All categories

3 years ago

Can you help me understand force=mass x accleration

1 answer:

6 0

Force = mass * acceleration

Here is an equation for force, used to calculate force.

Look at it this way:-

Joules is used to measure force, while mass is measured by kg. Acceleration is measured by m/s/s (meters per second per second, or meters per second squared.)

joules = m/s/s * kg

Which makes everything fit in just right! :D

You might be interested in

An electron traverses a vacuum tube with a length of 2 m in 2 X 10- 4

Maslowich

Answer:

Average speed = 10,000 m/s

Explanation:

Given the following data;

Distance = 2m

Time = 0.0002secs

To find the average speed;

Average speed = distance/time

Average speed = 2/0.0002

Average speed = 10,000 m/s

Therefore, the average speed of the

electron is 10,000 meters per seconds.

7 0

3 years ago

Certain neutron stars (extremely dense stars) are believed to be rotating at about 10 rev/s. If such a star has a radius of 18 k

Aleks [24]

Answer:

mass of the neutron star =3.45185×10^26 Kg

Explanation:

When the neutron star rotates rapidly, a material on its surface to remain in place, the magnitude of the gravitational acceleration on the central material must be equal to magnitude of the centripetal acc. of the rotating star.

That is

$\frac{GM_{ns}}{R^2}= \omega^2 R$

M_ns = mass odf the netron star.

G= gravitational constant = 6.67×10^{-11}

R= radius of the star = 18×10^3 m

ω = 10 rev/sec = 20π rads/sec

therefore,

$M_{ns}= \frac{\omega^2R^3}{G} = \frac{4\pi^2\times(18\times10^3)^3}{6.67\times10^{-11}}$

= 3.45185... E26 Kg

= 3.45185×10^26 Kg

4 0

3 years ago

A4) A straight wire that is 0.60 m long is carrying a current of 2.0 A. It is placed in a uniform magnetic field of strength 0.3

Klio2033 [76]

Answer:

$\theta=30^{\circ}$

Explanation:

It is given that,

Length of the wire, L = 0.6 m

Current flowing inside the wire, I = 2 A

Uniform magnetic field, B = 0.3 T

Force experienced by the wire in the magnetic field, F = 0.18 N

To find,

The angle made by the wire with the magnetic field.

Solve,

We know that the magnetic force acting on the wire inside the magnetic field is given by :

$F=ILB\ sin\theta$

$sin\theta=\dfrac{F}{ILB}$

$sin\theta=\dfrac{0.18}{2\times 0.6\times 0.3}$

$\theta=30^{\circ}$

Therefore, the wire makes an angle of 30 degrees with respect to magnetic field.

5 0

3 years ago

According to Newton’s Second Law of Motion, an object will accelerate if which kind of force is applied?

Kobotan [32]

Answer:

unbalanced force

Explanation:

this is a guess so just look it up

7 0

3 years ago

Read 2 more answers

A parallel-plate capacitor is charged by connecting it to a battery. If the battery is disconnected and then the separation betw

TEA [102]

Answer:

The charge stored in the capacitor will stay the same. However, the electric potential across the two plates will increase. (Assuming that the permittivity of the space between the two plates stays the same.)

Explanation:

The two plates of this capacitor are no longer connected to each other. As a result, there's no way for the charge on one plate to move to the other. $Q$ , the amount of charge stored in this capacitor, will stay the same.

The formula $\displaystyle Q = C\, V$ relates the electric potential across a capacitor to:

$Q$ , the charge stored in the capacitor, and
$C$ , the capacitance of this capacitor.

While $Q$ stays the same, moving the two plates apart could affect the potential $V$ by changing the capacitance $C$ of this capacitor. The formula for the capacitance of a parallel-plate capacitor is:

$\displaystyle C = \frac{\epsilon\, A}{d}$ ,

where

$\epsilon$ is the permittivity of the material between the two plates.
$A$ is the area of each of the two plates.
$d$ is the distance between the two plates.

Assume that the two plates are separated with vacuum. Moving the two plates apart will not affect the value of $\epsilon$ . Neither will that change the area of the two plates.

However, as $d$ (the distance between the two plates) increases, the value of $\displaystyle C = \frac{\epsilon\, A}{d}$ will become smaller. In other words, moving the two plates of a parallel-plate capacitor apart would reduce its capacitance.

On the other hand, the formula $\displaystyle Q = C\, V$ can be rewritten as:

$V = \displaystyle \frac{Q}{C}$ .

The value of $Q$ (charge stored in this capacitor) stays the same. As the value of $C$ becomes smaller, the value of the fraction will become larger. Hence, the electric potential across this capacitor will become larger as the two plates are moved away from one another.

3 0

3 years ago