1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
morpeh [17]
2 years ago
5

Two football players run towards each other along a straight path in Penrith Park in the clash between the Melbourne storms and

the Penrith Panthers a month ago. Melbourne's Justin Olam who is about 95kg and ran towards Viliame Kikau at 3.75m/s. Viliame Kikau is 111kg and moves towards Justin Olam at 4.10m/s. They end up in a head-on collision and are stuck together.
A) What is their velocity immediately after the collision?

B) What are the initial and final kinetic energies of the system?​
Physics
1 answer:
Agata [3.3K]2 years ago
3 0

Answer:

a) v = 0.4799 m / s,  b)  K₀ = 1600.92 J,    K_f = 5.46 J

Explanation:

a) How the two players collide this is a momentum conservation exercise. Let's define a system formed by the two players, so that the forces during the collision are internal and also the system is isolated, so the moment is conserved.

Initial instant. Before the crash

        p₀ = m v₁ + M v₂

where m = 95 kg and his velocity is v₁ = -3.75 m / s, the other player's data is M = 111 kg with velocity v₂ = 4.10 m / s, we have selected the direction of this player as positive

Final moment. After the crash

       p_f = (m + M) v

as the system is isolated, the moment is preserved

       p₀ = p_f

       m v₁ + M v₂ = (m + M) v

       v =\frac{m v_1 + M v_2}{m+M}

let's calculate

        v = \frac{ -95 \ 3.75 \ + 111 \ 4.10}{95+111}

        v = 0.4799 m / s

b) let's find the initial kinetic energy of the system

         K₀ = ½ m v1 ^ 2 + ½ M v2 ^ 2

         K₀ = ½ 95 3.75 ^ 2 + ½ 111 4.10 ^ 2

         K₀ = 1600.92 J

the final kinetic energy

         K_f = ½ (m + M) v ^ 2

         k_f = ½ (95 + 111) 0.4799 ^ 2

         K_f = 5.46 J

You might be interested in
Develop a hypothesis for why one of the two types of soup should indeed be rolling down faster than the other. This hypothesis s
Eva8 [605]

Answer:

Assume two identical cans filled with two types of soup having same mass are rolling down on an inclined plane in same conditions. In terms of inertia different types of soup will indicate different viscosity. The higher viscosity fillings indicates more part of the soup mass is rotating together with the can’s body. This means that for the can with lower viscosity soup has a lower moment of inertia and the can with higher viscosity has higher moment of inertia while the same gravity makes them to roll.

incline angle = θ ; can's mass =  m ; Radius of the can's = R , Angular acceleration for Can 1 = α1 ; Angular acceleration for Can 2 = α2

T1 = Inertia of Can with high viscosity soup

T2 = Inertia of Can with low viscosity soup

M1 rolling moment of Can 1

M2 rolling moment of Can 2

equation is given by

T1*α1 = M1   - (a)

T2*α2 = M2 - (b)

M1 = M2 = m*g*R*sin(θ). (c)

as assumed T1 > T2

from the three equation (a), (b) & (c)

the α2 > α1

Angular acceleration of Can 2 is higher than Can 1. Already stated that Can 1 has more viscous soup as compared to Can 2.

7 0
3 years ago
Explain how the mass of a planet affects the motion of the planet around the sun?
vodomira [7]

Answer:

The gravitational attraction of the Sun is what holds the planets in their elliptical orbits. So to explain this the mass effects the motion of the planets because the strength of gravitational force depends of the mass.

Explanation:

6 0
3 years ago
(1 point) A projectile is fired from ground level with an initial speed of 600 m/sec and an angle of elevation of 30 degrees. Us
konstantin123 [22]

Answer:

(a) The range of the projectile is 31,813.18 m

(b) The maximum height of the projectile is 4,591.84 m

(c) The speed with which the projectile hits the ground is 670.82 m/s.

Explanation:

Given;

initial speed of the projectile, u = 600 m/s

angle of projection, θ = 30⁰

acceleration due to gravity, g = 9.8 m/s²

(a) The range of the projectile in meters;

R = \frac{u^2sin \ 2\theta}{g} \\\\R = \frac{600^2 sin(2\times 30^0)}{9.8} \\\\R = \frac{600^2 sin (60^0)}{9.8} \\\\R = 31,813.18 \ m

(b) The maximum height of the projectile in meters;

H = \frac{u^2 sin^2\theta}{2g} \\\\H = \frac{600^2 (sin \ 30)^2}{2\times 9.8} \\\\H = \frac{600^2 (0.5)^2}{19.6} \\\\H = 4,591.84 \ m

(c) The speed with which the projectile hits the ground is;

v^2 = u^2 + 2gh\\\\v^2 = 600^2 + (2 \times 9.8)(4,591.84)\\\\v^2 = 360,000 + 90,000.064\\\\v = \sqrt{450,000.064} \\\\v = 670.82 \ m/s

5 0
2 years ago
In a jump spike, a volleyball player slams the ball from overhead and toward the opposite floor. Controlling the angle of the sp
MAVERICK [17]

Answer:

The ball would have landed 3.31m farther if the downward angle were 6.0° instead.

Explanation:

In order to solve this problem we must first start by doing a drawing that will represent the situation. (See picture attached).

We can see in the picture that the least the angle the farther the ball will go. So we need to find the A and B position to determine how farther the second shot would go. Let's start with point A.

So, first we need to determine the components of the velocity of the ball, like this:

V_{Ax}=V_{A}cos\theta

V_{Ax}=(21m/s)cos(-14^{o})

V_{Ax}=20.38 m/s

V_{Ay}=V_{A}sin\theta

V_{Ay}=(21m/s)sin(-14^{o})

V_{Ay}=-5.08 m/s

we pick the positive one, so it takes 0.317s for the ball to hit on point A.

so now we can find the distance from the net to point A with this time. We can find it like this:

x_{A}=V_{Ax}t

x_{A}=(20.38m/s)(0.317s)

x_{A}=6.46m

Once we found the distance between the net and point A, we can similarly find the distance between the net and point B:

V_{Bx}=20.88 m/s

V_{By}=-2.195 m/s

y_{Bf}=y_{B0}+V_{0}t-\frac{1}{2}at^{2}

0=2.1m+(-2.195m/s)t-\frac{1}{2}(-9.8m/s^{2})t^{2}

-4.9t^{2}-2.195t+2.1=0

t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}

t=\frac{-(-2.195)\pm\sqrt{(-2.195)^{2}-4(-4.9)(2.1)}}{2(-4.9)}

t= -0.9159s    or   t=0.468s

we pick the positive one, so it takes 0.468s for the ball to hit on point B.

so now we can find the distance from the net to point B with this time. We can find it like this:

x_{B}=V_{Bx}t

x_{B}=(20.88m/s)(0.468s)

x_{B}=9.77m

So once we got the two distances we can now find the difference between them:

x_{B}-x_{A}=9.77m-6.46m=3.31m

so the ball would have landed 3.31m farther if the downward angle were 6.0° instead.

7 0
2 years ago
Determine tge energy in joules of a photon whose frequency is 3.55x10
JulijaS [17]
E = hf
E = 6.63×10^-34 × 3.55×10 eV
1 eV = 1.60×10^-19 J
E = 6.63×10^-34 × 3.55×10 × 1.60×10^-19
E = 3.77×10^-51 J

Hope it helped!
6 0
3 years ago
Other questions:
  • How can air particles be trapped for investigation? (Subject: Science)
    13·1 answer
  • If a freely falling rock were equipped with a speedometer, by how much would itd speed readings increase with each second if it
    6·1 answer
  • Suppose a single light bulb burns out. How do you think this will affect lights that are strung
    13·2 answers
  • The temperature is cold if
    5·1 answer
  • What factors affect where a plant or animal lives
    13·1 answer
  • A boat is travelling down stream in a river at 20 m/s south. A passenger on the boat walks toward the back of the boat at 10 m/s
    15·1 answer
  • 4. You run from your house to a friend's house that is 3 miles away. You then walk
    12·1 answer
  • The 2nd maximum of a double slit diffraction pattern makes an angle 20 degrees when a wavelength of 600 nm is used. What is the
    10·1 answer
  • Three liters (3000 cm3) of oil is spilled onto a smooth lake. If the oil spreads out uniformly until it makes a circular oil sli
    6·1 answer
  • All the following are tiny particles that are parts of atoms EXCEPT:
    14·2 answers
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!