Does a physical change affect the identity of a substance? Does a chemical change affect the identity of a substance?

Car A is traveling at 22.0 m/s and car B at 29.0 m/s. Car A is 300 m behind car B when the driver of car A accelerates his car w

Len [333]

Answer:

(c) 19.0s

Explanation:

Given Data

Car A speed v=22.0 m/s

Car B speed v=29.0 m/s

Car A distance S=300 m behind Car B

Car A acceleration a=2.40 m/s²

To find

Time required For Car A to take over Car B

Solution

We can represent Car A Coordinate by using equation of simple motion

$X_{A} =vt+1/2at^{2}\\X_{A} =22t+(1/2)(2.40)t^{2}$

And Coordinates of car B equals

$X_{B}=300+29t\\$

Car A is overtake car B when:

$X_{A}=X_{B}\\ 22t+(1/2)(2.4)t^{2}=300+29t\\1.2t^{2}-7t-300=0\\ time=19.0s$

Option (C) 19.0s is correct one

6 0

4 years ago

The resting heart rate will increase quickly in a person who has a high level of cardiovascular fitness. True or false

mestny [16]

False . It won't increase, take an athlete for example . They don't get tired as easily because their cardiovascular fitness is strong .

6 0

3 years ago

Read 2 more answers

A gas in a closed container is heated with 10J of energy, causing the lid of the container to rise 2m with 3N of force. What is

AveGali [126]

Explanation:

For this problem, use the first law of thermodynamics. The change in energy equals the increase in heat energy minus the work done.

ΔU=Q−W

We are not given a value for work, but we can solve for it using the force and distance. Work is the product of force and displacement.

W=FΔx

W=3N×2m

W=6J

Now that we have the value of work done and the value for heat added, we can solve for the total change in energy.

ΔU=Q−W

ΔU=10J−6J

ΔU=4J

Answer is 4J

i think this may help you very much

3 0

3 years ago

A wave pulse travels along a string at a speed of 200 cm/s. What will be the speed if:

34kurt

Answer:

a) $v_f = \sqrt{\frac{2TL}{m}} = \sqrt{2} v = \sqrt{2}2m/s =2.83 m/s$

The velocity increase by a factor of $\sqrt{2}$

b) $v_f = \sqrt{\frac{TL}{4m}} = \frac{1}{2} v = \frac{1}{2} *2m/s =1 m/s$

The velocity decrease by a factor of 2.

c) $v_f = \sqrt{\frac{4TL}{m}} = 2} v = 2 *2m/s =4m/s$

The velocity increase by a factor of 2

d) $v_f = \sqrt{\frac{4TL}{4m}} = v = 2m/s$

The velocity not changes.

Explanation:

For this case we know that the velocity is $v = 200 cm/s = 2m/s$

$v_f$ represent the final velocity after the changes specified,

Part a

The formula for the speed of a wave in a string is given by:

$v = \sqrt{\frac{T}{\rho}}$

And the linear density is defined as:

$\rho = \frac{m}{L}$

And if we replace this we got:

$v = \sqrt{\frac{TL}{m}}$

If the tension mass is doubled we have this:

$v_f = \sqrt{\frac{2TL}{m}} = \sqrt{2} v = \sqrt{2}2m/s =2.83 m/s$

The velocity increase by a factor of $\sqrt{2}$

Part b

If we mass is quadrupled we have this:

$v_f = \sqrt{\frac{TL}{4m}} = \frac{1}{2} v = \frac{1}{2} *2m/s =1 m/s$

The velocity decrease by a factor of 2.

Part c

If the length is quadrupled we have this:

$v_f = \sqrt{\frac{4TL}{m}} = 2} v = 2 *2m/s =4m/s$

The velocity increase by a factor of 2

Part d

For this case we know that the mass and the length are both quadrupled and we got:

$v_f = \sqrt{\frac{4TL}{4m}} = v = 2m/s$

The velocity not changes.

7 0

3 years ago

Suppose the ski patrol lowers a rescue sled and victim, having a total mass of 90.0 kg, down a p slope at constant speed, as sho

kondor19780726 [428]

The work done by friction to move the sled is - 1,323 J.

<h3>What is Coefficient of friction?</h3>

The friction coefficient is the ratio of the normal force pressing two surfaces together to the frictional force preventing motion between them.
Typically, it is represented by the Greek letter µ. In terms of math, is equal to F/N, where F stands for frictional force and N for normal force.
The coefficient of friction has no dimensions because both F and N are measured in units of force (such as newtons or pounds). For both static and kinetic friction, the coefficient of friction has a range of values.
When an object experiences static friction, the frictional force resists any applied force, causing the object to stay at rest until the static frictional force is removed. The frictional force opposes an object's motion in kinetic friction.

Solution:

Given that

Coefficient of friction (µ) = 0.10

Mass (m) = 90kg

distance covered (d) = 30m

We use the formula:

friction work = -µmgdcos∅

friction work = -0.100 × 90 kg × 9.8 m/s² × 30 m × cos 60°

friction work = - 1,323 J

Know more about Coefficient of friction numerical brainly.com/question/19308401

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5 0

2 years ago