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2 years ago

Why does a solid have a definite shape and volume

Physics

2 answers:

spayn [35]2 years ago

6 0

Particles in a solid are arranged in a regular, tightly packed configuration.

Lady_Fox [76]2 years ago

4 0

A solid has a definite shape and volume, since its particles are held closely together leaving no room.

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Please can someone solve this physics question with a good explenation.

zimovet [89]

Answer:

The coefficient of dynamic friction is 0.025.

Explanation:

Given:

Initial speed after the push is 'v' as seen in the graph.

Final speed of the stone is 0 m/s as it comes to rest.

Total distance traveled is, $D=29.8\ m$

Total time taken is, $t_{total}=17.5\ s$

Time interval for deceleration is 3.5 to 17.5 s which is for 14 s.

Now, average speed of the stone is given as:

$v_{avg}=\frac{D}{t_{total}}=\frac{29.8}{17.5}=1.703\ m/s$

Now, we know that, average speed can also be expressed as:

$v_{avg}=\frac{v_i+v_f}{2}\\1.703=\frac{v+0}{2}\\v=2\times 1.703=3.41\ m/s$

Now, from the graph, the vertical height of the triangles is, $v=3.41\ m/s$

The deceleration is given as the slope of the line from time 3.5 s to 17.5 s.

Therefore, deceleration is:

$a=\frac{\textrm{Vertical height}}{\textrm{Time interval}}\\a=\frac{v-0}{17.5-3.5}\\a=\frac{3.41}{14}=0.244\ m/s^2$

Frictional force is the net force acting on the stone. Frictional force is given as:

$f=\mu_dN\\Where, \mu_d\rightarrow \textrm{coefficient of dynamic friction}\\N\rightarrow \textrm{Normal force}\\N=mg\\\therefore f=\mu_dmg$

Now, from Newton's second law, net force is equal to the product of mass and acceleration.

Therefore,

$\mu_dmg=ma\\\mu_d=\frac{a}{g}$

Plug in 0.244 for 'a' and 9.8 for 'g'. This gives,

$\mu_d=\frac{a}{g}=\frac{0.244}{9.8}=0.025$

Therefore, the coefficient of dynamic friction is 0.025.

3 0

2 years ago

Two positive charges q1 = q2 = 2.0 μC are located at x = 0, y = 0.30 m and x = 0, y = -0.30 m, respectively. Third point charge

Wittaler [7]

Answer:

F = 0.111015 N

Explanation:

For this exercise the force is given by Coulomb's law

F = k q₁q₂ / r₂₁²

we calculate the electric force of the other two particles on the charge q1

Charges q₁ and q₂

the distance between them is

r₁₂ = y₁ -y₂

r₁₂ = 0.30 + 0.30

r₁₂ = 0.60 m

let's calculate

F₁₂ = 9 10⁹ 2 10⁻⁶ 2 10⁻⁶ / 0.60 2

F₁₂ = 1 10⁻¹ N

directed towards the positive side of the y-axis

Charges 1 and 3

Let's find the distance using the Pythagorean Theorem

r₁₃ = RA [(0.40-0) 2 + (0-0.30) 2]

r₁₃ = 0.50 m

F₁₃ = 9 10⁹ 2 10⁻⁶ 4 10⁻⁶ / 0.50²

F₁₃ = 1.697 10⁻² N

The direction of this force is on the line that joins the two charges (1 and 3), let's use trigonometry to find the components of this force

tan θ = y / x

θ = tan⁻¹ y / x

θ = tan⁻¹ 0.3 / 0.4

tea = 36.87º

The angle from the positive side of the x-axis is

θ ’= 180 - θ

θ ’= 180 - 36.87

θ ’= 143.13º

sin143.13 = F_13y / F₁₃

F_13y = F₁₃ sin 143.13

F{13y} = 1.697 10⁻² sin 143.13

F_13y = 1.0183 10⁻² N

cos 143.13 = F_13x / F₁₃

F₁₃ₓ = F₁₃ cos 143.13

F₁₃ₓ = 1.697 10⁻² cos 143.13

F₁₃ₓ = -1.357 10-2 N

Now we can find the components of the resultant force

Fx = F13x + F12x

Fx = -1,357 10-2 +0

Fx = -1.357 10-2 N

Fy = F13y + F12y

Fy = 1.0183 10-2 + 1 10-1

Fy = 0.110183 N

We use the Pythagorean theorem to find the modulus

F = Ra (Fx2 + Fy2)

F = RA [(1.357 10-2) 2 + 0.110183 2]

F = 0.111015 N

Let's use trigonometry for the angles

tan tea = Fy / Fx

tea = tan-1 (0.110183 / -0.01357)

tea = 1,448 rad

to find the angle about the positive side of the + x axis

tea '= pi - 1,448

Tea = 1.6936 rad

6 0

3 years ago

A 81 kg person sits on a 3.8 kg chair. Each leg of the chair makes contact with the floor in a circle that is 1.2 cm in diameter

tankabanditka [31]

Answer:

1.9 MPa

Explanation:

Mass of person = 81 kg

Mass of chair = 3.8 kg

Diameter of contact point = 1.2 cm = D

Radius of contact point = 1.2/2 = 0.6 cm

Total mass of chair and person = 81 + 3.8 = 84.8 kg = M

Acceleration due to gravity = 9.81 m/s²

Force acting on the floor

F = Mg

⇒F = 84.8×9.81

⇒F = 831.888 N

Area of the contact point

A = πR²

⇒A = π0.006²

⇒A = π0.000036 m²

Area of the four points is

4A = 0.000144π m²

Pressure

$p=\frac{F}{A}\\\Rightarrow p=\frac{831.888}{0.000144\pi}\\\Rightarrow p=1838876.21\ Pa=1.83887621\times 10^6\ Pa=1.9 MPa$

Pressure exerted on the floor by each leg of the chair is 1.9 MPa

5 0

2 years ago

A 5 kg mass is oscillating on a spring with a time period of 2.8 seconds. What is the spring constant k of the spring?

katrin [286]

Answer:

k = 25.18 N/m

Explanation:

Simple Harmonic Oscillator

It consists of a weight attached to one end of a spring being allowed to move forth and back.

If m is the mass of the weight and k is the constant of the spring, the period of the oscillation is given by:

$\displaystyle T=2\pi {\sqrt {\frac {m}{k}}}$

If the period is known, we can find the value of the constant by solving for k:

$\displaystyle k=m\left(\frac{2\pi}{T}\right)^2$

Substituting the given values m=5 Kg and T=2.8 seconds:

$\displaystyle k=5 \left(\frac{2\pi}{2.8}\right)^2$

k = 25.18 N/m

5 0

2 years ago

A wire carries a steady current of 2.60 A. A straight section of the wire is 0.750 m long and lies along the x axis within a uni

ad-work [718]

Answer:

The magnetic force on the section of wire is $-2.925\hat{j}\ N$ .

Explanation:

Given that,

Current $I = 2.60\hat{i}\ A$

Length = 0.750 m

Magnetic field $B = 1.50\hat{k}\ T$

We need to calculate the magnetic force on the section of wire

Using formula of magnetic force

$\vec{F}=l\vec{I}\times\vec{B}$

$\vec{F}=0.750\times2.60\hat{i}\times1.50\hat{k}$

Since, $\hat{i}\times\hat{k}=-\hat{j}$

$\vec{F}=-2.925\hat{j}\ N$

Hence, The magnetic force on the section of wire is $-2.925\hat{j}\ N$ .

7 0

2 years ago