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Masteriza [31]
2 years ago
12

Please help which one !!

Physics
1 answer:
Nostrana [21]2 years ago
6 0

Answer:

4th one

Explanation:

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Which of these phase changes requires energy?
frutty [35]

Answer:

A

Explanation:

You have to give energy away for B. You have to think about that carefully. The CO2 starts out with a great deal of energy (mostly KE) and has to slow down to go from gas to a solid. Not B

In general C is the same way. Water has more energy than ice. Not C

Same principle in D.  Not D.

So it's A

7 0
2 years ago
The force involves the attraction between objects with mass. Strong nuclear. Gravitational. Electromagnetic. Weak nuclear
Viktor [21]
The answer is Gravitational.
4 0
3 years ago
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Two horses are side by side on a carousel. Which has a greater tangential speed the one closer to the center or the one farther
photoshop1234 [79]

Answer:

The horse father from the center has a greater tangential speed. Although both horses complete one circle in the same time period, the one farther from the center covers a greater distance during that same period.

Explanation:

8 0
3 years ago
Problem: Hooke's law states that the force on a spring varies directly with the distance that it is stretched. If a spring has a
Scorpion4ik [409]

Answer:

Restoring force of the spring is 50 N.

Explanation:

Given that,

Spring constant of the spring, k = 100 N/m

Stretching in the spring, x = 0.5 m

We need to find the restoring force of the spring. It can be calculated using Hooke's law as "the force on a spring varies directly with the distance that it is stretched".

F=kx

F=100\ N/m\times 0.5\ m

F = 50 N

So, the restoring force of the spring is 50 N. Hence, this is the required solution.

7 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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