All categories

3 years ago

Which statement about force is incorrect

Physics

1 answer:

garik1379 [7]3 years ago

4 0

Answer:

What are the options?

Explanation:

You might be interested in

If an object goes through a

aivan3 [116]

Answer:

A, The same amount of gravity

Explanation:

4 0

2 years ago

Problem PageQuestion A cylinder measuring 2.3cm wide and 2.7cm high is filled with gas. The piston is pushed down with a steady

Alexandra [31]

Jehedhdhsh dbdbdbdbdbdbdbebddnsbdb
Dhdhdbdbdbdbdhfhbdbddbdbdbdbdbdbsbsbsbdbdbdb d s DVD’s

8 0

3 years ago

A point charge q1 = 1.0 µC is at the origin and a point charge q2 = 6.0 µC is on the x axis at x = 1 m.

iris [78.8K]

To solve this problem we will apply the concepts related to the Electrostatic Force given by Coulomb's law. This force can be mathematically described as

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

Here

k = Coulomb's Constant

$q_{1,2} =$ Charge of each object

d = Distance

Our values are given as,

$q_1 = 1 \mu C$

$q_2 = 6 \mu C$

d = 1 m

$k = 9*10^9 Nm^2/C^2$

a) The electric force on charge $q_2$ is

$F_{12} = \frac{ (9*10^9 Nm^2/C^2)(1*10^{-6} C)(6*10^{-6} C)}{(1 m)^2}$

$F_{12} = 54 mN$

Force is positive i.e. repulsive

b) As the force exerted on $q_2$ will be equal to that act on $q_1$ ,

$F_{21} = F_{12}$

$F_{21} = 54 mN$

Force is positive i.e. repulsive

c) If $q_2 = -6 \mu C$ , a negative sign will be introduced into the expression above i.e.

$F_{12} = \frac{(9*10^9 Nm^2/C^2)(1*10^{-6} C)(-6*10^{-6} C)}{(1 m)^{2}}$

$F_{12} = F_{21} = -54 mN$

Force is negative i.e. attractive

6 0

3 years ago

Mike is a runner. Mike runs 200m in 25s; what is his average velocity during the run? assume that units are in m/s. (asap, will

maw [93]

Velocity is displacement/time
(Displacement is the overall change in distance)

So you’ll want to divide 200 by 25, which should give you:

8 m/s

5 0

3 years ago

A drop

exis [7]

Answer:

$27.5\ m$

Explanation:

As we know that volume of cylinder is

$v=\pi r^{2} *h$

Where v=volume , h= height or thickness and r= radius

Here,

$v= 10 m ,\ diameter= 10, \ r=\frac{diameter}{2} \ r=\frac{10}{2}\\ r=5$

Putting these values in the previous equation , we get

$10\ = \frac{22}{7} *5 *5*h\\ 14\ =\ 110*h\\h=\frac{110}{14} \\h=\frac{55}{2} \\\\h=27.5\ m$

Therefore thickness is 27.5 m

7 0

3 years ago