Which is a substance that takes the shape and volume of its container?

8. Fig. 4.1 shows a heavy ball B of weight W suspended from a fixed beam by two ropes P and Q.

mart [117]

Answer:

The resultant tension of the two ropes is approximately 42.4 N

The length of the line representing the resultant tension is approximately 8.48 cm

Please find included with the answer the scale drawing created with Microsoft Word

Explanation:

The given parameters are;

The tension in rope P, $T_P$ = 30 N

The tension in rope Q, $T_Q$ = 30 N

The angle the rope, 'P', makes with the horizontal = 45°

The angle the rope, 'Q', makes with the horizontal = 45°

The scale factor of the scale diagram, S.F. = 5.0 N/cm

By the resolution of forces at equilibrium, we have;

The sum of the vertical forces, $\Sigma F_y$ = $T_P_y$ + $T_Q_y$ + W = 0

∴ W = -( $T_P_y$ + $T_Q_y$ )

W = -(30 × sin(45°) + 30 × sin(45°)) = -42.4264068712

The weight of the heavy ball, W ≈ 42.4 N acting downwards

The sum of the horizontal forces, $\Sigma F_x$ = $T_P_x$ + $T_Q_x$ = 0

The length of the resultant force, W = W/(S.F.) ≈ 42.4 N/(5.0 N/cm) = 8.48 cm

The drawing of the vectors using the scale factor of 5.0 N/cm is created using Microsoft Word is included

3 0

3 years ago

An infinite sheet of charge, oriented perpendicular to the x-axis, passes through x = 0. It has a surface charge density σ1 = -2

pishuonlain [190]

Answer:

$E_{total}=4.82*10^6N/C$

vector with direction equal to the axis X.

Explanation:

We use the Gauss Law and the superposition law in order to solve this problem.

Superposition Law: the Total Electric field is the sum of the electric field of the first infinite sheet and the Electric field of the second infinite sheet:

$E_{total}=E_1+E_2$

Thanks Gauss Law we know that the electric field of a infinite sheet with density of charge σ is:

$E=\sigma/(2\epsilon_o)$

Then:

$E_{total}=(\sigma_1+\sigma_2)/(2\epsilon_o)=(-2.7*10^{-6}+88*10^{-6})/(2*8.85*10^{-12})=4.82*10^6N/C$

This electric field has a direction in the axis perpendicular to the sheets, that means it has the same direction as the axis X.

7 0

3 years ago

Read 2 more answers

A uniform log of length L is inclined 30° from the horizontal when supported by a frictionless rock located 0.6L from its left e

mafiozo [28]

Answer:

x = 0.974 L

Explanation:

given,

length of inclination of log = 30°

mass of log = 200 Kg

rock is located at = 0.6 L

L is the length of the log

mass of engineer = 53.5 Kg

let x be the distance from left at which log is horizontal.

For log to be horizontal system should be in equilibrium

∑ M = 0

mass of the log will be concentrated at the center

distance of rock from CM of log = 0.1 L

now,

∑ M = 0

$m_{log} g \times 0.1 L = m_{engineer} g \times (x - 0.6 L)$

$200 \times 0.1 L = 53.5 \times (x - 0.6 L)$

$0.374 L =x - 0.6 L$

x = 0.974 L

hence, distance of the engineer from the left side is equal to x = 0.974 L

7 0

3 years ago

How much power does a 2000 kg car need to accelerate from 20 m/s to 35 m/s in 7 seconds?

Alexus [3.1K]

firstly you get your acceleration with the formula, a=v-u/t. Then you use the formula for kinetic energy 1/2mv^2

then you can finally get the answer for power by dividing your previous answer by the time

3 0

3 years ago

A motorboat accelerates uniformly from a velocity of 6.5m/s to the west to a velocity of 1.5m/s to the west. if its accelerate w

grigory [225]

A motorboat accelerates uniformly from a velocity of 6.5m/s to the west to a velocity of 1.5m/s to the west. if its accelerate was 2.7m/s2 to the east , how far did it travel during the accelration? Give your answer in units of kilometers per hour/sec. To find the acceleration of the car we have to 
1. First determine the suitable formula for this word problem.
Which is a. A=vf-vi/t which will be
Given are: Vi= 6.5 m/s Vf= 1.5 m/s a= 2.7 m/sec2 t=1.85s
Solution: 
x = v0t + ½at2
x = 16.645375 m

7 0

3 years ago