Ask question

Ask question

All categories

3 years ago

10

A child standing on a bus remains still when the bus is at rest. When the bus moves forward and then slows down, the child conti

nues moving forward at the original speed. This is an example of

2 answers:

expeople1 [14]3 years ago

3 0

Newton's first law of motion

dlinn [17]3 years ago

3 0

The best answer would be, <u><em>Newton's first law of motion. </em></u>

You might be interested in

A 15.0-μF capacitor is charged by a 130.0-V power supply, then disconnected from the power and connected in series with a 0.280-

SVETLANKA909090 [29]

The resonant frequency of a circuit is the frequency $\omega_0$ at which the equivalent impedance of a circuit is purely real (the imaginary part is null).

Mathematically this frequency is described as

$f = \frac{1}{2\pi}(\sqrt{\frac{1}{LC}})$

Where

L = Inductance

C = Capacitance

Our values are given as

$C = 15*10^{-6}\mu F$

$L = 0.280*10^{-3}mH$

Replacing we have,

$f = \frac{1}{2\pi}(\sqrt{\frac{1}{LC}})$

$f = \frac{1}{2\pi}(\sqrt{\frac{1}{(15*10^{-6})(0.280*10^{-3})}})$

$f= 2455.81Hz$

From this relationship we can also appreciate that the resonance frequency infers the maximum related transfer in the system and that therefore given an input a maximum output is obtained.

For this particular case, the smaller the capacitance and inductance values, the higher the frequency obtained is likely to be.

7 0

3 years ago

A charged box ( m = 495 g, q = + 2.50 μ C ) is placed on a frictionless incline plane. Another charged box ( Q = + 75.0 μ C ) is

Rudik [331]

Answer:

$a=16.2m/s^{2}$

Explanation:

From the attached file diagram, the total force acting on the charged box is the downward weight and the repulsive force acting in opposite to the weight force . Hence we can write the total force as

$F=masin\alpha -\frac{kQq}{r^{2}} \\\alpha =35^{0}, m=0.495kg, r=0.61m, Q=2.5*10^{-6}, q=75.0*10^{-6}\\$

When fixed,F=o

Hence

$masin\alpha =\frac{kQq}{r^{2}}\\0.495kg*asin35=\frac{9*10^{9}*2.5*10^{-6}*75.0*10^{-6}}{0.61^{2}} \\0.28a=4.5351\\a=\frac{4.5351}{0.28}\\\\ a=16.2m/s^{2}$

The value of the acceleration is 16.2m/s^2

7 0

3 years ago

A 4113 N piano is to be pushed up a(n) 3.99 mfrictionless plank that makes an angle of 25.5â—¦with the horizontal.Calculate the

Serhud [2]

Answer:

W = 14.8 kJ

Explanation:

W = F S cos ∅

W = 4113 x 3.99 x cos 25.5

W = 16410.87 x 0.9025 = 14810.8 J or 14.8 kJ

3 0

3 years ago

If each washer has a mass of 4.9 g, then the mass of two washers in kilograms would be Given that F = ma, and if 4 washers are a

zysi [14]

Given that,

Mass of each washer = 4.9 g

We need to calculate the mass of two washers in kg

Using conversion of unit

Mass of each washer $m= 4.9\times10^{-3}\ Kg$

So, Mass of two washers is

$M=2\times m$

Put the value of m

$M=2\times4.9\times10^{-3}\ k$

$M=0.0098\ Kg$

If 4 washer are attached to the spring

We need to calculate the applied force on the car

Using formula of force

$F=mg$

Put the value into the formula

$F=4\times4.9\times10^{-3}\times9.8$

$F=0.192\ N$

Hence, (i), The mass of two washers is 0.0098 kg.

(ii). The applied force on the car is 0.192 N.

6 0

3 years ago

What is the area of a circle of a radius 4.1×10^4 cm

dmitriy555 [2]

Answer:

128805.2988 cm squared or 128000 to 3 significant figures.

Explanation:

Area of a circle : pi times radius squared or $r^{2} \pi$

pi times 4.1×10^4 cm = 41000 pi or 128805.2988

4 0

3 years ago