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JulijaS [17]
3 years ago
10

Which statement describes an example of destructive interference?

Physics
1 answer:
a_sh-v [17]3 years ago
4 0

Answer:

B. Light shining on a surface has gaps after passing through two

slits.

Explanation:

Destructive interference is said to occur when two waves cancel out each other. The occurrence of this phenomenon implies that the waves have equal amplitude in opposite directions.

When light shinning on a surface has two gaps after passing through two slits, then the waves have interfered destructively which led to the gaps.

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A tree loses water to the air by the process of transpiration at the rate of 110 g/h. This water is replaced by the upward flow
Ksivusya [100]

Answer:

The answer is 1.87nm/s.

Explanation:

The 110g/hr  water loss must be replaced by 110g/hr of sap. 110g of sap corresponds to a volume of  

110g \div \dfrac{1040*10^3g}{1*10^6cm^3}  = 106cm^3

thus rate of sap replacement is

106cm^3/hr = 106*10^{-6}m^3/3600s  = 2.94*10^{-8}m^3/s

The volume of sap in the vessel of length x is

V = Ax,

where A is the cross sectional area of the vessel.

For 2000 such vessels, the volume is

V = 2000Ax

taking the derivative of both sides we get:

\dfrac{dV}{dt} = 2000A \dfrac{dx}{dt}

on the left-hand-side \dfrac{dx}{dt} is the velocity v of the sap, and on right-hand-side \dfrac{dV}{dt}  = 2.94*10^{-8}m^3/s; therefore,

2.94*10^{-8}m^3/s=2000Av

and since the cross-sectional area is

A = \pi (\dfrac{100*10^{-3}m}{2} )^2 = 7.85*10^{-3}m^2;

therefore,

2.94*10^{-8}m^3/s =2000(7.85*10^{-3}m^2)v

solving for v we get:

v = \dfrac{2.94*10^{-8}m^3/s}{2000(7.85*10^{-3}m^2)}

\boxed{v =1.875*10^{-9}m/s = 1.875nm/s}

which is the upward speed of the sap in each vessel.

8 0
3 years ago
Write the following number in scientific notation 156.60
Liula [17]

Answer:

1.566 x 10^2

Move the decimal to where the number being multiplied by 10^x is greater than 1 but less than 10. Then multiply it by 10^x

X is the number of times you moved the decimal, so in this case it would be 10^2

8 0
3 years ago
Suppose you want to design an air bag system that can protect the driver at a speed 100 km/h (60 mph) if the car hits a brick wa
34kurt

When solving question that contains equations and the use mathematical computations, It is always ideal to list the parameters given.

Now, given that:

  • the speed of the car which is the initial velocity (u) = 100 km/h before it hits the wall.
  • after hitting the wall, the final velocity will be (v) = 0 km/h

Assumptions:

  • Suppose we make an assumption that the distance travelled during the collision of the car with the brick wall (S) = 1 m
  • That the car's acceleration is also constant.

∴

For a motion under constant acceleration, we can apply the kinematic equation:

\mathsf{v^2 = u^2 + 2as}

where;

v = final velocity

u = initial velocity

a = acceleration

s = distance

From the above equation, making acceleration (a) the subject of the formula:

\mathsf{v^2 - u^2 =2as }

\mathsf{a = \dfrac{v^2 - u^2 }{2s}}

The initial velocity (u) is given in km/h, and we need to convert it to m/s as it has an effect on the unit of the acceleration.

since 1 km/h = 0.2778 m/s

100 km/h = 27.78 m/s

\mathsf{a = \dfrac{(0)^2 - (27.78)^2 }{2(1)}}

\mathsf{a = \dfrac{- 771.7284 }{2}}

a = - 385.86 m/s²

Similarly, from the kinematic equation of motion, the formula showing the relation between time, acceleration and velocity is;

v = u + at

where;

v = 0

-u = at

\mathsf{t = \dfrac{-u}{a}}

\mathsf{t = \dfrac{-27.78}{-385.86}}

t = 0.07 seconds

An airbag is designed in such a way as to prevent the driver from hitting on the steering wheel or other hard substance that could damage the part of the body. The use of the seat belt is to keep the driver in shape and in a balanced position against the expansion that occurred by the airbag during the collision on the brick wall.

Thus, we can conclude that the airbag must be inflated at 0.07 seconds faster before the collision to effectively protect the driver.

Learn more about the kinematic equation here:

brainly.com/question/11298125?referrer=searchResults

3 0
3 years ago
After landing on an unfamiliar planet, a space explorer constructs a simple pendulum of length 49.0 cm . The explorer finds that
iren2701 [21]

Answer: g = 10.0 m/s/s

Explanation:

For a simple pendulum, provided that the angle between the lowest and highest point of  his trajectory be small, the oscillation period is given by the following expression:

T = 2π √L/g , where L = pendulum length, g= accelleration of gravity.

We can also define the period, as the time needed to complete a full swing, so from the measured  values, we can conclude the following :

T = 140 sec/ 101 cycles = 1.39 sec

Equating both definitions for T, we can solve for g, as follows:

g = 4 π² L / T² = 4π². 0.49 m / (1.39)² = 10.0 m/s/s

7 0
3 years ago
Two clouds collide and form another, more massive cloud. One cloud is stationary, while the other is traveling at 1 m/s. After t
sweet [91]

Answer: 3

Explanation:

Given

One cloud is traveling at rate of u_2=1\ m/s

combined velocity of the two is v=0.25\ m/s

Suppose the masses of the clouds be m_1,m_2

Conserving momentum

\Rightarrow m_1u_1+m_2u_2=\left(m_1+m_2\right)v\\\text{Divide whole equation by }m_2\\\Rightarrow \dfrac{m_1}{m_2}u_1+u_2=\left(\dfrac{m_1}{m_2}+1\right)v\\\\\Rightarrow 0+1=0.25\dfrac{m_1}{m_2}+0.25\\\\\Rightarrow \dfrac{m_1}{m_2}=\dfrac{0.75}{0.25}\\\\\Rightarrow \dfrac{m_1}{m_2}=3

6 0
3 years ago
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