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Levart [38]
3 years ago
6

A car is traveling with a constant speed when the driver suddenly applies the brakes, causing the car to slow down with a consta

nt acceleration of magnitude 3.50 m/s2. if the car comes to a stop in a distance of 30.0 m, what was the car's original speed?
Physics
1 answer:
Margaret [11]3 years ago
4 0
Consider a car<span> that travels between points A and B. The </span>car's<span> average </span>speed<span> can be ..... the </span>car<span> to </span>slow down<span> with a </span>constant acceleration<span> of </span>magnitude 3.50 m/s2<span>. </span>If<span> the </span>car comes<span> to a </span>stop<span> in a </span>distance<span> of</span>30.0 m<span>, what was the </span>car's original speed<span>? ... A </span>car<span> is </span>traveling<span> at 26.0 </span>m<span>/s when the </span>driver suddenly applies<span> the </span>brakes<span>, ...</span>
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The first car has twice the mass of a second car, but only half as much kinetic energy. When both cars increase their speed by 9
kirill [66]

Answer:

Speed of the car 1 =V_1=8.98m/s

Speed of the car 2 =V_2=17.96m/s

Explanation:

Given:

Mass of the car 1 , M₁ = Twice the mass of car 2(M₂)

mathematically,

M₁ = 2M₂

Kinetic Energy of the car 1 = Half the kinetic energy of the car 2

KE₁ = 0.5 KE₂

Now, the kinetic energy for a body is given as

KE =\frac{1}{2}mv^2

where,

m = mass of the body

v = velocity of the body

thus,

\frac{1}{2}M_1V_1^2=0.5\times \frac{1}{2}M_2V_2^2

or

\frac{1}{2}2M_2V_1^2=0.5\times \frac{1}{2}M_2V_2^2

or

2M_2V_1^2=0.5\times M_2V_2^2

or

2V_1^2=0.5\times V_2^2

or

4V_1^2= V_2^2

or

2V_1= V_2  .................(1)

also,

\frac{1}{2}M_1(V_1+9.0)^2=\frac{1}{2}M_2(V_2+9.0)^2

or

\frac{1}{2}2M_2(V_1+9.0)^2=\frac{1}{2}M_2(2V_1+9.0)^2

or

2(V_1+9.0)^2=(2V_1+9.0)^2

or

\sqrt{2}(V_1+9.0)=(2V_1+9.0)

or

(\sqrt{2}V_1+ \sqrt{2}\times 9.0)=(2V_1+9.0)

or

(\sqrt{2}V_1+ 12.72)=(2V_1+9.0)

or

(2V_1-\sqrt{2}V_1)=(12.72-9.0)

or

(0.404V_1)=(3.72)

or

V_1=8.98m/s

and, from equation (1)

V_2=2\times 8.98m/s = 17.96m/s

Hence,

Speed of car 1 =V_1=8.98m/s

Speed of car 2 =V_2=17.96m/s

8 0
2 years ago
Read 2 more answers
Consider the accuracy of recording muscle electrical activity using electrodes placed on the skin surface, compared to directly
OverLord2011 [107]

Answer: The common difference between surface EMG and intramuscular EMG is that that former is non-invasive while the later is an invasive method

Explanation:

Electromyography (EMG) is used clinically for the examination of muscle excitations (muscle electrical activity) in both normal or abnormal conditions. There are two forms of EMG includes:

--> Surface EMT and

--> Intramuscular EMT

Surface EMT is a non invasive method of examination of muscle excitations for superficial and easily accessible muscles.

Intramuscular EMT is the invasive method of examination of muscle excitations usually for deep muscles.

The difference between the two forms of EMT includes:

- surface EMT is non- invasive while intramuscular EMT is invasive

- surface EMT is used to access superficial muscle while intramuscular EMT is used to access deep muscles.

- surface EMT requires less skill and time to carry out while intramuscular EMT requires special skills and takes more time while carrying out the procedure.

8 0
3 years ago
Can someone help with this???
Rus_ich [418]

no BECQUSE POSUM BROOB SHSHSJ

8 0
3 years ago
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Complete the sentence-
trasher [3.6K]

Answer:

Friction always acts opposite to the motion.

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How far does a ca4 go
miv72 [106K]

Answer:

as far as the car can withstand

Explanation:

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