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4 years ago

What is reflection? Reflection is light that has struck a surface and has _______.

Physics

1 answer:

andrew11 [14]4 years ago

5 0

1.)bounced off...in pretty sure(sorry if wrong)

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A current of 16.0 mA is maintained in a single circular loop of 1.90 m circumference. A magnetic field of 0.790 T is directed pa

pav-90 [236]

Answer

given,

current (I) = 16 mA

circumference of the circular loop (S)= 1.90 m

Magnetic field (B)= 0.790 T

S = 2 π r

1.9 = 2 π r

r = 0.3024 m

a) magnetic moment of loop

M= I A

M= $16 \times 10^{-3} \times \pi \times r^2$

M= $16 \times 10^{-3} \times \pi \times 0.3024^2$

M= $4.59 \times 10^{-3}\ A m^2$

b) torque exerted in the loop

$\tau = M\ B$

$\tau = 4.59 \times 10^{-3}\times 0.79$

$\tau = 3.63 \times 10^{-3} N.m$

8 0

3 years ago

Three objects are positioned along the x axis as follows: 4.4 kg at x = + 1.1 m, 3.7 kg at x = +0.80 m, and 2.9 kg at x = +1.6 m

maxonik [38]

Answer:

the distance from the location of the center of gravity to the location of the center o mass for this system is 1.13m

Explanation:

Given that

m₁=4.4kg

x₁=+1.1m

m₂=3.7kg

x₂=+0.80m

m₃=2.9kg

x₃=+1.6m

The position of the center of mass is

Xcm = [m₁x₁ +m₂x₂ +m₃x₃]/(m₁+m₂+m₃)

= [(4.40kg)(1.1 m)+(3.70 kg)(0.80 m)+(2.90 kg)(1.60 m)]/(4.4 kg + 3.70 kg+2.90 kg)

= 1.13 m

The position of the center of gravity is 1.13m

Therefore, the distance from the location of the center of gravity to the location of the center o mass for this system is 1.13m

4 0

3 years ago

Water at the top of a slope has potential energy. please select the best answer from the choices provided: t f

prisoha [69]

True because it have a lot of energy at the top because of gravity and air restrictsence

4 0

4 years ago

Read 2 more answers

When an object is moving with uniform circular motion, the centripetal acceleration of the object a. is circular. b. is perpendi

koban [17]

When an object is moving with uniform circular motion, the centripetal acceleration of the object d. is directed toward the center of motion.

7 0

3 years ago

Read 2 more answers

For fully developed laminar pipe flow in a circular pipe, the velocity profile is u(r) = 2(1-r2 /R2 ) in m/s, where R is the inn

sdas [7]

Answer:

a) $v_{max} = 2\ \textup{m/s}$

b) $v_{avg} = 1\ \textup{m/s}$

c) Q = 1.256 × 10⁻³ m³/s

Explanation:

Given:

The velocity profile as:

$u(r) = 2(1-\frac{r^2}{R^2} )$

Now, the maximum velocity of the flow is obtained at the center of the pipe

i.e r = 0

thus,

$v_{max}=u(0) = 2(1-\frac{0^2}{R^2} )$

$v_{max} = 2\ \textup{m/s}$

Now,

$v_{avg} = \frac{v_{max}}{2}\ \textup{m/s}$

$v_{avg} = \frac{2}}{2}\ \textup{m/s}$

$v_{avg} = 1\ \textup{m/s}$

Now, the flow rate is given as:

Q = Area of cross-section of pipe × $v_{avg}$

Q = $\frac{\pi D^2}{4}\times v_{avg}$

Q = $\frac{\pi 0.04^2}{4}\times 1$

Q = 1.256 × 10⁻³ m³/s

7 0

3 years ago