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3 years ago

A bicycle accelerates from rest to 6 m/s in 2 s. What is the bicycle's acceleration?

Physics

2 answers:

USPshnik [31]3 years ago

5 0

So the right answer is 3m/s^2.

Look at the attached picture

hope it will help you

Good luck on your assignment

Angelina_Jolie [31]3 years ago

4 0

Answer:

Explanation:I don't say you have to mark my ans as brainliest but if you think it has really helped you plz don't forget to thank me...

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If velocity is decreasing,then acceleration:

ziro4ka [17]

If velocity is decreasing, then acceleration is in the direction
opposite to the velocity.

If the object is moving in the direction that you call 'positive',
then acceleration is negative.

5 0

3 years ago

1-A car moves toward east 12km is represented as A and it turns towards south 16km is represented as B. What is the resultant ve

hjlf

1. A-20 km south east

The car's displacement consists of two components into two different directions. Using a system of coordinates in which x represents the east direction and y represents the south direction, the two displacements are:

$d_x = 12 km$ east

$d_y = 16 km$ south

Since the two components are orthogonal to each other, we can find the resultant displacement by using Pythagorean's theorem:

$d=\sqrt{d_x^2+d_y^2}=\sqrt{(12 km)^2+(16 km)^2}=\sqrt{400}=20 km$

and the direction is between the two original directions, so south-east.

2. D. 10 m/s

First of all, we need to calculate the total time the stone took to hit the ground. Since the vertical distance covered is S = 78.4 m, and since the motion is an accelerated motion with constant acceleration g=9.8 m/s^2, we have

$S=\frac{1}{2}gt^2$

From which we find the total time of the fall, t:

$t=\sqrt{\frac{2S}{g}}=\sqrt{\frac{2(78.4 m)}{9.8 m/s^2}}=4 s$

Now we can consider the horizontal motion of the stone: we know that the stone travels for d = 40 m in a time of t = 4 s, therefore the horizontal velocity of the stone is

$v=\frac{d}{t}=\frac{40 m}{4 s}=10 m/s$

3. B=32.32 m

As in the previous problem, we have to calculate the total time it takes for the stone to reach the river first. Since the vertical distance covered is S = 20 m, we have

$t=\sqrt{\frac{2S}{g}}=\sqrt{\frac{2(20 m)}{9.8 m/s^2}}=2.0 s$

And since the stone is traveling horizontally at v = 16 m/s, the horizontal distance covered is

$d=vt=(16 m/s)(2 s)=32 m$

So, the closest answer is B.

5 0

3 years ago

Why won’t anyone help me please anybody help me I really need help .

irga5000 [103]

Answer:

1➡️ this is the method of decomposition

2➡️ H2 and O2

3➡️ b

sorry if I am wrong

8 0

3 years ago

A thin soap bubble of index of refraction 1.33 is viewed with light of wavelength 550.0nm and appears very bright. Predict a pos

faust18 [17]

Answer:

The possible thickness of the soap bubble = $1.034\times 10^{-7}\ m.$

Explanation:

Given:

Refractive index of the soap bubble, $\mu=1.33.$
Wavelength of the light taken, $\lambda = 550.0\ nm = 550.0\times 10^{-9}\ m.$

Let the thickness of the soap bubble be $t$ .

It is given that the soap bubble appears very bright, it means, there is a constructive interference takes place.

For the constructive interference of light through a thin film ( soap bubble), the condition of constructive interference is given as:

$2\mu t=\left ( m+\dfrac 12 \right )\lambda.$

where $m$ is the order of constructive interference.

Since the soap bubble is appearing very bright, the order should be 0, as $0^{th}$ order interference has maximum intensity.

Thus,

$2\mu t=\left (0+\dfrac 12\right )\lambda\\t=\dfrac{\lambda}{4\mu}\\\ \ = \dfrac{550\times 10^{-9}}{4\times 1.33}\\\ \ = 1.034\times 10^{-7}\ m.$

It is the possible thickness of the soap bubble.

6 0

3 years ago

A pickup truck and a hatchback car start at the same position. If the truck is moving at a constant 33.2m/s and the hatchback ca

love history [14]

Answer:

The pickup truck and hatchback will meet again at 440.896 m

Explanation:

Let us assume that both vehicles are at origin at the start means initial position is zero i.e. $s_{o}$ = 0. Both the vehicles will cross each other at same time so we will make equations for both and will solve for time.