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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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Explanation:

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A heat pump is to be used for heating a house in winter. The house is to be maintained at 70°F at all times. When the temperatur
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Answer:

\dot{W_{H} } = 4244.48 Btu/h

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Convert to rankine, T_{H} = 70^{0}+ 460 = 530 R

Heat is extracted at 40°F i.e T_{L} = 40^{0}F  = 40 + 460 = 500 R

Calculate the coefficient of performance of the heat pump, COP

COP = \frac{T_{H} }{T_{H} - T_{L}  } \\COP = \frac{530 }{530 - 500  }\\ COP = \frac{530}{30} \\COP = 17.67

The minimum power required to run the heat pump is given by the formula:

\dot{W_{H} } = \frac{\dot{Q_{H} }}{COP} \\...............(*)

Where the heat losses from the house, \dot{Q_{H} } = 75,000 Btu/h

Substituting these values into * above

\dot{W_{H} } = \frac{75000}{17.67} \\ \dot{W_{H} } = 4244.48 Btu/h

3 0
2 years ago
A Neglecting air resistance, a ball projected straight upward so it remains in the air for 10 seconds needs an initial speed of
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Answer:

The initial velocity is 50 m/s.

(C) is correct option.

Explanation:

Given that,

Time = 10 sec

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We need to calculate the height

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v^2=u^2+2gh

h =\dfrac{v^2}{2g}....(I)

For second half,

We need to calculate the time

Using equation of motion

h =ut+\dfrac{1}{2}gt_{2}^2

h=0+\dfrac{1}{2}gt_{2}^2

t_{2}=\sqrt{\dfrac{2h}{g}}

Put the value of h from equation (I)

t_{2}=\sqrt{\dfrac{2\times v^2}{g^2}}

t_{2}=\dfrac{v}{g}

According to question,

t_{1}+t_{2}=10

t_{1}=t_{2}

Put the value of t₁ and t₂

\dfrac{v}{g}+\dfrac{v}{g}=10

\dfrac{2v}{g}=10

v=\dfrac{10\times g}{2}

Here, g = 10

The initial velocity is

v=\dfrac{10\times10}{2}

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2 years ago
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