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Ostrovityanka [42]
3 years ago
6

Calculate the magnitude of the total impulse applied to the car to bring it to rest

Physics
1 answer:
n200080 [17]3 years ago
5 0

Answer:

Total impulse = mv = Initial momentum of the car

Explanation:

Let the mass of the car be 'm' kg moving with a velocity 'v' m/s.

The final velocity of the car is 0 m/s as it is brought to rest.

Impulse is equal to the product of constant force applied to an object for a very small interval. Impulse is also calculated as the total change in the linear momentum of an object during the given time interval.

The magnitude of impulse is the absolute value of the change in momentum.

|J|=|p_f-p_i|

Momentum of an object is equal to the product of its mass and velocity.

So, the initial momentum of the car is given as:

p_i=mv

The final momentum of the car is given as:

p_f=m(0)=0

Therefore, the impulse is given as:

|J|=|p_f-p_i|=|0-mv|=|-mv|=mv

Hence, the magnitude of the impulse applied to the car to bring it to rest is equal to the initial momentum of the car.

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Three heavy rods are all made of the same uniform material. These rods have lengths 3 m, 4 m, and 5 m, and compose the sides of
Sonbull [250]

Answer:

[ 2.67 , 1 ] m

Explanation:

Given:-

- The side lengths of the rods are as follows:

                             a = 4 m , b = 4 m , c = 5 m

                             a = Base , b = Perpendicular , c = Hypotenuse

- All rods are made of same material with uniform density. With  

Find:-

Find the coordinates of the center of mass of the triangle.

Solution:-

- The center of mass of any triangle is at the intersection of its medians.

- So let’s say we have a triangle with vertices at points (0,0) , (a,0) , and (0,b).

  • Median from (0,0) to midpoint (a/2,b/2) of opposite side has equation:

                                       bx−ay=0

  • Median from (a,0) to midpoint (0,b/2) of opposite side has equation:

                                      bx+2ay=ab

  • Median from (0,b) to midpoint (a/2,0) of opposite side has equation:

                                     2bx+ay=ab

  • Solve all three equations simultaneously:

                                     bx−ay=0  , bx = ay

                                     ay + 2ay = ab , 3ay = ab , y = b/3

                                     bx = b/3

                                     x = a / 3

  • So the distance from the median to each leg of the triangle is 1/3 length of other leg.

- So the coordinates of the centroid for right angle triangle would be:

                                   [ 2a/3 , b/3 ]

                                   [ 2.67 , 1 ] m

                                                         

                                 

3 0
3 years ago
A generator with �# ' = 300 V and Zg = 50 Ω is connected to a load ZL = 75 Ω through a 50-Ω lossless line of length l = 0.15λ. (
ki77a [65]

Answer:

a. Zin = 41.25 - j 16.35 Ω

b. V₁ = 143. 6 e⁻ ¹¹ ⁴⁶

c.  Pin = 216 w

d. PL = Pin = 216 w

e. Pg = 478.4 w , Pzg = 262.4 w

Explanation:

a.

Zin = Zo * [ ZL + j Zo Tan (βl) ] / [ Zo + j ZL Tan (βl) ]  

βl = 2π / λ * 0.15 λ = 54 °

Zin = 50 * [ 75 + j 50 Tan (54) ] / [ 50 + j 75 Tan (54) ]

Zin = 41.25 - j 16.35 Ω

b.

I₁ = Vg / Zg + Zin ⇒ I₁ = 300 / 41.25 - j 16.35 = 3.24 e ¹⁰ ¹⁶

V₁ = I₁ * Zin = 3.24 e ¹⁰ ¹⁶ * ( 41.25 - j 16.35)

V₁ = 143. 6 e⁻ ¹¹ ⁴⁶

c.

Pin = ¹/₂ * Re * [V₁ * I₁]

Pin = ¹/₂ * 143.6 ⁻¹¹ ⁴⁶ * 3.24 e ⁻ ¹⁰ ¹⁶ = 143.6 * 3.24 / 2 * cos (21.62)

Pin = 216 w

d.

The power PL and Pin are the same as the line is lossless input to the line ends up in the load so

PL = Pin

PL = 216 w

e.

Pg Generator

Pg = ¹/₂ * Re * [ V₁ * I₁ ] = 486 * cos (10.16)

Pg = 478.4 w

Pzg dissipated

Pzg = ¹/₂ * I² * Zg = ¹/₂ * 3.24² * 50

Pzg = 262.4 w

4 0
3 years ago
The acceleration of an object as a function of time is given by a(t) = (1.00 m/s2)t2. If displacement of the object between time
jolli1 [7]

not enough information is given to determine the velocity of the object at time to=0.00s

3 0
3 years ago
If a galaxy has an apparent velocity of 2300 km/s, what is its distance if the Hubble constant is assumed to be 70 km/s/Mpc
prohojiy [21]

The distance of the galaxy is 32.86 Mpc.

Using the hubble law, v = H₀D where v = apparent velocity of galaxy = 2300 km/s, H = hubble constant = 70 km/s/Mpc and D = distance of galaxy.

Since we require the distance of the galaxy, we make D subject of the formula in the equation. So, we have

D = v/H₀

Substituting the values of the variables into the equation, we have

D = 2300 km/s ÷ 70 km/s/Mpc

D = 32.86 Mpc

So, the distance of the galaxy is 32.86 Mpc

Learn more about hubble law here:

brainly.com/question/18484687

4 0
2 years ago
A light ray travels in the +x direction and strikes a slanted surface with an angle of 62° between its normal and the ty axis. T
Elena-2011 [213]

Answer:

- 0.6

Explanation:

Given that angle between normal y axis is 62° so angle between  normal

and x axis will be 90- 62 = 28 °. Since incident ray is along x axis , 28 ° will be the angle between incident ray and normal ie it will be angle of incidence

Angle of incidence = 28 °

angle of reflection = 28°

Angle between incident ray and reflected ray = 28 + 28 = 56 °

Angle between x axis and reflected ray = 56 °

x component of reflected ray

= - cos 56 ( it will be towards - ve x axis. )

- 0.6

6 0
2 years ago
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