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saveliy_v [14]
2 years ago
15

What does a snow leopard eat?

Physics
1 answer:
Rom4ik [11]2 years ago
4 0

Answer:

They eat Himalayan blue sheep, boar, deers, wild mountain goats, pikas etc.

Explanation:

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To calculate how much rest energy an object contains, you multiply
Rina8888 [55]
This is what e=mc² means.
So rest energy is RH side of equation that is its mass times speed of light squared.
5 0
3 years ago
Calculate the KE of a car which has a mass of 1000 kg and is moving at the rate of 20 m/s
Elan Coil [88]

Answer:

The kinetic energy of the car is 2.0\cdot 10^5 J

Explanation:

3 0
2 years ago
Gayle runs at a speed of 3.85 m/s and dives on a sled, initially at rest on the top of a frictionless snow-covered hill. After s
enot [183]

Answer:

Final velocity at the bottom of hill is 15.56 m/s.

Explanation:

The given problem can be divided into four parts:

1. Use conservation of momentum to determine the speed of the combined mass (Gayle and sled)

From the law of conservation of momentum (perfectly inelastic collision), the combined velocity is given as:  

p_i = p_f  

m_1u_1 + m_2v_2 = (m_1 + m_2)v

v = \frac{(m_1u_1 + m_2v_2)}{(m_1 + m_2)}

v=\frac{[50.0\ kg)(3.85\ m/s) + 0]}{(50.0\ kg + 5.00\ kg)}= 3.5\ m/s  

2. Use conservation of energy to determine the speed after traveling a vertical height of 5 m.

The velocity of Gayle and sled at the instant her brother jumps on is found from the law of conservation of energy:  

E(i) = E(f)  

KE(i) + PE(i) = KE(f) + PE(f)  

0.5mv^2(i) + mgh(i) = 0.5mv^2(f) + mgh(f)  

v(f) = \sqrt{[v^2(i) + 2g(h(i) - h(f))]}

Here, initial velocity is the final velocity from the first stage. Therefore:  

v(f) = \sqrt{[(3.5)^2+2(9.8)(5.00-0)]}= 10.5\ m/s

3. Use conservation of momentum to find the combined speed of Gayle and her brother.  

Given:

Initial velocity of Gayle and sled is, u_1(i)=10.5 m/s

Initial velocity of her brother is, u_2(i)=0 m/s

Mass of Gayle and sled is, m_1=55.0 kg

Mass of her brother is, m_2=30.0 kg

Final combined velocity is given as:

v(f) = \frac{[m_1u_1(i) + m_2u_2(i)]}{(m_1 + m_2)}  

v(f)=\frac{[(55.0)(10.5) + 0]}{(55.0+30.0)}= 6.79 m/s  

4. Finally, use conservation of energy to determine the final speed at the bottom of the hill.

Using conservation of energy, the final velocity at the bottom of the hill is:  

E(i) = E(f)  

KE(i) + PE(i) = KE(f) + PE(f)  

0.5mv^2(i) + mgh(i) = 0.5mv^2(f) + mgh(f)  

v(f) = \sqrt{[v^2(i) + 2g(h(i) - h(f))]} \\v(f)=\sqrt{[(6.79)^2 + 2(9.8)(15 - 5.00)]}\\v(f)= 15.56\ m/s

6 0
3 years ago
⬇️⬇️⬇️⬇️⬇️⬇️⬇️⬇️⬇️⬇️⬇️⬇️⬇️⬇️⬇️⬇️⬇️⬇️⬇️⬇️⬇️⬇️⬇️⬇️⬇️⬇️
zloy xaker [14]

Answer:

its all black I cant see the image can you re upload it

Explanation:

6 0
2 years ago
A 36-cm-circumference loop of wire is placed between the poles of an electromagnet. When a field of 0.76 T is switched on in a t
Marrrta [24]

Answer:

R = 25.3ohms

Explanation:

See attachment below.

5 0
3 years ago
Read 2 more answers
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