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Lera25 [3.4K]
3 years ago
6

A boy is exerting a force of 70 N at a 50-degree angle on a lawn mower. He is accelerating at 1.8 m/s2. Round the answers to the

nearest whole number.
What is the mass of the lawn mower?

What is the normal force exerted on the lawn mower?
Physics
2 answers:
kumpel [21]3 years ago
5 0

The mass is 25kg and the normal force is 299N

Crazy boy [7]3 years ago
3 0
The mass of the lawn mower can be calculated from the expression force is equal to the product of mass and acceleration. We do as follows:

F = ma
70 = m(1.8)
m = 38.89 kg

The normal force is the upward force perpendicular with the object. For this case it is equal with Wcos(50). We calculate as follows:

Fn = Wcos50 = 38.89(9.81)(cos50) = 245.23 N
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Answer:

F=9/5(°C)+32

Explanation:

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A ball is bouncing to a height that is 1/3 of the previous height after every bounce. What kind of sequence is this? How could y
tamaranim1 [39]

Answer and Explanation:

The ball is bouncing to a height of 1/3 of its previous height this is a type of geometric sequence the total distance can be found by the sum of geometric sequence

For example let the initial height is 243 fit

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After second bounce 81/3=27 feet

After third bounce 27/3 =9 feet

After fourth bounce 9/3 =3 feet

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\frac{1}{10^{10}}\\

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Block A of mass M is on a horizontal surface of negligible friction. An identical block B is attached to block A by a light stri
miv72 [106K]

Answer:

T’= 4/3 T  

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Explanation:

For this problem let's use Newton's second law applied to each body

Body A

X axis

      T = m_A a

Axis y

     N- W_A = 0

Body B

Vertical axis

     W_B - T = m_B a

In the reference system we have selected the direction to the right as positive, therefore the downward movement is also positive. The acceleration of the two bodies must be the same so that the rope cannot tension

We write the equations

    T = m_A a

    W_B –T = M_B a

We solve this system of equations

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Let's find the tension

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Now we change the mass of the second block

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We seek tension for this case

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Let's look for the relationship between the tensions of the two cases

   T’/ T = 2/3 M g / (½ M g)

   T’/ T = 4/3

   T’= 4/3 T

The new tension is 4/3 = 1.33 of the previous tension the answer  e

4 0
3 years ago
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