2 years ago

Which instrument should, ideally, have zero resistance?

Physics

2 answers:

Temka [501]2 years ago

6 0

The answer is C voltmeter

MrMuchimi2 years ago

5 0

I think the answer is D) ammeter, because the resistance of the ammeter has to zero, to ensure there is no error readings during measurements.

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A car turns a certain curve of radius 24.98 m with constant linear speed of

Anastaziya [24]

Answer:

3525.19 kg

Explanation:

The computation of the mass of the car is shown below:

As we know that

Fc = m × V^2 ÷ R

m = Fc × R ÷ V^2

Provided that:

Fc = 34.652 kN = 34652 N

R = Radius = 24.98 m

V = speed = 15.67 m/s

So,

m = 34652 × 24.98 ÷ 15.67^2

= 3525.19 kg

7 0

3 years ago

Objects in our solar system, including planets and their moons, stay in orbit because of gravity and inertia. Draw a model to sh

coldgirl [10]

Answer:

You cant draw on brainly

Explanation:

6 0

2 years ago

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How many sandwiches were made in the hour?

S_A_V [24]

Answer:

Explanation:

simple facts.

7 0

3 years ago

A typical wall outlet voltage is 120 volts in the United States. Personal MP3 players require much smaller voltages, typically 2

butalik [34]

Answer:

Number of turns in secondary will be 7

Explanation:

We have given primary voltage $V_p=120volt$

Number of turns in the primary is $N_p=3575$

Secondary voltage is given $V_s=235mV=0.235volt$

We have to find the number of turns in secondary

We know that $\frac{N_p}{N_s}=\frac{V_p}{V_s}$

So $\frac{3575}{N_s}=\frac{120}{0.235}$

$N_s=6.60$

As the number of turns can not be in decimal so number of turns will be 7

6 0

3 years ago

A small airplane takes on 302 l of fuel. If the density of the fuel is 0.821 g/ml, what mass of fuel has the airplane taken on?

ella [17]

To calculate the mass of the fuel, we use the formula

$m = V \times \rho$

Here, m is the mass of fuel, V is the volume of the fuel and its value is $V =302 \ L = 302 \ L \times \frac{10^{3}m L }{L} = 302 \times 10^{3} \ mL$ and $\rho$ is the density and its value of 0.821 g/mL.

Substituting these values in above relation, we get

$m = 302 \times 10^{3} \ mL \times 0.821 g/mL = 247942 g \\\\ m = 247. 94 \ kg$

Thus, the mass of the fuel 247 .94 kg.

8 0

3 years ago