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2 years ago

Help me do this question

Physics

1 answer:

Zepler [3.9K]2 years ago

3 0

Answer:

c20800

Explanation:

go to bear khana bro your book looks cheap go and study in durbar kanda school and take somee cash and do ash ah boy

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An airplane starts at rest and accelerates down the runway for 20 s. At the end of the runway, its velocity is 80 m/s north. Wha

goldenfox [79]

Vi = vf + ( a.t)
0 m/s (rest) = 80 m/s + (a.20s.-1)

a.20.-1= -80, we check if this is true; = 0 m/s= 80 + -80 = 0 ITS TRUE.

so a.20.-1= -80, a= 80/20 , Answers a= 4 m/s

5 0

2 years ago

What should be the spring constant k of a spring designed to bring a 1200 kg car to rest from a speed of 85 km/h so that the occ

borishaifa [10]

Answer:

k = 5178.8 N/m

Explanation:

As we know that spring mass system will oscillate at angular frequency given as

$\omega = \sqrt{\frac{k}{m}}$

now we have

$\omega = \sqrt{\frac{k}{1200}}$

now the maximum acceleration of the spring block system is at its maximum compression state which is given as

$a = \omega^2 A$

here A= maximum compression of the spring

so here in order to find maximum compression of the spring we will use energy conservation as we know that initial total kinetic energy of the car will convert into spring potential energy

$\frac{1}{2}mv^2 = \frac{1}{2}kA^2$

here we know that

v = 85 km/h

$v = 85 \times\frac{1000}{3600} = 23.61 m/s$

now we have

$(1200)(23.61^2) = kA^2$

$A^2 = \frac{6.68 \times 10^5}{k}$

now from above equation of acceleration we have

$5.0 g = (\frac{k}{m})\sqrt{\frac{6.68 \times 10^5}{k}}$

$5.0(9.81) = \sqrt{k}(0.68)$

$k = 5178.8 N/m$

6 0

3 years ago

If a car is moving on a straight line with a velocity of 40 m/s and it changes its velocity to 60 m/s in 4 seconds, calculate it

Bogdan [553]

Answer:

5m/s²

Explanation:

Initial Velocity= 40m/s

Final Velocity=60m/s

Time = 4s

a=(v-u)/t

a=60-40/4

a=20/4

a=5m/s²

8 0

2 years ago

Traveling in a circle requires a net force

sergij07 [2.7K]

I don’t think I understand the question but true?

4 0

2 years ago

When devising a model, scientists can only use the information available during their lifetime. This means that the current mode

11111nata11111 [884]

no, it not useless. we still learn Bohr's model in HS n dats almost 200 yr old! while there may be new models, previous one is good for explaining the basics. it is also useful to learn previous model n see how our understanding improves over time.

6 0

2 years ago

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