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Setler79 [48]
3 years ago
11

Marys airplane trip took 5.8 hours for one-half of that time, the airplane flew at a constant speed of 640 miles per hour and fo

r the rest of the time, it flew at a speed of 580 mph. what distance did mary travel?
Physics
1 answer:
Olenka [21]3 years ago
7 0
Distance is speed x time.  Half of the trip is 5.8/2 = 2.9hrs.
640 x 2.9 = 1856mi
580 x 2.9 = 1682mi
1856mi+1682mi=3538mi.

You could also calculate her average speed.  This is easy since it was divided in two equal time slices.  Average Speed = (640+580)/2 = 610mi/hr
Now 610mi/hr x 5.8hrs = 3538mi
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What physical property makes the red light different from blue light, or radio waves different from microwaves?​
inna [77]

Only their wavelength does.

Blue light waves have only roughly half the wavelength of red light waves, and the so-called "microwaves" are the radio waves with the shortest wavelengths.

4 0
3 years ago
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A 30.0-μF capacitor is connected to a 49.0-Ω resistor and a generator whose rms output is 30.0 V at 60.0 Hz. (a) Find the rms
Natali5045456 [20]

Explanation:

Given that,

Capacitor = 30μC

Resistor = 49.0Ω

Voltage = 30.0 V

Frequency = 60.0 Hz

We need to calculate the impedance

Using formula of impedance

Z=\sqrt{R^2+X_{c}^2}.....(I)

We need to calculate the value of X_{c}

Using formula of X_{c}

X_{c}=\dfrac{1}{2\pi f c}

X_{c}=\dfrac{1}{2\times\pi\times60.0\times30\times10^{-6}}

X_{c}=88.42\ \Omega

Put the value of X_{c} into the formula of impedance

Z=\sqrt{(49.0)^2+(88.42)^2}

Z=101.08\ \Omega

(a). We need to calculate the rms current in the circuit

Using formula of rms current

I_{rms}=\dfrac{V}{Z}

I_{rms}=\dfrac{30.0}{101.08}

I_{rms}=0.30\ A

The rms current in the circuit is 0.30 A.

(b). We need to calculate the rms voltage drop across the resistor

Using formula of rms voltage

V_{rms}=I_{rms}\times R

Put the value into the formula

V_{rms}=0.30\times49.0

V_{rms}=14.7\ V

The rms voltage drop across the resistor is 14.7 V

(c). We need to calculate the rms voltage drop across the capacitor

Using formula of rms voltage

V_{rms}=I_{rms}\times X_{C}

V_{rms}=0.30\times88.42

V_{rms}=26.53\ V

The rms voltage drop across the capacitor is 26.53 V.

Hence, This is the required solution.

4 0
3 years ago
Two tiny beads are 25 cm apart with no other charges or fields present. Bead A carries 10 µC of charge and bead B carries 1 µC.
steposvetlana [31]
The electric forces on both beads are equal.
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3 years ago
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The SAF operates the M113 Ultra APC.
HACTEHA [7]

The volume of the object must be no larger than 11.15 m^3.

Explanation:

In order for an object to be able to float in water, its density must be equal or smaller than the water density.

The density of water is:

\rho = 1000 kg/m^3

This means that the density of the object must be no larger than this value.

We also know that the density of an object is given by

\rho = \frac{m}{V}

where

m is the mass of the object

V is its volume

For the object in this problem, the mass is

m=1.115\cdot 10^4 kg

Therefore, we can re-arrange the equation to find its volume:

V=\frac{m}{\rho}=\frac{1.115\cdot 10^4}{1000}=11.15 m^3

So, the volume of the object must be no larger than 11.15 m^3.

Learn more about density:

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8 0
3 years ago
Two rigid tanks of equal size and shape are filled with different gases. The tank on the left contains oxygen, and the tank on t
fredd [130]

Answer:

The number of oxygen molecules in the left container greater than the number of hydrogen molecules in the right container.

Explanation:

Given:

Molar mass of oxygen, M_O=32

Molar mass of hydrogen, M_H=2

We know ideal gas law as:

PV=nRT

where:

P = pressure of the gas

V = volume of the gas

n= no. of moles of the gas molecules

R = universal gs constant

T = temperature of the gas

∵n=\frac{m}{M}

where:

m = mass of gas in grams

M = molecular mass of the gas

∴Eq. (1) can be written as:

PV=\frac{m}{M}.RT

P=\frac{m}{V}.\frac{RT}{M}

        as: \frac{m}{V}=\rho\ (\rm density)

So,

P=\rho.\frac{RT}{M}

Now, according to given we have T,P,R same for both the gases.

P_O=P_H

\rho_O.\frac{RT}{M_O}=\rho_H.\frac{RT}{M_H}

\Rightarrow \frac{\rho_O}{32}=\frac{\rho_H}{2}

\rho_O=16\rho_H

∴The molecules of oxygen are more densely packed than the molecules of hydrogen in the same volume at the same temperature and pressure. So, <em>the number of oxygen molecules in the left container greater than the number of hydrogen molecules in the right container.</em>

5 0
3 years ago
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