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kow [346]
3 years ago
14

A cart loaded with bricks has a total mass of 20.4 kg and is pulled at constant speed by a rope. The rope is inclined at 26.1 ◦

above the horizontal and the cart moves 20.1 m on a horizontal floor. The coefficient of kinetic friction between ground and cart is 0.6 . The acceleration of gravity is 9.8 m/s 2 . What is the normal force exerted on the cart by the floor? Answer in units of N.
Physics
1 answer:
Readme [11.4K]3 years ago
4 0

Answer:

Normal force, N = 154.5 N

Explanation:

Given that,

Total mass of the cart, m = 20.4 kg

Angle of inclination, \theta=26.1^{\circ}

Distance moved, d = 20.1 m

The coefficient of kinetic friction between ground and cart is 0.6

The value of acceleration due to gravity, g=9.8\ m/s^2

Let N is the normal force exerted on the cart by the floor. It is given by :

N=mg-T\ sin\theta

T\ cos\theta=\mu N

T\ cos\theta=\mu (mg-T\ sin\theta)

T\ cos(26.1)=0.6 (20.4\times 9.8-T\ sin(26.1))

T = 103.23 N

So, the normal force is :

N=20.4\times 9.8-103.23\ sin(26.1)

N = 154.5 N

So, the normal force exerted on the cart by the floor is 154.5 N. Hence, this is the required solution.

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galina1969 [7]

Answer:

μ =tanθ

Explanation:=

The ratio of the force of static friction and the normal reaction is equal to tanθ. F=mgsinθ. R = mgcosθ.

μ=tanθ

6 0
2 years ago
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What is the thinnest soap film (excluding the case of zero thickness) that appears black when illuminated with light with a wave
Firlakuza [10]

Answer:

thinnest soap film is  206.76 nm

Explanation:

Given data

wavelength = 550 nm

index of refraction n  = 1.33

to find out

What is the thinnest soap film

solution

we have wavelength  λ = 550 nm

that is  λ = 550 × 10^{-9} m

and n = 1.3

we will find the thickness of soap film as given by formula that is

thickness = λ/2n

thickness = 550 × 10^{-9}  / 2(1.33)

thickness = 206.76 × 10^{-9}  m

thinnest soap film is  206.76 nm

7 0
3 years ago
An object is thrown straight up with an initial velocity of 10 m/s, and there is an air resistance force causing an acceleration
lana [24]

Answer:

Vf= 7.29 m/s

Explanation:

Two force act on the object:

1) Gravity

2) Air resistance

Upward motion:

Initial velocity = Vi= 10 m/s

Final velocity = Vf= 0 m/s

Gravity acting downward =  g = -9.8 m/s²

Air resistance acting downward = a₁ = - 3 m/s²

Net acceleration = a = -(g + a₁ ) = - ( 9.8 + 3 ) = - 12.8 m/s²

( Acceleration is consider negative if it is in opposite direction of velocity )

Now

2as = Vf² - Vi²

⇒ 2 * (-12.8) *s = 0 - 10²

⇒-25.6 *s = -100

⇒ s = 100/ 25.6

⇒ s = 3.9 m

Downward motion:

Vi= 0 m/s

s = 3.9 m

Gravity acting downward =  g = 9.8 m/s²

Air resistance acting upward = a₁ = - 3 m/s²

Net acceleration = a = g - a₁  =  9.8 - 3  = 6.8 m/s²

Now

2as = Vf² - Vi²

⇒ 2 * 6.8 * 3.9 = Vf² - 0

⇒ Vf² = 53. 125

⇒ Vf= 7.29 m/s

8 0
3 years ago
PHYSICS!
Allushta [10]

Answer:

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Explanation:

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6 0
2 years ago
The density of a block of wood is 694 kg/m3. Its mass is 689 g. We tie the block to the bottom of a swimming pool using a single
Serhud [2]

Answer:

<em>The tension in the string = 2.065 N</em>

Explanation:

From Archimedes principle,

R.d = density of the wood block/density of water = weight of the wood block/Upthrust of the wood block in water.

R.d = D₁/D₂ = W/U

W/U =D₁/D₂.................................. Equation 1

Where W = weight of the wood block, U = upthrust of the wood block in water, D₁ = Density of the wood block, D₂ = Density of water.

Making U the subject of the equation,

U = WD₂/D₁........................... Equation 2

Given: W =  mg = (689/1000)9.8 = 6.75 N,  D₁ = 694 kg/m³, D₂ = 1000 kg/m³.

Substituting these values into equation 2,

U = 6.75(694)/1000

U = 4684.5/1000

U = 4.685 N.

Note: Three forces act on the wood block in the pool. and they are

(i) The weight(W) acting downs

(ii) The upthrust (U) acting upwards,

(iii) The Tension (T) in the string, acting upwards.

Thus,

W = U+T

T = W - U ................................. Equation 3

Where W = 6.75 N, U = 4.685 N

T = 6.75 - 4.685

T = 2.065 N.

T = 2.065 N

<em>Thus the tension in the string = 2.065 N</em>

7 0
2 years ago
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