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2 years ago

In the diagram below, a 10-kilogram ball is fired with a

Physics

1 answer:

Tcecarenko [31]2 years ago

3 0

Answer:

A) 5m/s

Explanation:

Momentum of ball in one direction=momentum of cannon in other direction

mv = mv

(10)(500)=(1000)v

5000=1000v

∴v=5m/s

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A +2.00nc point charge is at the origin, and a second -5.00nc point charge is on the x-axis at x = 0.800m find the magnitude of

Damm [24]

You have to reduce 2.00 an5.00 I order to use the×that=0.800

3 0

2 years ago

An ideal gas, initially at a pressure of 11.2 atm and a temperature of 299 K, is allowed to expand adiabatically until its volum

Tju [1.3M]

Answer:

The pressure is $P_2 = 4.25 \ a.t.m$

Explanation:

From the question we are told that

The initial pressure is $P_1 = 11.2\ a.t.m$

The temperature is $T_1 = 299 \ K$

Let the first volume be $V_1$ Then the final volume will be $2 V_1$

Generally for a diatomic gas

$P_1 V_1 ^r = P_2 V_2 ^r$

Here r is the radius of the molecules which is mathematically represented as

$r = \frac{C_p}{C_v}$

Where $C_p \ and\ C_v$ are the molar specific heat of a gas at constant pressure and the molar specific heat of a gas at constant volume with values

$C_p=7 \ and\ C_v=5$

=> $r = \frac{7}{5}$

=> $11.2*( V_1 ^{\frac{7}{5} } ) = P_2 * (2 V_1 ^{\frac{7}{5} } )$

=> $P_2 = [\frac{1}{2} ]^{\frac{7}{5} } * 11.2$

=> $P_2 = 4.25 \ a.t.m$

8 0

3 years ago

Suppose the magnitude of the proton charge differs from the magnitude of the electron charge by a mere 1 part in 109

Oksana_A [137]

Answer:

A) F = 1.09 10 5 N, b) Yes

Explanation:

Part A

For this exercise we need the number of free electrons in copper, as the valence of copper +1 there is a free electron for each atom. Let's use the concept of density to find the mass of copper in the sphere

ρ = m / V

.m = ρ V = ρ 4/3 π r³

The radius is half the diameter

r = 1.9 10⁻² / 2 = 0.95 10⁻² m

ρ = 8960 kg / m3

m = 8960 4/3 π (0.95 10⁻²)³

m = 3.2179 10⁻² kg

The molecular weight of copper is 63,546 g / mol which has 6,022 10²³ atoms

With this we can use a rule of proportions to enter the number of atom is this mass

#_atom = 6.022 10²³ 3.2179 10⁻² / 63.546 10⁻³

#_atom = 3,049 10²³

Therefore there is the same number of electrons, as they indicate that the charge of the protone and the electon differs by 1/10⁹ the total charge for each spherical is

q = e / 10⁹ #_atom

q = e / 10⁹ 3,049 1023

q = 3,049 10⁴ (-1.6 10⁻¹⁹)

q = -4,878 10-5 C

Electric force is

F = k q₁q₂ / r²

F = k q² / r²

Let's calculate

F = 8.99 10⁹ (4.878 10⁻⁵)²2 / (1.4 10⁻²)²

F = 1.09 10 5 N

This is a force of repulsion.

Part B

The magnitude of this force is in very easy to detect

4 0

3 years ago

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

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