Ask question

Ask question

All categories

3 years ago

10

A bicycle takes 8.0 seconds to accelerate at a constant rate from rest to a speed of 4.0 m/s. If the mass of the bicycle and rid

er together is 85 kg, what is the net force acting on the bicycle

1 answer:

yanalaym [24]3 years ago

6 0

11.0 second hope i helped

You might be interested in

A sinusoidal wave travels along a string. if the time for a particular point to move from maximum displacement to zero displacem

Lisa [10]

The time it takes for a point to reach 0 displacement from maximum displacement is $\frac{\pi}{2}$ of a cycle. Thus, one period ( $2\pi$ ) will take $T=0.17*4=0.68s$ .

The frequency is simply $f=\frac{1}{T}=\frac{1}{0.68}=1.47Hz$ .

The speed is given by: $v=f\lambda=1.47*1.4 = 2.06m/s$

7 0

3 years ago

Which of the following would demonstrate an attraction between magnetic poles?

vredina [299]

"C. north pole to a south pole" would demonstrate an attraction between magnetic poles, since in the field of magnetics, opposites attract and similar forces repel one another.

5 0

3 years ago

Read 2 more answers

Evelyn meets her friends at a local restaurant 5 km away. Afterwards she stops at the post office which is another 3 km away, bu

marissa [1.9K]

16km because she is 8km away and if she goes back it’s gonna be 8km more so it’s 16km

8 0

3 years ago

At what constant rate of acceleration will a car starting from rest cover 18m in the first 3 seconds?

irga5000 [103]

69 hehehehe hehehehe here

7 0

3 years ago

How much extra water does a 147-lb concrete canoe displace compared to an ultralightweight 38-lb kevlar canoe of the same size c

romanna [79]

We use the formula, to calculate the volume of water displaced by concrete canoe,

$V =\frac{W}{\gamma }$

Here, W is the weight of concrete canoe and $\gamma$ is the specific weight of water and its value is $62.4\ lb/ft^3$ .

So,

$V =\frac{ 147\ lb}{62.4\ lb/ft^3}=2.356\ ft^3$ .

Now the volume of water occupied in ultra lightweight kevlar canoe,

$v=\frac{w}{\gamma}$

Here, w is weight of kevlar canoe.

So,

$v=\frac{38\ lb}{62.4\ lb/ft^3} =0.6089\ ft^3$

Thus, the volume of water displaced,

$=V-v=2.356\ ft^3-0.6089\ ft^3=2.19\ ft^3$ .

Hence, the volume of water displaced canoe compared to an ultra-lightweight kevlar canoe is $2.19\ ft^3$

5 0

3 years ago