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2 years ago

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Bianca is standing at x =600m. Firecracker 1, at the origin, and firecracker 2, at x =900m, explode simultaneously. The flash fr

om firecracker 1 reaches Bianca's eye at 4.0 μs . Part A At what time does she see the flash from firecracker 2?

1 answer:

agasfer [191]2 years ago

4 0

Answer: $3 \mu s$

Explanation:

Given

Bianca is at $x=600 m$

i.e. distance between origin and Bianca is $600 m$

time taken to reach Bianca eyes is

$t=\frac{600}{speed\ of\ light}$

$t=\frac{600}{3\times 10^8}$

$t=2\times 10^{-6} s$

$t=2 \mu s$

i.e. Cracker exploded at $t=2\mu s$ because it is observed at $t=4\mu s$

Time taken by second cracker flash to reach Bianca eyes

$t_2=\frac{300}{3\times 10^8}$

$t_2=10^{-6}$

$t_2=1 \mu s$

Therefore it will be observed at $t=3 \mu s$

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A baseball pitcher throws a ball at 90.0 mi/h in the horizontal direction. How far does the ball fall vertically by the time it

Lisa [10]

Answer:

Vertical distance= 3.3803ft

Explanation:

First with the speed of the ball and the distance traveled horizontally we can determine the flight time to reach the plate:

Velocity= (90 mi/h) × (1 mile/5280ft) = 475200ft/h

Distance= Velocity × time⇒ time= 60.5ft / (475200ft/h) = 0.00012731h

time= 0.00012731h × (3600s/h)= 0.458316s

With this time we can determine the distance traveled vertically taking into account that its initial vertical velocity is zero and its acceleration is that of gravity, 9.81m/s²:

Vertical distance= (1/2) × 9.81 (m/s²) × (0.458316s)²=1.0303m

Vertical distance= 1.0303m × (1ft/0.3048m) = 3.3803ft

This is the vertical distance traveled by the ball from the time it is thrown by the pitcher until it reaches the plate, regardless of air resistance.

3 0

2 years ago

A conducting sphere of radius 5.0 cm carries a net charge of 7.5 µC. What is the surface charge density on the sphere?

11111nata11111 [884]

Answer:

$\sigma=0.014\ C/m^2$

Explanation:

Given that,

The radius of sphere, r = 5 cm = 0.05 m

Net charge carries, q = 7.5 µC = 7.5 × 10⁻⁶ C

We need to find the surface charge density on the sphere. Net charge per unit area is called the surface charge density. So,

$\sigma=\dfrac{7.5\times 10^{-6}}{\dfrac{4}{3}\pi \times (0.05)^3}\\\\=0.014\ C/m^2$

So, the surface charge density on the sphere is $0.014\ C/m^2$ .

7 0

2 years ago

Two lasers, one red (with wavelength 633.0 nm) and the other green (with wavelength 532.0 nm), are mounted behind a 0.150-mm sli

Orlov [11]

Answer:

a.3.20m

b.0.45cm

Explanation:

a. Equation for minima is defined as: $sin \theta=\frac{m\lambda}{\alpha}$

Given $m=3$ , $\lambda=6.33\times 10^-^7$ and $\alpha=0.00015$ :

#Substitute our variable values in the minima equation to obtain $\theta$ :

$\theta=sin^-^1 (\frac{3\times 6.33\times 10^-^7}{0.00015})\\\\\theta=0.01266rad$

#draw a triangle to find the relationship between $\theta, y \$ and $L$ .

$tan(\theta)=y/L$ #where $y=4.05cm$

$L=y/tan(\theta)=3.20$

Hence the screen is 3.20m from the split.

b. To find the closest minima for green(the fourth min will give you the smallest distance)

#Like with a above, the minima equation will be defined as:

$sin \theta=\frac{m\lambda}{\alpha}$ , where $m=4$ given that it's the minima with the smallest distance.

$sin \theta=\frac{4\lambda}{\alpha}\\\theta=sin^-^1 (\frac{4\times 6.33\times 10^-^7}{0.00015})\\\\\theta=0.01688rad$

#we then use $tan(\theta)=y/L$ to calculate $L$ =4.5cm

Then from the equation subtract $y_3$ from $y$ :

$4.50cm-4.05cm=0.45cm$

Hence, the distance $\bigtriangleup y$ is 0.45cm

8 0

2 years ago

3. A snail crawls 5 inches in 15 minutes. What is its speed in in./min?

suter [353]

Answer:

3.0.33in/min...(c)

4.40m/min....(b)

5.10m/s....(a)

6.20mph...(b)

7.4m/s...(a)

3 0

2 years ago

Assume that a uniform magnetic field is directed into thispage. If an electron is released with an initial velocity directedfrom

Feliz [49]

Answer:

B). to the right

Explanation:

Since the direction of magnetic field is into the page

So here we know that

$B = B_o(-\hat k)$

now the velocity is from bottom to top

so we have

$v = v_o \hat j$

now the force on the moving charge is given as

$\vec F = q(\vec v \times \vec B)$

now we have

$\vec F = (-e)(v_o \hat j \times B_o(-\hat k))$

$\vec F = e v_o B \hat i$

so force will be towards Right

7 0

2 years ago