How many centimeters are in 4 inches?

A ball thrown with 50N of force accelerates at 25 m/s2. What is the mass of the ball?

MrRissso [65]

F=ma
Force is 50N. Acceleration is 25 m/s^2.
50N=m*25 m/s^2
Divide both sides by 25.
mass=2 kg

7 0

3 years ago

How is the electrostatic force affected when the magnitude of a charge is doubled?

BaLLatris [955]

The magnitude of the electrostatic force between two charges is given by:
$F=k_e \frac{q_1 q_2}{r^2}$
where
ke is the Coulomb's constant
q1 and q2 are the two charges
r is the separation between the two charges

We can see that the magnitude of the force is directly proportional to the charges. This means that when one of the charges is doubled, the magnitude of the electrostatic force will double as well, so the correct answer is
A) The magnitude of the electrostatic force doubles

4 0

3 years ago

1. An electric iron has a rating of 750W, 220V. Calculate:

horrorfan [7]

The equation for electrical power isP=VIwhere V is the voltage and I is the current. This can be rearranged to solve for I in 6(a). 6(b) can be solved with Ohm's LawV=IRor if you'd like, from power, after substituting Ohm's law in for IP=V2R For 7, realize that because they are in parallel, their voltages are the same. We can find the resistance of each lamp fromP=V2RThen the equivalent resistance as1R∗=1R1+1R2Then the total power asPt=V2R∗However, this will reveal that (with a bit of algebra)Pt=P1+P2 For 8, again the resistance can be found asP=V2RThe energy usage is simplyE=P⋅t

6 0

3 years ago

A note of frequency 200Hz has a velocity of 400m/s. what is the wavelength of the note

Xelga [282]

Answer:

$\huge\boxed{\sf \lambda = 2 m}$

Explanation:

<h3>Given data:</h3>

Frequency = f = 200 Hz

Velocity = v = 400 m/s

<h3>Required:</h3>

Wavelength = λ = ?

<h3>Formula:</h3>

v = fλ

<h3>Solution:</h3>

Put the givens in the formula

400 = (200)λ

Divide 200 to both sides

400/200 = λ

2 m = λ

λ = 2 m

$\rule[225]{225}{2}$

8 0

1 year ago

The pressure drop needed to force water through a horizontal 1-in diameter pipe if 0.60 psi for every 12-ft length of pipe. Dete

oksian1 [2.3K]

Answer:

The shear stress at a distance 0.3-in away from the pipe wall is 0.06012lb/ft²

The shear stress at a distance 0.5-in away from the pipe wall is 0

Explanation:

Given;

pressure drop per unit length of pipe = 0.6 psi/ft

length of the pipe = 12 feet

diameter of the pipe = 1 -in

Pressure drop per unit length in a circular pipe is given as;

$\frac{\delta P}{L} = \frac{2 \tau}{r} \\\\$

make shear stress (τ) the subject of the formula

$\frac{\delta P}{L} = \frac{2 \tau}{r} \\\\\tau = \frac{\delta P *r}{2L}$

Where;

τ is the shear stress on the pipe wall.

ΔP is the pressure drop

L is the length of the pipe

r is the distance from the pipe wall

Part (a) shear stress at a distance of 0.3-in away from the pipe wall

Radius of the pipe = 0.5 -in

r = 0.5 - 0.3 = 0.2-in = 0.0167 ft

ΔP = 0.6 psi/ft

ΔP, in lb/ft² = 0.6 x 144 = 86.4 lb/ft²

$\tau = \frac{\delta P *r}{2L} = \frac{86.4 *0.0167}{2*12} =0.06012 \ lb/ft^2$

Part (b) shear stress at a distance of 0.5-in away from the pipe wall

r = 0.5 - 0.5 = 0

$\tau = \frac{\delta P *r}{2L} = \frac{86.4 *0}{2*12} =0$

3 0

3 years ago