If you measured 25 g jelly beans and 25 g of gumdrops, the number of jellybeans would be

An isolated conducting sphere has a 17 cm radius. One wire carries a current of 1.0000020 A into it. Another wire carries a curr

notsponge [240]

14 ms is required to reach the potential of 1500 V.

Explanation:

The current is measured as the amount of charge traveling per unit time. So the charge of electrons required for each current is determined as the product of current with time.

$Charge = Current \times Time$

As two different current is passing at two different times, the net charge will be the different in current. So,

$\text { Charge }=(1.0000020-1.0000000) \times t=2 \times 10^{-6} \times t$

The electric voltage on the surface of cylinder can be obtained as the ratio of charge to the radius of the cylinder.

$V=\frac{k q}{R}$

Here $k = 9 * 10^9$ , q is the charge and R is the radius. As $q=2 \times 10^{-6} \times t$ and R =17 cm = 0.17 m, then the voltage will be

$V=\frac{9 \times 10^{9} \times 2 \times 10^{-6} \times t}{0.17}$

The time is required to find to reach the voltage of 1500 V, so

$1500 =\frac{9 \times 10^{9} \times 2 \times 10^{-6} \times t}{0.17}$

$\begin{aligned}&t=\frac{1500 \times 0.17}{\left(9 \times 10^{9} \times 2 \times 10^{-6}\right)}\\&t=14.1666 \times 10^{-3} s=14\ \mathrm{ms}\end{aligned}$

So, 14 ms is required to reach the potential of 1500 V.

3 0

3 years ago

In a region of space where gravitational forces can be neglected, a sphere is accelerated by a uniform light beam of intensity 8

BlackZzzverrR [31]

Answer:

The correct answer is B

Explanation:

To calculate the acceleration we must use Newton's second law

F = m a

a = F / m

To calculate the force we use the defined pressure and the radiation pressure for an absorbent surface

P = I / c absorbent surface

P = F / A

F / A = I / c

F = I A / c

The area of area of a circle is

A = π r²

We replace

F = I π r² / c

Let's calculate

F = 8.0 10⁻³ π (1.0 10⁻⁶)²/3 10⁸

F = 8.375 10⁻²³ N

Density is

ρ = m / V

m = ρ V

m = ρ (4/3 π r³)

m = 4500 (4/3 π (1 10⁻⁶)³)

m = 1,885 10⁻¹⁴ kg

Let's calculate the acceleration

a = 8.375 10⁻²³ / 1.885 10⁻¹⁴

a = 4.44 10⁻⁹ m/s² absorbent surface

The correct answer is B

4 0

3 years ago

What is the difference between a graph of linear motion and a graph of harmonic motion?

fredd [130]

Hope this helps :)

When describing linear motion, you need only one graph representing each of the three terms, while projectile motion requires a graph of the x and y axes. Graphs of simple harmonic motion are sine curves. Circular motion is different from other forms of motion because the speed of the object is constant.

5 0

3 years ago

A car and a motorcycle start from rest at the same time on a straight track, but the motorcycle is 25.0 m behind the car. The ca

sergeinik [125]

Answer:

t = 8.45 sec

car distance d = 132.09 m

bike distance d = 157.08 m

Explanation:

GIVEN :

motorcycle is 25 m behind the car , therefore distance need to covered by bike to overtake car is 25+ d, when car reache distance d at time t

for car

by equation of motion

$d = ut + \frac{1}{2}at^2$

u = 0 starting from rest

$d = \frac{1}{2}at^2$

$t^2 = \frac{2d}{a}$

for bike

$d+25 = 0 + \frac{1}{2}*4.40t^2$

$t^2= \frac{d+25}{2.20}$

equating time of both

$\frac{2d}{a} = \frac{d+25}{2.20}$

solving for d we get

d = 132 m

therefore t is $= \sqrt{\frac{2d}{a}}$

$t = \sqrt{\frac{2*132}{3.70}}$

t = 8.45 sec

each travelled in time 8.45 sec as

for car

$d = \frac{1}{2}*3.70 *8.45^2$

d = 132.09 m

fro bike

$d = \frac{1}{2}*4.40 *8.45^2$

d = 157.08 m

7 0

3 years ago

Refer to a long, straight wire carrying constant current I. What can be concluded about the magnitude of the magnetic field at d

sergij07 [2.7K]

Answer:

"the magnitude of the magnetic field at a point of distance a around a wire, carrying a constant current I, is inversely proportional to the distance a of the wire from that point"

Explanation:

The magnitude of the magnetic field from a long straight wire (A approximately a finite length of wire at least for close points around the wire.) decreases with distance from the wire. It does not follow the inverse square rule as is the electric field from a point charge. We can then say that "the magnitude of the magnetic field at a point of distance a around a wire, carrying a constant current I, is inversely proportional to the distance a of the wire from that point"

From the Biot-Savart rule,

B = μI/2πR

where B is the magnitude of the magnetic field

I is the current through the wire

μ is the permeability of free space or vacuum

R is the distance between the point and the wire, in this case is = a

5 0

3 years ago