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Charra [1.4K]
3 years ago
7

You can jump 3 m (10 feet) off of a diving board and into a swimming pool and be uninjured. However, if you jumped from the same

height onto concrete it would be very bad. Use the principles of impulse and momentum to explain this.
2) A 125 kg linebacker running to the right at 2.5 m/s and an 80 kg quarterback running to the left at 3 m/s collide in midair. The linebacker holds on to the quarterback and they move off together. What is their velocity (magnitude and direction) after this collision?
Physics
1 answer:
Mrac [35]3 years ago
7 0

Answer:

1) this is due to force due to momentum change of a body falling from such a height.

2) they move with velocity of 2.69 m/s towards the right.

Explanation:

1) in the case where one jumps 3 meters into water, the water deform on impact extending the time of impact for your body's momentum to get to zero, so that the time it takes for the momentum of your body due to your mass and your velocity to get to zero is relatively long lessening the force felt due to the impact.

For a fall from the same height onto a concrete, the concrete does not deform at all. This means that there is a very short time for your momentum to get to zero and hence the force is huge. In order to cushion this a little, your body deforms in a bid to reduce the force and this is very deadly.

This is in line with Newton's second law of motion which states that the rate of change of momentum is proportional to the force produced.

2)

M1 = 125 kg

M2 = 80 kg

V1 = 2.5 m/s

V2 = 3 m/s

From momentum conservation, total initial momentum must be equal to the total momentum after collision.

Momentum = mass x velocity.

That is,

M1V1 + M2V2 = (M1 + M2) Vf

Where Vf is their final velocity together.

Note that the masses stick together because the collision is inelastic.

Substituting and solving, we have

(125 x 2.5) + (80 x 3) = (125 + 80)Vf

312.5 + 240 = 205Vf

552.5 = 205Vf

Vf = 552.5/205 = 2.69 m/s

Note that the linebacker runner has the greater momentum 125 x 2.5 = 312.15

This means they both move to the right at velocity of 2.69 m/s

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You watch a distant lady driving nails into her front porch at a regular rate of 1 stroke per second. You hear the sound of the
Igoryamba

Answer:

λ = 1360 m

Explanation:

Given data:

frequency of driving nails is given as 1 stroke per second mean at every 0.25 sec she hit the nails

speed of sound is given as 340 m/s

we know that the wave equation is given as

Speed = frequency × wavelength,

v = f × λ

where,

v = speed in meters/second (m/s)

f = frequency in Hertz (Hz)

substituing value to get wavelength  of her driving nails

340 m/s = (1Hz)\times  \lambda

\lambda = \frac{340}{0.25}

λ = 1360 m

4 0
3 years ago
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How much work must be done to bring three electrons from a great distance apart to 3.0×10−10 m from one another (at the corners
Natasha_Volkova [10]

Answer:

Potential\ Energy=Work \ Done=2.301*10^{-18} J

Work  done to bring three electrons from a great distance apart to 3.0×10−10 m from one another (at the corners of an equilateral triangle) is 2.301*10^{-18} Joules

Explanation:

The potential energy is given by:

U=Q*V

where:

Q is the charge

V is the potential difference

Potential Difference=V=\frac{kq}{r}

So,

Potential\ Energy=\frac{Qkq}{r} \\Q=q\\Potential\ Energy=\frac{kq^2}{r}

Where:

k is Coulomb Constant=8.99*10^9 Nm^2/C^2

q is the charge on electron=-1.6*10^-19 C

r is the distance=3.0*10^{-10}m

For 3 Electrons Potential Energy or work Done is:

Potential\ Energy=3*\frac{kq^2}{r}

Potential\ Energy=3*\frac{(8.99*10^9)(-1.6*10^{-19})^2}{3*10^{-10}}\\Potential\ Energy=2.301*10^{-18} J

Work  done to bring three electrons from a great distance apart to 3.0×10−10 m from one another (at the corners of an equilateral triangle) is 2.301*10^{-18} Joules

7 0
3 years ago
A container in the shape of a cube 10.0 cm on each edge contains air (with equivalent molar mass 28.9 g/mol) at atmospheric pres
Vikentia [17]

Answer:

a) m = 1.174 grams

b) F_g = 0.01151 N

c) F_c = 1013 N

Explanation:

Given:

- The length of a cube L = 10.0 cm

- The molar mass of air M = 28.9 g/mol

- Pressure of air P = 101.3 KPa

- Temperature of air T = 300 K

- Universal Gas constant R = 8.314 J/kgK

Find:

(a) the mass of the gas

(b) the gravitational force exerted on it

(c) the force it exerts on each face of the cube

(d) Why does such a small sample exert such a great force? (6%)

Solution:

- Compute the volume of the cube:

                               V = L^3  = 0.1^3 = 0.001 m^3

- Use Ideal gas law equation and compute number of moles of air n:

                               P*V = n*R*T

                                n = P*V / R*T

                                n = 101.3*10^3 * 0.001 / 8.314*300

                                n = 0.04061 moles

- Compute the mass of the gas:

                                m = n*M

                                m = 0.04061*28.9

                                m = 1.174 grams

- The gravitational force exerted on the mass of gas is due to its weight:

                                F_g = m*g

                                F_g = 1.174*9.81*10^-3

                               F_g = 0.01151 N

- The force exerted on each face of cube is due its surface area:

                                F_c = P*A

                                F_c = (101.3*10^3)*(0.1)^2

                                F_c = 1013 N

- The molecules of a gas have high kinetic energy; hence, high momentum. When they collide with the walls they transfer momentum per unit time as force. Higher the velocity of the particles higher the momentum higher the force exerted.

4 0
3 years ago
9. To completely describe the motion of an object, you need
pav-90 [236]
I think it’s D) all of the above
4 0
3 years ago
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In the diagram, q1 = +6.60*10^-9 C and q2 = +3.10*10^-9 C. Find the magnitude of the total electric field at point P. pls help?
kirza4 [7]

Answer:

|E(t)| = 1258,46 [N/C]

α = 67,5⁰  (angle with respect x-axis)

Explanation:

E(t)  Electric Field is a vector, so we need to determine module and direction

E(t)  =  E(q₁)  + E (q₂)  Where E(q₁) and E (q₂) are electric fields due to electric charge q₁ and q₂  respectively.

E(q₁) = K * q₁/ (d₁)²         K = 9 *10⁹   [N*m²/C²]    d₁ = 0,350 m

E(q₁) = 9 *10⁹ * 6,6*10⁻⁹ / 0,1225      [N*m²/C²] *C/m²

E(q₁) = 484,9 [N/C]

E(q₂) =  9 *10⁹ * 3,1*10⁻⁹ / 0,024025

E(q₂) = 1161,29

Then

|E(t)| = √ |Eq₁|² + |Eq₂|²

|E(t)| = √ ( 484,9)² +( 1161,29)²

|E(t)| = √ 235128 + 1348594,46

|E(t)| = 1258,46 [N/C]

And tanα = 1161,29/484,9        tanα =  2,395      α = 67,5⁰

The angle of the vector electric field with the x-axis

3 0
3 years ago
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