Our solar system is made up of the Sun, 8 planets, and other bodies such as

A 150 kg car hits a wall with 45 N of force, what was its acceleration?

alisha [4.7K]

Answer:

a = 0.3 m/s²

Explanation:

Given: 45 N, 150 kg

To find: a

Formula: $a = \frac{F}{m}$

Solution: To find a, divide the force by the weight

A = F ÷ m

= 45 ÷ 150

= 0.3 m/s²

Newtons are derived units, equal to 1 kg-m/s². In other words, a single Newton is equal to the force needed to accelerate one kilogram one meter per second squared.

4 0

2 years ago

A slender, uniform metal rod of mass M and length lis pivoted without friction about an axis through its midpoint andperpendicul

lys-0071 [83]

Answer:

Explanation:

Moment of inertia of the metal rod pivoted in the middle

= M l² / 12

If the spring is compressed by small distance x twisting the rod by angle θ

restoring force by spring

= k x

moment of torque about axis

= k x l /2

= k θ( l /2 )² ( x / .5 l = θ )

=

moment of torque = moment of inertia of rod x angular acceleration

k θ( l /2 )² = M l² / 12 d²θ/dt²

d²θ/dt² = 3 k/M θ

acceleration = ω² θ

ω² = 3 k/M

ω = √ 3 k / M

8 0

2 years ago

Read 2 more answers

Describe the movement of the man when the resultant horizontal force is ON

vagabundo [1.1K]

Answer:

Explanation:

hope this helps <3

6 0

2 years ago

Students in an introductory physics lab are performing an experiment with a parallel-plate capacitor made of two circular alumin

Zarrin [17]

Answer:

$\rm 9.186\times 10^{-7}\ C.$

Explanation:

<u>Given:</u>

Diameter of the plates of the capacitor, D = 21 cm = 0.21 m.
Distance of separation between the plates, d = 1.0 cm = 0.01 m.
Minimum value of electric field that produces spark, $\rm E=3\times 10^6\ N/C.$

When the dimensions of the plate of the capacitor is comparatively much larger than the distance of separation between the plates, then, according to the Gauss' law of electrostatics, the value of the electric field strength in the region between the plates of the capacitor is given by

$\rm E=\dfrac{\sigma}{\epsilon_o}.$

where,

$\rm \sigma$ = surface charge density of the plate of the capacitor = $\dfrac qA$ .

$\rm q$ = magnitude of the charge on each of the plate.
$\rm A$ = surface area of each of the plate = $\rm \pi \times (Radius)^2=\pi \times\left ( \dfrac{D}{2}\right )^2= \pi \times \left ( \dfrac{0.21}{2}\right )^2=3.46\times 10^{-2}\ m^2.$
$\epsilon_o$ = electrical permittivity of free space, having value = $8.85\times 10^{-12}\rm \ C^2N^{-1}m^{-2}.$

For the minimum value of electric field that produces spark,

$\rm E = \dfrac{q}{A\epsilon_o}\\\Rightarrow q = E\ A\epsilon_o\\=3\times 10^6\times 3.46\times 10^{-2}\times 8.85\times 10^{-12}\\=9.186\times 10^{-7}\ C.$

It is the maximum value of the magnitude of charge which can be added up to each of the plates of the capacitor.

4 0

2 years ago

Which statement explains how it is possible to carry books to school without changing the kinetic or potential energy of the boo

Mamont248 [21]

Answer:

a. by moving the book without acceleration and keeping the height of the book constant

Explanation:

FOR CONSTANT KINETIC ENERGY:

The kinetic energy of a body depends upon its speed according to its formula:

ΔK.E = (1/2)mΔv²

So, for Δv = 0 m/s

ΔK.E = 0 J

So, for keeping kinetic energy constant, the books must be moved at constant speed without acceleration.

FOR CONSTANT POTENTIAL ENERGY:

The potential energy of a body depends upon its height according to its formula:

ΔP.E = mgΔh

So, for Δh = 0 m/s

ΔP.E = 0 J

So, for keeping potential energy constant, the books must be moved at constant height.

So, the correct option is:

<u>a. by moving the book without acceleration and keeping the height of the book constant</u>

8 0

2 years ago