A rabbit is trying to cross the street. Its velocity v as a function of time t is given in the graph below where

If F1 is the force on q due to Q1 and F2 is the force on q due to Q2, how do F1 and F2 compare? Assume that n=2.

Anna35 [415]

This question is incomplete

Complete Question

Three equal point charges are held in place as shown in the figure below

If F1 is the force on q due to Q1 and F2 is the force on q due to Q2, how do F1 and F2 compare? Assume that n=2.

A) F1=2F2

B) F1=3F2

C) F1=4F2

D) F1=9F2

Answer:

D) F1=9F2

Explanation:

We are told in the question that there are three equal point charges.

q, Q1, Q2 ,

q = Q1 = Q2

From the diagram we see the distance between the points d

q to Q1 = d

Q1 to Q2 = nd

Assuming n = 2

= 2 × d = 2d

Sum of the two distances = d + 2d = 3d

F1 is the force on q due to Q1 and

F2 is the force on q due to Q2,

Since we have 3 equal point charges and a total sum of distance which is 3d

Hence,

F1 = 9F2

6 0

2 years ago

Suppose the gas resulting from the sublimation of 1.00 g carbon dioxide is collected over water at 25.0◦c into a 1.00 l containe

AlexFokin [52]

Answer:

0.56 atm

Explanation:

First of all, we need to find the number of moles of the gas.

We know that

m = 1.00 g is the mass of the gas

$Mm=44.0 g/mol$ is the molar mass of the carbon dioxide

So, the number of moles of the gas is

$n=\frac{m}{M_m}=\frac{1.00 g}{44.0 g/mol}=0.023 mol$

Now we can find the pressure of the gas by using the ideal gas equation:

$pV=nRT$

where

p is the pressure

$V=1.00 L = 0.001 m^3$ is the volume

n = 0.023 mol is the number of moles

$R=8.314 J/mol K$ is the gas constant

$T=25.0^{\circ}+273=298 K$ is the temperature of the gas

Solving the equation for p, we find

$p=\frac{nRT}{V}=\frac{(0.023 mol)(8.314 J/mol K)(298 K)}{0.001 m^3}=5.7 \cdot 10^4 Pa$

And since we have

$1 atm = 1.01\cdot 10^5 Pa$

the pressure in atmospheres is

$p=\frac{5.7\cdot 10^4 Pa}{1.01\cdot 10^5 Pa/atm}=0.56 atm$

5 0

2 years ago

Henrietta is going off to her physics class, jogging down the sidewalk at a speed of 4.15 m/s . Her husband Bruce suddenly reali

aleksandr82 [10.1K]

Answer:

a 15.22 m/s

b 45.65 m

Explanation:

Using the same formula,

x = vt, where

x is now 45.65, and

t is 3 s, then

45.65 = 3v

v = 45.65/3

v = 15.22 m/s

See the attachment for the part b. We used the distance gotten in part B, to find question A

5 0

2 years ago

A cyclist traveling at 5m/s uniformly accelerates up to 10 m/s in 2 seconds. Each tire of the bike has a 35 cm radius, and a sma

konstantin123 [22]

Answer:

$a=2.5\ m/s^2$

Explanation:

Given that,

Initial speed, u = 5 m/s

Final speed, v = 10 m/s

Time, t = 2 s

The radius of the tire of the bike, r = 35 cm

We need to find the angular acceleration of the pebble during those two seconds. It can be calculated as follows.

$a=\dfrac{v-u}t{}\\\\a=\dfrac{10-5}{2}\\\\a=2.5\ m/s^2$

So, the required angular acceleration of the pebble is equal to $2.5\ m/s^2$ .

8 0

1 year ago

Alex throws a 0.15-kg rubber ball down onto the floor. The ball’s speed just before impact is 6.5 m/s, and just after is 3.5 m/s

Jet001 [13]

Answer: Change in ball's momentum is 1.5 kg-m/s.

Explanation: It is given that,

Mass of the ball, m = 0.15 kg

Speed before the impact, u = 6.5 m/s

Speed after the impact, v = -3.5 m/s (as it will rebound)

We need to find the change in the magnitude of the ball's momentum. It is given by :

So, the change in the ball's momentum is 1.5 kg-m/s. Hence, this is the required solution.

Read more on Brainly.com - brainly.com/question/12946012#readmore

7 0

2 years ago