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3 years ago

If I walk 30 m north and turn around and walk 10 m to the south, what is the magnitude

Physics

2 answers:

Karo-lina-s [1.5K]3 years ago

8 0

Answer:

40m

Explanation:

Your total displacement would be 40m! Let me explain.

Going 30m north adds 30 to 0... so your total right now is 30m.

if you turn around and walk 10 m you have to add that 10 to the 30.

so your total displacement is 40m.

miss Akunina [59]3 years ago

5 0

Answer:

Hey

it would be 20 m

If you walk away 30 m then come back 10 you are now 20 m away from your starting point.

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Question No 1 Find the voltage drop across 24 ohm resistor and current flowing through 22 ohm resistor in the given circuit as s

lord [1]

Answer:

8.25 V

Explanation:

We can ignore the 22Ω and 122Ω resistors at the bottom. Since there's a short across those bottom nodes, any current will go through the short, and none through those two resistors.

The 2Ω resistor and the 44Ω resistor are in parallel. The equivalent resistance is:

1 / (1 / (2Ω) + 1 / (44Ω)) = 1.913Ω

This resistance is in series with the 12Ω resistor. The equivalent resistance is:

1.913Ω + 12Ω = 13.913Ω

This resistance is in parallel with the 24Ω resistor. The equivalent resistance is:

1 / (1 / (13.913Ω) + 1 / (24Ω)) = 8.807Ω

Finally, this resistance is in series with the 4Ω resistor. The equivalent resistance of the circuit is:

8.807Ω + 4Ω = 12.807Ω

The current through the battery is:

12 V / 12.807Ω = 0.937 A

The voltage drop across the 4Ω resistor is:

(0.937 A) (4Ω) = 3.75 V

So the voltage between the bottom nodes and the top nodes is:

12 V − 3.75 V = 8.25 V

4 0

3 years ago

A person might be using social media in an unhealthy way if:

Rus_ich [418]

Answer:

they are comparing themselves to other ppl

5 0

2 years ago

Read 2 more answers

an object is moving in a straight line with a constant acceleration initially it is traveling at 16 meters per second. 3 seconds

VLD [36.1K]

A)It moved 6 meters.

B)It would 3 meters.

I think this is the answer.

Have a good day.

4 0

3 years ago

Certain insects can achieve seemingly impossible accelerations while jumping. the click beetle accelerates at an astonishing 400

hichkok12 [17]

(a) The launching velocity of the beetle is 6.4 m/s

(b) The time taken to achieve the speed for launch is 1.63 ms

(a) The beetle starts from rest and accelerates with an upward acceleration of 400 g and reaches its launching speed in a distance 0.53 cm. Here g is the acceleration due to gravity.

Use the equation of motion,

$v^2=u^2+2as$

Here, the initial velocity of the beetle is u, its final velocity is v, the acceleration of the beetle is a, and the beetle accelerates over a distance s.

Substitute 0 m/s for u, 400 g for a, 9.8 m/s² for g and 0.52×10⁻²m for s.

$v^2=u^2+2as\\ = (0 m/s)^2+2 (400)(9.8 m/s^2)(0.52*10^-^2 m)\\ =40.768 (m/s)^2\\ v=6.385 m/s$

The launching speed of the beetle is 6.4 m/s.

(b) To determine the time t taken by the beetle for launching itself upwards is determined by using the equation of motion,

$v=u+at$

Substitute 0 m/s for u, 400 g for a, 9.8 m/s² for g and 6.385 m/s for v.

$v=u+at\\ 6.385 m/s = (0 m/s) +400(9.8 m/s^2)t\\ t = \frac{6.385 m/s}{3920 m/s^2} = 1.63*10^-^3s=1.63 ms$

The time taken by the beetle to launch itself upwards is 1.62 ms.

(c) After the beetle launches itself upwards, it is acted upon by the earth's gravitational force, which pulls it downwards towards the earth with an acceleration equal to the acceleration due to gravity g. Its velocity reduces and when it reaches the maximum height in its path upwards, its final velocity becomes equal to zero.

Use the equation of motion,

$v^2=u^2+2as$

Substitute 6.385 m/s for u, -9.8 m/s² for g and 0 m/s for v.

$v^2=u^2+2as\\ (0m/s)^2=(6.385 m/s)^2+2(-9.8m/s^2)s\\ s=\frac{(6.385 m/s)^2}{2(9.8m/s^2)} =2.08 m$

The beetle can jump to a height of 2.1 m

7 0

2 years ago

The length of a certain wire is kept same while its radius is doubled. what is the new resistivity of this wire?

anastassius [24]

The text does not specify whether the resistance R of the wire must be kept the same or not: here I assume R must be kept the same.

The relationship between the resistance and the resistivity of a wire is
$\rho = \frac{AR}{L}$
where
$\rho$ is the resistivity
A is the cross-sectional area
R is the resistance
L is the wire length

the cross-sectional area is given by
$A=\pi r^2$
where r is the radius of the wire. Substituting in the previous equation ,we find
$\rho = \frac{\pi r^2 R}{L}$

For the new wire, the length L is kept the same (L'=L) while the radius is doubled (r'=2r), so the new resistivity is
$\rho' = \frac{\pi r'^2 R}{L'}= \frac{\pi (2r)^2 R}{L}=4 \frac{\pi r^2 R}{L} = 4 \rho$
Therefore, the new resistivity must be 4 times the original one.

5 0

3 years ago

Read 2 more answers