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3 years ago

8

A bicyclist slow with a force of 3.5 × 10² N. if the bicyclist and bicycle have a total mass of 1.0 × 10² kg, what is the accele

ration

1 answer:

Aleonysh [2.5K]3 years ago

7 0

Answer:

3.5 m/s²

Step-by-step explanation:

We are given;

The force of a bicycle as 3.5 × 10² N
Mass of the bicycle as 1.0 × 10² kg

We are required to determine the acceleration of the bicycle;

From Newton's second law of motion;

F = ma

Rearranging this;

a = F÷ m

Therefore;

Acceleration = 3.5 × 10² N ÷ 1.0 × 10² kg

= 3.5 m/s²

Thus, the acceleration of the bicycle is 3.5 m/s²

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What values of y satisfy this equation?<br><br> 3y - 14 = 16

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2 years ago

Read 2 more answers

The position of an object moving along an x axis is given by x = 3.24 t - 4.20 t2 + 1.07 t3, where x is in meters and t in secon

Pavel [41]

Answer and explanation:

Given : The position of an object moving along an x axis is given by $x=3.24t-4.20t^2+1.07t^3$ where x is in meters and t in seconds.

To find : The position of the object at the following values of t :

a) At t= 1 s

$x(t)=3.24t-4.20t^2+1.07t^3$

$x(1)=3.24(1)-4.20(1)^2+1.07(1)^3$

$x(1)=3.24-4.20+1.07$

$x(1)=0.11$

b) At t= 2 s

$x(t)=3.24t-4.20t^2+1.07t^3$

$x(2)=3.24(2)-4.20(2)^2+1.07(2)^3$

$x(2)=6.48-16.8+8.56$

$x(2)=-1.76$

c) At t= 3 s

$x(t)=3.24t-4.20t^2+1.07t^3$

$x(3)=3.24(3)-4.20(3)^2+1.07(3)^3$

$x(3)=9.72-37.8+28.89$

$x(3)=0.81$

d) At t= 4 s

$x(t)=3.24t-4.20t^2+1.07t^3$

$x(4)=3.24(4)-4.20(4)^2+1.07(4)^3$

$x(4)=12.96-67.2+68.48$

$x(4)=14.24$

(e) What is the object's displacement between t = 0 and t = 4 s?

At t=0, x(0)=0

At t=4, x(4)=14.24

The displacement is given by,

$\triangle x=x(4)-x(0)$

$\triangle x=14.24-0$

$\triangle x=14.24$

(f) What is its average velocity from t = 2 s to t = 4 s?

At t=2, x(2)=-1.76

At t=4, x(4)=14.24

The average velocity is given by,

$\triangle x=x(4)-x(2)$

$\triangle x=14.24-(-1.76)$

$\triangle x=14.24+1.76$

$\triangle x=16$

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2 years ago

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sattari [20]

8y - 11y + 6

Combine like terms

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