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goldfiish [28.3K]

1 year ago

14

The drawing shows the top view of a door that is 1.68 m wide. two forces are applied to the door as indicated. what is the magni

tude of the net torque on the door with respect to the hinge?

1 answer:

jekas [21]1 year ago

8 0

First, torque is equal to force times the distance. for the first force that is applied, the torque is zero because is applied at the hinge. so the net torque:
t = ( 12 N ) ( 0 m ) ( cos 30 ) + ( 12 N ) ( 1.68 m ) cos 45
t = 14.26 Nm is the torque with respect to the hinge

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A 2.0 g metal cube and a 4.0 g metal cube are 6.0 cm apart, measured between their centers, on a horizontal surface. For both, t

emmasim [6.3K]

Answer:

a) t=10.2s

b) The 2g-cube moves first

Explanation:

Since the electric force is the same on both cubes and so is the coefficient of static friction, the first one to move will be the one with less mass.

So, on the 2g-cube the sum of forces are:

$\left \{ {{Ff-Fe=0} \atop {N-m*g=0}} \right.$

Replacing the friction on the first equation:

$\mu*m*g-Fe=0$ Thus $Fe=\mu*m*g=12.74*10^{-3}N$

The electric force is:

$Fe = \frac{K*q^2}{d^2}$ Solving for q:

q=71.44nC

This amount divided by the rate at which they are being charged:

t = 71.44nC / 7nC/s = 10.2s

7 0

1 year ago

An object is dropped from rest into a pit, and accelerates due to gravity at roughly 10 m/s2. It hits the ground in 5 seconds. A

vitfil [10]

Answer:

Second pit is 375 m deeper compared to first pit.

Explanation:

We have equation of motion s = ut + 0.5at²

First object hits the ground after 5 seconds,

Initial velocity, u = 0 m/s

Acceleration, a = 10 m/s²

Time, t = 5 s

Substituting,

s = ut + 0.5 at²

s = 0 x 5 + 0.5 x 10 x 5²

s = 125 m

Depth of pit 1 = 125 m

Second object hits the ground after 10 seconds,

Initial velocity, u = 0 m/s

Acceleration, a = 10 m/s²

Time, t = 10 s

Substituting,

s = ut + 0.5 at²

s = 0 x 10 + 0.5 x 10 x 10²

s = 500 m

Depth of pit 2 = 500 m

Difference in depths = 500 - 125 = 375 m

Second pit is 375 m deeper compared to first pit.

7 0

1 year ago

The δe of a system that releases 12.4 j of heat and does 4.2 j of work on the surroundings is __________ j.

Vedmedyk [2.9K]

The answer for this problem is clarified through this, the system is absorbing (+). And now see that it uses that the SURROUNDINGS are doing 84 KJ of work. Any time a system is overshadowing work done on it by the surroundings the sign will be +. So it's just 12.4 KJ + 4.2 = 16.6 KJ.

5 0

2 years ago

In lab, your instructor generates a standing wave using a thin string of length L = 1.65 m fixed at both ends. You are told that

mars1129 [50]

Answer:

The maximum transverse speed of the bead is 0.4 m/s

Explanation:

As we know that the Amplitude of the travelling wave is

A = 3.65 mm

Now the speed of the travelling wave is

$v_x = 13.5 m/s$

now we know that distance of first antinode from one end is 27.5 cm

so length of the loop of the standing wave is given as

$\frac{\lambda}{4} = 27.5 cm$

$\lambda = 110 cm$

now we have

$N = \frac{2L}{\lambda}$

$N = \frac{2(1.65)}{1.10}$

$N = 3$

now we have

$R = 2A sin(kx)$

$R = 2(3.65) sin(\frac{2\pi}{1.10}x)$

$R = 7.3 sin(1.82 \pi x)$

now at x = 13.8 cm

$R = 7.3 sin(1.82 \pi (0.138))$

$R = 5.18 mm$

now we have

$f = \frac{v}{\lambda}$

$f = \frac{13.5}{1.1}$

$f = 12.27 Hz$

now maximum speed is given as

$v_y = R\omega$

$v_y = (5.18 \times 10^{-3})(2\pi(12.27))$

$v_y = 0.4 m/s$

4 0

1 year ago

Solve A and B using energy considerations.

Alisiya [41]

<span>Use the kinematic equation vf^2 = vi^2 + 2ad where;
vf = ?
vi = 0 m/s
a = 9.8 m/s^2
d1 = 10 m
d2 = 25 m

final velocity at the ground (d1): vf = sqrt(2)(9.8)(10) = 14 m/s
final velocity to the bottom of the cliff (d2): vf = sqrt(2)(9.8)(25) = 22.14 m/s
</span>

7 0

2 years ago