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kondaur [170]
3 years ago
6

Can you please help - the answer I got so far is 7.624 but when I enter it in my homework in Quest, it i stating that this answe

r is wrong. Can you please let me know what the correct answer is? You are driving at the speed of 33.5 m/s (74.9534 mph) when suddenly the car in front of you (previously traveling at the same speed) brakes and begins to slow down with the largest deceleration possible without skidding. Considering an average human reaction, you press your brakes 0.549 s later. You also brake and decelerate as rapidly as possible without skidding. Assume that the coefficient of static friction is 0.778 between both cars’ wheels and the road. The acceleration of gravity is 9.8 m/s 2 . Calculate the acceleration of the car in front of you when it brakes. Answer in units of m/s 2 .
Physics
1 answer:
torisob [31]3 years ago
6 0

Answer:

-7.04

Explanation:

9.8 multipled by -0.719 b

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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
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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

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Given a force of 88N and an acceleration of 4m/s2 what is the mass
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Use newtons second law F=ma, plug in the given values which gives us the answer of 22 kg for the mass
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