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koban [17]
3 years ago
5

a particle with charge Q is on the y axis a distance a from the origin and a particle with charge qi is on the x axis at a dista

nce d from the origin. The value of d for which the x component of the force on the second particle is the greatest is
Physics
1 answer:
vova2212 [387]3 years ago
3 0

Answer:

Explanation:

Force on Q due to q₁

= k Q.q₁ / r² , r is distance between two.

X component

= k Q.q₁ / r²  x cos θ ,  θ is angle r makes with x - axis.

= k Q.q₁ / r²  x cos θ

r = a / sinθ

X component

= k Q.q₁ sin²θ . cosθ/ a²  

= k Q.q₁ / 2 a² x sin2θ . sinθ

for maximum value of it ,

sin2θ = 1  , θ = 45°

a / d = tan45

a = d .

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To practice Problem-Solving Strategy 30.1: Inductors in Circuits. A circuit has a 1 V battery connected in series with a switch.
ki77a [65]

Answer:

0.0133A

Explanation:

Since we have two sections, for the Inductor region there would be a current i_1. In the case of resistance 2, it will cross a current i_2

Defined this we proceed to obtain our equations,

For i_1,

\frac{di_1}{dt}+i_1R_1 = V

I_1 = \frac{V}{R_1} (1-e^{-\frac{R_1t}{L}})

For i_2,

I_2R_2 =V

I_2 = \frac{V}{R_2}

The current in the entire battery is equivalent to,

i_t = I_1+I_2

i_t = \frac{V}{R_2}+\frac{V}{R_1} (1-e^{-\frac{R_1t}{L}})

Our values are,

V=1V

R_1 = 95\Omega

L= 1.5*10^{-2}H

R_2 =360\Omega

Replacing in the current for t= 0.4m/s

i=\frac{1}{360}+\frac{1}{95}(1-e^{-\frac{95*0.4}{1.5*10^{-2}}})

i= 0.0133A

i_1 = 0.01052A

3 0
3 years ago
A basketball is tossed up into the air, falls freely, and bounces from the wooden floor. From the moment after the player releas
Tju [1.3M]

Answer:

Tha ball- earth/floor system.

Explanation:

The force acting on the ball is the force of gravity when ignoring air resistance. At the moment the player releases the ball, until it reaches the top of its bounce, the small system for which the momentum is conserved is the ball- floor system. The balls exerts and equal and opposite force on the floor. <u>Here the ball hits the floor, because in any collision the momentum is conserved. Moment of the ball -floor system is conserved</u>. Mutual gravitation bring the ball and floor together in one system. As the ball moves downwards, the earth moves upwards, although with an acceleration on the order of 1025 times smaller than that of the ball. The two objects meet, rebound and separate.

5 0
3 years ago
An elevator (mass 4100 kg) is to be designed so that the maximum acceleration is 0.0400g. what is the maximum force the motor sh
melomori [17]

Answer:

The maximum force on the supporting cable is 80688 N.

The minimum force on the supporting cable is -164 N.

Explanation:

For maximum force movement of elevator is in upward direction. Thus, equation of motion is given by,

ma = T - mg

where m is the mass of elevator

a is acceleration of elevator

g is acceleration due to gravity

T is the maximum tension in the supporting cable

T = ma + mg

T = m (a + g)

T = 4100 ( 0.04g + 9.8)

T = 80688 N

This is the maximum force on the supporting cable.

For minimum force movement of elevator is in downward direction. Thus, equation of motion is given by,

ma = T - mg

where m is the mass of elevator

a= -0.04g is acceleration of elevator because elevator is moving downward

g is acceleration due to gravity

T is the minimum tension in the supporting cable

T = ma + mg

T = m (a + g)

T = 4100 ( 9.8 - 0.04g)

T = -164 N

This is the minimum force on the supporting cable.


7 0
3 years ago
Find the density of seawater at a depth where the pressure is 500 atm if the density at the surface is 1100 kg/m^3 . Seawater ha
mixer [17]

The density of seawater at a depth where the pressure is 500 atm is 1124kg/m^3

Explanation:

The relationship between bulk modulus and pressure is the following:

B=\rho_0 \frac{\Delta p}{\Delta \rho}

where

B is the bulk modulus

\rho_0 is the density at surface

\Delta p is the variation of pressure

\Delta \rho is the variation of density

In this problem, we have:

B=2.3\cdot 10^9 N/m^2 is the bulk modulus

\rho_0 =1100 kg/m^3

\Delta p = p-p_0 = 500 atm - 1 atm = 499 atm = 5.05\cdot 10^7 Pa is the change in pressure with respect to the surface (the pressure at the surface is 1 atm)

Therefore, we can find the density of the water where the pressure is 500 atm as follows:

\rho = \rho_0 + \Delta \rho = \rho_0+\frac{\rho_0 \Delta p}{B}=\rho_0 (1+\frac{\Delta p}{B})=(1100)(1+\frac{5.05\cdot 10^7}{2.3\cdot 10^9})=1124kg/m^3

Learn more about pressure in a fluid:

brainly.com/question/9805263

#LearnwithBrainly

7 0
3 years ago
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