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dalvyx [7]
2 years ago
5

7. A mother pushes her 9.5 kg baby in her 5kg baby carriage over the grass with a force of 110N @ an angle

Physics
1 answer:
jasenka [17]2 years ago
3 0

Weight of the carriage =(m+M)g =142.1\ N

Normal force =Fsin(\theta) + W = 197.1\ N

Frictional force =\mu N=27.59\ N

Acceleration =4.66\ m\ s^{-2}

Explanation:

We have to look into the FBD of the carriage.

Horizontal forces and Vertical forces separately.

To calculate Weight we know that both the mass of the baby and the carriage will be added.

  • So Weight(W) =(m+M)\times g =(9.5+5)\ kg \times 9.8 =142.1\ Newton\ (N)

To calculate normal force we have to look upon the vertical component of forces, as Normal force is acting vertically.We have weight which is a downward force along with F_x, force of 110\ N acting vertically downward.Both are downward and Normal is upward so Normal force =Summation\ of\ both\ forces

  • Normal force (N) = Fsin(\theta)+W=110sin(30) + 142.1 =197.1\ N
  • Frictional force (f) =\mu N=0.14\times 197.1 =27.59\ N

To calculate acceleration we will use Newtons second law.

That is Force is product of mass and acceleration.

We can see in the diagram that F_y=Horizontal and F_x=Vertical component of forces.

So Fnet = Fy(Horizontal) - f(friction) = m\times a

  • Acceleration (a) =\frac{Fcos(\theta)-\mu N}{mass(m)} =\frac{(95.26-27.59)}{14.5}= 4.66\ m\ s{^2 }

So we have the weight of the carriage, normal force,frictional force and acceleration.

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A. A ball is thrown directly up with an initial speed of 4.00 m/s at y = 0. What is the maximum height h that it achieves, and w
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Answer:

A)     t = 0.40816 s , y = 0.916 m

Explanation:

A) For this problem we use the kinematic relations

           v = v₀ - g t

the highest point zero velocities (v = 0)

           t = (v₀-v) / g

           t = (4 - 0) / 9.8

           t = 0.40816 s

to calculate the height let's use

          v² = v₀² - 2 g y

          y = vo2 / 2g

           y = 4 2 / (2 9.8)

          y = 0.916 m

To find the number of photos, we can use a direct proportions rule, if you take 30 photos in a second in 0.40816 s how many photos does it take

          # _photos1 = 0.40916 (30/1)

          # _photos1 = 12

yes i take 120 fps

          #_fotod = 0.40916 (120/1)

          #photos = 5.87 10³

 

B) The ball is released from a latura h how long it takes to reach the floor

           v² = v₀² + 2 g y

where the initial velocity is zero and the velocity with which the expert leaves is equal to the velocity with which v = 4 m / s leaves

            v² = 2gy

             v = √ (2 9.8 0.916)

             v = √ (2.1397 101)

             v = 4.6257 m / s

c) we ask us for the time for latura

             y = L / 2

             y = 0.916 / 2

             y = 0.458 m

now we can use the formula

             y = v₀ t - ½ g t²

           0.458 = 4.00 t - ½ 9.8 t²

            4.9 t² - 4t + 0.458 = 0

            t² -0.8163 t +0.09346 = 0

we solve second degree execution

           t = [0.8163 ±√ (0.8163² - 4 0.09346)] / 2

           t = [0.8163 ± 0.540] / 2

           t₁ = 0.678 m

           t₂ = 0.2763 m

the shortest time is for when the ball goes up and the longest when it goes down

D) the graph of vs Vs is expected to be a closed line

and the graph of position versus time a parabola

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3 years ago
Determine the magnitude of the component of F directed along the axis of AB. Set F = 520 N .
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Answer:

The component of F along AB is equal to Fcos45

F = 520N

Component along AB = 520cos45

= 367.7N

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Explanation:

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charle [14.2K]

Answer:

a = -0.33 m/s² k^

Direction: negative

Explanation:

From Newton's law of motion, we know that;

F = ma

Now, from magnetic fields, we know that;. F = qVB

Thus;

ma = qVB

Where;

m is mass

a is acceleration

q is charge

V is velocity

B is magnetic field

We are given;

m = 1.81 × 10^(−3) kg

q = 1.22 × 10 ^(−8) C

V = (3.00 × 10⁴ m/s) ȷ^.

B = (1.63T) ı^ + (0.980T) ȷ^

Thus, since we are looking for acceleration, from, ma = qVB; let's make a the subject;

a = qVB/m

a = [(1.22 × 10 ^(−8)) × (3.00 × 10⁴)ȷ^ × ((1.63T) ı^ + (0.980T) ȷ^)]/(1.81 × 10^(−3))

From vector multiplication, ȷ^ × ȷ^ = 0 and ȷ^ × i^ = -k^

Thus;

a = -0.33 m/s² k^

7 0
2 years ago
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