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3 years ago

Define temperature

Physics

2 answers:

Whitepunk [10]3 years ago

8 0

Please mark me as Brainliest ........

ladessa [460]3 years ago

7 0

Answer:

temperature is the degree or intensity of heat present in a substance or object.

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3 years ago

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What happens when a negatively charged object A is brought near a neutral object B?

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3 years ago

A wheel of diameter 40.0 cm starts from rest and rotates with a constant angular acceleration of 3.00 rad/s^2. At the instant th

xxMikexx [17]

Answer:

ac= 15.07 m/s²

Explanation:

The Wheel rotates with a constant angular acceleration.:

Centripetal acceleration is calculated as follows:

ac =ω² *R Formula (1)

ac =v² / R Formula (2)

Kinematics of the wheel

We apply the equations of circular motion uniformly accelerated :

ωf²= ω₀² + 2αθ Formula (3)

v = ω* R Formula (4)

Where:

θ : angle that the wheel has rotated in a given time interval (rad)

α : angular acceleration (rad/s²)

ω₀ : initial angular speed ( rad/s)

ωf : final angular speed ( rad/s)

v: tangential velocity of a point on the rim ( m/s)

R : radius of wheel (m)

ac: centripetal acceleration, (m/s²)

Data:

D = 40.0 cm : diameter of the wheel

R = D/2= 40.0 cm/ 2 = 20 cm = 0.2m

α = 3.00 rad/s^2

ω₀ = 0

n = 2 revolutions : number of revolutions

θ =2πn (rad) = 2π*2 (rad) = 4π rad

Calculate of the ωf

We replace data in the formula (3)

ωf²= ω₀² + 2αθ

ωf²= 0 + 2(3)(4π)

ωf²= 24π

$w_{f} = \sqrt{24\pi }$

ωf = 8.68 rad/s

Calculate of the v

We replace data in the formula (4)

v = ω*R

v = (8.68)*(0.2)

v = 1.736 m/s

Calculate of the ac

We replace data in the formula (1)

ac = ( ω)²*(R)

ac = (8.68)²*(0.2)

ac = 15.07 m/s²

We replace data in the formula (2)

ac = v²/ R

ac = ( 1.736 )²/(0.2)

ac = 15.07 m/s²

7 0

4 years ago

A massless string connects a 1.00 kg mass to a 3.00 kg cart which is resting on a frictionless horizontal surface. The mass hang

Romashka-Z-Leto [24]

Both masses will have the same acceleration. The cart accelerates to the right with a magnitude of 4.9 m/ $s^{2}$ . The correct answer is 4.90 m/ $s^{2}$

Given that a massless string connects a 1.00 kg mass to a 3.00 kg cart which is resting on a frictionless horizontal surface.

Let M = 1kg and m = 3 kg

Since the horizontal surface is frictionless, the tension in the string will be the same. when the mass is hanged over a frictionless pulley, the tension will also be the same.

When the mass is released, the cart accelerates to the right can be calculated from Newton' second law of motion. That is,

M( g + a) = m(g - a)

1(9.8 + a) = 3( 9.8 - a)

9.8 +a = 29.4 - 3a

collect the like terms

4a = 19.6

a = 19.6/4

a = 4.9 m/ $s^{2}$

Therefore, the cart accelerates to the right with a magnitude of 4.9 m/ $s^{2}$ . The correct answer is 4.90 m/ $s^{2}$

Learn more about dynamics here: brainly.com/question/24994188

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3 years ago