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MrRissso [65]
4 years ago
14

g A rod is 2.0 m long and lies along the x-axis, with one end at the origin. A force of 25 N is applied at the point x = 1.2 m,

and is directed 30° above the x-axis. What is the torque on the rod?
Physics
1 answer:
Nataly [62]4 years ago
6 0

Answer:

τ = 15N.m

Explanation:

Torque on the rod is given by:

\tau=F*sin(30\°)*d

\tau=25*0.5*1.2

\tau=15N.m

If you were told that gravity contributes as well, and you were given the mass of the rod, you would have to subtract the torque of the weight to this value.

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what would you want the after life to be like. examples are heaven and hell ,reincarnation ,eternal darkness , reliving your liv
NeTakaya

Answer:

i would want to be a dog or a cat

Explanation:

there just funny

4 0
3 years ago
An ion in a mass spectrometer follows a semicircular path of radius 14.8. What is the distance it travels?
aleksklad [387]

The circumference of a circle is (2π · the circle's radius).

The length of a semi-circle is  (1π · the circle's radius) =

                                                 (π · 14.8) = 46.5 (rounded)

(The unit is the same as whatever the unit of the  14.8  is.)

7 0
3 years ago
Read 2 more answers
Which is the result of using a machine?
Eddi Din [679]
Correct Answers is A.

The machines gives us some mechanical advantage. This means the mechanical average makes the work output greater than the work input
Simple most example is a lever. The force applied is smaller and the output work is larger as compared to input.

Option B cannot be true, as there must be a force to get some work done.
Option C and D are inverse of what a machine is designed for. A small force can be exerted through a large distance to have a large force exerted through a small distance. Common Example of this principle is a screw opener. 
3 0
3 years ago
A horizontal spring with stiffness 0.4 N/m has a relaxed length of 11 cm (0.11 m). A mass of 21 grams (0.021 kg) is attached and
riadik2000 [5.3K]

Answer:

0.6983 m/s

Explanation:

k = spring constant of the spring = 0.4 N/m

L₀ = Initial length = 11 cm = 0.11 m

L = Final length = 27 cm = 0.27 m

x = stretch in the spring = L - L₀ = 0.27 - 0.11 = 0.16 m

m = mass of the mass attached = 0.021 kg

v = speed of the mass

Using conservation of energy

Kinetic energy of mass = Spring potential energy

(0.5) m v² = (0.5) k x²

m v² = k x²

(0.021) v² = (0.4) (0.16)²

v = 0.6983 m/s

5 0
4 years ago
A spring on Earth has a 0.500 kg mass suspended from one end and the mass is displaced by 0.3 m. What will the displacement of t
julia-pushkina [17]

To solve this problem we will apply the concepts related to the Force of gravity given by Newton's second law (which defines the weight of an object) and at the same time we will apply the Hooke relation that talks about the strength of a body in a system with spring.

The extension of the spring due to the weight of the object on Earth is 0.3m, then

F_k = F_{W,E}

kx_1 = mg

The extension of the spring due to the weight of the object on Moon is a value of x_2, then

kx_2 = mg_m

Recall that gravity on the moon is a sixth of Earth's gravity.

kx_2 = m\frac{g}{6}

kx_2 = \frac{1}{6} mg

kx_2 = \frac{1}{6} kx_1

x_2 = \frac{1}{6} x_1

We have that the displacement at the earth was x_1 = 0.3m, then

x_2 = \frac{1}{6} 0.3

x_2 = 0.05m

Therefore the displacement of the mass on the spring on Moon is 0.05m

6 0
3 years ago
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