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3 years ago

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An object at rest stays at rest and an object in stays in motion with the same speed and in the same direction unless acted upon

by an unbalanced force is Newton’s 1st law of motion
True or false

1 answer:

IrinaVladis [17]3 years ago

3 0

Answer:

Hi I'm happy to meet you and the answer for your question is true. If you need me to state the law ask me

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Two infinite wires 20 cm apart each carry a current of 3 A into the paper. d I I d/2 d/2 At a distance d 2 below their midpoint,

Rzqust [24]

<u>Answer:</u>

<h3>As electric current is carried in a cable, around it, a magnetic field is created. The lines of the magnetic fields form concentric circles around the wire. The direction of the magnetic field hinges on the direction of the current. It can be calculated by pointing the thumb of your right hand in the direction of the moment, using the "right hand law." The position of your curled fingers is in the magnetic field lines. The magnetic field magnitude depends on the sum of current, and the distance from the wire carrying the charge.</h3>

<u></u>

<u>Explanation:</u>

Determine the direction of vector B magnitude B: $B: B=\mu_{0} * 1 /(2 \pi r): r=d / 2 * \sqrt{2}$

$\cos \alpha=1 / 2 \Rightarrow \alpha=450$

Resultant magnitude strength: $B=2 B^{*} \cos \alpha=B=u_{0}^{*} 1 /(2 \pi r)=4.24^{*} 10^{-6} T=4.24 u T$ its direction is pointing to the left.

Note: Refer the image attached below

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3 years ago

The tension in the horizontal cord is 30N. Find the weight of the object.

vazorg [7]

Making a drawing of the system, we will have two forces which are tension and the weight of the object. Balancing the forces present, we do as follows:

T = W
W = 30 N

Therefore, weight is equal to 30 N. Hope this answers the question. Have a nice day. Feel free to ask more questions.

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3 years ago

All the organelles in a cell work together to help the cell and the whole organism function. Some diseases are caused by the fai

soldier1979 [14.2K]

The correct answer is cell membrane

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3 years ago

Read 2 more answers

For the image of the overhead projector to be in focus, the distance from the projector lens to the image, <img src="https://tex

rjkz [21]

Given:
distance from the projector lens to the image, di
projector lens focal length, f
distance from the transparency to the projector lens, do

thin lens equation: 1/f = 1/di + 1/do
do = 4 inches
di = 8 feet

convert feet to inches, for uniformity.
1 foot = 12 inches
8 feet * 12 inches/ft = 96 inches

1/f = 1/96 inches + 1/4 inches

Adding fractions, denominator must be the same.

1/f = (1/96 * 1/1) + (1/4 * 24/24)
1/f = 1/96 + 24/96
1/f = 25/96

to find the value of f, do cross multiplication
1*96 = f * 25
96 = 25f
96/25 = f
3.84 = f

The focal length of the project lens is 3.84 inches

4 0

3 years ago

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