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2 years ago

If light moves at a speed of 299,792,458 m/s, how long will it take light to move a distance of 1,000 km

Physics

1 answer:

astra-53 [7]2 years ago

6 0

Answer:

change 1000km into Metre

1000km=1000000 M

speed= distance/time

time=distance/speed

time=1000000/299792458

time= 0.0033second

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A lawyer drives from her home, located 1 mile east and 8 miles north of the town courthouse, to her office, located 4 miles w

givi [52]

Answer:

d = 13 miles

Explanation:

Lets say the position of court house is origin in this case

her office is located at 4 miles west and 4 miles south of court house

so here we have coordinate of the office with respect to court house is given as

$r_1 = (-4\hat i - 4\hat j)$

now the position of her home is located at 1 miles east and 8 miles north of the court house

so the coordinates of her home is given as

$r_2 = (1\hat i + 8 \hat j)$

now the change in the position is given as the distance between office and home

$d = r_2 - r_1$

$d = 5 \hat i + 12 \hat j$

$d = \sqrt{5^2 + 12^2} = 13 miles$

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3 years ago

Technician A says that hill assist and hill descent controls are added features to some electronic stability control systems. Te

Pani-rosa [81]

Answer:

Both technicians A and B

Explanation:

Both trailer sway control, hill assist and hill descent controls are additional featires that enhance stability of electronics and their control systems. Majorly, these features track and reduce skidding in electronics, therfore, enhancing electronic system stability. During the process, these newly added features help to automatically apply brakes and direct the sytem where the controller wants to take it.

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3 years ago

All moving objects don’t have momentum<br> A. True<br> B. False

mestny [16]

Answer:

Explanation:

Some objects gain momentum.

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2 years ago

Read 2 more answers

A thin spherical spherical shell of radius R which carried a uniform surface charge density σ. Write an expression for the volum

ozzi

Answer:

Explanation:

From the given information:

We know that the thin spherical shell is on a uniform surface which implies that both the inside and outside the charge of the sphere are equal, Then

The volume charge distribution relates to the radial direction at r = R

∴

$\rho (r) \ \alpha \ \delta (r -R)$

$\rho (r) = k \ \delta (r -R) \ \ at \ \ (r = R)$

$\rho (r) = 0\ \ since \ r< R \ \ or \ \ r>R---- (1)$

To find the constant k, we examine the total charge Q which is:

$Q = \int \rho (r) \ dV = \int \sigma \times dA$

$Q = \int \rho (r) \ dV = \sigma \times4 \pi R^2$

∴

$\int ^{2 \pi}_{0} \int ^{\pi}_{0} \int ^{R}_{0} \rho (r) r^2sin \theta \ dr \ d\theta \ d\phi = \sigma \times 4 \pi R^2$

$\int^{2 \pi}_{0} d \phi* \int ^{\pi}_{0} \ sin \theta d \theta * \int ^{R}_{0} k \delta (r -R) * r^2dr = \sigma \times 4 \pi R^2$

$(2 \pi)(2) * \int ^{R}_{0} k \delta (r -R) * r^2dr = \sigma \times 4 \pi R^2$

Thus;

$k * 4 \pi \int ^{R}_{0} \delta (r -R) * r^2dr = \sigma \times R^2$

$k * \int ^{R}_{0} \delta (r -R) r^2dr = \sigma \times R^2$

$k * R^2= \sigma \times R^2$

$k = R^2 --- (2)$

Hence, from equation (1), if k = $\sigma$

$\mathbf{\rho (r) = \delta* \delta (r -R) \ \ at \ \ (r=R)}$

$\mathbf{\rho (r) =0 \ \ at \ \ rR}$

To verify the units:

$\mathbf{\rho (r) =\sigma \ * \ \delta (r-R)}$

↓ ↓ ↓

c/m³ c/m³ × 1/m

Thus, the units are verified.

The integrated charge Q

$Q = \int \rho (r) \ dV \\ \\ Q = \int ^{2 \ \pi}_{0} \int ^{\pi}_{0} \int ^R_0 \rho (r) \ \ r^2 \ \ sin \theta \ dr \ d\theta \ d \phi \\ \\ Q = \int ^{2 \pi}_{0} \ d \phi \int ^{\pi}_{0} \ sin \theta \int ^R_{0} \rho (r) r^2 \ dr$