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3 years ago

In which of the following situations are there a force on the object balanced

Physics

1 answer:

ser-zykov [4K]3 years ago

3 0

Where’s the rest at????

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What factors cause a rock to become metamorphic

Lynna [10]

Pressure and heat. I hope this helps

3 0

3 years ago

Read 2 more answers

A student (m = 68 kg) falls freely from rest and strikes the ground. During the collision with the ground, he comes to rest in a

Gnesinka [82]

Answer:

5.7141 m

Explanation:

Here the potential and kinetic energy will balance each other

$mgh=\frac{1}{2}mv^2\\\Rightarrow v=\sqrt{2gh}$

This is the initial velocity of the system and the final velocity is 0

t = Time taken = 0.04 seconds

F = Force = 18000 N

a = Acceleration

g = Acceleration due to gravity = 9.81 m/s²

Equation of motion

$v=u+at\\\Rightarrow a=\frac{v-u}{t}$

From Newton's second law

$F=ma\\\Rightarrow F=m\frac{v-u}{t}\\\Rightarrow 18000=68\frac{0-\sqrt{2gh}}{0.04}\\\Rightarrow \frac{18000}{68}\times 0.04=-\sqrt{2\times 9.81\times h}\\\Rightarrow 10.58823=-\sqrt{2\times 9.81\times h}$

Squarring both sides

$112.11061=2\times 9.81\times h\\\Rightarrow h=\frac{112.11061}{2\times 9.81}\\\Rightarrow h=5.7141\ m$

The height from which the student fell is 5.7141 m

5 0

3 years ago

Never mind i dont need help anymore

Usimov [2.4K]

Yay good for you!!!!

8 0

3 years ago

En un m.A.S. La amplitud tiene un valor de 10 centimetros y el periodo es de 2 segundos calcular el valor de la velocidad de 0.8

Ivanshal [37]

Answer:

v1=18.46m/s

v2=29.8cm/s

Explanation:

We know that

$A=10cm\\T=2s$

the equation of the motion is

$x=Acos(\omega t)\\$

we can calculate w by using

$\omega=\frac{2\pi}{T}=\frac{2\pi}{2}=\pi$

Hence, we have that

$x=10cm*cos(\pi t)\\$

the speed will be

$v=-\omega*Asin(\omega t)\\|v(0.8)|=|\pi*10cm*sin(\pi *0.8)|=18.46\frac{cm}{s}\\|v(1.4)|=|\pi*10cm*sin(\pi *1.4)|=29.8\frac{cm}{s}$

hope this helps!

6 0

3 years ago

(a) In what direction would the ship in Exercise 3.57 have to travel in order to have a velocity straight north relative to the

shtirl [24]

Answer:

Direction of ship: 9.45° West of North

Ship's relative speed: 7.87m/s

Explanation:

A. Direction of ship: since horizontal of the velocity of boat relative to the ground is 0

Vx=0

Therefore, -VsSin∅+VcCos∅40°

Sin∅ = Vc/Vs × Cos 40°

Sin∅ = 1.5/7 ×Cos40°

Sin∅= 0.164

∅= Sin-¹ (0.164)

∅= 9.45° W of N

B. Ship's relative speed:

Vy= VsCos∅ + Vcsin40°

= 7Cos9.45° + 1.5sin40°

= 7×0.986 + 1.5×0.642

= 7.865

= 7.87m/s

4 0

3 years ago