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BaLLatris [955]

3 years ago

13

Which of the following is a legal requirement for boat operation?

1 answer:

Korvikt [17]3 years ago

3 0

Answer:

B

Explanation:

it's iSpy because negative it's always see does not match what he was trying to say orgy but I ate would say that but no it's not I'mma teach about the way

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What is the smallest possible piece an element in broken down into my normal laboratory processes

lozanna [386]

answer would be: A. atom

8 0

3 years ago

Two identical small charged spheres hang in equilibrium with equal masses (0.02kg). The length of the strings is equal (0.18m) a

Yuliya22 [10]

Answer:

The value is $q = 3.4 *10^{-6} \ C$

Explanation:

From the question we are told that

The mass of each sphere is $m_1 = m_2 = m = 0.020 \ kg$

The length of the string is $l = 0.18 \ m$

The angle of with the vertical is $\theta = 7^o$

The acceleration due to gravity is $g = 9.8 \ m/s^2$

Generally the force acting between the forces is mathematically represented as

$F = T cos \theta = \frac{k* q^2}{ r^2}$

=> $T cos \theta = \frac{k* q^2}{ r^2}$

Generally from Pythagoras theorem the radius of the circular curve created by the force is

$r = 2 L sin (\theta )$

=> $r = 2* 0.180 sin (7)$

=> $r = 0.043 \ m$

=> $q = tan \theta * \frac{m * g * r^2 }{k}$

=> $q = tan(7)* \frac{ 0.02 * 9.8 * 0.043^2 }{9*10^{9}}$

=> $q = 3.4 *10^{-6} \ C$

7 0

3 years ago

A person of mass 70 kg stands at the center of a rotating merry-go-round platform of radius 2.9 m and moment of inertia 900 kg⋅m

Cloud [144]

Explanation:

It is given that,

Mass of person, m = 70 kg

Radius of merry go round, r = 2.9 m

The moment of inertia, $I_1=900\ kg.m^2$

Initial angular velocity of the platform, $\omega=0.95\ rad/s$

Part A,

Let $\omega_2$ is the angular velocity when the person reaches the edge. We need to find it. It can be calculated using the conservation of angular momentum as :

$I_1\omega_1=I_2\omega_2$

Here, $I_2=I_1+mr^2$

$I_1\omega_1=(I_1+mr^2)\omega_2$

$900\times 0.95=(900+70\times (2.9)^2)\omega_2$

Solving the above equation, we get the value as :

$\omega_2=0.574\ rad/s$

Part B,

The initial rotational kinetic energy is given by :

$k_i=\dfrac{1}{2}I_1\omega_1^2$

$k_i=\dfrac{1}{2}\times 900\times (0.95)^2$

$k_i=406.12\ rad/s$

The final rotational kinetic energy is given by :

$k_f=\dfrac{1}{2}(I_1+mr^2)\omega_1^2$

$k_f=\dfrac{1}{2}\times (900+70\times (2.9)^2)(0.574)^2$

$k_f=245.24\ rad/s$

Hence, this is the required solution.

5 0

3 years ago

During a braking test, a car is brought to rest beginning from an initial speed of 60 mi/hr in a distance of 120 ft. With the sa

maw [93]

Answer:

Explanation:

Given

Initial speed $u=60\ mi/hr\approx 88\ ft/s$

distance traveled before coming to rest $d_1=120\ ft$

using equation of motion

$v^2-u^2=2as$

where v=final velocity

u=initial velocity

a=acceleration

s=displacement

$0-(88)^2=2\times a\times 120---1$

for $u_2=80\ mi/hr\approx 117.33\ ft/s$

using same relation we get

$0-(117.33)^2=2\times a\times (d_2)----2$

divide 1 and 2 we get

$(\frac{88}{117.33})^2=\frac{120}{d_2}$

$d_2=213.32\ ft$

So a distance if 213.32 ft is required to stop the vehicle with 80 mph speed

8 0

3 years ago

There are two tuning forks struck at the same time. One tuning fork is tuned to a frequency note of 500 Hz. The other is tuned t

fomenos

Answer: 26 beats in 2 seconds

Explanation:

The number of beats per second = frequency of tuning fork 1 - frequency of tuning fork 2.

Given :

- frequency of tuning fork 1 = 500Hz

- frequency of tuning fork 2 = 487Hz

Thus,

Beats per second = 500 - 487

= 13 beats per second

Therefore in two(2) seconds, you will have 2 x 13 = 26 seconds.

6 0

3 years ago