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Sati [7]
1 year ago
11

A child jumps from a moving sled with a speed of 2.2 m/s and in the

Physics
1 answer:
butalik [34]1 year ago
4 0

The initial velocity of the sled will be 7.34 m/sec. V is the initial velocity of the sled.

<h3>What is the law of conservation of momentum?</h3>

According to the law of conservation of momentum, the momentum of the body before the collision is always equal to the momentum of the body after the collision.

The given data in the problem is;

(m₁)  mass of child = 38 kg

(u₁) is the initial velocity child = 2.2 m/s

(m₂) is the mass of sled = 68 kg

(u₂) is the initial velocity of sled = ?

(v) is the velocity after collision = 5.5 m/s

According to the law of conservation of momentum;

Momentum before collision =Momentum after collision

\rm  m_1u_1 + m_2u_2 = v(m_1 + m_2)\\\\(38 \times  2.2) + (68 \times V) = 5.5 \times (38+68) \\\\ 83.6 + 68V = 5.5 \times 106 \\\\  83.6 + 68V =583 \\\\ \rm 68\ V= 583 -83.6 \\\\ \rm 68 \ V= 499.4\\\\ V= 7.34 \ m/sec

Hence,the initial velocity of the sled will be 7.34 m/sec.

To learn more about the law of conservation of momentum refer;

brainly.com/question/1113396

#SPJ1

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10. A worker uses a pulley system to raise a 24.0 kg carton 16.5 m. A force of 129 N is exerted and the rope is pulled 33.0 m. a
lozanna [386]

Answer: Machanical advantage of the machine is 1.86

Explanation: Machanical advantage of a machine is the ratio of the Force to overcome which is the load in this case 24kg * 10= 240N to the force exerted(Effort) to overcome the load in this case 129N.

So, we have

MA = load/effort

= 240N/129N

= 1.86.

5 0
2 years ago
When a 3.0 kg mass is hung from a vertical massless spring, the spring is stretched 40 cm. What is the spring constant of the sp
Dmitry_Shevchenko [17]

Answer:

0.74 N/cm

Explanation:

The following data were obtained from the question:

Mass (m) = 3 Kg

Extention (e) = 40 cm

Spring constant (K) =?

Next, we shall determine the force exerted on the spring.

This can be obtained as follow:

Mass (m) = 3 Kg

Acceleration due to gravity (g) = 9.8 m/s²

Force (F) =?

F = mg

F = 3 × 9.8

F = 29.4 N

Finally, we shall determine the spring constant of the spring. This can be obtained as follow:

Extention (e) = 40 cm

Force (F) = 29.4 N

Spring constant (K) =?

F = Ke

29.4 = K × 40

Divide both side by 40

K = 29.4 / 40

K = 0.74 N/cm

Therefore, the spring constant of the spring is 0.74 N/cm

5 0
2 years ago
Need help on this,<br>will mark brainliest if you answer.<br>​
Alecsey [184]

Answer:

(1) tropical storm

(2) Severe tropical storm

(3) 305 kmh

5 0
2 years ago
How fast is a 90kg man running if his kinetic energy is 720 J?
zheka24 [161]
Answer:
4 m/s
Explanation:
KE=1/2mv^2
720=1/2(90)
720=45v^2
divide by 45
16= v^2
over the square root
4=v
v= 4m/s
6 0
2 years ago
The index of refraction for a certain type of glass is 1.641 for blue light and 1.603 for red light. When a beam of white light
exis [7]

Answer:

The angle between the emergent blue and red light is 0.566^{o}

Explanation:

We have according to Snell's law

n_{1}sin(\theta _{i})=n_{2}sin(\theta _{r})

Since medium from which light enter's is air thus n_{1}=1

Thus for blue incident light we have

1\times sin(40.05)=1.641\times sin(\theta _{rb})\\\\\therefore \theta _{rb}=sin^{-1}(\frac{sin(40.05)}{1.64})\\\\\theta _{rb}=23.10

Similarly using the same procedure for red light we have

1\times sin(40.05)=1.603\times sin(\theta _{rr})\\\\\therefore \theta _{rr}=sin^{-1}(\frac{sin(40.05)}{1.603})\\\\\theta _{rr}=23.66^{o}

Thus the absolute value of angle between the refracted blue and red light is

|23.66-23.10|=0.566^{o}

6 0
3 years ago
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