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Marta_Voda [28]

2 years ago

11

A boat's capacity plate gives the maximum weight and/or number of people the boat can carry safely in certain weather conditions

. what are these conditions?

1 answer:

erastova [34]2 years ago

5 0

•wind
•snow
•high tide/low tide
•thunder/lightning storms

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For the meter stick shown in figure 10-4, the force F1 10.0 N acts at 10.0 cm. What is the magnitude of torque due to F1 about a

Phantasy [73]

Torque is equal position vector times (r) times force vector (F). Since F= 10 N and r = 0.1 m, so the torque is equal to (10 N) x ( 0.1 m) = 1Nm. The direction of the torque would be into the screen, clockwise rotation.

8 0

1 year ago

Focus groups are one of the most widely used ________ methods to gain greater understanding of a current problem or to develop p

Zigmanuir [339]

Answer:

Exploratory

Explanation:

<u>Focus groups</u>

It is a small group of 8-12 respondents guided by a moderator through a thorough debate on a specific subject or idea.It's great for generation of ideas, brainstorming, insight into motives, attitudes, and perceptions. it can show likes, dislikes,emotional requirements and prejudices.

Exploratory methods are used to gain initial insights that could pave the way for further investigation.

Some of the exploratory methods are focus groups, Key informant,case studies,secondary data and observational data.

8 0

1 year ago

A turntable that is initially at rest is set in motion with a constant angular acceleration α. What is the magnitude of the angu

bekas [8.4K]

Explanation:

If the turntable starts from rest and is set in motion with a constant angular acceleration α. Let $\omega$ is the angular velocity of the turntable. We know that the rate of change of angular velocity is called the angular acceleration of an object. Its formula is given by :

$\alpha =\dfrac{\omega_f-\omega_i}{t}$

$\alpha =\dfrac{\omega-0}{t}$

$\alpha =\dfrac{\omega}{t}$

$t=\dfrac{\omega}{\alpha }$ ............(1)

Using second equation of kinematics as :

$\theta=\omega_i t+\dfrac{1}{2}\alpha t^2$

$\theta=\dfrac{1}{2}\alpha t^2$

Using equation (1) in above equation

$\theta=\dfrac{1}{2}\times \dfrac{\omega^2}{\alpha }$

In one revolution, $\theta=4\pi$ (in 2 revolutions)

$4\pi =\dfrac{1}{2}\times \dfrac{\omega^2}{\alpha }$

$\omega=\sqrt{8\pi \alpha}$

$\omega=2\sqrt{2\pi \alpha}$

Hence, this is the required solution.

6 0

2 years ago

Read 2 more answers

Which statements best describe x-rays? check all that apply. x-rays are electromagnetic waves. x-rays are longitudinal waves. x-

laiz [17]

X ray is one of the electromagnetic waves.

As per Clark Maxwell's electromagnetic theory, all the electromagnetic waves move with the velocity of light i.e c= 3×10^8 m/s

In case of electromagnetic waves,the electric field and magnetic field are perpendicular to each other as well as perpendicular to the direction of propagation.The electromagnetic waves exhibit the property of polarisation. Hence they are transverse in nature.

Hence the best statements about X- ray will be-

1- X -rays are electromagnetic waves

2-X-rays are transverse transverse waves

3- X- rays travel at the speed of light.

5 0

1 year ago

Read 2 more answers

The same physics student jumps off the back of her Laser again, but this time the Laser is

soldi70 [24.7K]

a) The speed of the student after the jump is 1.07 m/s

b) The final speed of the laser is 10.4 m/s

Explanation:

a)

We can solve this problem by applying the law of conservation of momentum: if there are no external forces acting on the system, the total momentum of the student+Laser system must be constant. Therefore, we can write:

$p_i = p_f\\0=mv+MV$

where

The initial momentum is zero

m = 42 kg is the mass of the Laser

v = 1.5 m/s is the final velocity of the Laser

M = 59 kg is the mass of the student

V is the final velocity of the student

Solving the equation for V, we find the velocity of the student:

$V=-\frac{mv}{M}=-\frac{(42)(1.5)}{59}=-1.07 m/s$

So, the final speed of the student is 1.07 m/s.

b)

In this case, the laser and the student are travelling at 3.1 m/s before the student jumps off: therefore, the total momentum before the jump is not zero.

So, the equation of the conservation of momentum is

$(m+M)u=mv+MV$

where

m = 42 kg is the mass of the Laser

M = 59 kg is the student's mass

u = 3.1 m/s is the initial velocity of the student and the Laser

V = -2.1 m/s is the velocity of the student after the jump (she jumps backward)

v is the final velocity of the Laser

And solving for v, we find

$v=\frac{(m+M)u-MV}{m}=\frac{(42+59)(3.1)-(59)(-2.1)}{42}=10.4 m/s$

Learn more about momentum:

brainly.com/question/7973509

brainly.com/question/6573742

brainly.com/question/2370982

brainly.com/question/9484203

#LearnwithBrainly

3 0

1 year ago

Read 2 more answers