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3 years ago

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When jogging outside you accidently bump into a curb. Your feet stop but your body continues to move forward and you end up on t

he ground. Which law of motion is being described in this scenario?

1 answer:

oee [108]3 years ago

6 0

Inertia I think because I've heard it around school and in science

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The revolution of the earth around the sun demonstrate what motion?

olga2289 [7]

Answer:

Anticlockwise directions

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1 year ago

to measure the static friction coefficient between a block and a vertical wall, a spring is attached to the block, is pushed on

Stolb23 [73]

Answer:

μ = mg/kx

Explanation:

Since the bock does not slip, the frictional force equals the weight of the block. So, F = mg. Now, the frictional force, F = μN where μ = coefficient of static friction and N = Normal force.

Now, the normal force equals the spring force F' = kx where k = spring constant and x = compression of spring.

N = F' = kx

So, F = μN = μkx

μkx = mg

So, μ = mg/kx

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3 years ago

The moon's orbit around the Earth will advance in one day:<br><br> 1°<br> 13°<br> 27°<br> 29°

Sever21 [200]

<span>The moon's orbit around the Earth will advance in one day:

1°
13° correct answer
27°
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3 years ago

Read 2 more answers

Which best describes how energy changes form in a car's engine?

andrezito [222]

Answer:

Multiple transformations occur because the chemical energy of the fuel is changed to several forms of energy

Explanation:

In a car engine, multiple energy transformation takes place. The chemical energy storef in fuel is transformed into mechanical energy which helps move the wheels of the vehicle.

The mechanical energy can also be transformed into electrical energy through a sort of dynamo system in vehicles. Stereo players use the electrical energy to produce sound.

We see that multiple energy conversions are common in a motor car.

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3 years ago

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Cars A and B are racing each other along the same straight road in the following manner: Car A has a head start and is a distanc

4vir4ik [10]

The question is incomplete. Here is the complete question.

Cars A nad B are racing each other along the same straight road in the following manner: Car A has a head start and is a distance $D_{A}$ beyond the starting line at t = 0. The starting line is at x = 0. Car A travels at a constant speed $v_{A}$ . Car B starts at the starting line but has a better engine than Car A and thus Car B travels at a constant speed $v_{B}$ , which is greater than $v_{A}$ .

Part A: How long after Car B started the race will Car B catch up with Car A? Express the time in terms of given quantities.

Part B: How far from Car B's starting line will the cars be when Car B passes Car A? Express your answer in terms of known quantities.

Answer: Part A: $t=\frac{D_{A}}{v_{B}-v_{A}}$

Part B: $x_{B}=\frac{v_{B}D_{A}}{v_{B}-v_{A}}$

Explanation: First, let's write an equation of motion for each car.

Both cars travels with constant speed. So, they are an uniform rectilinear motion and its position equation is of the form:

$x=x_{0}+vt$

where

$x_{0}$ is initial position

v is velocity

t is time

Car A started the race at a distance. So at t = 0, initial position is $D_{A}$ .

The equation will be:

$x_{A}=D_{A}+v_{A}t$

Car B started at the starting line. So, its equation is

$x_{B}=v_{B}t$

Part A: When they meet, both car are at "the same position":

$D_{A}+v_{A}t=v_{B}t$

$v_{B}t-v_{A}t=D_{A}$

$t(v_{B}-v_{A})=D_{A}$

$t=\frac{D_{A}}{v_{B}-v_{A}}$

Car B meet with Car A after $t=\frac{D_{A}}{v_{B}-v_{A}}$ units of time.

Part B: With the meeting time, we can determine the position they will be:

$x_{B}=v_{B}(\frac{D_{A}}{v_{B}-v_{A}} )$

$x_{B}=\frac{v_{B}D_{A}}{v_{B}-v_{A}}$

Since Car B started at the starting line, the distance Car B will be when it passes Car A is $x_{B}=\frac{v_{B}D_{A}}{v_{B}-v_{A}}$ units of distance.

5 0

3 years ago