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atroni [7]
3 years ago
13

Suppose that a constant force is applied to an object. newton's second law of motion states that the acceleration of the object

varies inversely with its mass. a certain force acting upon an object with mass 2 kg produces an acceleration of 38 /ms2 . if the same force acts upon another object whose mass is 19 kg , what would this object's acceleration be?
Physics
1 answer:
amid [387]3 years ago
3 0

Newton’s second law states that force is the product of mass and acceleration. This is expressed mathematically as:

F = m * a

Where F = force, m = mass, a = acceleration

Since in the given problem, the force is constant or same force is acting upon two objects. Therefore we can simply equate the expression m * a of the 2 objects.

m1 * a1 = m2 * a2

Where m1 = 2 kg, a1 = 38 m/s^2, m2 = 19 kg. Therefore finding for a2:

2 kg * 38 m/s^2 = 19 kg * a2

<span>a2 = 4 m/s^2     (ANSWER)</span>

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Two forces, F⃗ 1F→1F_1_vec and F⃗ 2F→2F_2_vec, act at a point,F⃗ 1F→1F_1_vec has a magnitude of 8.80 NN and is directed at an an
castortr0y [4]

Answer:

  • Fx = -9.15 N
  • Fy = 1.72 N
  • F∠γ ≈ 9.31∠-10.6°

Explanation:

You apparently want the sum of forces ...

  F = 8.80∠-56° +7.00∠52.8°

Your angle reference is a bit unconventional, so we'll compute the components of the forces as ...

  f∠α = (-f·cos(α), -f·sin(α))

This way, the 2nd quadrant angle that has a negative angle measure will have a positive y component.

  = -8.80(cos(-56°), sin(-56°)) -7.00(cos(52.8°), sin(52.8°))

  ≈ (-4.92090, 7.29553) +(-4.23219, -5.57571)

  ≈ (-9.15309, 1.71982)

The resultant component forces are ...

  • Fx = -9.15 N
  • Fy = 1.72 N

Then the magnitude and direction of the resultant are

  F∠γ = (√(9.15309² +1.71982²))∠arctan(-1.71982/9.15309)

  F∠γ ≈ 9.31∠-10.6°

4 0
3 years ago
Jen pushed a box for a distance of 80m with 20 N of force. How much work did she do?
Drupady [299]
The answer is 1,600 J.

A work (W) can be expressed as a product of a force (F) and a distance (d):
W = F · d<span>

We have:
W = ?
F = 20 N = 20 kg*m/s</span>²
d = 80 m
_____
W = 20 kg*m/s² * 80 m
W = 20 * 80 kg*m/s² * m
W = 1600 kg*m²/s²
W = 1600 J
8 0
3 years ago
A student pushes on a 20.0 kg box with a force of 50 N at an angle of 30° below the horizontal. The box accelerates at a rate of
PtichkaEL [24]

Answer:

225 N

Explanation:

"Below the horizontal" means he's pushing down at an angle.

Draw a free body diagram of the box.  There are three forces: normal force N pushing up, weight force mg pulling down, and the applied force F at an angle θ.

Sum of forces in the y direction:

∑F = ma

N − mg − F sin θ = 0

N = F sin θ + mg

Plug in values:

N = (50 N) (sin 30°) + (20.0 kg) (10 m/s²)

N = 225 N

8 0
3 years ago
The work-energy theorem states that the work done on an object is equal to a change in which quantity?
Fynjy0 [20]

The work-energy theorem states that the net work done by the forces on an object equals the change in its kinetic energy.

8 0
2 years ago
A particle moving along the x-axis has a position given by m, where t is measured in s. What is the magnitude of the acceleratio
8090 [49]

Question:

A particle moving along the x-axis has a position given by x=(24t - 2.0t³)m, where t is measured in s. What is the magnitude of the acceleration of the particle at the instant when its velocity is zero

Answer:

24 m/s

Explanation:

Given:

x=(24t - 2.0t³)m

First find velocity function v(t):

v(t) = ẋ(t) = 24 - 2*3t²

v(t) = ẋ(t) = 24 - 6t²

Find the acceleration function a(t):

a(t) = Ẍ(t) = V(t) = -6*2t

a(t) = Ẍ(t) = V(t) = -12t

At acceleration = 0, take time as T in velocity function.

0 =v(T) = 24 - 6T²

Solve for T

T = \sqrt{\frac{-24}{6}} = \sqrt{-4} = -2

Substitute -2 for t in acceleration function:

a(t) = a(T) = a(-2) = -12(-2) = 24 m/s

Acceleration = 24m/s

4 0
3 years ago
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