=\rho \frac{L}{A}$

where

$\rho$ is the resistivity of the material

L is the length of the wire

A is the cross-sectional area of the wire

In this problem, we have a wire of platinoid, whose resistivity is

$\rho = 3.3\cdot 10^{-7} \Omega m$

The length of the wire is

L = 7.0 m

And its radius is

$r=\frac{0.14 cm}{2}=0.07 cm = 7\cdot 10^{-4} m$ , so the cross-sectional area is

$A=\pi r^2=\pi(7\cdot 10^{-4})^2=1.54\cdot 10^{-6}m^2$

Solving for R, we find the resistance of the wire:

$R=(3.3\cdot 10^{-7})\frac{7.0}{1.54\cdot 10^{-6}}=1.5 \Omega$

3 0

3 years ago

In a head-on collision, a ball of mass 0.3 kg travelling with velocity 2.8 m/s in the positive x-direction hits a stationary sec

Pie

Answer:

The final velocity of the first ball (0.3 kg) is 0.4 m/s in negative x direction.

Explanation:

Given;

mass of the first object, m₁ = 0.3 kg

initial velocity of the first ball, u₁ = 2.8 m/s

mass of the second ball, m₂ = 0.4 kg

initial velocity of the second ball, u₂ = 0

let the final velocity of the first ball, = v₁

let the final velocity of the second ball, = v₂

Apply the principle of conservation of linear momentum;

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

(0.3 x 2.8) + (0.4 x 0) = 0.3v₁ + 0.4v₂

0.84 = 0.3v₁ + 0.4v₂

2.8 = v₁ + 1.333v₂ -------equation (1)

Apply one-direction velocity;

u₁ + v₁ = u₂ + v₂

2.8 + v₁ = 0 + v₂

v₂ = 2.8 + v₁

substitute the value of v₂ into equation (1)

2.8 = v₁ + 1.333v₂

2.8 = v₁ + 1.333(2.8 + v₁)

2.8 = v₁ + 3.732 + 1.333v₁

2.8 - 3.732 = v₁ + 1.333v₁

-0.932 = 2.333v₁

v₁ = -0.932 / 2.333

v₁ = -0.4 m/s

Therefore, the final velocity of the first ball (0.3 kg) is 0.4 m/s in negative x direction.

8 0

3 years ago

To run the physics cart, the fan speed of the cart is manipulated. This is the (blank) variable. The cart accelerates due to the

vlabodo [156]

Answer:

A) Independent

B) Dependent

C) Mass

Explanation:

The speed applied to the physics cart is manipulated. This doesn't depend on another variable. Thus, speed is the (independent) variable.

The cart accelerates due to the speed applied to the cart. Acceleration depends on the speed applied. Thus, acceleration is therefore the (dependent) variable.

A “constant” is a parameter that remains the same regardless of the variables.

One parameter of the cart that is held constant is the (mass).

6 0

3 years ago

Design a tension member and slip-critical splice to carry a factored load of 500 kips. Please use a wide-flange section for the

Svetlanka [38]

Answer:

Kindly check the explanation section.

Explanation:

For the design we are asked for in this question/problem there is the need for us to calculate or determine the strength in fracture and that of the yield. Also, we need to calculate for the block shear strength.

From the question, we have that the factored load = 500kips. Also, note that the tension splice must not slip.

Also, the shear force are resisted by friction, that is to say shear resistance = 1.13 × Tb × Ns.

Assuming our db = 3/4 inches, then the slip critical resistance to shear service load = 18ksi(refer to AISC manual for the table).

If db = 7/8 inches, then the shear force resistance for n bolt = 10.2kips, n > 49.6.

The yielding strength = 0.9 × Aj × Fhb= 736 kips > 500

The fracture strength = .75 × Ah × Fhb = 309 kips.

The bearing strength of 7/8 inches bolt at the edge hole and other holes = 46 kips and 102 kips.

4 0

4 years ago