Which of the following material offers the highest resistivity?

A metal smith pours 3.00 kg of lead shot at 99oc into 1.00 kg of water at 25oc in an isolated container. what is the final tempe

cluponka [151]

The answer is in attachment.

7 0

4 years ago

The knee extensors insert on the tibia at an angle of 30 degrees (from the longitudinal axis of the tibia), at a distance of 3 c

Reika [66]

Answer:

the knee extensors must exert 15.87 N

Explanation:

Given the data in the question;

mass m = 4.5 kg

radius of gyration k = 23 cm = 0.23 m

angle ∅ = 30°

∝ = 1 rad/s²

distance of 3 cm from the axis of rotation at the knee r = 3 cm = 0.03 m

using the expression;

ζ = I∝

ζ = mk²∝

we substitute

ζ = 4.5 × (0.23)² × 1

ζ = 0.23805 N-m

so

from; ζ = rFsin∅

F = ζ / rsin∅

we substitute

F = 0.23805 / (0.03 × sin( 30 ° )

F = 0.23805 / (0.03 × 0.5)

F F = 0.23805 / 0.015

F = 15.87 N

Therefore, the knee extensors must exert 15.87 N

7 0

3 years ago

Estimate the order of magnitude of the length in meters of each of the following: a. a ladybug b. your leg c. your school buildi

sukhopar [10]

The order of the magnitude of the length in meters is estimated based on the average length of the object: if it is a small object then the unit would be cm and if it is a long object (like a road or something) the distance can be measured in km. Then we convert the unit we measured in into the SI unit of the meter.

Based on this, for the mentioned objects, the estimated length would be as follows:
a- ladybug: 10^-2 meters
b- your leg : 10^0 meters
c- your school building : 10^1 to 10^2 meters
d- a giraffe: 10^0 meters
e- city block: 10^2 meters

4 0

3 years ago

1. What is work done in holding a 15kg suitcase while waiting for a bus for 15 minutes?

Tems11 [23]

Work is (force applied) x (distance through which the force moves).

Since the suitcase doesn't move up or down during the 15 minutes,
no work is done ... zero, zip, nada ... according to the real Physics
definition of 'work'.

6 0

3 years ago

Consider the cylindrical weir of diameter 3 m and length 6m. If the fluid on the left has a specific gravity of 0.8, find the ma

sladkih [1.3K]

This question is incomplete, the complete question is;

Consider the cylindrical weir of diameter 3m and length 6m. If the fluid on the left has a specific gravity of 1.6 and on the right has a specific gravity of 0.8, Find the magnitude and direction of the resultant force.

Answer:

- the magnitude of the resultant force is 557.32 kN

- the direction of resultant force is 48.29°

Explanation:

Given the data in the question and the diagram below,

First we work on the force on the left hand side.

Left Horizontal

$F_{LH$ = βgAr

here, h = 3/2 = 1.5 m, β = 1.6, g = 9.81 m/s², A = 3 m × 6 m = 18 m²

we substitute

$F_{LH$ = βgAh = ( 1.6 × 1000 ) × 9.81 × 18 × 1.5 = 423792 N

Left Vertical

$F_{LV$ = ( βgπh² / 2 ) × W

we substitute

$F_{LV$ = [ ( ( 1.6 × 1000 ) × 9.81 × π(1.5)² ) / 2 ] × 6 = 332845.458 N

Now we go to the right hand side

Right Horizontal

$F_{RH$ = βgAh

here, h' = 1.5/2 = 0.75 m, β = 0.8, g = 9.81 m/s², A = 1.5 m × 6 m = 9 m²

we substitute

$F_{RH$ = ( 0.8 × 1000 ) × 9.81 × 9 × 0.75 ) = 52974 N

Right Vertical

$F_{RV$ = ( βgπh² / 4 ) × W

we substitute

$F_{RV$ = [ ( ( 0.8 × 1000 ) × 9.81 × π(1.5)² ) / 4 ] × 6 = 83211.36 N

Hence

Fx = $F_{LH$ - $F_{RH$ = 52974 N - 423792 N = 370818 N

Fy = $F_{LV$ + $F_{RV$ = 332845.458 N + 83211.36 N = 416056.818 N

R = √( Fx² + Fy² ) = √[ (370818 N)² + (416056.818 N)² ] = 557323.3 N

R = 557.32 kN

Therefore, the magnitude of the resultant force is 557.32 kN

Direction of resultant force;

tanθ = Fy / Fx

we substitute

tanθ = 416056.818 N / 370818 N

tanθ = 1.121997

θ = tan⁻¹( 1.121997 )

θ = 48.29°

Therefore, the direction of resultant force is 48.29°

4 0

3 years ago