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2 years ago

11

What is the weight of 5kgs of apples on Earth?

1 answer:

gregori [183]2 years ago

3 0

Answer:5kgs

Explanation:

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A 16.2 kg person climbs up a uniform ladder with negligible mass. The upper end of the ladder rests on a frictionless wall. The

S_A_V [24]

To solve this problem we will apply the concepts related to the balance of forces. We will decompose the forces in the vertical and horizontal sense, and at the same time, we will perform summation of torques to eliminate some variables and obtain a system of equations that allow us to obtain the angle.

The forces in the vertical direction would be,

$\sum F_x = 0$

$f-N_w = 0$

$N_w = f$

The forces in the horizontal direction would be,

$\sum F_y = 0$

$N_f -W =0$

$N_f = W$

The sum of Torques at equilibrium,

$\sum \tau = 0$

$Wdcos\theta - N_wLsin\theta = 0$

$WdCos\theta = fLSin\theta$

$f = \frac{Wd}{Ltan\theta}$

The maximum friction force would be equivalent to the coefficient of friction by the person, but at the same time to the expression previously found, therefore

$f_{max} = \mu W=\frac{Wd}{Ltan\theta}$

$\theta = tan^{-1} (\frac{d}{\mu L})$

Replacing,

$\theta = tan^{-1} (\frac{0.9}{0.42*2})$

$\theta = 46.975\°$

Therefore the minimum angle that the person can reach is 46.9°

8 0

3 years ago

Two identical conducting spheres are placed 80.0 cm apart. One is given a charge of 5.8 C and the other is given a charge of 6.4

Zepler [3.9K]

Answer: $5.214(10)^{11} N$

Explanation:

According to <u>Coulomb's Law:</u>

<em>"The electrostatic force $F_{E}$ between two point charges $q_{1}$ and $q_{2}$ is proportional to the product of the charges and inversely proportional to the square of the distance $d$ that separates them, and has the direction of the line that joins them".</em>

<em />

Mathematically this law is written as:

$F_{E}=K\frac{q_{1}.q_{2}}{d^{2}}$

Where:

$F_{E}$ is the electrostatic force

$K=8.99(10)^{9} Nm^{2}/C^{2}$ is the Coulomb's constant

$q_{1}=5.8 C$ and $q_{2}=6.4 C$ are the electric charges

$d=80 cm \frac{1 m}{100 cm}=0.8 m$ is the separation distance between the charges

Solving:

$F_{E}=8.99(10)^{9} Nm^{2}/C^{2}\frac{(5.8 C)(6.4 C)}{(0.8 m)^{2}}$

$F_{E}=5.214(10)^{11} N$

7 0

3 years ago

In which layer of the sun is energy transfferred between atoms?

sasho [114]

Answer:

Core is the layer of the sun is energy transferred between atoms.

Explanation:

4 0

2 years ago

A dust particle with mass of 5.4×10−2 g and a charge of 2.3×10−6 C is in a region of space where the potential is given by V(x)=

user100 [1]

Answer:

1.6 m/s^2

Explanation:

Hello!

To calculate the acceleration we must know the electric field. The electric field and the potential are related by:

$E = -\frac{dV}{dx} =- 2.6(\frac{V}{m^{2}})x + 8.1(\frac{V}{m^{3}})x^{2}$

If the particle starts at 2.3m, the electric field is:

E = 36.869 V/m = 36.869 N/C

So, the force on the particle is:

F = q E = 2.3×10^−6 C * 36.869 N/C = 8.48 x 10^-5 N

And its acceleration is :

a = F/m = 8.48 x 10^-5 N / 5.4×10−5 kg = 1.57 m/s^2

Rounded to two significant figures:

1.6 m/s^2

6 0

3 years ago

Some enterprising physics students working on a catapult decide to have a water balloon fight in the school hallway. The ceiling

sergejj [24]

Answer:

$\alpha =54.7$ º

Explanation:

From the exercise we have our initial information

$y=3.4m\\v_{o}=10m/s\\g=-9.8m/s^2$

When the balloon gets to the ceiling its velocity at that moment is 0 m/s. Being said that we can calculate velocity at the vertical direction

$v_{y}^2=v_{oy}^2+ag(y-y_{o})$

Since $v_{y}=0$ and $y_{o}=0$

$0=v_{oy}^2-2(9.8m/s^2)(3.4m)$

$v_{oy}=\sqrt{2(9.8m/s^2)(3.4m)}=8.16m/s$

Knowing that

$v_{oy}=v_{o}sin\alpha$

$sin\alpha =\frac{v_{oy} }{v_{o} }$

$\alpha =sin^{-1}(\frac{v_{oy}}{v_{o}})=sin^{-1}(\frac{8.16m/s}{10m/s})=54.7$ º

8 0

3 years ago