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Roman55 [17]
3 years ago
12

Ajar can hold 60 mL of liquid. It is 10 cm high and 2 cm wide. The length of the jar is:

Physics
2 answers:
ZanzabumX [31]3 years ago
8 0
60:2=30 despues 30:10= 3
marin [14]3 years ago
5 0

Answer:

3cm

Explanation:

From the question, we obtained the following:

Volume = 60mL = 60cm^3

Height = 10cm

Width = 2m

Length =?

Volume = length x width x height

Length = volume / width x height

Length = 60/(2 x 10) = 60/ 20

Length = 3cm

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Beginning at the NW corner of the intersection of Pine &amp; 675, thence north 950 feet, thence west 380 feet, thence south 950
Serjik [45]

Answer:

this description is valid for mediadle displacement, bone is an acceptable description

Explanation:

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Each displacement occurs with respect to the previous one, indicating the magnitude of the displacement and its direction.

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6 0
3 years ago
The wheel having a mass of 100 kg and a radius of gyration about the z axis of kz=300mm, rests on the smooth horizontal plane.a.
pickupchik [31]

Answer:

a) 20 rad/s

b) 6 m/s

Explanation:

b) Force acting on the wheel is 200 N

mass of the wheel is 100 kg

From Newton's second law of motion, F = m × a

Where F is the net force acting on the body

m is mass of the body

a is the acceleration of the body

By substituting the values we get, a = 2 m/s²

As acceleration is constant, we can use the below formula for calculating the final velocity of the object

v = u + a × t

Where v is the final velocity

u is the initial velocity

a is the acceleration

t is the time taken

u = 0 (∵ it starts from rest)

By substituting the values we get

v = 0 + 2 × 3 = 6 m/s

∴ Speed of center of mass after 3 seconds = 6 m/s

a) As the wheel rotates about z-axis, radius of gyration will be the radius of wheel

∴ Radius of the wheel = 300 mm

Torque acting on the wheel about axis of rotation = 300 mm × 200 N =

60 N·m

Torque = (Moment of inertia) × (angular acceleration)

Assuming that the mass of spokes of the wheel to be negligible,

Moment of inertia of the wheel about axis of rotation = 100 × 300² × 10^{-6} = 9 kg·m²

Then,

60 = 9 × (angular acceleration)

∴ angular acceleration ≈ 6·67 rad/s²

As angular acceleration of the wheel is constant, we can use the below formula for calculation of final angular speed

w_{f} = w_{i} + α × t

Where

w_{f} is the final angular velocity

w_{i} is the initial angular velocity

α is the angular acceleration

t is the time taken

w_{i} is 0 (∵ initially it starts from rest)

By substituting the values we get

w_{f} = 6·67 × 3 = 20 rad/s

∴ Angular velocity of the wheel after three seconds = 20 rad/s

3 0
3 years ago
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