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3 years ago

Mary cycled at an average speed of 8 km/h. How far has she traveled if she rides for 4 hours?

Physics

2 answers:

Sati [7]3 years ago

4 0

Answer:

Which sentence from the passage shows that the function of the river depicted here has carried through to modern times?

Explanation:

Which sentence from the passage shows that the function of the river depicted here has carried through to modern times?

Lana71 [14]3 years ago

3 0

If it’s 8km/h =8x4km/1x4h= 32km/4h

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the coefficient of static friction between a 40 kg picnic table and the ground below is .43. what is the greatest horizontal for

Gre4nikov [31]

The force equals the coefficient of static friction times the weight. Use gravity g=9.8 m/s^2
0.43*40*9.8=16.856 N

4 0

4 years ago

Read 2 more answers

In which electric circuit would the voltmeter read 10 volts ?

ki77a [65]

Given that,

Voltage = 10 volt

Suppose, The three resistance is connected in parallel and each resistance is 12 Ω. find the current in the electric circuit.

We need to calculate the equivalent resistance

Using formula of parallel

$\dfrac{1}{R}=\dfrac{1}{R_{1}}+\dfrac{1}{R_{2}}+\dfrac{1}{R_{3}}$

Put the value into the formula

$\dfrac{1}{R}=\dfrac{1}{12}+\dfrac{1}{12}+\dfrac{1}{12}$

$\dfrac{1}{R}=\dfrac{1}{4}$

$R=4\ \Omega$

We need to calculate the current in the circuit

Using ohm's law

$V=IR$

$I=\dfrac{V}{R}$

Where, V = voltage

R = resistance

Put the value into the formula

$I=\dfrac{10}{4}$

$I=2.5\ A$

Hence, The current in the circuit is 2.5 A

4 0

3 years ago

PLEASE HELP MEEEEEE!!!!!!!!!!!!!!!!!!

Simora [160]

You would need to freeze it in a freezer. Hope this helps if it does could I have brainlist thanks

5 0

3 years ago

Read 2 more answers

single goose sounds a loud warning when an intruder enters the farmyard. Some distance from the goose, you measure the sound lev

Goryan [66]

Answer:

The sound level of the 26 geese is $Z_{26}= 96.15 dB$

Explanation:

From the question we are told that

The sound level is $Z_1 = 81.0 \ dB$

The number of geese is $N = 26$

Generally the intensity level of sound is mathematically represented as

The intensity of sound level in dB for one goose is mathematically represented as

$Z_1 = 10 log [\frac{I}{I_O} ]$

Where I_o is the threshold level of intensity with value $I_o = 1*10^{-12} \ W/m^2$

$I$ is the intensity for one goose in $W/m^2$

For 26 geese the intensity would be

$I_{26} = 26 * I$

Then the intensity of 26 geese in dB is

$Z_{26} = 10 log[\frac{26 I }{I_o} ]$

$Z_{26} = 10 log (\ \ 26 * [\frac{ I }{I_o} ]\ \ )$

$Z_{26} = 10 log (\ \ 26 \ \ ) * (\ \ 10 log [\frac{ I }{I_o} ]\ \ )$

From the law of logarithm we have that

$Z_{26} = 10 log 26 + 10 log [\frac{I}{I_0} ]$

$= 14.15 + 82$

$Z_{26}= 96.15 dB$

4 0

3 years ago

An object that weighs 2.450 N is attached to an ideal massless spring and undergoes simple harmonic oscillations with a period o

Viktor [21]

Answer:

Spring constant, k = 24.1 N/m

Explanation:

Given that,

Weight of the object, W = 2.45 N

Time period of oscillation of simple harmonic motion, T = 0.64 s

To find,

Spring constant of the spring.

Solution,

In case of simple harmonic motion, the time period of oscillation is given by :

$T=2\pi\sqrt{\dfrac{m}{k}}$

m is the mass of object

$m=\dfrac{W}{g}$

$m=\dfrac{2.45}{9.8}$

m = 0.25 kg

$k=\dfrac{4\pi^2m}{T^2}$

$k=\dfrac{4\pi^2\times 0.25}{(0.64)^2}$

k = 24.09 N/m

k = 24.11 N/m

So, the spring constant of the spring is 24.1 N/m.

6 0

3 years ago