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miss Akunina [59]
3 years ago
10

If I hold a ball on my hand. Which force is stronger, my hand or gravity?

Physics
2 answers:
tatuchka [14]3 years ago
8 0
Your hand because if gravity was stronger you would not be able to hold the ball :)
docker41 [41]3 years ago
7 0

Answer: Your hand i think

Explanation: because if grvity was stronger then you would not be able to lift your hand and the ball.

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A 29.0 kg beam is attached to a wall with a hi.nge while its far end is supported by a cable such that the beam is horizontal.
Elanso [62]

The tension in the cable is 169.43 N and the vertical component of the force exerted by the hi.nge on the beam is 114.77 N.

<h3>Tension in the cable</h3>

Apply the principle of moment and calculate the tension in the cable;

Clockwise torque = TL sinθ

Anticlockwise torque = ¹/₂WL

TL sinθ  =  ¹/₂WL

T sinθ  =  ¹/₂W

T = (W)/(2 sinθ)

T = (29 x 9.8)/(2 x sin57)

T = 169.43 N

<h3>Vertical component of the force</h3>

T + F = W

F = W - T

F = (9.8 x 29) - 169.43

F = 114.77 N

Thus, the tension in the cable is 169.43 N and the vertical component of the force exerted by the hi.nge on the beam is 114.77 N.

Learn more about tension here: brainly.com/question/24994188

#SPJ1

6 0
2 years ago
What is the work energy transfer equation?
rosijanka [135]

Answer:

The equation used to calculate the work done is: work done = force × distance. W = F × d. This is when: work done (W) is measured in joules (J)

8 0
3 years ago
Read 2 more answers
3) What is the weight of an object with a mass of 5.2 kilograms?​
inna [77]

Answer:

50.96 N

Explanation:

weight = mass x gravity

We know that gravity = 9.8 m/s^2 and mass = 5.2 kg.

w = m x g

w = 5.2 kg x 9.8 m/s^2

w = 50.96 N

The weight of the object is 50.96 N (newtons). Hope this helps, thank you !!

8 0
3 years ago
Place several E-Field Sensors at a few points on different equipotential lines, and look at the relationship between the electri
Vlada [557]

Answer:

2.

Explanation:

  • If a charge is moved along a equipotential line, no work is done on the charge.
  • If we remember that the work done by an external force, is just the product of the component of the force parallel to the displacement, if the force produces no work, this means that is perpendicular to the displacement.
  • So, as the electric field is just the force per unit charge, and has the same direction as the force (for a positive charge), it must be perpendicular to any equipotential line.
  • As the electric field (by convention) has the same direction as it would be taken by a positive test charge, and positive charges move from higher voltages to lower ones, the electric field is directed toward lines of lower voltages (like it happens between the plates of a capacitor).
3 0
3 years ago
The cabinet is mounted on coasters and has a mass of 45 kg. The casters are locked to prevent the tires from rotating. The coeff
stira [4]

Answer:

the force P required for impending motion is 132.3 N

the largest value of "h" allowed if the cabinet is not to tip over is 0.8 m

Explanation:

Given that:

mass of the cabinet  m = 45 kg

coefficient of static friction μ =  0.30

A free flow body diagram illustrating what the question represents is attached in the file below;

The given condition from the question let us realize that ; the casters are locked to prevent the tires from rotating.

Thus; considering the forces along the vertical axis ; we have :

\sum f_y =0

The upward force and the downward force is :

N_A+N_B = mg

where;

\mathbf { N_A  \ and  \ N_B} are the normal contact force at center point A and B respectively .

N_A+N_B = 45*9.8

N_A+N_B = 441    ------- equation (1)

Considering the forces on the horizontal axis:

\sum f_x = 0

F_A +F_B  = P

where ;

\mathbf{ F_A \ and \ F_B } are the static friction at center point A and B respectively.

which can be written also as:

\mu_s N_A + \mu_s N_B  = P

\mu_s( N_A +  N_B)  = P

replacing our value from equation (1)

P = 0.30 ( 441)    

P = 132.3 N

Thus; the force P required for impending motion is 132.3 N

b) Since the horizontal distance between the casters A and B is 480 mm; Then half the distance = 480 mm/2 = 240 mm = 0.24 cm

the largest value of "h" allowed for  the cabinet is not to tip over is calculated by determining the limiting condition  of the unbalanced torque whose effect is canceled by the normal reaction at N_A and it is shifted to N_B:  

Then:

\sum M _B = 0

P*h = mg*0.24

h =\frac{45*9.8*0.24}{132.3}

h = 0.8 m

Thus; the largest value of "h" allowed if the cabinet is not to tip over is 0.8 m

6 0
3 years ago
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