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LenaWriter [7]
3 years ago
12

What is force x time is called

Physics
1 answer:
Law Incorporation [45]3 years ago
4 0

Answer:

In words, it could be said that the force times the time equals the mass times the change in velocity. In physics, the quantity Force • time is known as impulse. And since the quantity m•v is the momentum, the quantity m•Δv must be the change in momentum. The equation really says that the Impulse = Change in momentum.

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Two charges are located in the x – y plane. If ????1=−4.10 nC and is located at (x=0.00 m,y=0.600 m) , and the second charge has
faust18 [17]

Answer:

The x-component of the electric field at the origin = -11.74 N/C.

The y-component of the electric field at the origin = 97.41 N/C.

Explanation:

<u>Given:</u>

  • Charge on first charged particle, q_1=-4.10\ nC=-4.10\times 10^{-9}\ C.
  • Charge on the second charged particle, q_2=3.80\ nC=3.80\times 10^{-9}\ C.
  • Position of the first charge = (x_1=0.00\ m,\ y_1=0.600\ m).
  • Position of the second charge = (x_2=1.50\ m,\ y_2=0.650\ m).

The electric field at a point due to a charge q at a point r distance away is given by

\vec E = \dfrac{kq}{|\vec r|^2}\ \hat r.

where,

  • k = Coulomb's constant, having value \rm 8.99\times 10^9\ Nm^2/C^2.
  • \vec r = position vector of the point where the electric field is to be found with respect to the position of the charge q.
  • \hat r = unit vector along \vec r.

The electric field at the origin due to first charge is given by

\vec E_1 = \dfrac{kq_1}{|\vec r_1|^2}\ \hat r_1.

\vec r_1 is the position vector of the origin with respect to the position of the first charge.

Assuming, \hat i,\ \hat j are the units vectors along x and y axes respectively.

\vec r_1=(0-x_1)\hat i+(0-y_1)\hat j\\=(0-0)\hat i+(0-0.6)\hat j\\=-0.6\hat j.\\\\|\vec r_1| = 0.6\ m.\\\hat r_1=\dfrac{\vec r_1}{|\vec r_1|}=\dfrac{0.6\ \hat j}{0.6}=-\hat j.

Using these values,

\vec E_1 = \dfrac{(8.99\times 10^9)\times (-4.10\times 10^{-9})}{(0.6)^2}\ (-\hat j)=1.025\times 10^2\ N/C\ \hat j.

The electric field at the origin due to the second charge is given by

\vec E_2 = \dfrac{kq_2}{|\vec r_2|^2}\ \hat r_2.

\vec r_2 is the position vector of the origin with respect to the position of the second charge.

\vec r_2=(0-x_2)\hat i+(0-y_2)\hat j\\=(0-1.50)\hat i+(0-0.650)\hat j\\=-1.5\hat i-0.65\hat j.\\\\|\vec r_2| = \sqrt{(-1.5)^2+(-0.65)^2}=1.635\ m.\\\hat r_2=\dfrac{\vec r_2}{|\vec r_2|}=\dfrac{-1.5\hat i-0.65\hat j}{1.634}=-0.918\ \hat i-0.398\hat j.

Using these values,

\vec E_2= \dfrac{(8.99\times 10^9)\times (3.80\times 10^{-9})}{(1.635)^2}(-0.918\ \hat i-0.398\hat j) =-11.74\ \hat i-5.09\ \hat j\  N/C.

The net electric field at the origin due to both the charges is given by

\vec E = \vec E_1+\vec E_2\\=(102.5\ \hat j)+(-11.74\ \hat i-5.09\ \hat j)\\=-11.74\ \hat i+(102.5-5.09)\hat j\\=(-11.74\ \hat i+97.41\ \hat j)\ N/C.

Thus,

x-component of the electric field at the origin = -11.74 N/C.

y-component of the electric field at the origin = 97.41 N/C.

4 0
3 years ago
A box is sitting on a 2 m long board at one end. A worker picks up the board at the end with the box so it makes an angle with t
Vlada [557]

Answer:

Explanation:

When the box is on the ramp , component of its weight along the ramp

= mg sinθ

Friction force acting on it in upward direction

=μ mg cosθ

For sliding

μ mg cosθ < mg sinθ

μ cosθ < sinθ

.5 x cos35 < sin35

.41 < .57

So the box will slide

When sliding starts , kinetic friction acts

Net force in downward direction

mgsinθ - μ mg cosθ

acceleration

= gsinθ - μ g cosθ

= 5.62 - .3 x 9.8 x cos35

= 5.62 - 2.4

= 3.22 m /s²

3 0
2 years ago
A car is running at a velocity of 50 miles/hour and the driver accelerates the car by 10miles/hour.How far the car travels from
dezoksy [38]
Initial velocity u = 50 miles/hour
acceleration a = 10 miles/hour
Time t = 2 hours
Distance travelled S = ut + (at^2)/2
Substituting the values in the second equation of motion,
S = 50*2 + (10 * 2 *2)/2
S = 100 + 20
S = 120 miles
Therefore the distance travelled by the car in the next two hours is 120 miles
4 0
3 years ago
Read 2 more answers
Conductors of large instrumental ensembles use thin stick called a __________ to help performers keep time.
Dmitry [639]
I think the word is "baton"
6 0
3 years ago
Read 2 more answers
Water is leaking out of an inverted conical tank at a rate of 10,500 cm3/min at the same time that water is being pumped into th
satela [25.4K]

The tank has a volume of \dfrac\pi3R^2H, where H=6\,\rm m is its height and R=\dfrac d2=2\,\rm m is its radius.

At any point, the water filling the tank and the tank itself form a pair of similar triangles (see the attached picture) from which we obtain the following relationship:

\dfrac26=\dfrac rh\implies r=\dfrac h3

The volume of water in the tank at any given time is

V=\dfrac\pi3r^2h

and can be expressed as a function of the water level alone:

V=\dfrac\pi3\left(\frac h3\right)^2h=\dfrac\pi{27}h^3

Implicity differentiating both sides with respect to time t gives

\dfrac{\mathrm dV}{\mathrm dt}=\dfrac\pi9h^2\,\dfrac{\mathrm dh}{\mathrm dt}

We're told the water level rises at a rate of \dfrac{\mathrm dh}{\mathrm dt}=20\,\frac{\rm cm}{\rm min} at the time when the water level is h=2\,\mathrm m=200\,\mathrm{cm}, so the net change in the volume of water \dfrac{\mathrm dV}{\mathrm dt} can be computed:

\dfrac{\mathrm dV}{\mathrm dt}=\dfrac\pi9(200\,\mathrm{cm})^2\left(20\,\dfrac{\rm cm}{\rm min}\right)=\dfrac{800,000\pi}9\,\dfrac{\mathrm{cm}^3}{\rm min}

The net rate of change in volume is the difference between the rate at which water is pumped into the tank and the rate at which it is leaking out:

\dfrac{\mathrm dV}{\mathrm dt}=(\text{rate in})-(\text{rate out})

We're told the water is leaking out at a rate of 10,500\,\frac{\mathrm{cm}^3}{\rm min}, so we find the rate at which it's being pumped in to be

\dfrac{800,000\pi}9\,\dfrac{\mathrm{cm}^3}{\rm min}=(\text{rate in})-10,500\,\dfrac{\mathrm{cm}^3}{\rm min}

\implies\text{rate in}\approx289,753\,\dfrac{\mathrm{cm}^3}{\rm min}

4 0
3 years ago
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