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Nesterboy [21]
3 years ago
5

Which universal force acts on protons and neutrons in an atom's nucleus?

Physics
2 answers:
ser-zykov [4K]3 years ago
7 0
The strong nuclear force
quester [9]3 years ago
6 0

The strong nuclear force

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Every winter i fly home to Michigan. it takes 5 hours. what is my average speed?
Alisiya [41]

It depends how far you travel.

If for instance you travel 1000 miles, then your speed is distance/time = 1000/5 = 200 mph

8 0
2 years ago
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An 12 N force is applied to a 1 kg object. What is the magnitude of the objects acceleration?
Sauron [17]

Answer:

a=12 m/s²

Explanation:

Newton's second law of motion states that the acceleration of a body is directly proportional to the force applied and takes place in the direction of force.

This can be summarized as: F=ma, where m is the mass of the object on which force F acts. a is the acceleration due to the force applied.

12N= 1kg×a

a=12N/1kg

a=12m/s²

6 0
3 years ago
Please help!!!! Will mark brainliest.
julia-pushkina [17]

Answer:

Approximately 8.4 \times 10^{2}\; \rm N, assuming that g = 9.8\; \rm m \cdot s^{-2}.

Explanation:

Let m and a denote the mass and acceleration of Spiderman, respectively.

There are two forces on Spiderman:

  • Downward gravitational attraction from the earth: W = m \cdot g.
  • Upward tension force from the strand of web F(\text{tension}).

The directions of these two forces are exactly opposite of one another. Besides, because Spiderman is accelerating upwards, the magnitude of F(\text{tension}) (which points upwards) should be greater than that of W (which points downwards towards the ground.)

Subtract the smaller force from the larger one to find the net force on Spiderman:

(\text{Net Force}) = F(\text{tension}) - W.

On the other hand, apply Newton's Second Law of motion to find the value of the net force on Spiderman:

(\text{Net Force}) = m \cdot a.

Combine these two equations to get:

m \cdot a = (\text{Net Force}) = F(\text{tension}) - W.

Therefore:

\begin{aligned}& F(\text{tension})\\ &= m \cdot a + W \\ &= m \cdot (a + g)\\ &= 76\; \rm kg \times \left(1.3\; \rm m \cdot s^{-2} + 9.8\; \rm m \cdot s^{-2}\right)\\ &\approx 8.4\times 10^{2}\; \rm N\end{aligned}.

By Newton's Third Law of motion, Spiderman would exert a force of the same size on the strand of web. Hence, the size of the force in the strand of the web should be approximately 8.4\times 10^{2}\; \rm N (downwards.)

4 0
3 years ago
Anna and Jon sit on a seesaw. Anna has a mass of 60 kg and sits 2 m from the center. Jon has a mass of 70 kg. How far from the c
Ronch [10]

Answer:

dJ = 1.7 m

Explanation:

The Equation of the Balancing the moments in the center of the seesaw  is like this:

∑Mo = 0

Mo = F*d

Where:

∑Mo : Algebraic sum of moments in the center(o) of the balance

Mo : moment in the o point ( N*m)

F  : Force ( N)

d  : distancia of the force to the the o point  ( N*m)

Data

mA = 60 kg : mass of the Anna

mJ = 70 kg :  mass of theJon

dA = 2 m : Distance from Anna to the center of the seesaw

g: acceleration due to gravity

Calculation of the distance from Jon to the center of the seesaw  (dJ)

∑Mo = 0   WA : Ana's weight   , WJ : Jon's weight

W = m*g

(WA)(dA) - (WJ) (dJ) = 0

(mA*g)(dA) - (mJ*g)(dJ) = 0

We divide by g the equation:

(mA)(dA) - (mJ)(dJ)= 0

(mA)(dA) = (mJ)(dJ)

d_{J} = \frac{m_{A} *d_{A}}{m_{J}}

d_{J} = \frac{60 kg *2 m}{70 kg}

dJ = 1.7 m

5 0
3 years ago
As you are trying to move a heavy box of mass m, you realize that it is too heavy for you to lift by yourself.There is no one ar
Ira Lisetskai [31]

Answer: magnitude of applied force is FA = mg + F

Where F is the resultant force downward that the rope moves with

Explanation:

Force downwards F is,

F = FA - T

T is the upwards tension force on the rope

FA is the actual applied force in pulling the rope down.

Therefore, T = FA - F .....equ. (1)

For the box to move up with force ma ( it's mass times its acceleration upwards) upwards tension on the roap must exceed its own weight mg ( it's mass times acceleration due to gravity 9.8m/s^2)

Therefore, ma = T - mg

T = ma + mg ..... equ. (2)

Equating equ. 1 and 2

T = FA - F = ma + mg

Therefore FA = ma + mg + F

But at constant velocity a = 0

Magnitude of applied force becomes

FA = mg + F

See image below

5 0
3 years ago
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