All categories

2 years ago

Pls help I will mark brainliest

Physics

2 answers:

earnstyle [38]2 years ago

7 0

It will be unstable system because it will not be able to recover from the disturbance

anzhelika [568]2 years ago

5 0

8). first choice choice

9). 7.0 ohms

You might be interested in

If an orange falls off the table, what causes it to stop?

emmasim [6.3K]

The floor or chair absorbing the force of the impact from the orange causes the orange to stop.

7 0

2 years ago

A portable x-ray unit has a step-up transformer. The 120 V input is transformed to the 100 kV output needed by the x-ray tube. T

slega [8]

Answer:

$N_s\approx41667 \hspace{3}lo ops$

Explanation:

In an ideal transformer, the ratio of the voltages is proportional to the ratio of the number of turns of the windings. In this way:

$\frac{V_p}{V_s} =\frac{N_p}{N_s} \\\\Where:\\\\V_p=Primary\hspace{3} Voltage\\V_s=V_p=Secondary\hspace{3} Voltage\\N_p=Number\hspace{3} of\hspace{3} Primary\hspace{3} Windings\\N_s=Number\hspace{3} of\hspace{3} Secondary\hspace{3} Windings$

In this case:

$V_p=120V\\V_s=100kV=100000V\\N_p=50$

Therefore, using the previous equation and the data provided, let's solve for $N_s$ :

$N_s=\frac{N_p V_s}{V_p} =\frac{(50)(100000)}{120} =\frac{125000}{3} \approx41667\hspace{3}loo ps$

Hence, the number of loops in the secondary is approximately 41667.

3 0

3 years ago

Electrical current is defined as

Dmitry_Shevchenko [17]

Explanation:

The flow of electrical charge per unit time

8 0

2 years ago

Car and a train move together along straight, parallel paths with the same constant cruising speed v0. At t=0 the car driver not

musickatia [10]

Answer:

Part A: $t = v_0/a_0$

Part B: $t = v_0/a_0$

Part C: $v_0^2/a_0$

Explanation:

Part A:

We will use the following kinematics equation:

$v = v_0 + at\\0 = v_0 - a_0t\\t = \frac{v_0}{a_0}$

Part B:

We will use the same kinematics equation:

$v = v_0 + at\\v_0 = 0 + a_0t\\t = \frac{v_0}{a_0}$

Part C:

The total time takes is 2t.

So the train moves a distance of

$x = v_0(2t) = 2v_0(\frac{v_0}{a_0}) = \frac{2v_0^2}{a_0}$

And the car moves a distance in Part A and in Part B:

$d_A = v_0t + \frac{1}{2}at^2 = v_0(\frac{v_0}{a_0}) - \frac{1}{2}a_0(\frac{v_0^2}{a_0^2}) = \frac{v_0^2}{a_0} - \frac{v_0^2}{2a_0} = \frac{v_0^2}{2a_0}\\d_B = v_0t + \frac{1}{2}at^2 = 0 + \frac{1}{2}a_0(\frac{v_0^2}{a_0^2}) = \frac{v_0^2}{2a_0}$

So the total distance that the car traveled is $d = \frac{v_0^2}{a_0}$

The difference between the train and the car is

$x - d = \frac{2v_0^2}{a_0} - \frac{v_0^2}{a_0} = \frac{v_0^2}{a_0}$

8 0

2 years ago

A tightrope walker is walking between two buildings holding a pole with length L=14.0 m, and mass mp=17.5 kg. The daredevil grip

Artemon [7]

Answer:

F = 32.28 N

Explanation:

For this exercise we must use the rotational equilibrium relation

Σ τ = 0

In the initial configuration it is in equilibrium, for which all the torque and forces are compensated. By the time the payment lands on the bar, we assume that the counter-clockwise turns are positive.

W_bird L / 2 - F_left 0.595 - F_right 0.595 = 0

we assume that the magnitude of the forces applied by the hands is the same

F_left = F_right = F

W_bird L / 2 - 2 F 0.595 = 0

F = $\frac{m_{bird} \ g L} { 4 \ 0.595}$

we calculate

F = 0.560 9.8 14.0 /2.38

F = 32.28 N

7 0

3 years ago