How many cubic feet are there in a gallon?

A strip of copper 150 µm thick and 4.50 mm wide is placed in a uniform magnetic field of magnitude B = 0.74 T, that is perpendic

Sophie [7]

Answer:

V = 6.55*10^{-6} v

Explanation:

The number density can be determined by using below formula:

$n = \frac{Bi}{Vle}$

where,

B is uniform magnetic field 0.74

i is current 18 A

V is hall potential difference

l is thickness 150 MICRO METER

e is electron charge 1.6 *10^{-19} C

therefore V can be determined as

$V = \frac{iB}{nle}$

$V = \frac{18*0.74}{8.47*10^{28}*150*10^{-6}*1.6 *10^{-19}}$

V = 6.55*10^{-6} v

8 0

3 years ago

For a person with a Near Point (NP) = 45 cm, what would be his prescription's lens power = _________ diopters A) 1.08 diopters

natita [175]

Answer:

The power of lens is 2.22 D.

Explanation:

Given that,

Distance of image v= 45 cm

We know that,

Distance of object u= ∞

Power of lens is the reciprocal of focal length.

$P=\dfrac{1}{f}$

We need to calculate the power of lens

Using formula of lens

$\dfrac{1}{f}=\dfrac{1}{v}+\dfrac{1}{u}$

Where, f = focal length

v = image distance

u = object distance

Put the value into the formula

$\dfrac{1}{f}=\dfrac{1}{45}+\dfrac{1}{\infty}$

$\dfrac{1}{f}=\dfrac{1}{0.45}+0$

$P=\dfrac{1}{f}=2.22$

Hence, The power of lens is 2.22 D.

7 0

3 years ago

Which event signals the birth of a star?

oksano4ka [1.4K]

When the density and temperature at the core of the gravitationally collapsing nebula reaches values when nuclear fusion is triggered and sustained, that marks the birth of a star

4 0

3 years ago

A block of mass 200g is oscillating on the end of a horizontal spring of spring constant 100 N/m and natural length 12 cm. When

malfutka [58]

In order to determine the acceleration of the block, use the following formula:

$F=ma$

Moreover, remind that for an object attached to a spring the magnitude of the force acting over a mass is given by:

$F=kx$

Then, you have:

$ma=kx$

by solving for a, you obtain:

$a=\frac{kx}{m}$

In this case, you have:

k: spring constant = 100N/m

m: mass of the block = 200g = 0.2kg

x: distance related to the equilibrium position = 14cm - 12cm = 2cm = 0.02m

Replace the previous values of the parameters into the expression for a:

$a=\frac{(\frac{100N}{m})(0.02m)}{0.2\operatorname{kg}}=10\frac{m}{s^2}$

Hence, the acceleration of the block is 10 m/s^2

8 0

1 year ago

1.

Gwar [14]

Answer:

For 1: The correct option is Option C.

For 3: The final velocity of the opponent is 1m/s

Explanation:

During collision, the energy and momentum remains conserved. The equation for the conservation of momentum follows:

$m_1u_1+m_2u_2=m_1v_1+m_2v_2$ ...(1)

where,

$m_1,u_1\text{ and }v_1$ are the mass, initial velocity and final velocity of first object

$m_2,u_2\text{ and }v_2$ are the mass, initial velocity and final velocity of second object

For 1:

We are Given:

$m_1=150g=0.15kg\\u_1=?m/s\\v_1=0.85m/s\\m_2=3500g=3.5kg\\u_2=0m/s\\v_2=0.85m/s$

Putting values in equation 1, we get:

$(0.15\times u_1)+(3.5\times 0)=(3.5+0.15)\times 0.85\\\\u_1=20.683\approx 21m/s$

Hence, the correct answer is Option C.

For 2:

Impulse is defined as the product of force applied on an object and time taken by the object.

Mathematically,

$J=F\times t$

where,

F = force applied on the object

t = time taken

J = impulse on that object

Impulse depends only on the force and time taken by the object and not dependent on the surface which is stopping the object.

Hence, the impulse remains the same.

For 3:

Let the speed in right direction be positive and left direction be negative.

We are Given:

$m_1=240kg\\u_1=0m/s\\v_1=-1m/s\\m_2=80kg\\u_2=-2m/s\\v_2=?m/s$

Putting values in equation 1, we get:

$(240\times 0)+(80\times (-2))=(240\times (-1))+(80\times v_2)\\\\v_2=1m/s$

Hence, the final velocity of the opponent is 1m/s and has moved backwards to its direction of the initial velocity.

4 0

3 years ago