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mixas84 [53]
4 years ago
12

Initially a car accelerates at 2 m/s2 for x seconds. The car then travels at a velocity of -6 m/s for x seconds. If the car disp

laced a total of 16 m, how long did it travel?
Physics
1 answer:
Luda [366]4 years ago
5 0

Answer:

The time travel is

t=8 s

Explanation:

a= 2 \frac{m}{s^{2} } \\v=-6 \frac{m}{s} \\x=16m

x_{f}=x_{o}+v_{o}*t+\frac{1}{2} *a*t^{2}

x_{f}=0-6*t+\frac{1}{2} *2*t^{2}

t^{2}-6*t-16=0\\ using :\\\frac{-b+/-\sqrt{b^{2}-4*c*a } }{2} \\\frac{-(-6)+/-\sqrt{(-6)^{2}-4*(-16)*(1) } }{2}=\frac{3}{2} +/- \frac{10}{2} \\t_{1} = 2s \\t_{2} = 8s

Check

t_{2}=8s

x_{f}=x_{o}+v_{o}*t+\frac{1}{2} *a*t^{2}

x_{f}=0-6*+\frac{1}{2} *2*8^{2}

x_{f}=-48+64\\x_{f}=16

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A person tries to lift each object with a force of 25 N, upward. Which
Vlad1618 [11]

Answer:

Option C. Objects 1 and 3 will not move, and objects 2 and 4 will accelerate

upward.

Explanation:

The following data were obtained from the question:

OBJECT >>>>>>>>> WEIGHT (N)

1 >>>>>>>>>>>>>>>> 35

2 >>>>>>>>>>>>>>>> 23

3 >>>>>>>>>>>>>>>> 26

4 >>>>>>>>>>>>>>>> 18

Force (F) applied = 25 N

From the above, the force applied to each object is 25N. Thus the following can be concluded based on the data given above:

For object 1:

Weight = 35 N

Force applied = 25 N

Thus, the object will not move since the weight of the object is greater than the force applied

For object 2:

Weight = 23 N

Force applied = 25 N

Thus, the object will move since the force applied is greater than the weight of the object.

For object 3:

Weight = 26 N

Force applied = 25 N

Thus, the object will not move since the weight of the object is greater than the force applied.

For object 4:

Weight = 18 N

Force applied = 25 N

Thus, the object will move since the force applied is greater than the weight of the object.

From the above illustrations, Object 1 and 3 will not move, and objects 2 and 4 will accelerate i.e move

4 0
3 years ago
Read 2 more answers
A 3.35 kg object initially moving in the positive x direction with a velocity of 4.90 m s collides with and sticks to a 1.88 kg
ahrayia [7]

Answer:

The final components of velocity of the composite object is 3.33 m/s.

Explanation:

Given;

mass of the first object, m₁ = 3.35 kg

initial velocity of the first object, u₁ = 4.90 m/s in positive x-direction

mass of the second object, m₂ = 1.88 kg

initial velocity of the second object, u₂ = 3.12 m/s in negative y-direction

initial momentum of the first object, P₁ = 3.35 x 4.9 = 16.415 kgm/s

initial momentum of the second object, P₂ = 1.88 x 3.12 = 5.8656 kgm/s

The resultant velocity of the two objects is given by;

R² = 16.415² + 5.8656²

R² = 303.858

R = √303.858

R = 17.432 kgm/s

Apply the principle of conservation of linear momentum for inelastic collision;

total initial momentum before = total final momentum after collision

P₁(x) + P₂(y) = Pf

R = Pf

R = v(m₁ + m₂)

17.432 = v(m₁ + m₂)

where;

v is the final components of velocity of the composite object

v = \frac{17.432}{m_1 + m_2} \\\\v = \frac{17.432}{3.35+1.88} \\\\v = 3.33 \ m/s

Therefore, the final components of velocity of the composite object is 3.33 m/s.

8 0
3 years ago
A waitperson carrying a tray with a platter on it tips the tray at an angle of 12° below the horizontal. If the gravitational fo
Nadusha1986 [10]

Answer:

The answer is 1.0 N

Explanation:

inclination of tray=12^{\circ}

gravitational Force=5 N

Now this gravitational force has two component i.e.

5\sin \theta  is parallel to the tray =1.039 N

5\cos \theta  is perpendicular to the tray =4.890 N

7 0
3 years ago
An object is dropped from a platform 100 feet high. Ignoring wind resistance, what will its speed be when it reaches the ground?
Scorpion4ik [409]

Answer:

80 ft/s

Explanation:

Use III equation of motion

V^2 = U^2 + 2g h

Here, U = 0, g = 32 ft/s^2, h = 100 ft

V^2 = 0 + 2 × 32 ×100

V^2 = 6400

V = 80 ft/s

8 0
3 years ago
A spherical Christmas tree ornament is 8.00 cm in diameter. What is the magnification of an object placed 12.0 cm away from the
LiRa [457]

The magnification of the ornament is 0.25

To calculate the magnification of the ornament, first, we need to find the image distance.

Formula:

  • 1/f = u⁻¹+v⁻¹.................... Equation 1

Where:

  • f = Focal length of the ornament
  • u = image distance
  • v = object distance.

make u the subject of the equation

  • u = fv/(f+v)................ Equation 2

From the question,

Given:

  • f = 8/2 = 4 cm
  • v = 12 cm

Substitute these values into equation 2

  • u = (12×4)/(12+4)
  • u = 48/16
  • u = 3 cm.

Finally, to get the magnification of the ornament, we use the formula below.

  • M = u/v.................. Equation 3

Where

  • M = magnification of the ornament.

Substitute these values above into equation 3

  • M = 3/12
  • M = 0.25.

Hence, The magnification of the ornament is 0.25

8 0
3 years ago
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