All categories

3 years ago

How to solve part (b)?

Physics

1 answer:

maria [59]3 years ago

8 0

Answer:

1.11m

Explanation:

From the diagram we are given the following forces;

F1 = 24.3N

F3 = 30N

Since the sum of upward forces is equal to that of downward force, then;

F2 = F1 + F3

F2 = 24.3N + 30N

F2 = 54.3N

Required

Distance between B and C

First we need to get Length of AC

Take moment about A

Anticlockwise moment = F3 cos20 * AC

Anticlockwise moment = 30ACcos 20

Clockwise moment = 1.2 * F2

Clockwise moment = 1.2(54.3) = 65.16Nm

Applying the principle of moment;

Sum of ACW moment = Sum of CW moments

30ACcos 20 = 65.16

AC = 65.16/30cos20

AC = 65.16/28.19

AC = 2.31m

Get the distance BC

AC = AB + BC

BC = AC-AB

BC = 2.31 - 1.2

BC = 1.11m

Hence the separation between B and C is 1.11m

Note that the force F1 got in (a) was the value used in the calculation.

You might be interested in

What specific space object is shown in the picture below?

PIT_PIT [208]

From context clues, I believe the correct answer is B) Red Dwarf!

3 0

3 years ago

6. According to Ohm's Law, current (I), voltage (V), and resistance (R)

SIZIF [17.4K]

Answer: C

Explanation: according to the law voltages = to current and resistance

7 0

3 years ago

What energy change takes place when the wrecking ball strikes the wall?

Ilia_Sergeevich [38]

Answer:

Explanation:

The wrecking ball transfers kinetic energy to the wall.

3 0

2 years ago

A proton travels with a speed of 4.2×106 m/s at an angle of 30◦ west of north. A magnetic field of 2.5 T points to the north. Fi

Arada [10]

Answer:

$8.4\cdot 10^{-13} N$

Explanation:

The magnitude of the magnetic force on the proton is given by:

$F=qvB sin \theta$

where:

$q=1.6\cdot 10^{-19} C$ is the proton charge

$v=4.2\cdot 10^6 m/s$ is the proton velocity

$B=2.5 T$ is the magnetic field

$\theta=30^{\circ}$ is the angle between the direction of v and B

Substituting into the formula, we find

$F=(1.6\cdot 10^{-19}C)(4.2\cdot 10^6 m/s)(2.5 T) sin 30^{\circ}=8.4\cdot 10^{-13} N$

6 0

3 years ago

A kangaroo jumps up with an initial velocity of 36 feet persecond from the ground (assume its starting height is 0 feet).Use the

kkurt [141]

Given

Initial velocity:

36 ft/s

Initial height:

0 ft

Vertical motion model:

h(t) = -16t^2 + ut + s

v = initial velocity

s = is the height

Procedure

We are going to use the model provided for the vertical motion.

$\begin{gathered} h(t)=-16t^2+36t+0 \\ h(t)=-16t^2+36t \end{gathered}$

We know that at the maximum height the final velocity is 0.

Then we will use the following expression to calculate the maximum height:

$\begin{gathered} v^2_f=v^2_o-2ah_{\max } \\ 0=v^2_o-2ah_{\max } \\ 2ah_{\max }=v^2_o \\ h_{\max }=\frac{v^2_o}{2a} \\ h_{\max }=\frac{(36ft/s)^2}{2\cdot32ft/s^2} \\ h_{\max }=20.25\text{ ft} \end{gathered}$

Now for time:

$\begin{gathered} 20.25=-16t^2+36t \\ 16t^2-36t+20.25=0 \end{gathered}$

Solving for t,

$\begin{gathered} t_1=2.25 \\ t_2=0 \end{gathered}$

The total time the kangaroo takes in the air is 2.3s.

3 0

1 year ago

How to solve part (b)?​

How to solve part (b)?