Which of the following would not make a good insulator

A weightlifter deadlifts a 300 kg weight. If the weightlifter has a mass of 100 kg, what is the force acting on his legs?

sergeinik [125]

300 + 100 = 400

(Im not 100% sure of this answer)

3 0

3 years ago

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An object is launched at a velocity of 20 m/s in a direction making an angle of 25° upward with the horizontal.

Hitman42 [59]

Answer:

(a) max. height = 3.641 m

(b) flight time = 1.723 s

(c) horizontal range = 31.235 m

(d) impact velocity = 20 m/s

Above values have been given to third decimal. Adjust significant figures to suit accuracy required.

Explanation:

This problem requires the use of kinematics equations

v1^2-v0^2=2aS .............(1)

v1.t + at^2/2 = S ............(2)

where

v0=initial velocity

v1=final velocity

a=acceleration

S=distance travelled

SI units and degrees will be used throughout

Let

theta = angle of elevation = 25 degrees above horizontal

v=initial velocity at 25 degrees elevation in m/s

a = g = -9.81 = acceleration due to gravity (downwards)

(a) Maximum height

Consider vertical direction,

v0 = v sin(theta) = 8.452 m/s

To find maximum height, we find the distance travelled when vertical velocity = 0, i.e. v1=0,

solve for S in equation (1)

v1^2 - v0^2 = 2aS

S = (v1^2-v0^2)/2g = (0-8.452^2)/(2*(-9.81)) = 3.641 m/s

(b) total flight time

We solve for the time t when the vertical height of the object is AGAIN = 0.

Using equation (2) for vertical direction,

v0*t + at^2/2 = S substitute values

8.452*t + (-9.81)t^2 = 3.641

Solve for t in the above quadratic equation to get t=0, or t=1.723 s.

So time for the flight = 1.723 s

(c) Horiontal range

We know the horizontal velocity is constant (neglect air resistance) at

vh = v*cos(theta) = 25*cos(25) = 18.126 m/s

Time of flight = 1.723 s

Horizontal range = 18.126 m/s * 1.723 s = 31.235 m

(d) Magnitude of object on hitting ground, Vfinal

By symmetry of the trajectory, Vfinal = v = 20, or

Vfinal = sqrt(v0^2+vh^2) = sqrt(8.452^2+18.126^2) = 20 m/s

7 0

4 years ago

Whoever answers correctly gets brainlist!

UNO [17]

Infared = used by police

gamma = short wavelength

radio = largest wavelength

visible = only ones we can see

5 0

3 years ago

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If the real gi tract is 9 meters long, the model you created is what fraction of the real one

Sedbober [7]

Answer:

a

Explanation:

a

7 0

4 years ago

Suppose that the loudspeaker in the problem had a mass of 500 kg and the ropes hung 20∘ from the vertical. Into which of the fol

nexus9112 [7]

The following intervals would you expect T(tension) to fall : <u>2000 to 4000 N </u>

<h3>Further explanation</h3>

The force acting on a system with static equilibrium is 0

$\large{\boxed{\bold{\sum F=0}}$

(forces acting as translational motion only, not including rotational forces)

$\displaystyle \sum F_x = 0\\\\ \sum F_y = 0$

We complete the questions:

A 20-kg loudspeaker is suspended 2.0 m below the ceiling by two ropes that are each 30? from vertical.

Find the value of T, the magnitude of the tension in either of the ropes.

Express your answer in newtons.

Suppose that the loudspeaker in the problem has a mass of 500 kg and the ropes hung 20? from the vertical. Into which of the following intervals would you expect T to fall? You don't have to calculate Texactly to answer this question; just make an estimate.

500 to 1000 N

1000 to 2000 N

2000 to 4000 N

4000 to 8000 N

8000 to 16,000 N

In a 500-kg loudspeaker system and two ropes that are each 20° from vertical, we see the forces acting on the y axis (vertical)

$\displaystyle \sum F_y = 0\\\\T1_y+T2_y-w=0\\\\T1~cos~20^o+T2~cos~20^o=m\times g$

we assume g = 10 m/s², then :

$\displaystyle 2T~cos~20=500\times 10\\\\T~cos~20=2500\\\\\boxed{\bold{T=26592 N}}}$

So the value of T is between 2000 and 4000 N

<h3>Learn more</h3>

Newton's Law

brainly.com/question/13725525

Keywords : the loudspeaker, ropes, Tension, mass, intervals

#LearnWithBrainly

4 0

4 years ago