Calculate the current in a circuit whose total resistance is 4 ohms and whose emf is 12 volts.

A 500 kg crate is lifted by a crane and accelerated upward at 2 m/s/s. What is the tension force in the cable pulling up on the

Likurg_2 [28]

Answer = 1000N
F = ma
M = 500kg
a= 2 m/s/s
Solve for F

6 0

3 years ago

The greek letter (delta) indicates a(n)

iragen [17]

A
The delta indicates a difference or change. such as delta y is the change in y position

6 0

4 years ago

Please Please I only need help in Questions 4 and 5!

sdas [7]

I'm not sure but for the first one, if there were more electrons than protons that would mean the object would have a positive charge so you could put another object that is positively charged near it, to see if it would attract and if it would it would mean it's negatively charged and if it wouldn't it would mean it's positively charged. (not sure)

for the second one, after you've rubbed the balloon oh hair, the electrons from your hair have transferred onto the balloon, meaning that the balloon is now negatively charged. because the wall is neutral, it means that it has the SAME number of both protons and electrons ( positive and negative charges cancel out to create a neutral charge). because the protons are positively charged, and the balloon is negatively charged, the two objects will attract because opposite charges attract. this happens because of static electricity.

hope this helps :)

4 0

3 years ago

A 0.26 kg rod of length 80 cm is suspended by a frictionless pivot at one end. It is held horizontal and released.

Daniel [21]

Answer:

a) $a_{center} = 7.38 ~m/s^2$

b) $a_{end} = 14.77 ~m/s^2$

c) $v_{center} = 2.43~m/s$

Explanation:

a) Immediately after the rod is released, <u>the rod is still horizontal but now subject to gravity.</u> Since one end of the rod is fixed, then the weight of the rod applies a torque. Then by Newton's Second Law, the acceleration can be found.

$\tau =I\alpha$

where I is the moment of inertia of the rod with respect to its fixed end, and α is the angular acceleration.

The net torque of the rod is

$\vec{\tau} = \vec{r} \times \vec{F}\\\tau = rF\sin(90) = rF$

where r is the distance from center of the mass to the fixed end, so r = 0.4 m.

The weight of the rod is w = mg = 0.26 x 9.8 = 2.54 N.

So the net torque is τ = 1.01 Nm.

The moment of inertia of the rod is

$I = \frac{1}{3}mL^2 = \frac{1}{3}(0.26)(0.8)^2 = 0.055~kg m^2$

So, the Newton's Second Law yields

$\tau = I\alpha\\\alpha = \frac{\tau}{I} = \frac{1.01}{0.055} = 18.47$

<u>The relation between angular acceleration and linear acceleration is a = αr </u>

So, the linear acceleration of the rod is

$a = \alpha r = 7.38~m/s^2$

b) Using the same relationship between angular acceleration and linear acceleration, the linear acceleration of the end of the rod can be found.

$a = \alpha L = 14.77~m/s^2$

c) The conservation of energy can be used to find the velocity when the rod is vertical.

$K_1 + U_1 = K_2 + U_2\\0 + mg(L/2) = \frac{1}{2}I\omega^2 + 0\\(0.26)(9.8)(0.4) = \frac{1}{2}(0.055)\omega^2\\\omega = 6.08~rad/s$

The linear velocity is v = ωr, so

v = 2.43 m/s.

4 0

3 years ago

How is gravity simulated on the hermes spacecraft?

gulaghasi [49]

Hermes has a rotating wheel, or centrifuge, to provide artificial gravity for the crew. Spaceships in “2001: A Space Odyssey” and “Mission to Mars” were similarly equipped. Although it is not shown in the film, the astronauts would have used a Mars descent vehicle (MDV) to get down from Hermes to the surface.

3 0

3 years ago