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IRINA_888 [86]
3 years ago
15

Two forces Upper FSubscript Upper A Baseline Overscript right-arrow EndScripts and Upper FSubscript Upper B Baseline Overscript

right-arrow EndScripts are applied to an object whose mass is 10.6 kg. The larger force is Upper FSubscript Upper A Baseline Overscript right-arrow EndScripts. When both forces point due east, the object's acceleration has a magnitude of 0.554 m/s2. However, when Upper FSubscript Upper A Baseline Overscript right-arrow EndScripts points due east and Upper FSubscript Upper B Baseline Overscript right-arrow EndScripts points due west, the acceleration is 0.313 m/s2, due east. Find (a) the magnitude of Upper FSubscript Upper A Baseline Overscript right-arrow EndScripts and (b) the magnitude of Upper FSubscript Upper B Baseline Overscript right-arrow EndScripts.
Physics
1 answer:
Pavel [41]3 years ago
5 0

Answer:

Part a)

F_A = 4.59 N

Part B)

F_B = 1.28 N

Explanation:

As we know that when both the forces are acting on the object in same direction then we will have

F_A + F_B = ma

as we know that

a = 0.554 m/s^2

m = 10.6 kg

now we will have

F_A + F_B = 10.6(0.554)

F_A + F_B = 5.87 N

Now two forces are in opposite direction then we have

F_A - F_B = 10.6(0.313)

F_A - F_B = 3.32 N

Part A)

Now we will have from above two equation

F_A = 4.59 N

Part B)

Similarly for other force we have

F_B = 1.28 N

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Alex Ar [27]

Answer:

v = 1.25 m/s

Explanation:

We have,

Distance covered by a person is 10 meters

Time taken by him to cover that distance is 8 seconds.

If we want to find the speed of a person, we must know distance covered by it and taken. In this case, we know both distance and time. His speed is given by :

v=\dfrac{d}{t}\\\\v=\dfrac{10\ m}{8\ s}\\\\v=1.25\ m/s

So, the speed of the person is 1.25 m/s.

4 0
3 years ago
Two objects with masses m and 3m are connected by a compressed spring so that the energy stored in the spring is 375 joules. If
cluponka [151]

Answer:

281.25 J

Explanation:

We are told that the two objects with masses m and 3m.

Also that energy stored in the spring is 375 joules.

Now, initially the centre of mass of the system took place at rest, it means v1 = v and v2 = v/3

Thus, from principle of conservation of energy, we have;

½mv² + ½(3m)(v/3)² = 375J

(m + 3m/9)½v² = 375

(4/3)m × ½v² = 375

Multiply both sides by ¾ to get;

½mv² = 375 × ¾

½mv² = 281.25 J

Therefore, energy of lighter body is 281.25 J

7 0
2 years ago
According to meet one law of universal gravitation, gravity increases when?​
mina [271]

Newton's law of universal gravitation states that every point mass in the universe attracts every other point mass with a force that is directly proportional to the product of their masses, and inversely proportional to the square of the distance between them. Newton's law of universal gravitation states that every point mass in the universe attracts every other point mass with a force that is directly proportional to the product of their masses, and inversely proportional to the square of the distance between them.

5 0
3 years ago
A 4.00 kg ball is swung in a circle on the edge of a 1.50 m rope. The time it takes for the ball to complete one rotation is 3.4
lorasvet [3.4K]

Answer:

The answer is below

Explanation:

The length of the rope is equal to the radius of the circle formed by the complete rotation of the rope. Therefore the radius = 1.50 m.

a) The distance covered by the rope when completing one rotation is the same as the perimeter of the circle. Hence:

Distance covered in one rotation = 2π * radius = 2π * 1.5 = 3π meters

The velocity of the ball = Distance / time = 3π meters / 3.4 seconds = 2.77  m/s

b) The initial velocity (u) is 0 m/s, the final velocity is 2.77 m/s during time (t) = 3.4 s. Hence acceleration (a):

v = u + at

2.77 = 3.4a

a = 0.82 m/s²

c) Force on ball = mass * acceleration = 4 * 0.82 = 3.28 N

8 0
3 years ago
An object is pulled up an incline plane at a constant velocity as the picture above shows. Calculate the tension on the rope.
WARRIOR [948]

Hi there!

\large\boxed{545N}

Since the object is being pulled at a constant velocity, the forces must be balanced.

Since there is no movement vertically, we must take into account the horizontal forces. We can also assume a positive acceleration to be in the direction of motion.

The acceleration and force due to gravity on an incline is:

a = gsinФ

F = MgsinФ

∑F = -MgsinФ + T

Since it is getting pulled at a constant velocity, ∑F = 0. So:

0  = -MgsinФ + T

MgsinФ = T

Solve for T by plugging in values. Let g = 10 m/s²

T = (120)(10)sin(27) ≈ 545 N

6 0
3 years ago
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