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Dahasolnce [82]
3 years ago
7

A force of F = (2.00ˆi + 3.00ˆj) N is applied to an object that is pivoted about a fixed axle aligned along the z coordinate axi

s. The force is applied at the point r = (4.00ˆi + 5.00ˆj) m.
Find:

(a) the magnitude of the net torque about the z axis

(b) the direction of the torque vector τ
Physics
1 answer:
Vladimir [108]3 years ago
5 0

Explanation:

It is given that,

Force applied to object, F=(2i+3j)\ N

Position, r=(4i+5j)\ m  

(b) The cross product of force and position vector is used to find the net torque about the z axis. It is given by :

\tau=F\times r

\tau=(2i+3j) \times (4i+5j)

\tau=\begin{pmatrix}0&0&-2\end{pmatrix}

or

\tau=(-2k)\ N-m

The torque is acting in -z axis.

(a) The magnitude of torque is given by :

|\tau|=\sqrt{(-2)^2}

|\tau|=2\ N-m

Hence, this is the required solution.

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A wave is travelling at 3000 m/s has a wavelength of 1 m.
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Answer:

a] 3000hz

b]3.33 × 10⁻⁴

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3 0
3 years ago
A parachutist falls 50.0 m without friction. When the parachute opens, he slows down at a rate of 67 m/s*2. If he reaches the gr
KIM [24]

Answer:

3.49 seconds

3.75 seconds

-43200 ft/s²

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

s=ut+\frac{1}{2}at^2\\\Rightarrow 50=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{50\times 2}{9.81}}\\\Rightarrow t=3.19\ s

Time the parachutist falls without friction is 3.19 seconds

v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 9.81\times 50+0^2}\\\Rightarrow v=31.32\ m/s

Speed of the parachutist when he opens the parachute 31.32 m/s. Now, this will be considered as the initial velocity

v=u+at\\\Rightarrow 11=31.32+9.81t\\\Rightarrow t=\frac{11-31.32}{-67}=0.3\ s

So, time the parachutist stayed in the air was 3.19+0.3 = 3.49 seconds

s=ut+\frac{1}{2}at^2\\\Rightarrow \frac{s}{2}=0t+\frac{1}{2}\times a\times t^2\\\Rightarrow \frac{s}{2}=\frac{1}{2}at^2

s=ut+\frac{1}{2}at^2\\\Rightarrow \frac{s}{2}=u1.1+\frac{1}{2}\times a\times 1.1^2

Now the initial velocity of the last half height will be the final velocity of the first half height.

v=u+at\\\Rightarrow v=at

Since the height are equal

\frac{1}{2}at^2=u1.1+\frac{1}{2}\times a\times 1.1^2\\\Rightarrow \frac{1}{2}at^2=at1.1+\frac{1}{2}\times a\times 1.1^2\\\Rightarrow 0.5t^2-1.1t-0.605=0\\\Rightarrow 500t^2-1100t-605=0

t=\frac{11\left(1+\sqrt{2}\right)}{10},\:t=\frac{11\left(1-\sqrt{2}\right)}{10}\\\Rightarrow t=2.65, -0.45

Time taken to fall the first half is 2.65 seconds

Total time taken to fall is 2.65+1.1 = 3.75 seconds.

When an object is thrown with a velocity upwards then the velocity of the object at the point to where it was thrown becomes equal to the initial velocity.

v^2-u^2=2as\\\Rightarrow a=\frac{v^2-u^2}{2s}\\\Rightarrow a=\frac{0^2-240^2}{2\times \frac{8}{12}}\\\Rightarrow a=-43200\ ft/s^2

Magnitude of acceleration is -43200 ft/s²

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What is 0 point energy times negative 0 point energy ????????
elixir [45]
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