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Roman55 [17]
3 years ago
8

How to solve this.

Physics
1 answer:
viva [34]3 years ago
5 0

Answer:

8.1 s

Explanation:

Draw a free body diagram of the small block.  There are four forces acting on the block:

Applied force F pushing to the right,

Weight force mg pulling down,

Normal force N pushing up,

and friction force Nμ pushing left.

Sum of forces in the y direction:

∑F = ma

N − mg = 0

N = mg

Sum of forces in the x direction:

∑F = ma

F − Nμ = ma

F − mgμ = ma

a = (F − mgμ) / m

Plug in values:

a = (12.2 N − (3.0 kg) (9.8 m/s²) (0.320)) / (3.0 kg)

a = 0.931 m/s²

Next, draw a free body diagram of the larger block.  There are four forces:

Normal force N pushing down (equal and opposite),

Friction force Nμ pushing right (equal and opposite),

Weight force Mg pulling down,

and normal force N₂ pushing up.

Sum of forces in the x direction:

∑F = ma

Nμ = Ma

mgμ = Ma

a = mgμ / M

Plug in values:

a = (3.0 kg) (9.8 m/s²) (0.320) / (11.0 kg)

a = 0.855 m/s²

So the acceleration of the smaller block relative to the larger block is 0.931 m/s² − 0.855 m/s² = 0.0754 m/s².

Given:

Δx = 2.5 m

v₀ = 0 m/s

a = 0.0754 m/s²

Find: t

Δx = v₀ t + ½ at²

2.5 m = (0 m/s) t + ½ (0.0754 m/s²) t²

t = 8.14 s

Rounded to 2 significant figures, it takes 8.1 seconds.

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Consider an elevator with a table and a book on top of the table. The mass of the table is 10kg and the mass of the book is 2kg.
Step2247 [10]

Newton's second law allows us to find the force of the block on the table is 126 N

Newton's second law says that the net force is proportional to the product of the mass and the acceleration of the body

                Σ F = m a

Where the bold letters indicate vectors, m is the mass and the acceleration of the body

A free body diagram is a diagram where the forces are represented without the details of the bodies, in the attached we can see a free body diagram of the system.

Let's start by finding the acceleration of the elevator with kinematics  

                 v = v₀ + a t

                 a = \frac{v-v_o}{a}  

Where v and v₀ are the current and initial velocity, respectively, at acceleration and t is the time

                 a = \frac{8-1}{2}

                 a = 3.5 m / s²

Let's write Newton's second law for each body

The book

                N₂ - W₂ + N₁ = m a

               

Table

                N₁ - W₁ - W₂ = M a

                W₁ = Mg

                W₂ = mg

                N₁ = (M + m) g + M a

                N₁ = (10 + 2) 9.8 + 10 3.5

                N₁ = 152.6 N

This is the reaction of the earth to the support of the block and the table

               N₂ = ma + m g  - N₁

               N₂ = m ( a +g) - N₁  

               N₂ = 2 (3.5 + 9.8) - 152.6

               N₂ = 26.6 - 152.6

               N₂ = -126 N

The negative sign indicates that the direction is opposite to the one assigned, this is the action of the block on the table.

In conclusion using Newton's second law we can find the forces of the block on the table is 126 N

Learn more here: brainly.com/question/19860811

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3 0
3 years ago
Please help me with this question
valkas [14]

Answer: m∠P ≈ 46,42°

because using the law of sines in ΔPQR

=> sin 75°/ 4 = sin P/3

so ur friend is wrong due to confusion between edges

+) we have: sin 75°/4 = sin P/3

=> sin P = sin 75°/4 . 3 = (3√6 + 3√2)/16

=> m∠P ≈ 46,42°

Explanation:

4 0
3 years ago
The height of a typical playground slide is about 1.8 m and it rises at an angle of 30 ∘ above the horizontal.
Salsk061 [2.6K]

Answer:

5.94\ \text{m/s}

1.7

0.577

Explanation:

g = Acceleration due to gravity = 9.81\ \text{m/s}^2

\theta = Angle of slope = 30^{\circ}

v = Velocity of child at the bottom of the slide

\mu_k = Coefficient of kinetic friction

\mu_s = Coefficient of static friction

h = Height of slope = 1.8 m

The energy balance of the system is given by

mgh=\dfrac{1}{2}mv^2\\\Rightarrow v=\sqrt{2gh}\\\Rightarrow v=\sqrt{2\times 9.81\times 1.8}\\\Rightarrow v=5.94\ \text{m/s}

The speed of the child at the bottom of the slide is 5.94\ \text{m/s}

Length of the slide is given by

l=h\sin\theta\\\Rightarrow l=1.8\sin30^{\circ}\\\Rightarrow l=0.9\ \text{m}

v=\dfrac{1}{2}\times5.94\\\Rightarrow v=2.97\ \text{m/s}

The force energy balance of the system is given by

mgh=\dfrac{1}{2}mv^2+\mu_kmg\cos\theta l\\\Rightarrow \mu_k=\dfrac{gh-\dfrac{1}{2}v^2}{gl\cos\theta}\\\Rightarrow \mu_k=\dfrac{9.81\times 1.8-\dfrac{1}{2}\times 2.97^2}{9.81\times 0.9\cos30^{\circ}}\\\Rightarrow \mu_k=1.73

The coefficient of kinetic friction is 1.7.

For static friction

\mu_s\geq\tan30^{\circ}\\\Rightarrow \mu_s\geq0.577

So, the minimum possible value for the coefficient of static friction is 0.577.

8 0
3 years ago
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