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3 years ago

Suppose a 1.2 m antenna is installed on top of a building that is 225 m tall,

Physics

1 answer:

vazorg [7]3 years ago

7 0

Answer: 226.2 m

Explanation:

If we want to find the new total height $H$ of the building, we have to add to its initial height $h_{B}=225m$ the height of the antenna $h_{A}=1.2m$ , as follows:

$H=h_{B}+h_{A}$

$H=225m+1.2m$

Then:

$H=226.2m$ This is the new total height of the building.

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The earth exerts a gravitational force on the moon. the moon exerts a gravitational force on the earth. how do the magnitudes of

Sonbull [250]

The forces are the same like the third law of Newton states.I think

3 0

3 years ago

A current-carrying wire oriented north-south and laid over a compass deflects the compass 8° to the east. What is the magnitude

zhenek [66]

$2.8 \times 10^{-6}\ T$ is the magnitude of the magnetic field made by the current

<u>Explanation:</u>

Given data:

$\theta=8^{\circ}$

Magnetic field of earth, $B_{\text {earth}}=2 \times 10^{-5}\ \text {tesla}$

We need to find the magnetic field of wire, $B_{\text {wire }}$

The compass needle moves toward a direction of magnetic field. The current in wire makes a magnetic field in available space where the compass is on the ground. The vector sum of the Earth's magnetic field and the wire's magnetic field represents the net magnetic field, as shown in the attached drawing, expressing the angle:

$\tan \theta=\frac{B_{\text {wire}}}{B_{\text {earth}}}$

By substituting the given values, we get

$B_{\text {wire}}=\tan \theta \times B_{\text {earth}}=\tan 8^{\circ} \times 2 \times 10^{-5}=0.1405 \times 2 \times 10^{-5}$

$B_{\text {wire}}=0.28 \times 10^{-5}=2.8 \times 10^{-6}\ T$

4 0

3 years ago

An ultrasound unit is being used to measure a patient's heartbeat by combining the emitted 2.0 MHz signal with the sound waves r

alexandr402 [8]

Hi there,
for this question we have:
Signal 2.0 MHz = Emitted so we can call it $f_e$
and we need the Reflected = $f_{r}$
In this question, we have a source which goes to the heart and a reflected which comes back from the heart and we need the speed of the reflected.
So you should know that the speed of reflected is lower than the source(Emitted).
we also know: ΔBeat frequency(max) = 560 Hz = $f_{b}$
so we have:
$f_{e}$ - $f_{r}$ = $f_{b}$
so frequency of Reflected is:
2.0 × 10^6 Hz - 560 Hz = 1.99 × 10^6 Hz = $f_{r}$
now you know that Lambda = v/f
so if we find the lambda with our Emitted then we can find v with the Reflected:
Lambda = 1540(m/s) / 2.0 × 10^6 Hz = 7.7 × 10^-4 m
=> $v_{max}$ = (lambda)( $f_{r}$
=> 7.7 × 10^-4m (1.99 × 10^6Hz) = 1532 m/s
so the $v_{max}$ is equal to 1532 m/s :)))
This question is solved by two top teachers as fast as they could :))
I hope this is helpful
have a nice day

8 0

4 years ago

An engine absorbs 1.69 kJ from a hot reservoir at 277°C and expels 1.25 kJ to a cold reservoir at 27°C in each cycle.

Anna35 [415]

Answer

Given,

Energy absorbed, $Q_H = 1.69\ kJ$