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3 years ago

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A car of mass 500 kg increases its velocity from 40 metre per second to 60 metre per second in 10 second find the distance trave

lled and amount of force applied

1 answer:

Oksana_A [137]3 years ago

5 0

Answer:

it is answer of u are question

You might be interested in

Which of the following would decrease in size during the contraction of a sarcomere? The width of the I-bands The width of the A

ANEK [815]

Hi!

The correct answer would be: the width of I-bands

The sacromere is the smallest contractile unit of striated muscles. These units comprise of filaments (fibrous proteins) that, upon muscle contraction or relaxation, slide past each other. The sacromere consists of thick filaments (myosin) and thin filaments (actin).

<em>Refer to the attached picture to clearly see the structure of a sacromere.</em>

<u>When a sacromere contracts, a series of changes take place which include:</u>

<em>- Shortening of I band, and consequently the H zone</em>

<em>- The A line remains unchanged</em>

<em>- Z lines come closer to each other (and this is due to the shortening of the I bands) </em>

The only changes that take place occur in the zones/areas in the sacromere (as mentioned), not in the filaments (actin and myosin) that make the up the sacromere; hence all other options are wrong.

Hope this helps!

8 0

3 years ago

A horizontal 953 N merry-go-round of radius 1.68 m is started from rest by a constant horizontal force of 73.9 N applied tangent

solniwko [45]

Answer:

K.E=365.2 J

Explanation:

Given data

Weight w =953 N

radius r=1.68 m

F=73.9 N

t=2.55 s

g=9.8 m/s²

To find

Kinetic Energy K.E

Solution

From the moment of inertia

$I=(1/2)MR^{2}\\ as \\W=mg\\So\\I=(1/2)(W/g)R^{2}\\I=(1/2)(953/9.8)(1.68)^{2}\\I=137.232kg.m^{2}$

The angular acceleration is given as

$a=T/I\\a=\frac{FR}{I}\\ a=\frac{(73.9)(1.68)}{137.232}\\a=0.905rad/s^{2}$

The angular velocity is given as

$w=at\\w=(0.905)(2.55)\\w=2.31rad/s$

So the Kinetic Energy is given as

$K.E=(1/2)Iw^{2}\\ K.E=(1/2)(137.232)(2.31)^{2}\\ K.E=365.2J$

3 0

3 years ago

A 2-kg box is pushed to the right by a force of 4 N for a distance of 32 m. It has an initial velocity of 4 m/s to the right. NO

rewona [7]

Answer: a) 8 Kg m/s b) 16 Kg m/s c) 24 Kg m/s d) 16 J e) 128 J f) 144 J

g) 4 s

Explanation:

a) As momentum by definition is the product of mass times the velocity (is a vector quantity), we can write in this case the following:

pi = m. v₀ = 2 Kg . 4 m/s = 8 Kg. m/s

b) In order to get the change in momentum, we need to get first the final speed of the object.

As we have the total distance travelled, and we could find the acceleration, we could use a kinematic equation to solve the question, but later we will need the kinetic energy, it would be better to apply the work-energy theorem, and calculate ΔK as the work done by external force F, as follows:

ΔK = F . d = 1/2 m (vf² - v₀²)

As we know F, d, m, and v₀, we can solve the equation above for vf:

vf = 12 m/s

So, we can compute the final momentum as follows:

pf = m. vf = 2 Kg. 12 m/s = 24 Kg. m/s

Finally, we can find the change in momentum, as the difference between the final momentum and the initial one, calculated in a):

Δp = pf - pi = 24 Kg. m/s - 8 Kg. m/s = 16 Kg. m/s

c) As we have already found, final momentum is as follows:

pf = m . vf = 2 Kg. 12 m/s = 24 Kg. m/s

d) By definition the initial kinetic energy of the box is as follows:

Ki = 1/2 m v₀² = 1/2. 2 Kg .4² m²/s² = 16 J

e) We can find the change in the kinetic energy taking directly the difference between the final and initial ones, as follows:

ΔK = Kf - Ki = 1/2. 2 Kg (12² - 4²) m²/s² = 128 J

f) From above, we have Kf = 1/2 m. vf² = 1/2 . 2 Kg. 12² m²/s² = 144 J

g) As we know the magnitude of F, and the value of m, we can find the acceleration (assumed constant) , applying Newton's Second Law, as follows:

Fext = m .a ⇒ a = F/m = 4 N / 2 Kg = 2 m/s²

Appying the definition of acceleration, we can solve for t, as follows:

t = (vf-v₀) / a = (12 m/s - 4 m/s) / 2 m/s² = 4 s

6 0

3 years ago

Your in an escape room. Only its real and your oxygen runs out in 3 minutes unless you solve this problem, The problem :a ball i

alex41 [277]

The sum of the series allows to find the result for the total distance that the ball bounces is:

total distance = 59.52 in

A series is a set of things or numbers related by a specific operation.

They indicate that the ball falls from an initial height y₀ = 30 in. and it bounces 50% of the height and the process is repeated until it stops, see attached.

Let's build a table to observe the sequence.

drop height rebound

1 30 15

2 15 7.5

3 7.5 3.75

If we call the first term y₀

The first bounce can be found.

y₁ = $\frac{y_o}{2}$

The second bounces.

$y_2 = \frac{y_1}{2} \\y_2 = \frac{y_o}{4}$

The third bounce.

$y_3 = \frac{y_2}{2} \\y_3 = \frac{y_0}{ 8}$

By observing this table we can construct a series of the form

Total distance = $y_o \ ( 1 + \frac{1}{2} + \frac{1}{4}+ \frac{1}{8} + ... +\frac{1}{2n} )$

The sum of the serie has a result of

sum = 127/64 = 1,984

Let's calculate

distance total = 30 1,984

Distance total = 59.52 cm

In conclusion, using the sum of the series we can find the result for the total distance that the ball bounces is:

total distance = 59.52 in

Learn more here: brainly.com/question/8879163

4 0

3 years ago

Read 2 more answers

We repeat the experiment from the video, but this time we connect the wires in parallel rather than in series. Which wire will n

Fed [463]

Answer: B. The copper wire (resistance 0.1)

Explanation: When resistance is in parallel, voltage (V) is the same but current is different for every resistance. Current (i) is related to voltage and resistance (R) by <em>Ohm's Law</em>

i = $\frac{V}{R}$

So, since both wires are in parallel, they have the same voltage but because the copper wire resistance is smaller than Nichrome wire, the first's current will be bigger.

Every resistor in a circuit dissipates electrical power (P) that is converted into heat energy. The dissipation can be found by:

P = $i^{2}*R$

As current for copper wire is bigger than nichrome, power will be bigger and it will dissipate more heat.

In conclusion, the copper wire will dissipate more heat when connected in parallel.

4 0

3 years ago