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ollegr [7]
2 years ago
5

A 75 kg object is moving west at 5.6 m/s. What is the object's momentum? (p = mv)

Physics
1 answer:
Licemer1 [7]2 years ago
7 0

Explanation:

p=mv

p=5.6×75

p= 420

<em>hope</em><em> it</em><em> was</em><em> helpful</em><em> to</em><em> you</em>

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Which property of a star can help us deduce its surface temperature?
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I think I has to be C

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A steam engine absorbs 4 x 105 J and expels 3.5 x 105 J in each cycle. What is its efficiency?
Wewaii [24]
Input heat, Qin = 4 x 10⁵ J
Output heat, Qout = 3.5 x 10⁵ J

From the first Law of thermodynamics, obtain useful work performed as
W = Qin  -  Qout
     = 0.5 x 10⁵ J

By definition, the efficiency is
η = W/Qin
   = 100*(0.5 x 10⁵/4 x 10⁵)
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3 0
2 years ago
15.One girl was involved in a bus crash. She told the police that she was seated in the last line of seats and that when the bus
Alona [7]

Answer: It is not likely.

Explanation:

When the bus is moving forward, all the objects inside of it also are moving forward.

Now, as the objects inside the buss are not fixed to the bus, if the bus suddenly stops the objects inside of it will keep moving forward, because of the conservation of the momentum, defined as the quantity of motion (Similar to when you are in a car and it suddenly stops, you can feel the forward impulse).

Then is not likely that, in a case where the bus stops suddenly, an object inside the bus flies backward in opposite direction to the previous movement of the bus.

5 0
2 years ago
Swinging a golf club or baseball bat are examples of ______________ stretching.
kari74 [83]

Answer:

Static stretching is the answer.

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6 0
3 years ago
Comets travel around the sun in elliptical orbits with large eccentricities. If a comet has speed 1.6×104 m/s when at a distance
xz_007 [3.2K]

Answer:

v₂ = 7.6 x 10⁴ m/s

Explanation:

given,

speed of comet(v₁) = 1.6 x 10⁴ m/s

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to find the speed when he is at distance of(d₂) 4.8 × 10¹⁰ m

v₂ = ?

speed of planet can be determine using conservation of energy

K.E₁ + P.E₁ = K.E₂ + P.E₂

\dfrac{1}{2}mv_1^2-\dfrac{GMm}{r_1} = \dfrac{1}{2}mv_2^2-\dfrac{GMm}{r_2}

\dfrac{1}{2}v_1^2-\dfrac{GM}{r_1} = \dfrac{1}{2}v_2^2-\dfrac{GM}{r_2}

v_2^2= v_1^2 + \dfrac{2GM}{r_2}-\dfrac{2GM}{r_1}

v_2= \sqrt{v_1^2 +2GM(\dfrac{1}{r_2}-\dfrac{1}{r_1})}

v_2= \sqrt{(1.6\times 10^4)^2 +2\times 6.67 \times 10^{-11}\times 1.99 \times 10^{30}(\dfrac{1}{4.8\times 10^{10}}-\dfrac{1}{2.7\times 10^{11}})}

v₂ = 7.6 x 10⁴ m/s

3 0
2 years ago
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