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Dimas [21]
3 years ago
11

A bug walking on the circular rim of a flower

Physics
1 answer:
Mekhanik [1.2K]3 years ago
8 0

Given parameters;

Time taken to complete a lap  = 8.667s

Radius of flower = 13.9cm

    convert to SI unit of m,  100cm = 1m

            13.9cm gives \frac{13.9}{100}   = 0.139m

Unknown  = speed

To solve this problem, we need to first find the circumference of the flower.

 Circumference of the circular flower  = 2 π r

    where r is the radius of the flower;

           Circumference  = 2 x 3.142 x 0.139 = 0.87m

Now to find the how fast the bug is travelling,

   Speed  = \frac{distance}{time}

Since the bug covered 1 lap, the distance is 0.87m

Now input the parameters and solve for speed;

    Speed  = \frac{0.87}{8.667}  = 0.1m/s

The bug is travelling at a speed of  0.1m/s

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The main driver of climate change is the greenhouse effect. Some gases in the Earth's atmosphere

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2 years ago
A 120 kg box is on the verge of slipping down an inclined plane with an angle of inclination of 47º. What is the coefficient of
Alex_Xolod [135]

Given :

A 120 kg box is on the verge of slipping down an inclined plane with an angle of inclination of 47º.

To Find :

The coefficient of static friction between the box and the plane.

Solution :

Vertical component of force :

mg\ sin\ \theta =  120\times 10 \times sin\ 47^\circ{}=877.62 \ N

Horizontal component of force(Normal reaction) :

mg\ cos\ \theta =  120\times 10 \times cos\ 47^\circ{}=818.40 \ N

Since, box is on the verge of slipping :

mg\ sin\ \theta= \mu(mg \ cos\ \theta)\\\\\mu = tan \ \theta\\\\\mu = tan\ 47^o\\\\\mu = 1.07

Therefore, the coefficient of static friction between the box and the plane is 1.07.

Hence, this is the required solution.

7 0
3 years ago
I have an astronomy question... Spinning up the solar nebula. The orbital speed of the material in the solar nebula at Pluto's a
attashe74 [19]
<span>The angular momentum of a particle in orbit is 

l = m v r 

Assuming that no torques act and that angular momentum is conserved then if we compare two epochs "1" and "2" 

m_1 v_1 r_1 = m_2 v_2 r_2 

Assuming that the mass did not change, conservation of angular momentum demands that 

v_1 r_1 = v_2 r_2 

or 

v1 = v_2 (r_2/r_1) 

Setting r_1 = 40,000 AU and v_2 = 5 km/s and r_2 = 39 AU (appropriate for Pluto's orbit) we have 

v_2 = 5 km/s (39 AU /40,000 AU) = 4.875E-3 km/s

Therefore, </span> the orbital speed of this material when it was 40,000 AU from the sun is <span>4.875E-3 km/s.

I hope my answer has come to your help. Thank you for posting your question here in Brainly.
</span>
3 0
3 years ago
a horse began running due east and covered 25km in 4.0 hours what is the average velocity of the horse
Komok [63]

25km / 4hr = <em>6.25 km/hr east</em>

6 0
3 years ago
Read 2 more answers
A 1200 kg car accelerates from 13 m/s to 17 m/s. find the change in momentum of the car.
mixas84 [53]
P=4800kgm/s
As
p=mΔv
where p is momentum, m is mass and v is velocity
Given values is
m =1200kg
Δv= 17m/s-13m/s=4m/s
Now
p=mΔv
p=(1200kg)*(4m/s)
p=4800kgm/s


8 0
2 years ago
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