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Brut [27]
3 years ago
8

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?

⇒ 25 kg

What is the normal force exerted on the lawn mower?

⇒ 299 N
Physics
1 answer:
LenaWriter [7]3 years ago
4 0

Explanation:

Given that,

Force, F = 70 N

Angle, \theta=50^{\circ}

Acceleration of the boy, a = 1.8 m/s²

(a) Net force is equal to the product of mass and acceleration. So,

m=\dfrac{F}{a}\\\\m=\dfrac{70}{1.8}\\\\m=38.89\ kg

(2) Normal force = upward force

So,

F=mg\cos\theta\\\\F=38.89\times 9.81\times \cos(50)\\F=245.23\ N

Hence, mass is 38.89 kg and the normal force is 245.23 N.

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A 0.520-kg object attached to a spring with a force constant of 8.00 N/m vibrates in simple harmonic motion with an amplitude of
Veseljchak [2.6K]

Answer:

Explanation:

Given that,

Mass attached to spring

M = 0.52kg

Force constant K = 8N/m

Amplitude A = 11.6 cm

a. Maximum speed?

Angular velocity is calculated using

w = √k/m

w = √8/0.52

w = √15.385

w = 3.922rad/s

Then, the relation ship between angular velocity and linear velocity is given as

v = - w•A

v = - 3.922 × 11.6

v = - 45.5 cm/s

Then, the maximum velocity is

vmax = |v|= 45.5cm/s

b. Acceleration a?

Acceleration can be determine using the formula

a = -w²• A

a = -3.922² × 11.6

a = -178.46 cm/s²

Magnitude of the acceleration is 178.46cm/s²

c. Speed when the object is at 9.6cm from equilibrium position?

Generally,

The position of the object at equilibrium is

x(t) = A•Cos(wt)

x(t) = 11.6 Cos (3.922t)

Then, when x(t) = 9.6cm

9.6 = 11.6 Cos(3.92t)

Cos(3.922t) = 9.6/11.6

Cos(3.922t) = 0.8276

3.922t = ArcCos(0.8276)

Note: the angle is in radiant

3.922t = 0.596

t = 0.596/3.922

t = 0.152 second

Then, v(t) at that time is

v(t) = x'(t) = -11.6×3.92Sin(3.922t)

v(t) = -45.5Sin(3.922t)

Now, when t =0.152

v(t) = -45.5 Sin(3.922×0.152)

v(t) = -45.5Sin(0.596)

v(t) = -25.5 cm/s

Then, it's magnitude is 25.5cm/s

d. Acceleration at same position

t = 0.152s

a(t) = v'(t) = - 45.5×3.922Cos(3.922t)

a(t) = -178.46Cos(3.92t)

a(t) = -178.46 Cos(3.92×0.152)

a(t) = -178.46 Cos(0.596)

a(t) = -147.68 cm/s²

Magnitude of the acceleration is 147.68 cm/s²

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3 years ago
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Zina [86]

Elastic

Explanation is that it is force which is snapping back

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3 years ago
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You can increase the capacitance of a capacitor by A. Decreasing the plate spacing B. Increasing the plate spacing. ° C. Decreas
eimsori [14]

You can increase the capacitance of a capacitor by decreasing the plate spacing (A) or by increasing the area of the plates (D).

'A' and 'D' both do the job, so the correct choice is<em> (E)</em> .

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Explanation:

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The answer is D interferometry
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