Light passing through the center of a lens will carry on undeviated.

True or False:In a current, electrons will always flow from negative to positive.

DIA [1.3K]

It should be true: electrons low from negative toward positive to negative toward positive because opposite charges attract each other.

I hope this was correct

3 0

3 years ago

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A hot air balloon is traveling vertically upward at a constant speed of 4.5 m/s. When it is 28 m above the ground, a package is

ella [17]

Answer:

1.97 seconds

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.8 m/s²

$s=ut+\frac{1}{2}at^2\\\Rightarrow 21=4.5t+\frac{1}{2}\times -9.8\times t^2\\\Rightarrow 21=-4.5t-4.9t^2\\\Rightarrow 4.9t^2+4.5t-28=0\\\Rightarrow 49t^2+45t-280=0$

Solving the above equation we get

$t=\frac{-45+\sqrt{45^2-4\cdot \:49\left(-280\right)}}{2\cdot \:49}, \frac{-45-\sqrt{45^2-4\cdot \:49\left(-280\right)}}{2\cdot \:49}\\\Rightarrow t=1.97, -2.89$

So, the time the package was in the air is 1.97 seconds

3 0

2 years ago

Where is the frequency of ultrasound in relation to the range of human ability to hear

kykrilka [37]

Answer:

ultra sounds have frequency higher than the upper audible limit of human hearing, for healthy, young adults.

Explanation:

4 0

3 years ago

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Use the following half-life graph to answer the following question:

Temka [501]

Answer:

A 1.0 min

Explanation:

The half-life of a radioisotope is defined as the time it takes for the mass of the isotope to halve compared to the initial value.

From the graph in the problem, we see that the initial mass of the isotope at time t=0 is

$m_0 = 50.0 g$

The half-life of the isotope is the time it takes for half the mass of the sample to decay, so it is the time t at which the mass will be halved:

$m'=\frac{50.0 g}{2}=25.0 g$

We see that this occurs at t = 1.0 min, so the half-life of the isotope is exactly 1.0 min.

3 0

3 years ago

A space vehicle is traveling at 3760 km/h relative to Earth when the exhausted rocket motor is disengaged and sent backward. The

almond37 [142]

Answer:

3688 km/h

Explanation:

Given:-

- The speed of vehicle relative to earth, vs_e = 3760 km/h

- The relative speed of command and motor, v_c/m = 90 km/h

- The mass of command = m

- The mass of motor = 4m

Find:-

What is the speed of the command module relative to Earth just after the separation?

Solution:-

- Consider the space vehicle as a system that detaches itself into two parts ( command and motor ). We will assume that the gravitational pull due to Earth on the space vehicle is negligible. With that assumption we have our system in isolation. We will apply the principle of conservation of linear momentum on the system as follows:

Initial momentum = Final momentum

Pi = Pf

M*vs_e = m*vc_e + 4m*vm_e

Where,

M = m + 4m = 5m

vc_e = Velocity of command relative to earth

vm_e = Velocity of motor relative to earth

- We will develop a relation of velocities of command and motor in the frame of earth as follows:

vm_e = v_c/m + vc_e

- Substituting (vm_e) from Equation 2 into Equation 1, we have:

5m*vs_e = m*vc_e + 4m*(v_c/m + vc_e)

5m*vs_e = 5m*vc_e + 4m*(v_c/m)

- Solve for vc_e:

5m*vs_e - 4m*(v_c/m) = 5m*vc_e

vs_e - 0.8*(v_c/m) = vc_e

- Plug in values and evaluate vc_e:

vc_e = 3760 - 0.8*(90)

vc_e = 3,688 km/h

5 0

3 years ago

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