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leonid [27]
3 years ago
14

3. A 20 kg lawnmower is pushed by its handle, which makes a 40 degree angle with the horizontal. What

Physics
1 answer:
Mkey [24]3 years ago
5 0

<u>13788 J is required to finish the work.</u>

Explanation:

Given that,  

Mass of lawnmower is 20 kg  

Force = 30 N

\text { Angle }(\theta)=40^{\circ}

Displacement ( distance moved )= 20m

The force of 30N is applied along the direction of motion and was applied for the whole 20 m.

The formula of work done for one pass is

\text { Work done }=\text { force } \times \text { displacement } \times \cos \theta

Substituting the given data into the formula,

W=30 \times 20 \times \cos 40^{\circ}

\mathrm{W}=600 \times \cos 40^{\circ}

\mathrm{W}=600 \times 0.766

W = 459.6 J

Work done for one pass = 459.6J

Now, work done for 30 passes is calculated by multiplying the work done for one pass into 30.

\text { Then the work done for } 30 \text { passes is } 459.61 \times 30=13788 \mathrm{J}

Work is done by the person pushing the mower to finish the work is <u>13788 J</u>.

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Compared to its weight on Earth, a 5kg object on the moon will weigh
shutvik [7]

Answer:

8.1 N/49 N=0.1653  which means 16.53% of the weight of the object on Earth.

Explanation:

On the Moon, where the gravitational constant is 1.62 \frac{m}{s^2}, the weight of the 5 kg object will be: weight_M=m*g_M = 5 kg * 1.62 \frac{m}{s^2} =8.1 N

Where the answer is in Newtons (N) since all quantities are given in the SI system.

On Earth, on the other hand, the weight of the object is:

weight_E=m*g_E= 5 kg* 9.8 \frac{m}{s^2} = 49N

Therefore the object's weight on the Moon compared to that on Earth will be:

8.1N/49N=0.1653

That is, 16.53% of the weight the object has on Earth.

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4 years ago
What happens in the process of gravitational condensation?
marissa [1.9K]

Answer:

An object decreases in size due to the collision of materials. An object increases in size due to the addition of materials. Gas particles are formed from solar nebula materials.

6 0
3 years ago
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Which of the following concepts helps explain the modern technology used in 3D films?
stealth61 [152]

Answer:

a.

Explanation:

6 0
3 years ago
A horizontal 826 N merry-go-round of radius 1.17 m is started from rest by a constant horizontal force of 57.8 N applied tangent
Julli [10]

Answer:

The kinetic energy of the merry-go-round is \bf{475.47~J}.

Explanation:

Given:

Weight of the merry-go-round, W_{g} = 826~N

Radius of the merry-go-round, r = 1.17~m

the force on the merry-go-round, F = 57.8~N

Acceleration due to gravity, g= 9.8~m.s^{-2}

Time given, t=3.47~s

Mass of the merry-go-round is given by

m &=& \dfrac{W_{g}}{g}\\~~~~&=& \dfrac{826~N}{9.8~m.s^{-2}}\\~~~~&=& 84.29~Kg

Moment of inertial of the merry-go-round is given by

I &=& \dfrac{1}{2}mr^{2}\\~~~&=& \dfrac{1}{2}(84.29~Kg)(1.17~m)^{2}\\~~~&=& 57.69~Kg.m^{2}

Torque on the merry-go-round is given by

\tau &=& F.r\\~~~&=& (57.8~N)(1.17~m)\\~~~&=& 67.63~N.m

The angular acceleration is given by

\alpha &=& \dfrac{\tau}{I}\\~~~&=& \dfrac{67.63~N.m}{57.69~Kg.m^{2}}\\~~~&=& 1.17~rad.s^{-2}

The angular velocity is given by

\omega &=& \alpha.t\\~~~&=& (1.17~rad.s^{-2})(3.47~s)\\~~~&=& 4.06~rad.s^{-1}

The kinetic energy of the merry-go-round is given by

E &=& \dfrac{1}{2}I\omega^{2}\\~~~&=&\dfrac{1}{2}(57.69~Kg.m^{2})(4.06~rad.s^{-1})^{2}\\~~~&=& 475.47~J

5 0
3 years ago
The current in a resistor is 3.0 A, and its power is 60 W. What is the voltage?
Novosadov [1.4K]

Answer:

20 volts

Explanation:

Use the equation P=VI

60=V(3)

V=20

6 0
2 years ago
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