Which diagram represent light with the longest wavelength ?

Which of these letters is the symbol for current in equations A: c B: i C: r D: t

True [87]

Answer: [B]: The letter, "<em /> I " ; for current; in units of "Amps" .
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4 0

2 years ago

A plane is flying east when it drops some supplies to a designated target below. The supplies land after falling for 10 seconds.

Diano4ka-milaya [45]

solution:

As Given plane is flying in east direction.

It throws back some supplies to designated target.

Time taken by the supply to reach the target =10 seconds

g = Acceleration due to gravity = - 9.8 m/s²[Taken negative as object is falling Downwards]

As we have to find distance from the ground to plane which is given by d.

d = $\frac{1}{2}\times g\times t^2$

= $\frac{1}{2}\times (9.8) \times(100) =50\times 9.8=490$ meters

Distance from the ground where supplies has to be land to plane = Option B =490 meters

6 0

3 years ago

Read 2 more answers

Suppose the current in a conductor decreases exponentially with time according to the equation I(t) = I0e-t/τ, where I0 is the i

ELEN [110]

Answer:

Pls see attached file

Explanation:

5 0

3 years ago

Cesium-137 undergoes beta decay and has a half-life of 30.0 years. How many beta particles are emitted by a 14.0-g sample of ces

Mandarinka [93]

Answer: $0.81\times 10^{16}$ beta particles

Explanation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Given mass = 14.0 g

Molar mass = 137 g/mol

$\text{Number of moles of cesium}=\frac{14.0g}{137g/mol}=0.102moles$

According to avogadro's law, 1 mole of every substance weighs equal to its molecular mass and contains avogadro's number $6.023\times 10^{23}$ of particles.

1 mole of cesium contains atoms = $6.023\times 10^{23}$

0.102 moles of cesium contains atoms = $\frac{6.023\times 10^{23}}{1}\times 0.102=0.614\times 10^{23}$

The relation of atoms with time for radioactivbe decay is:

$N_t=N_0\times \frac{1}{2}^{\frac{t}{t_{\frac{1}{2}}}}$

Where $N_t$ =atoms left undecayed

$N_0$ = initial atoms

t = time taken for decay = 3 minutes

${t_{\frac{1}{2}}}$ = half life = 30.0 years = $1.577\times 10^7$ minutes

The fraction that decays : $1-(\frac{1}{2})^{\frac{3}{1.577\times 10^7}}=1.32\times 10^{-7}$

Amount of particles that decay is = $0.614\times 10^{23}\times 1.32\times 10^{-7}=0.81\times 10^{16}$

Thus $0.81\times 10^{16}$ beta particles are emitted by a 14.0-g sample of cesium-137 in three minutes.

7 0

2 years ago

A student drops an egg from the roof of their house onto a trampoline. The egg feels a change in momentum of 2.2 kg^ * m/s in 1.

geniusboy [140]

Answer:

F = m a = m v / t where v is the change in velocity in time t

F = p / t since m v is equal to p

F = 2.2 (kg m / s) / 1.1 s = 2 kg-m / s^2 = 2 N

Or you can use the impulse equation

8 0

2 years ago