Ask question

Ask question

All categories

4 years ago

15

Two 1.0 kg masses are 4.0 m apart on a frictionless table. Each has 1.0μC of charge. Part A What is the magnitude of the electri

c force on one of the masses? What is the initial acceleration of each mass if they are released and allowed to move?

2 answers:

9966 [12]4 years ago

8 0

Force between two charges is given by

$F =\frac{kq_1q_2}{r^2}$

$F =\frac{9*10^9* 1* 10^{-6}* 1 * 10^{-6}}{4^2}$

$F = 5.625 * 10^{-4} N$

Now in order to find the acceleration of each mass

we can use

F = ma

$5.625 * 10^{-4} = 1 * a$

$a= 5.625 * 10^{-4} m/s^2$

forsale [732]4 years ago

8 0

The force between two charged particles can be determined using the formula
F = k Q1 Q2 / d²

Using the given values
F = 9 x 10^9 (1.0 x 10^-9)(1.0 x 10^-9) / 4²
F = 5.625 x 10^-10 N

The acceleration is
ma = F
1 (a) = 5.625 x 10^-10
a = 5.625 x 10^-10 m/s²

You might be interested in

Which statement is the contrapositive of:

HACTEHA [7]

Answer:

E.two angles are vertical angles if, and only if they are not adjacent angles

5 0

3 years ago

Spring #1 has a force constant of k, and spring #2 has a force constant of 2k. Both springs are attached to the ceiling. Identic

Gre4nikov [31]

Answer:

The ratio of the energy stored by spring #1 to that stored by spring #2 is 2:1

Explanation:

Let the weight that is hooked to two springs be w.

Spring#1:

Force constant= k

let x1 be the extension in spring#1

Therefore by balancing the forces, we get

Spring force= weight

⇒k·x1=w

⇒x1=w/k

Energy stored in a spring is given by $\frac{1}{2}kx^{2}$ where k is the force constant and x is the extension in spring.

Therefore Energy stored in spring#1 is, $\frac{1}{2}k(x1)^{2}$

⇒ $\frac{1}{2}k(\frac{w}{k})^{2}$

⇒ $\frac{w^{2}}{2k}$

Spring #2:

Force constant= 2k

let x2 be the extension in spring#2

Therefore by balancing the forces, we get

Spring force= weight

⇒2k·x2=w

⇒x2=w/2k

Therefore Energy stored in spring#2 is, $\frac{1}{2}2k(x2)^{2}$

⇒ $\frac{1}{2}2k(\frac{w}{2k})^{2}$

⇒ $\frac{w^{2}}{4k}$

∴The ratio of the energy stored by spring #1 to that stored by spring #2 is $\frac{\frac{w^{2}}{2k}}{\frac{w^{2}}{4k}}=$ 2:1

4 0

3 years ago

2 Why <br> Force is needed

nirvana33 [79]

Answer:

The use of force in our everyday life is very common. We use force to walk on the road, to lift the objects, to throw a cricket ball, or to move a given body by some particular speed or direction. We are very familiar with the various effects of force. We can exert pull and push.

Explanation:

plzzz brainlist me

3 0

4 years ago

A jet liner must reach a speed of 82 m/s for takeoff. If the

SIZIF [17.4K]

Answer:

The acceleration that the jet liner that must have is 2.241 meters per square second.

Explanation:

Let suppose that the jet liner accelerates uniformly. From statement we know the initial ( $v_{o}$ ) and final speeds ( $v_{f}$ ), measured in meters per second, of the aircraft and likewise the runway length ( $d$ ), measured in meters. The following kinematic equation is used to calculate the minimum acceleration needed ( $a$ ), measured in meters per square second:

$a = \frac{v_{f}^{2}-v_{o}^{2}}{2\cdot d}$

If we know that $v_{o} = 0\,\frac{m}{s}$ , $v_{f} = 82\,\frac{m}{s}$ and $d = 1500\,m$ , then the acceleration that the jet must have is:

$a = \frac{\left(82\,\frac{m}{s} \right)^{2}-\left(0\,\frac{m}{s} \right)^{2}}{2\cdot (1500\,m)}$

$a = 2.241\,\frac{m}{s^{2}}$

The acceleration that the jet liner that must have is 2.241 meters per square second.

3 0

3 years ago

A jet plane flying 600 m/s experiences an acceleration of 4g when pulling out of the dive. What is the radius of curvature of th

Gnoma [55]

Answer:

91.84 m/s²

Explanation:

velocity, v = 600 m/s

acceleration, a = 4 g = 4 x 9.8 = 39.2 m/s^2

Let the radius of the loop is r.

he experiences a centripetal force.

centripetal acceleration,

a = v² / r

39.2 x r = 600 x 600

r = 3600 / 39.2

r = 91.84 m/s²

Thus, the radius of the loop is 91.84 m/s².

8 0

3 years ago