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-BARSIC- [3]
3 years ago
11

An unnamed professor is walking to class when an office chair (travelling in the same direction) collides with them. The profess

or fails into the chair and remains in the chair after collision. The professor, with a mass of 113kg, is travelling at 1.56 m/s. The chair has a velocity of 6.5 m/s, and a mass of 10kg. After the collision, and ignoring any effects of professor falling, what would be the speed of the chair professor system
Physics
1 answer:
ololo11 [35]3 years ago
5 0

Answer: 1.96 m/s

Explanation:

Given

Mass of Professor m_1=113\ kg

Velocity of professor u_1=1.56\ m/s

mass of chair m_2=10\ kg

velocity of chair u_2=6.5\ m/s

Suppose  after the collision, v is the common velocity

Conserving momentum

\Rightarrow m_1u_1+m_2u_2=(m_1+m_2)v\\\\\Rightarrow v=\dfrac{113\times 1.56+10\times 6.5}{113+10}=\dfrac{241.28}{123}\\\\\Rightarrow v=1.96\ m/s

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V = 1/3 Bh v = 1/3 (13 ac)(43560ft^2/ac)(481ft) v = 90793560 ft^3 * 0.3048m/ft * 0.3048m/ft * 0.3048m/ft = 2570987m^3
3 0
3 years ago
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Over 4 seconds, a car's momentum decreases by 1000 kg m/s how much force did it take to make this happen?
kotykmax [81]

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

6 0
3 years ago
Two point charges, a +45nC charge X and a +12nC charge Y are separated by a distance of 0.5m.
Gnoma [55]

A) Calculate the resultant electric field strength at the midpoint between the charges.

Qx is the charge at X and Qy is the charge at Y.

E at midpoint = k×Qx/0.25² - k×Qy/0.25²

k = 9×10⁹Nm²C⁻², Qx = 45nC, Qy = 12nC

E = 4752N/C

Well done.

B) Calculate the distance from X at which the electric field strength is zero.

Let D be some point between X and Y for which the net E field is 0.

Let d be the distance from X to D.

Set up the following equation:

E at D = k×Qx/d² - k×Qy/(0.5-d)² = 0

Do some algebra to solve for d:

k×Qx/d² = k×Qy/(0.5-d)²

Qx/d² = Qy/(0.5-d)²

Qx(0.5-d)² = Qyd²

(0.5-d)√Qx = d√Qy

0.5√Qx-d√Qx = d√Qy

d(√Qx+√Qy) = 0.5√Qx

d = (0.5√Qx)/(√Qx+√Qy)

Plug in Qx = 45nC, Qy = 12nC

d ≈ 330mm

C) Calculate the magnitude of the electric field strength at the point P on the diagram below.

First determine the angles of the triangle. The sides of the triangle are 0.3m, 0.4m, and 0.5m, so this is a right triangle where the angle between the 0.3m and 0.4m sides is 90°

∠Y = tan⁻¹(0.4/0.3) = 53.13°

∠X = 90-∠Y = 36.87°

Determine the horizontal component of E at P:

Ex = E from Qx × cos(∠X) - E from Qy × cos(∠Y)

Ex = k×Qx/0.4²×cos(36.87°) - k×Qy/0.3²×cos(53.13°)

Ex = 1305N/C

Determine the vertical component of E at P:

Ey = E from Qx × sin(∠X) - E from Qy × sin(∠Y)

Ey = k×Qx/0.4²×sin(36.87°) - k×Qy/0.3²×sin(53.13°)

Ey = 2479N/C

Use the Pythagorean theorem to determine the magnitude of E at P:

E = √(Ex²+Ey²)

E ≈ 2802N/C

4 0
3 years ago
A​ heavy-duty shock absorber is compressed 4 cm from its equilibrium position by a mass of 700nbspkg. How much work is required
tia_tia [17]

Answer:

Explanation:

A mass of 700 kg will exert a force of

700 x 9.8

= 6860 N.

Amount of compression x = 4 cm

= 4 x 10⁻² m

Force constant K = force of compression / compression

= 6860 / 4 x 10⁻²

= 1715 x 10² Nm⁻¹.

Let us take compression of r at any moment

Restoring force by spring

= k r

Force required to compress = kr

Let it is compressed  by small length dr during which force will remain constant.

Work done

dW =  Force x displacement

= -kr -dr

= kr dr

Work done to compress by length d

for it r ranges from 0 to -d

Integrating on both sides

W  = \int\limits^{-4}_0 {kr} \, dr

= [ kr²/2]₀^-4

= 1/2 kX16X10⁻⁴

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= 137.20 J

3 0
3 years ago
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Rasek [7]

Answer:

el plomo será el más largo

Explanation:

Dado que;

longitud inicial (l1) = 4m

Longitud final l2

aumento de temperatura (θ) = 10 ° C

Coeficiente de expansión lineal α

Ahora para el hierro;

α = 11,7 x 10-6

Desde;

l2-l / l1θ = α

l2 = α l1θ + l1

l2 = l1 (αθ + 1)

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l2 = 4.00044 m

Para el plomo

l2 = 4 ((27,3 x 10-6 * 10) + 1)

l2 = 4,00109 m

Para cobre

l2 = 4 ((16,7 x 10-6 * 10) + 1)

l2 = 4.000668 m

Por lo tanto, el plomo será el más largo

7 0
3 years ago
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