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Blababa [14]
2 years ago
7

a lady bug walks 10 cm forward then 5 cm backwards in 20 seconds. what is the average speed of the ladybug ?

Physics
1 answer:
igor_vitrenko [27]2 years ago
5 0

A lady bug moves 10 cm forward and 5 cm backwards

so total distance moved by lady bug = 10 + 5 = 15 cm

total time taken by the lady bug

t =  20 s

so the average speed is given as

v = \frac{d}{t}

v = \frac{15}{20}

v = 0.75 cm/s

so its average speed is 0.75 cm/s

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A piece of luggage is being loaded onto an airplane by way of an inclined conveyor belt. The bag, which has a mass of 15.0 kg, t
LenKa [72]

Answer:

a) W = - 318.26 J, b)  W = 0 , c) W = 318.275 J , d) W = 318.275 J , e) W = 0

Explanation:

The work is defined by

           W = F .ds = F ds cos θ

Bold indicate vectors

We create a reference system where the x-axis is parallel to the ramp and the axis and perpendicular, in the attached we see a scheme of the forces

Let's use trigonometry to break down weight

     sin θ = Wₓ / W

     Wₓ = W sin 60

     cos θ = Wy / W

      Wy = W cos 60

X axis

How the body is going at constant speed

    fr - Wₓ = 0

    fr = mg sin 60

    fr = 15 9.8 sin 60

    fr = 127.31 N

Y Axis  

    N - Wy = 0

    N = mg cos 60

    N = 15 9.8 cos 60

    N = 73.5 N

Let's calculate the different jobs

a) The work of the force of gravity is

     W = mg L cos θ

Where the angles are between the weight and the displacement is

      θ = 60 + 90 = 150

     W = 15 9.8 2.50 cos 150

     W = - 318.26 J

b) The work of the normal force

     From Newton's equations

          N = Wy = W cos 60

          N = mg cos 60

         W = N L cos 90

        W = 0

c) The work of the friction force

      W = fr L cos 0

      W = 127.31 2.50

      W = 318.275 J

d) as the body is going at constant speed the force of the tape is equal to the force of friction

      W = F L cos 0

      W = 127.31 2.50

       W = 318.275 J

e) the net force

    F ’= fr - Wx = 0

    W = F ’L cos 0

    W = 0

4 0
2 years ago
A quarterback throws a football at 40km/hr to a receiver 50yd away. How much time does it take the ball to reach the receiver
Akimi4 [234]

Given:

Distance = 50 yard = 45.72 meter

Speed = 40 km/hr = 11.11 m/s

To find:

Time required by ball to reach the receiver = ?

Formula used:

speed = \frac{distance}{time}

Solution:

The speed of the ball is given by,

speed = \frac{distance}{time}

Thus,

Time = \frac{distance}{speed}

Distance = 50 yard = 45.72 meter

Speed = 40 km/hr = 11.11 m/s

Time = 4.12 second

Hence, ball reaches the receiver in 4.12 second.

3 0
2 years ago
With your hand parallel to the floor and your palm upright, you lower a 3-kg book downward. If the force exerted on the book by
Stells [14]
For Newton's second law, the resultant of the forces acting on the book is equal to the product between the mass of the book and its acceleration:
\sum F = ma (1)

There are only two forces acting on the book:
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so, equation (1) becomes
mg - F = ma
from which we can calculate the book's acceleration, a:
a= g -  \frac{F}{m}= 9.81 m/s^2 - \frac{20 N}{3 kg}=3.14 m/s^2
7 0
2 years ago
Read 2 more answers
Two 8.0 Ω lightbulbs are connected in a 12 V series circuit. What is the power of both glowing bulbs?
V125BC [204]

Answer:

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Then,

R = R1+R2............. Equation 2

Where R1= Resistance of the first bulb, R2 = Resistance of the second bulb

Given: R1 = R2 = 8 Ω

Substitute into equation 1

R = 8+8

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Also Given: V = 12 V

Substitute into equation 1

P = 12²/8

P = 144/8

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7 0
2 years ago
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Inessa [10]
A = h / n => h = a*n

a = 0.290 hit / time
n = 300 times

=> h = 0.290 hit / time * 300 time = 87 hits

Answer: 87 hits
4 0
2 years ago
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