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3 years ago

Which of the following statements is correct?

1 answer:

6 0

These are the correct solutions:

It is 11 a.m. in the Eastern Time Zone; therefore, it is 8 a.m. in the Pacific Time Zone. (3 hrs behind)

It is 3 p.m. in the Central Time Zone; therefore, 2 p.m. in the Mountain Time Zone. (1 hr behind)

It is 6 p.m. in the Pacific Time Zone; therefore, it is 4 p.m in Hawaii. (2 or 3 hours behind depending on time of year)

It is 6 p.m. in Hawaii; therefore, it is 11 p.m. in the Eastern Time Zone (5 or 6 hours behind depending on time of year).

It is 3 p.m. in Hawaii; therefore, it is 6 p.m. in the Mountain Time Zone (3 or 4 hours behind depending on time of year).

You might be interested in

A car moves in a straight line at 22.0 m/s for 10.0miles, then at 30.0 m/s for another 10.0miles. Calculate the car’s average sp

maw [93]

Answer: 25.38 m/s

Explanation:

We have a straight line where the car travels a total distance $D$ , which is divided into two segments $d=10 miles$ :

$D=d+d=2d$ (1)

Where $d=10mi \frac{1609.34 m}{1 mi}=16093.4 m$

On the other hand, we know speed is defined as:

$S=\frac{d}{t}$ (2)

Where $t$ is the time, which can be isolated from (2):

$t=\frac{d}{S}$ (3)

Now, for the first segment $d=16093.4 m$ the car has a speed $S_{1}=22m/s$ , using equation (3):

$t_{1}=\frac{d}{S_{1}}$ (4)

$t_{1}=\frac{16093.4 m}{22m/s}$ (5)

$t_{1}=731.518 s$ (6) This is the time it takes to travel the first segment

For the second segment $d=16093.4 m$ the car has a speed $S_{1}=30m/s$ , hence:

$t_{2}=\frac{d}{S_{2}}$ (7)

$t_{2}=\frac{16093.4 m}{30m/s}$ (8)

$t_{2}=536.44 s$ (9) This is the time it takes to travel the secons segment

Having these values we can calculate the car's average speed $S_{ave}$ :

$S_{ave}=\frac{d + d}{t_{1} + t_{2}}=\frac{2d}{t_{1} + t_{2}}$ (10)

$S_{ave}=\frac{2(16093.4 m)}{731.518 s +536.44 s}$ (11)

Finally:

$S_{ave}=25.38 m/s$

3 0

3 years ago

what is the mass of an object that needs a force of 6600N to increase its speed from rest to 107 m/s in 2.3 seconds

Ray Of Light [21]

To answer this question do you need to know the formula to get the rate of change of acceleration (a=Δv/Δt; Δv= final velocity - initial velocity) and the formula to find the force of an object given a constant acceleration (F=m*a). Given these two formulas you can applicate them to solve for the mass of an object.

6 0

3 years ago

An automobile traveling 50km/hr. It decelerates at 5 meter per second square. How long will it take to stop the car? How far wil

const2013 [10]

We know, a = v/t
Here, a = 5 m/s²
v = 50 km/h= 13.88 m/s

Substitute their values into the expression:
5 = 13.88 / t
t = 13.88/5
t = 2.78 sec

Now, we know, v = d/t
13.88 = d/2.78
d = 13.88 * 2.78
d = 38.53 meter

In short, Your Answers would be:
i) It will take 2.78 sec
ii) It will travel for 38.53 m after a brakes applied.

Hope this helps!

3 0

3 years ago

25)

evablogger [386]

Answer:

radio waves electricity

5 0

2 years ago

he plates each have an area of 2 x 10-3 m2, and the spacing between the plates is 6 x 10-3 m. There is no dielectric between the

Serhud [2]

Complete question:

The left plate of a parallel plate capacitor carries a positive charge Q, and the right plate carries a negative charge -Q. The magnitude of the electric field between the plates is 100 kV/m. The plates each have an area of 2 x 10⁻³ m², and the spacing between the plates is 6 x 10⁻³ m. There is no dielectric between the plates. What is the charge on the capacitor?

Answer:

The charge on the capacitor is 1.77nC

Explanation:

Given;

magnitude of electric field between the plates, E = 100 kV/m

Area of each plate, A = 2 x 10⁻³ m²

Distance between the plates, d = 6 x 10⁻³ m

Charge on the capacitor is calculated as;

Q = CV

V = Ed

$C = \epsilon_o \frac{A}{d}$

$Q = CV = \epsilon_o \frac{A}{d}*Ed\\\\ Q = \epsilon_o AE\\\\Q = 8.85*10^{-12} *2*10^{-3}* 100,000\\\\Q = 1.77 *10^{-9} \ C\\Q = 1.77 \ nC$

Therefore, the charge on the capacitor is 1.77nC

8 0

3 years ago