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patriot [66]
3 years ago
8

HELP ASAP WITH number 6 plz

Physics
1 answer:
Alika [10]3 years ago
6 0
The answer would be c
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A circular saw blade with radius 0.175 m starts from rest and turns in a vertical plane with a constant angular acceleration of
adelina 88 [10]

Answer:

x = 11.23  m

Explanation:

For this interesting exercise, we must use angular kinematics, linear kinematics and the relationship between angular and linear quantities.

Let's reduce to SI system units

    θ = 155 rev (2pi rad / rev) = 310π rad

    α = 2.00rev / s2 (2pi rad / 1 rev) = 4π rad / s²

Let's look for the angular velocity at the time the piece is released, with starting from rest the initial angular velocity is zero (wo = 0)

    w² = w₀² + 2 α θ  

    w =√ 2 α θ

    w = √(2 4pi 310pi)

    w = 156.45  rad / s

The relationship between angular and linear velocity

    v = w r

    v = 156.45  0.175

    v = 27.38 m / s

In this part we have the linear speed and the height that it travels to reach the floor, so with the projectile launch equations we can find the time it takes to arrive

    y = v_{oy} t - ½ g t²

As it leaves the highest point its speed is horizontal

   y = 0 - ½ g t²

   t = √ (-2y / g)

   t = √ (-2 (-0.820) /9.8)

   t = 0.41 s

With this time we calculate the horizontal distance, because the constant horizontal speed

   x = vox t

   x = 27.38 0.41

   x = 11.23  m

5 0
3 years ago
How much work in joules is required to lift a 23 kg box up from the ground to your waist that is 1.0 meters high, carry it 6 met
PSYCHO15rus [73]

Answer:

2682

Explanation:

Work done is given by :

Work = Force x distance

         =  mg x d

So, work done in lifting the box of 23 kg up to my waist of 1 m high is :

W = mg x d

   = 23 x 9.18 x 1

   = 211.14

Now work done carrying the box horizontally 6 meters across the room is

W = mg x d

   = 23 x 9.18 x 6

   = 1266.84

Work done in placing the box on the shelf that is 5.7 m above the ground is

W = mg x d

   = 23 x 9.18 x 5.7

   = 1203.49

So the total work done is = 211.14 + 1266.84 + 1203.49

                                          = 2681.47

                                          = 2682 (rounding off)

5 0
2 years ago
The 68-kg crate is stationary when the force P is applied. Determine the resulting acceleration of the crate if (a) P = 0, (b) P
bogdanovich [222]

Explanation:

Mass of the crate, m = 68 kg

We need to find the resulting acceleration if :

(a) Force, P = 0

P = m a

⇒ a = 0

(b) P = 181 N

a=\dfrac{P}{m}

a=\dfrac{181\ N}{68\ kg}

a=2.67\ m/s^2

(c) P = 352 N

a=\dfrac{P}{m}

a=\dfrac{352\ N}{68\ kg}

a=5.17\ m/s^2

Hence, this is the required solution.

5 0
3 years ago
A tree loses water to the air by the process of transpiration at the rate of 110 g/h. This water is replaced by the upward flow
Ksivusya [100]

Answer:

The answer is 1.87nm/s.

Explanation:

The 110g/hr  water loss must be replaced by 110g/hr of sap. 110g of sap corresponds to a volume of  

110g \div \dfrac{1040*10^3g}{1*10^6cm^3}  = 106cm^3

thus rate of sap replacement is

106cm^3/hr = 106*10^{-6}m^3/3600s  = 2.94*10^{-8}m^3/s

The volume of sap in the vessel of length x is

V = Ax,

where A is the cross sectional area of the vessel.

For 2000 such vessels, the volume is

V = 2000Ax

taking the derivative of both sides we get:

\dfrac{dV}{dt} = 2000A \dfrac{dx}{dt}

on the left-hand-side \dfrac{dx}{dt} is the velocity v of the sap, and on right-hand-side \dfrac{dV}{dt}  = 2.94*10^{-8}m^3/s; therefore,

2.94*10^{-8}m^3/s=2000Av

and since the cross-sectional area is

A = \pi (\dfrac{100*10^{-3}m}{2} )^2 = 7.85*10^{-3}m^2;

therefore,

2.94*10^{-8}m^3/s =2000(7.85*10^{-3}m^2)v

solving for v we get:

v = \dfrac{2.94*10^{-8}m^3/s}{2000(7.85*10^{-3}m^2)}

\boxed{v =1.875*10^{-9}m/s = 1.875nm/s}

which is the upward speed of the sap in each vessel.

8 0
3 years ago
Emotions refer to the act or process of knowing. True or False​
vaieri [72.5K]

Answer:

True

Explanation:

Because they are central to the human experience and are related to everything we do .

4 0
3 years ago
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