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4 years ago

7

Which object would be considered to be in free fall? one that experiences both friction and centripetal force. one that is truly

weightless. one that is slowed down by air resistance. one with only gravity acting upon it

2 answers:

Vitek1552 [10]4 years ago

6 0

Answer:

one with only gravity acting upon it

Explanation:

edge 2020 took the quiz

Scorpion4ik [409]4 years ago

6 0

Answer:

D

Explanation:

Edge 2021

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Wan and Nurul sat on a see-saw. The see-saw is imbalanced because Wan’s mass is 45 kg while Nurul’s mass is only 30 kg. Suggest

Alex Ar [27]

We have that the see-saw will be balance if a weight of $x=15kg$ is added to Nural's side of the see-saw

$x=15kg$

From the Question we are told that

Wan’s mass is $M_w=45 kg$

Nurul’s mass is $M_n= 30 kg.$

Generally

The Will be balance when the weight on both sides of the see-saw are equal

$M_w=M_n+x$

$45=30+x$

$x=45-30$

$x=15kg$

In conclusion

The see-saw will be balance if a weight of $x=15kg$ is added to Nural's side of the see-saw

$x=15kg$

For more information on this visit

brainly.com/question/22255610

5 0

3 years ago

If a display has a dynamic range of 20 dB and the smallest voltage it can handle is 200 mV, then the largest voltage it can hand

DedPeter [7]

Answer:

The largest voltage is 0.02 V.

(d) is correct option.

Explanation:

Given that,

Range = 20 dB

Smallest voltage = 200 mV

We need to calculate the largest voltage

Using formula of voltage gain

$G_{dB}=10 log_{10}(\dfrac{V_{out}^2}{V_{in}^{2}})$

$20 =10 log_{10}(\dfrac{V_{out}^2}{V_{in}^{2}})$

$2=log_{10}(\dfrac{V_{out}^2}{V_{in}^{2}})$

$10^2=\dfrac{V_{out}^2}{V_{in}^{2}}$

$\dfrac{V_{out}}{V_{in}^}=10$

$V_{in}=\dfrac{V_{out}}{10}$

$V_{in}=\dfrac{200}{10}$

$V_{in}=20\ mV$

$V_{in}=0.02\ V$

Hence, The largest voltage is 0.02 V.

8 0

3 years ago

check the file.........................................................................................................

White raven [17]

What is meant by the number of complete oscillation made by an oscillating body in 10 seconds is 500 complete oscillations is that the frequency of the oscillating body is 50Hz

<h3>What is frequency?</h3>

Frequency of an oscillating body can be defined as the number of complete oscillations per unit time

Frequency is measured in hertz (Hz).

Also, the frequency of 1Hz is one oscillation per second.

frequency = number of oscillations/ time taken

frequency = 500/10 = 50 Hz

Thus, what is meant by the number of complete oscillation made by an oscillating body in 10 seconds is 500 complete oscillations is that the frequency of the oscillating body is 50Hz

Learn more about frequency here:

brainly.com/question/1199084

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6 0

2 years ago

A 1.5m wire carries a 6 A current when a potential difference of 61 V is applied. What is the resistance of the wire?

Alisiya [41]

Resistance = (voltage) / (current)

For this piece of wire . . .

Resistance = (61 volts) / (6 Amperes)

Resistance = (61/6) (V/A)

<em>Resistance = (10 and 1/6) ohms</em>

Since you know the voltage and current, the length doesn't matter.

7 0

3 years ago

Calculate the work performed by an ideal Carnot engine as a cold brick warms from 150 K to the temperature of the environment, w

olga nikolaevna [1]

To solve this problem, apply the concepts related to the calculation of the work performed according to the temperature change (in an ideal Carnot cycle), for which you have to:

$W = \int\limit_{T_c}^{T_H} C (1-\frac{T_H}{T})$

Where,

C = Heat capacity of the Brick

$T_C$ = Cold Temperature

$T_H$ = Hot Temperature

Integrating,

$W = C (T_H-T_C)- T_H C ln (\frac{T_H}{T_C})$

Our values are given as

$T_H= 300K$

$T_C = 150K$

Replacing,

$W = (1) (300-150)-300(1)ln(2)$

$W = 150-300ln2$

$W = -57.94kJ \approx 58kJ$

Therefore the work perfomed by this ideal carnot engine is 58kJ

5 0

4 years ago