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4 years ago

Forces that are equal in strength and opposite indirection are

Physics

2 answers:

blsea [12.9K]4 years ago

8 0

They are called balanced forces.

Otrada [13]4 years ago

4 0

Hey there! My name is Christy and I'm gladly to help you out!

The three main forces that stop moving objects are friction, gravity and wind resistance. Equal forces acting inopposite directions are called balanced forces. Balanced forcesacting on an object will not change the object's motion. When you add equal forces in opposite direction, the netforce is zero.

Hope this helped!

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Which system works with the respiratory system for the exchange of gases to take place?

4vir4ik [10]

Yep it is. Defiantly the circulatory system

6 0

3 years ago

A block is given a very brief push up a 20.0 degree frictionless incline to give it an initial speed of 12.0 m/s.(a) How far alo

Orlov [11]

Explanation:

(a) Net force acting on the block is as follows.

$F_{net} = -mg Sin (\theta)$

or, ma = -mg Sin (\theta)[/tex]

a = $-g Sin (\theta)$

= $-9.8 \times Sin (20^{o})$

= -3.35 $m/s^{2}$

According to the kinematic equation of motion,

$v^{2} - v^{2}_{o} = 2as$

Distance traveled by the block before stopping is as follows.

s = $\frac{v^{2} - v^{2}_{o}}{2a}$

= $\frac{(0)^{2} - (12.0)^{2}_{o}}{2 \times -3.35}$

= 21.5 m

According to the kinematic equation of motion,

v = $v_{o} + at$

0 = $12.0 m/s + \frac{1}{2} \times -3.35 m/s^{2} \times t$

$t_{1}$ = 7.16 sec

Therefore, before coming to rest the surface of the plane will slide the box till 7.16 sec.

(b) When the block is moving down the inline then net force acting on the block is as follows.

$F_{net} = -mg Sin (\theta)$

ma = $mg Sin (\theta)$

a = $g Sin (\theta)$

= $9.8 m/s^{2} \times Sin (20^{o})$

= 3.35 $m/s^{2}$

Kinematics equation of the motion is as follows.

s = $v_{o}t + \frac{1}{2}at^{2}$

21.5 m = $0 + \frac{1}{2} \times 3.35 m/s^{2} \times t^{2}$

$t_{2}$ = $\sqrt{\frac{2 \times 21.5 m}{3.35 m/s^{2}}}$

= 3.58 sec

Hence, total time taken by the block to return to its starting position is as follows.

t = $t_{1} + t_{2}$

= 7.16 sec + 3.58 sec

= 10.7 sec

Thus, we can conclude that 10.7 sec time it take to return to its starting position.

3 0

4 years ago

If an astronaut can throw a certain wrench 10.0 m vertically upward on earth, how high could he throw it on our moon if he gives

Jobisdone [24]

I think you forgot to include the acceleration due to gravity of astronauts. I assume that it is = 0.170 g. To get the answer we have to use the formula s = v0t – (1/2) At². Where s is the altitude, A is the acceleration of gravity, t is the time after throwing.

v = v0 –At

v = 0 at max altitude so v0 – At = 0

t = v0/A at max altitude

Using the formula above for the altitude:

s = v0t – (1/2) At²

s = v0(v0/A) – (1/2) A (v0/A)²

s = v0²/A – (1/2) v0²/A

s = (1/2) v0²/A

The earth: E = (1/2) v0²/g

The moon: M = (1/2)v0²(0.17g)

So, take the ratio of M/E = g/0.17g = 1/0.17 = 588

M = 5.88 E

He can throw the wrench 5.88 times higher on the moon

<span>M =5.88 (10 m) = 58.8 meters that the can throw the wrench a little over on the moon.</span>

7 0

4 years ago

To do work, this truck uses energy stored in chemical

GREYUIT [131]

800 J Got it right on edgenuity

4 0

3 years ago

you stretch a spring by a distance of 0.3 m. the spring has a spring constant of 440 n/m. when you release the spring, it snaps

Alina [70]

Answer:

19.8 J

Explanation:

According to the law of conservation of energy, the total mechanical energy of the spring (sum of kinetic energy and elastic potential energy) must be conserved:

$K_i + U_i = K_f + U_f$ (1)