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Ronch [10]
2 years ago
9

A 70.0Kg cyclist Rides a 15.0kg

Physics
1 answer:
Inessa [10]2 years ago
4 0

Answer:

1275     kg*m/s

Explanation:

We'll use the momentum equation:

p=mv

where:

p = momentum

m = mass

v = velocity

Since we're doing the magnitude of momentum of the system, we'll add the mass of the cyclist and the mountain bike together:

70.0 kg + 15.0kg=85.0 kg

Given that, we can now substitute our given values into the momentum equation:

p=mv\\p=85.0kg*15.0m/s

Our final answer is:

p= 1275 kg*m/s

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The third floor of a house is 8m above street level. How much work is needed to move a 136kg refrigerator to the third floor?
jonny [76]

m = Mass of the refrigerator to be moved to third floor = 136 kg

g = Acceleration due to gravity by earth on the refrigerator being moved = 9.8 m/s²

h = Height to which the refrigerator is moved  = 8 m

W = Work done in lifting the object

Work done in lifting the object is same as the gravitational potential energy gained by the refrigerator. hence

Work done = Gravitation potential energy of refrigerator

W = m g h

inserting the values

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8 0
3 years ago
A particle moves through an xyz coordinate system while a force acts on it. When the particle has the position vector r with arr
Paha777 [63]

Answer:

The question is incomplete, below is the complete question "A particle moves through an xyz coordinate system while a force acts on it. When the particle has the position vector r with arrow = (2.00 m)i hat − (3.00 m)j + (2.00 m)k, the force is F with arrow = Fxi hat + (7.00 N)j − (5.00 N)k and the corresponding torque about the origin is vector tau = (4 N · m)i hat + (10 N · m)j + (11N · m)k.

Determine Fx."

F_{x}=-1N.m

Explanation:

We asked to determine the "x" component of the applied force. To do this, we need to write out the expression for the torque in the in vector representation.

torque=cross product of force and position . mathematically this can be express as

T=r*F

Where

F=F_{x}i+(7N)j-(5N)k  and the position vector

r=(2m)i-(3m)j+(2m)k

using the determinant method to expand the cross product in order to determine the torque we have

\left[\begin{array}{ccc}i&j&k\\2&-3&2\\ F_{x} &7&-5\end{array}\right]\\\\

by expanding we arrive at

T=(18-14)i-(-12-2F_{x})j+(12+3F_{x})k\\T=4i-(-12-2F_{x})j+(12+3F_{x})k\\\\

since we have determine the vector value of the toque, we now compare with the torque value given in the question

(4Nm)i+(10Nm)j+(11Nm)k=4i-(-12-2F_{x})j+(12+3F_{x})k\\

if we directly compare the j coordinate we have

10=-(-12-2F_{x})\\10=12+2F_{x}\\ 10-12=2F_{x}\\ F_{x}=-1N.m

8 0
3 years ago
Newton’s law of universal gravitation a. is equivalent to Kepler’s first law of planetary motion. b. can be used to derive Keple
Finger [1]
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If the police are chasing a runaway car why is knowing the velocity of the car important
atroni [7]

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