When a positively charged body touches a neutral body, the neutral body will

All matter in the Universe consists of many substances called elements. <br><br> true or faluse

Mrrafil [7]

The answer to this statement is true!

8 0

3 years ago

A gymnast of mass 63.0 kg hangs from a vertical rope attached to the ceiling. You can ignore the weight of the rope and assume t

Sergio [31]

Answer:

Explanation:

A ) When gymnast is motionless , he is in equilibrium

T = mg

= 63 x 9.81

= 618.03 N

B )

When gymnast climbs up at a constant rate , he is still in equilibrium ie net force acting on it is zero as acceleration is zero.

T = mg

= 618.03 N

C ) If the gymnast climbs up the rope with an upward acceleration of magnitude 0.600 m/s2

Net force on it = T - mg , acting in upward direction

T - mg = m a

T = mg + m a

= m ( g + a )

= 63 ( 9.81 + .6)

= 655.83 N

D ) If the gymnast slides down the rope with a downward acceleration of magnitude 0.600 m/s2

Net force acting in downward direction

mg - T = ma

T = m ( g - a )

= 63 x ( 9.81 - .6 )

= 580.23 N

6 0

3 years ago

Which of the following is an example of a Cnidarian?

siniylev [52]

The answer to this is C jellyfish i am positive

hope this helps

5 0

4 years ago

Read 2 more answers

if a torque of 55.0 N/m is required and the largest force that can be exerted by you is 135 N what is th e length of the lever a

Whitepunk [10]

Answer:

$r=0.41m$

Explanation:

Torque is defined as the cross product between the position vector ( the lever arm vector connecting the origin to the point of force application) and the force vector.

$\tau=r\times F$

Due to the definition of cross product, the magnitude of the torque is given by:

$\tau=rFsin\theta$

Where $\theta$ is the angle between the force and lever arm vectors. So, the length of the lever arm (r) is minimun when $sin\theta$ is equal to one, solving for r:

$r=\frac{\tau}{F}\\r=\frac{55\frac{N}{m}}{135N}\\r=0.41m$

7 0

3 years ago

A person slaps her leg with her hand, which results in her hand coming to rest in a time interval of 2.65 ms from an initial spe

dezoksy [38]

Answer:

the magnitude of the average contact force exerted on the leg is 3466.98 N

Explanation:

Given the data in the question;

Initial velocity of hand v₀ = 5.25 m/s

final velocity of hand v = 0 m/s

time interval t = 2.65 ms = 0.00265 s

mass of hand m = 1.75 kg

We calculate force on the hand F $_{hand$

using equation for impulse in momentum

F $_{hand$ × t = m( v - v₀ )

we substitute

F $_{hand$ × 0.00265 = 1.75( 0 - 5.25 )

F $_{hand$ × 0.00265 = 1.75( - 5.25 )

F $_{hand$ × 0.00265 = -9.1875

F $_{hand$ = -9.1875 / 0.00265

F $_{hand$ = -3466.98 N

Next we determine force on the leg F $_{leg$

Using Newton's third law of motion

for every action, there is an equal opposite reaction;

so, F $_{leg$ = - F $_{hand$

we substitute

F $_{leg$ = - ( -3466.98 N )

F $_{leg$ = 3466.98 N

Therefore, the magnitude of the average contact force exerted on the leg is 3466.98 N

5 0

3 years ago