What is an object's acceleration that takes 550.0 meters to go from 6.0 m/s to 85.0 m/s?

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

Pavel [41]

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

$F_B = 1.28 N$

5 0

3 years ago

Two ropes are attached to either side of a 100.0 kg wagon as shown below. The rope on the right is pulled at an angle 40.0° rela

NikAS [45]

The acceleration of the wagon is found by applying Newton's Second Law of motion.

1. The responses for question 1 are;

x-component of the tension in the rope on the right is approximately 91.93 N

y-component of the tension in the rope on the right is approximately 71.135 N

x-component of the tension in the rope on the left is -80.0 N

y-component of the tension in the rope on the left is 0

2. The net force in the x-direction is approximately 11.93 N

3. The net acceleration of the wagon in the horizontal direction is approximately 0.1193 m/s².

Reasons:

The given parameters are;

Mass of the wagon, m = 100.0 kg

Angle of inclination to the horizontal of the rope to the right, θ = 40.0°

Tension in the rope on the right = 120.0 N

Direction in which the rope on the left is pulled = To the west

Tension in the rope on the left = 80.0 N

1. The x and y component of the tension in the rope on the right are;

x-component = 120.0 N × cos(40.0°) ≈ 91.93 N

y-component = 120.0 N × sin(40.0°) ≈ 77.135 N

The x and y component of the tension in the rope on the left are;

x-component = 80.0 N × cos(180°) = -80.0 N

y-component = 80.0 N × sin(180°) = 0.0 N

2. The net force in the horizontal direction, Fₓ, is found as follows;

Fₓ = The x-component of the rope on the left + The x-component of the rope on the right

Which gives;

Fₓ = 91.93 N - 80.0 N = 11.93 N

3. The net acceleration of the block is given as follows;

According to Newton's Second Law of motion, we have;

Force in the horizontal direction, Fₓ = Mass of wagon, m × Acceleration of the wagon in the horizontal direction, aₓ

Fₓ = m × aₓ

Therefore;

$\displaystyle a_x = \frac{F_x}{m} \approx \frac{11.93 \, N}{100.0 \, kg} = \mathbf{0.1193 \ m/s^2}$

The acceleration of the wagon in the horizontal direction, aₓ ≈ 0.1193 m/s².

Learn more here:

brainly.com/question/20357188

8 0

2 years ago

The position-time equation for a cheetah chasing an antelope is:

pochemuha

Answer:

x = 1.6 + 1.7 t^2 omitting signs

a) at t = 0 x = 1.6 m

b) V = d x / d t = 3.4 t

at t = 0 V = 0

c) A = d^2 x / d t^2 = 3.4 (at t = 0 A = 3.4 m/s^2)

d) x = 1.6 + 1.7 * (4.4)^2 = 34.5 (position at 4.4 sec = 34.5 m)

4 0

2 years ago

A 3.0kg weight W is initially at rest on incline AB, which is raised 40° above the horizontal. The effective coefficient is

lakkis [162]

(a) The acceleration of the system is determined as 1.58 m/s².

(b) The relative weight of P is pounds is determined as 0.14 lb.

<h3>Acceleration of the system</h3>

The acceleration of the system is calculated as follows;

W - T = m₂a --- (1)

T = m₁a ----(2)

μmgsinθ - m₁a = m₂a

(0.3 x 3 x 9.8 x sin40) - (0.4 + 0.2)a = 3a

5.67 - 0.6a = 3a

5.67 = 3.6a

a = 5.67/3.6

a = 1.58 m/s²

<h3> Relative Weight of P</h3>

W = ma

W = 0.4 x 1.58

W = 0.632 N = 0.14 lb

Learn more about weight here: brainly.com/question/2337612

#SPJ1

3 0

2 years ago

An octopus propels itself through the water by ejecting water through a funnel in its body. How does this demonstrate Newton’s t

valina [46]

Answer: The answer would be C. When Octopus eject the water out, the water acts with an equal force on the octopus in opposite direction.

Explanation:

An octopus propels itself forward in the water by ejecting water backward through a funnel in its body.

Hope this helps! :D

8 0

2 years ago

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