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miss Akunina [59]
3 years ago
15

If the time of impact in a collision is extended by 3 times, by hour much is

Physics
1 answer:
vodomira [7]3 years ago
4 0

Answer:

The force of the impact would be smaller

Explanation:                                                          Examples:

If the force is big then the time would be small (2500N of Force = 10 seconds)

If the force is small then the time would be big (250N of Force = 50 seconds)

Impulse/Collision -> [Ft] = [M (vf-vo)] <- Change in momentum

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A new landowner has a triangular piece of flat land she wishes to fence. Starting at the first corner, she measures the first si
Lera25 [3.4K]

Answer:

Length of third side = 3.97m

Explanation:

First of all, let we draw the triangle from the given information assuming our first corner is A. second corner is B and third corner is C.

From A-B we have distance = 5.5m = Say it c

From B-C we have distance = 4.3m = Say it a

From A-C we have distance = ? = Say it b which we have to find out.

Using the Law of Cosines: The square of the unknown side equals to the sum of squares of other 2 sides and subtracting 2*(Product of other sides)*(Cos(Angle opposite to the unknown side)

For our case it is:

b² = a² + c² - 2acCos(B)         -  Say it equation 1

From the attached triangle you may see that, a & c are our known sides and B is the angle opposite to the side b.

There values are:

a = 4.3m;  c = 5.5m ; Angle B = 0.8 rad = 0.8 * 57.3 = 45.84 degrees where 1 rad = 57.3 degrees

Now by putting the respectve values in equation 1 we have:

b² = (4.3)² + (5.5)² - 2*4.3*5.5*Cos(45.84)

b² = 18.49 + 30.25 - 32.95

b² = 15.79

b  = √15.79

b = 3.97m

Thus the length of third side is 3.97m.

PS: The picture of triangle is being attached for yours understanding.

6 0
3 years ago
Three point charges are placed on the x−y plane: a + 50.0-nC charge at the origin, a −50.0-nC charge on the x axis at 10.0 cm, a
butalik [34]

Answer:

(a) F = 0.00322i - 0.00793j with magnitude |F| = 0.00856N

(b) E = -42846.7 N/C

Explanation:

The diagram attached below explains some parameters.

Parameters given:

Charge Q1 = +50 nC at point (0, 0)

Charge Q2 = -50 nC at point (0.1, 0)

Charge Q3 = +150 nC at point (0.1, 0.08)

* The distances are in meters.

(a) The total electric force on the charge Q3 due to Q1 and Q2 is the vector sum of the forces due to Q1 and Q2. Mathematically,

F = F1 + F2

FORCE DUE TO Q1 i.e. F(Q1, Q3)

We have to find the x and y components.

From the diagram, we can find θ using SOHCAHTOA:

θ = tan⁻¹ (0.08/0.1)

θ = 38.66⁰

The distance between Q1 and Q3 can be found using Pythagoras theorem:

x² = 0.08² + 0.1²

x = 0.128 m

F1 = Fx(Q1, Q3)i + Fy(Q1, Q3)j

F1 = iF(Q1, Q3)cosθ + jF(Q1, Q3)sinθ

F(Q1, Q3) = (k * Q1 * Q3) / r²

k = Coulombs constant

F(Q1, Q3) = (9 * 10⁹ * 50 * 10⁻⁹ * 150 * 10⁻⁹) /(0.128)²

F(Q1, Q3) = 0.00412N

F1 = i0.00412 * cos38.66 + j0. 00412 * sin38.66

F1 = 0.00322i + 0.00257j N

FORCE DUE TO Q2 i.e. F(Q2, Q3)

We have to find the x and y components.

F2 = Fx(Q2, Q3)i + Fy(Q2, Q3)j

F2 = iF(Q2, Q3)cos90 + jF(Q2, Q3)cos0

F(Q2, Q3) = (k * Q2 * Q3) / r²

F(Q2, Q3) = (9 * 10⁹ * -50 * 10⁻⁹ * 150 * 10⁻⁹) /(0.08)²

F(Q2, Q3) = -0.0105N

F2 = -i0.0105 * cos90 - j0.0105 * cos0

F2 = - 0.0105j N

Hence, the total force will be

F = F1 + F2

F = 0.00322i + 0.00257j - 0.0105j

F = 0.00322i - 0.00793j N

The magnitude of this force is:

|F| = √(0.00322² + (-0.00793²)

|F| = 0.00856N

(b) The electric field at charge Q3 is the sum of the electric fields due to Q1 and Q2:

E = E1 + E2

E1, electric field due to Q1 = kQ1/r²

E1 = (9 * 10⁹ * 50 * 10⁻⁹) / (0.128²)

E1 = 27465.8 N/C

E2, electric field due to Q2 = (9 * 10⁹ * -50 * 10⁻⁹) / (0.08²)

E1 = -70312.5N/C

The total electric field:

E = E1 + E2

E = 27465.8 - 70312.5

E = -42846.7 N/C

3 0
3 years ago
If the sun in 148 million kilometers from the earth, how many minutes will it take the light from the sun to reach the earth?
Natasha2012 [34]

Answer:

about 8 mins and 20 secs

Explanation:

8 0
3 years ago
How to make snf62- lewis structure in aleks?
umka21 [38]
 Sn is in group 4 so has 4 electrons in the outer shell (draw<span> these as dots) Add 1 more electron from each bond to F (+6) </span>
4 0
3 years ago
A force parallel to the x-axis acts on a particle moving along the x-axis. This force produces potential energy U(x) given by U(
noname [10]

Answer:

Explanation:

U(x)=αx⁴

Force F = - dU / dx

= 4αx³

= 4 x .75 x³

= 3 x³

Force when x = .8 m

F =  3 (0.8)³

= 1.54 N

8 0
4 years ago
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