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2 years ago

13

The electric force is exerted by a 40-nC charged particle located at the origin of a Cartesian coordinate system on a 10-nC char

ged particle located at (2.0 m, 2.0 m ).
Required:
a. Determine the direction of the force.
b. Determine the magnitude of the force.

1 answer:

notsponge [240]2 years ago

5 0

Answer:

450 × 10 ⁻⁹ N

Explanation:

From the given:

The direction of force is;

$tan \ \theta = \dfrac{2}{2}$

$\theta =tan ^{-1} (1)$

$\theta = 45^0$

Hence, the angle is 45° counterclockwise from + x-axis

However;

the magnitude of the force is:

$F = \dfrac{kq_1q_2}{r^2}$

$F = \dfrac{(9*10^9 \ Nm^2/C^2 )*( 40 * 10^{-9} \ C ) * (10*10^{-9} \ C)}{(\sqrt{8})^2}$

$F = 450 \times 10^{-9} \ N$

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The wave length of violet light rounded to the nearest nanometer is a __ nm

nataly862011 [7]

Wavelength= speed / frequency

so.....3× 10^8 / 7.26×10^14
= .413× 10^(-6)

in scientific notation= 4.13×10^(-7)

in nanometer = 413 nm

6 0

2 years ago

Daniel takes his two dogs, Pauli the Pointer and Newton the Newfoundland, out to a field and lets them loose to exercise. Both d

DedPeter [7]

Answer:

Velocity of Pauli relative to Daniel = (-1.50ï + 3.90ĵ) m/s

x-component = -1.50 m/s

y-component = 3.90 m/s

Explanation:

Relative velocity of a body A relative to another body B, Vab, is given as

Vab = Va - Vb

where

Va = Relative velocity of body A with respect to another third body or frame of reference C

Vb = Relative velocity of body B with respect to that same third body or frame of reference C.

So, relative velocity can be given further as

Vab = Vac - Vbc

Velocity of Newton relative to Daniel = Vnd = 3.90 m/s due north = (3.90ĵ) m/s in vector form.

Velocity of Newton relative to Pauli = Vnp = 1.50 m/s due East = (1.50î) m/s in vector form

What is Pauli's velocity relative to Daniel?

Vpd = Vp - Vd

(Pauli's velocity relative to Daniel) = (Pauli's velocity relative to Newton) - (Daniel's velocity relative to Newton)

Vpd = Vpn - Vdn

Vpn = -Vnp = -(1.50î) m/s

Vdn = -Vnd = -(3.90ĵ) m/s

Vpd = -1.50î - (-3.90ĵ)

Velocity of Pauli relative to Daniel = (-1.50ï + 3.90ĵ) m/s

Hope this Helps!!!!

5 0

2 years ago

Block b rests upon a smooth surface. if the coefficients of static and kinetic friction between a and b are μs = 0.4 and μk = 0.

aliina [53]

Given

Weight of the block A, Wa = 20 lb, weight of block B Wb = 50 lb. Applied force to block A, P = 6lb, coefficient of static friction µs = 0.4, coefficient of kinetic friction µk = 0.3. If a force P is applied to the body, no relative motion will take place until the applied force is equal to the force of friction Ff, which is acting opposite to the direction of motion. Magnitude of static force of friction between block A and block B, Fs = µsN, where N is reaction force acting on block A. Now, resolve the forces Fx = max. P = (mA + mB)a,

6 = (20 / 32.2 + 50 / 32.2)a

2.173a = 6

A = 2.76 ft/s^2

To check slipping occurs between block A and block B, consider block A:

P – Ff = mAaA

6 – Ff = 1.71

Ff = 4.29 lb

And also,

N = wA. We know static friction,

Fs = µsN

Fs = 0.4 x 20

Fs = 8lb

Frictional force is less than static friction. Ff < Fs

<span>Therefors, acceleration of block A, aA = 2.76 ft/s^2, acceleration of block B aB = 2.76 ft/s^2</span>

6 0

2 years ago

05. The time required to complete one lap around a perfectly circular track having a radius of 1,835 meters is 86

likoan [24]

Answer:

v = 134.06 m/s

Explanation:

Given that,

Radius of a circular track is 1,835 m

Time required to complete one lap around a perfectly circular track is 86 seconds

We need to find the car's velocity. Velocity is equal to,

v=d/t

On circular path,

$v=\dfrac{2\pi r}{t}\\\\v=\dfrac{2\pi \times 1835}{86}\\\\v=134.06\ m/s$

So, car's velocity is 134.06 m/s.

7 0

3 years ago

A large helium filled balloon is used as the center piece for a graduation party. The balloon alone has a mass of 225 kg and it

Orlov [11]

To solve this problem it is necessary to apply the concepts related to Newton's second law, the definition of density and sum of forces in bodies.

From Newton's second law we understand that

$F= ma (\rightarrow$ Gravity at this case)

Where,

m = mass

a= acceleration

Also we know that

$\rho = \frac{m}{V} \Rightarrow m = \rho V$

Part A) The buoyant force acting on the balloon is given as

$F_b = ma$

As mass is equal to the density and Volume and acceleration equal to Gravity constant

$F_b = \rho V g$

$F_b = 1.2*323*9.8$

$F_b = 3798.5$

PART B) The forces acting on the balloon would be given by the upper thrust force given by the fluid and its weight, then

$F_{net} = F_b -W$

$F_{net} = F_b -(mg+\rho_H Vg)$

$F_{net} = 3798.5-(9.8*225*9.8*0.179*323)$

$F_{net} = 1030N$

PART C) The additional mass that can the balloon support in equilibrium is given as

$F_{net} = m' g$

$m' =\frac{F_{net}}{g}$

$m' = \frac{1030}{9.8}$

$m' = 105Kg$

4 0

3 years ago