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blondinia [14]
3 years ago
9

In straight line motion, if the velocity of an object is changing at a constant rate, then its position is _________ and its acc

eleration is___________: O changing: zero O changing; changing O constant and non-zero; constant and non-zero O None of the above
Physics
1 answer:
Serjik [45]3 years ago
8 0

Answer:

None of the above

It should be position is changing and acceleration is constant.

Explanation:

Since the velocity is changing, this means the object is moving, so the <u>position must also be changing</u>.

Acceleration is the change in velocity in time, if this change of velocity happens at a constant rate, the <u>acceleration must be constant</u> too.

So, for example, if the velocity were to stay the same (not changing), acceleration would be zero, because there wouldn't be a change in time on the velocity.

So in this case the answer sould be position is changing and acceleration is constant. But this isn't in the options so the correct answer is "None of the above"

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35 Ω

Explanation:

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If the pendulum is brought on the moon where the gravitational acceleration is about g/6, approximately what will its period now
Andru [333]

Answer:

The new period will be √6 *T

Explanation:

period ,T=2π√(L/g)       ................equation 1

where T is the period on earth

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T1 = 2π√[L/(g/6)]

T1=2π√(6L/g)               ...............equation 2

divide equation 2 by 1

T1/T =2π√(6L/g)÷2π√(L/g)

T1/T =√(6L/L)

T1/T =√6

T1 = √6 *T

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3 years ago
A kangaroo jumps straight up with an initial vertical velocity of 2.4m/s. We want to find the maximum height of the jump. We can
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Answer:

v  

2

=v  

0

2

​  

+2aΔxv, squared, equals, v, start subscript, 0, end subscript, squared, plus, 2, a, delta, x

Explanation:

8 0
3 years ago
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A rock is dropped from a vertical cliff. The rock takes 6.00 s to reach the ground below the cliff. What is the height of the cl
Lana71 [14]

Answer:

180 m

Explanation:

The rock follows a free-fall motion - so the vertical distance covered can be found by using the equation

h=\frac{1}{2}gt^2

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Substituting these data, we find the height of the cliff:

h=\frac{1}{2}(10 m/s^2)(6.00 s)^2=180 m

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Consider two objects (Object 1 and Object 2) moving in the same direction on a frictionless surface. Object 1 moves with speed v
d1i1m1o1n [39]

1) A) Object 1 has the greater momentum

The magnitude of the momentum of an object is given by:

p=mv

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v is its speed

Object 1 has a mass of m_1 = 2m and a speed of v_1 = v, so its momentum is

p_1 = m_1 v_1 = (2m)(v)=2mv

Object 2 has a mass of m_2 = m and a speed of v_2 = \sqrt{2} v, so its momentum is

p_2 = m_2 v_2 = (m)(\sqrt{2} v)=\sqrt{2}mv

So we see that p_1 > p_2, so object 1 has the greater momentum.

2) The objects have the same kinetic energy.

The kinetic energy of an object is given by

K=\frac{1}{2}mv^2

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m is the mass of the object

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Object 1 has a mass of m_1 = 2m and a speed of v_1 = v, so its kinetic energy is

K_1 = \frac{1}{2}m_1 v_1^2 = \frac{1}{2}(2m)(v)^2=mv^2

Object 2 has a mass of m_2 = m and a speed of v_2 = \sqrt{2} v, so its kinetic energy is

K_2 = \frac{1}{2}m_2 v_2^2 = \frac{1}{2}(m)(\sqrt{2} v)^2=mv^2

So we see that K_1 =K_2, so the objects have same kinetic energy

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3 years ago
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