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Marina86 [1]
3 years ago
9

Susan

Physics
1 answer:
maxonik [38]3 years ago
7 0
D. Was a leader in the woman's suffrage movement
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An object is tracked by a radar station and found to have a position vector given by r = (4570-160 t) i + 2930j + 130k , with r
Kipish [7]

Answer:

a)P= -4800 i

b)F= 0

Explanation:

Given that

r = (4570-160 t) i + 2930 j + 130 k

We know that velocity is rate of change of the space vector.

V= dr/dt

r = (4570-160 t) i + 2930 j + 130 k

dr/dt= -160 i + 0 + 0

dr/dt= -160 i

V= -160 i

It means that velocity is in only x-direction

We also know that acceleration is the rate of change of velocity .

a= dV/dt

V= -160 i

dV/dt=0

So we can say that acceleration is zero.

 a= 0

From Second law of Newton'

Force =  Mass x acceleration

F= 300 x 0

F= 0

We know that linear momentum P

P = m V

Given that m= 300 kg

P = 300 x (-160 i)

P= -4800 i

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3 years ago
How do you think the body deals with consuming a lot more<br> calories than are burnt?
natali 33 [55]

Answer:

The body stores it as fat

Explanation:

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When a species has no members left that are alive
denpristay [2]

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it means that species is extinct.

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have a good day :)

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How would a small bar magnet be oriented when placed at position x ? option c is wrong.
lubasha [3.4K]

Answer:

B

Explanation:

3 0
3 years ago
A box of mass m1 = 20.0 kg is released from rest at a warehouse loading dock and slides down a 3.0 m-high frictionless chute to
Harrizon [31]

´To develop this problem we will use the concepts related to the conservation of momentum and the application of energy conservation equations to find the velocity of the mass after the collision, like this:

Velocity of the mass m_1 just before the collision

v_1 = \sqrt{2gh}

v_1 = \sqrt{2(9.8)(3)}

v_1 = 7.67m/s

Therefore the momentum just before collision would be

p_2 = m_1v_1+40(0)\\p_1 = 20*7.67+40(0)\\p_1 = 153.36kg \cdot m/s

Momentum after the collision

p_1 = 20*u_1+40u_1\\p_1 = 60u_1

Since the momentum is conserved we have that

153.36= 60u_1

u_1 = \frac{153.36}{60}

u_1 = 2.56m/s

The velocity of mass m_2 after the collision is given by

v_2 = \frac{2m_1}{m_1+m_2} u_1

v_2 = \frac{2(20)}{20+40}(2.56)

v_2 = 1.71m/s

Therefore the change in momentum of mass 2 is

p_2 = m_2v_2

p_2 = 40*1.71

p_2 = 68.4kg\cdot m/s

Therefore the impulse acting on m2 during the collision between the two boxes is p = 68.4kg\cdot m/s

8 0
3 years ago
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