According to the law of conservation of energy, how will the sum of the kinetic

Lions can run at speeds up to approximately 80.0 km / h. A hungry 109 kg lion running northward at top speed attacks and holds o

sukhopar [10]

Answer:

17.34 m/s

Explanation:

Given:

Mass of lion (m₁) = 109 kg

Initial speed of lion (v₁) = 80.0 km/h (Northward direction)

Mass of gazelle (m₂) = 39.0 kg

Initial speed of gazelle (v₂) = 78.5 km/h (Eastward direction)

Final velocity of both lion and gazelle is, $v_f=?$

First, let us convert the speeds from km/h to m/s using the conversion factor.

We know that, 1 km/h = 5/18 m/s

Therefore,

$v_1= 80.0\ km/h=80\times \frac{5}{18}=22.22\ m/s\\\\v_2=78.5\ km/h=78.5\times \frac{5}{18}=21.81\ m/s$

Now, the concept of conservation of total momentum is used here as this is a case of perfectly inelastic collision. In inelastic collision, the masses move together with same velocity after collision.

Here, as the lion and gazelle are moving in directions at right angles to each other, the vector sum of their momentums will give the net initial momentum of the system.

So, initial momentum is given as:

$P_i=\sqrt{P_1^2+P_2^2}\\\\Where,\\\\P_1\to initial\ momentum\ of\ lion\\P_2\to initial\ momentum\ of\ gazelle$

Now, we calculate P₁ and P₂.

$P_1=m_1v_1=(109\ kg)(22.22\ m/s) = 2421.98\ Ns\\\\P_2=m_2v_2=(39\ kg)(21.81\ m/s) = 850.59\ Ns$

Therefore, the net initial momentum of the system is given as:

$P_i=\sqrt{(2421.98)^2+(850.59)^2}=2567\ Ns$

The final momentum of the system is given as:

$P_f=(m_1+m_2)(v_f)\\\\P_f=(109+39)v_f\\\\P_f=148v_f$

From the law of conservation of momentum, the final momentum is equal to the initial momentum. So,

$P_f=P_i\\\\148v_f=2567\\\\v_f=\frac{2567}{148}=17.34\ m/s$

Therefore, the final speed of the lion-gazelle system is 17.34 m/s

3 0

3 years ago

A proton travels at a velocity of 3.0x10^6 m/s in a direction perpendicular to a uniform magnetic field.

Virty [35]

Answer:

0.239 T

Explanation:

Applying,

F = Bvqsin∅................ Equation 1

Where F = magnetic force, B = magnetic Field, q = charge of a proton, v = velocity of proton, ∅ = angle between the velocity and the magnetic field.

make B the subject of the equation

B = F/(vqsin∅)................. Equation 2

From the question,

Given: F = 1.15×10⁻¹³ N, v = 3.0×10⁶ m/s, ∅ = 90°(perpendicular)

Constant: q = 1.602 x 10⁻¹⁹ C

Substitute into equation 2

B = 1.15×10⁻¹³ /(3.0×10⁶×1.602 x 10⁻¹⁹×sin90°)

B = 1.15×10⁻¹³/(4.806×10⁻¹³)

B = 0.239 T.

Hence the magnetic field = 0.239 T

7 0

3 years ago

(b) Figure 3.32 shows the orbit of a planet around the Sun. Compare the linear speed of the planet at positions X, Y and Z.

Viefleur [7K]

At Z ... slowest speed

At Y ... fastest speed

At X ... medium speed

Wherever it is in its orbit, the line from the planet to the Sun smears over the same amount of area every second.

That's Kepler's second law of planetary motion.

The reason this happens is: That's how gravity works. (A better explanation is available, but first you have to be able to twirl calculus and solid geometry in the air on long sticks.)

5 0

3 years ago

Can someone help me?

Alik [6]

Answer:

Resultant = 3.05N

Explanation:

$r = \sqrt{{5}^{2} + 5^{2} - 2(5)(5) \cos(120) }$

$r = \sqrt{25 + 25 - 50(0.8142)}$

$r = \sqrt{50 - 40.71} = \sqrt{9.29}$

$r = 3.05$

6 0

3 years ago

A 5.00 kg dog runs towards a skateboard with a speed of 2.50 m/s. The skateboard is initially at rest. The dog jumps onto the sk

ladessa [460]

Answer:

The mass of the skateboard is 1.25 kg.

Explanation:

We have,

Mass of dog, m = 5 kg

Initial speed of dog, u = 2.5 m/s

Initial speed of skateboard, u' = 0 (at rest)

The dog jumps onto the skateboard and both the dog and skateboard move with a speed of 2.00 m/s. It is the common speed of the dog +skateboard, V = 2 m/s. We need to find the mass of the skateboard. Using conservation of momentum as :

$mu+m'u'=(m+m')V$

m' is mass of skateboard

$5\times 2.5=(5+m')2\\\\m'=1.25\ kg$

So, the mass of the skateboard is 1.25 kg.

6 0

3 years ago