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-Dominant- [34]
3 years ago
13

Little Tammy lines up to tackle Jackson to (unsuccessfully) prove the law of conservation of momentum. Tammy’s mass is 34.0 kg a

nd Jackson’s is 54.0 kg. Jackson is running at a velocity of 3.0 m/s. What must Tammy’s minimum velocity be in order to stop forward momentum?
Physics
1 answer:
Naily [24]3 years ago
8 0

Answer:

So Tammy must move with speed 4.76 m/s in opposite direction of Jackson

Explanation:

As per law of conservation of momentum we know that there is no external force on it

So here we can say that initial momentum of the system must be equal to the final momentum of the system

now we have

m_1v_1 + m_2v_2 = 0

final they both comes to rest so here we can say that final momentum must be zero

now we have

34 v + 54 (3 m/s) = 0

v = -4.76 m/s

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Force F acts between a pair of charges, q1 and q2, separated by a distance d. For each of the statements, use the drop-down menu
lora16 [44]

The initial force between the two charges is given by:

F=k \frac{q_1 q_2}{d^2}

where k is the Coulomb's constant, q1 and q2 the two charges, d their separation. Let's analyze now the other situations:

1. F

In this case, q1 is halved, q2 is doubled, but the distance between the charges remains d.

So, we have:

q_1' = \frac{q_1}{2}\\q_2' = 2 q_2\\d' = d

So, the new force is:

F'=k \frac{q_1' q_2'}{d'^2}= k \frac{(\frac{q_1}{2})(2q_2)}{d^2}=k \frac{q_1 q_2}{d^2}=F

So the force has not changed.

2. F/4

In this case, q1 and q2 are unchanged. The distance between the charges is doubled to 2d.

So, we have:

q_1' = q_1\\q_2' = q_2\\d' = 2d

So, the new force is:

F'=k \frac{q_1' q_2'}{d'^2}= k \frac{q_1 q_2)}{(2d)^2}=\frac{1}{4} k \frac{q_1 q_2}{d^2}=\frac{F}{4}

So the force has decreased by a factor 4.

3. 6F

In this case, q1 is doubled and q2 is tripled. The distance between the charges remains d.

So, we have:

q_1' = 2 q_1\\q_2' = 3 q_2\\d' = d

So, the new force is:

F'=k \frac{q_1' q_2'}{d'^2}= k \frac{(2 q_1)(3 q_2)}{d^2}=6 k \frac{q_1 q_2}{d^2}=6F

So the force has increased by a factor 6.

8 0
3 years ago
Read 2 more answers
A 30kg boxed is pushed with a force of 20N. What is the boxes acceleration. Please show work
kozerog [31]

Answer:

<h3>The answer is 0.67 m/s²</h3>

Explanation:

The acceleration of an object given it's mass and the force acting on it can be found by using the formula

a =  \frac{f}{m}  \\

f is the force

m is the mass

From the question we have

a =  \frac{20}{30}  =  \frac{2}{3}  \\  = 0.6666666...

We have the final answer as

<h3>0.67 m/s²</h3>

Hope this helps you

8 0
3 years ago
Density of iron in CGS unit is 7.8 g/cm3. Its density is SI is
kumpel [21]

Answer:

7800kg/m³

Explanation:

Density of iron in CGS unit is 7.8 g/cm3. Its density is SI is

Given the density of iron = 7.8 g/cm3.

The SI units must be in kg/m³

7.8g = 7.8/1000 kg

7.8g = 0.0078kg

1cm³ = 0.000001m³

7.8g/cm³

= 0.0078/0.000001 kg/m³

= 7800kg/m³

Hence the density in SI unit is 7800kg/m³

4 0
3 years ago
2 pts
const2013 [10]

Answer:

I think the answer 1

Explanation:

im probably wrong too i dont know

5 0
3 years ago
C. A helium atom has a diameter of approximately 9.8 • 10-11 meters. What is the diameter of a helium atom in nanometers? Given
Studentka2010 [4]

1\ nm = 10^{-9} m\\\\1\ m =\frac{1\ nm}{10^{-9}}

Since the diameter of helium atom is approximately 9.8\times 10^{-11}\ m, therefore the diameter of helium atom in nano meter,

9.8\times 10^{-11}\ m=9.8\times 10^{-11} \times\frac{1\ nm}{10^{-9}}=0.098\ nm

4 0
4 years ago
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