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3 years ago

Over 4 seconds, a car's momentum decreases by 1000 kg m/s how much force did it take to make this happen?

Physics

1 answer:

kotykmax [81]3 years ago

6 0

Answer:

250N

Explanation:

Given parameters:

Time = 4s

Momentum = 1000kgm/s

Unknown:

Force = ?

Solution:

To solve this problem, we use Newton's second law of motion;

Ft = Momentum

F is the force

t is the time

So;

F x 4 = 1000kgm/s

F = 250N

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What is the overall charge of the wall?

NNADVOKAT [17]

Answer:

The overall charge on the wall is zero.

Explanation:

Because the atoms making the wall are neutral i.e. number of positive charges are equal to number of negative charges.

8 0

3 years ago

ASAP please!! <br>The force vectors on an aircraft are as shown. Find the net (resultant force).

ioda

Answer:

The magnitude of the net force is 5430N

Explanation:

I suggest to define the axes as aligned to the axis of the plane. This will require you to decompose only one vector, namely the Weight. We need two components of the W force: one in horizontal direction of the plane, the other perpendicular to it. Through a simple triangle argument you will se that the plane-horizontal component of W is

$W_D=3600 N\cdot\sin 27^\circ=1634N$

acting in the direction of the Drag, and the plane-perpendicular component is:

$W_L=-3600N\cdot\cos 27^\circ=-3208N$

with negative sign since it counteracts the Lift.

So the components of the netforce F are:

$F_h=T-D-W_D=(8000-1000-1634)N=5366N\\F_v=L+W_L=(4100-3208)N=829N$

The magnitude of the net force is:

$|F|=\sqrt{5366^2+829^2}N = 5430N$

6 0

3 years ago

An electric motor has an effective resistance of 22 22 and an inductive reactance of 72 12 when working under load. The voltage

Alecsey [184]

Answer:

Current amplitude, I = 5.57 A

Explanation:

It is given that,

Effective resistance, R = 22 ohms

Inductive reactance, L = 72 ohms

Voltage of the alternating source, V = 420 volt

The total impedance of the RL circuit is given by :

$Z=\sqrt{R^2+L^2}$

$Z=\sqrt{(22)^2+(72)^2}$

Z = 75.28 ohms

From Ohm's law,

$V=I\times Z$

$I=\dfrac{V}{Z}$

$I=\dfrac{420}{75.28}$

I = 5.57 A

So, the current amplitude of the electric motor is 5.57 A. Hence, this is the required solution.

6 0

4 years ago

Consider two point charges located on the x axis: one charge, q1= -11.0 nC, is located at x1= -1.675 m; the second charge, q2= 3

Mekhanik [1.2K]

Answer:

Please refer to the figure.

q1 is a negative charge, and q2 and q3 are positive charges. So, the force exerted by q1 on q3 is attractive, and the force exerted by q2 on q3 is repulsive, which means F13 is directed towards left, and F23 is also directed towards left.

$F_{13} = \frac{1}{4\pi \epsilon_0}\frac{q_1 q_3}{r_1^2} = \frac{1}{4\pi \epsilon_0}\frac{11\times 10^{-9}\times 47.5\times10^{-9}}{(-1.675 - (-1.18))^2} = 1.92\times 10^{-5}N$

$F_{23} = \frac{1}{4\pi\epsilon_0}\frac{q_2q_3}{r_2^2} = \frac{1}{4\pi\epsilon_0}\frac{31\times 10^{-9} \times 47.5\times 10^{-9}}{(-1.18 - 0)^2} = 9.5\times 10^{-6}$