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allsm [11]
4 years ago
8

На рычаге размещены два противовеса таким образом, что рычаг находится в состоянии равновесия. Вес расположенного слева противов

еса равен P1=13H.
Какой вес P2 расположенного справа противовеса, если все обозначенные на перекладине рычага участки имеют одинаковую длину?

P1=13H
P2=?
Physics
1 answer:
prisoha [69]4 years ago
8 0

Answer:

Вес противовеса, или сила нагрузки, составляет затем 100 000 фунтов-футов, разделенных на 20 футов, или 5000 фунтов.

Explanation:

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As a wave moves through a medium, particles are displaced and (2 points)
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As a wave moves through a medium, particles are displaced and return to their normal position after the wave passes.

Explanation:

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7 0
3 years ago
Two friends, Al and Jo, have a combined mass of 195 kg. At the ice skating rink, they stand close together on skates, at rest an
ddd [48]

Answer:

Al's mass is 102.92  kg  

Explanation:

As there are no external forces in the horizontal direction, the horizontal net force must be zero:

F_{net} = 0

As the force is the derivative in time of the momentum, this means that the horizontal momentum is constant:

F_{net} = \frac{dp_{horizontal}}{dt} = 0

p_{horizontal_i }= p_{horizontal_f}

where the suffix i and f means initial and final respectively.

The initial momentum will be:

p_{horizontal_}i = m_{Al} \ v_{Al_i} + m_{Jo} \ v_{Jo_i}

But, as they are at rest, initially

p_{horizontal_i} = m_{Al} * 0 + m_{Jo} * 0

p_{horizontal_i} = 0

So, this means:

p_{horizontal_f} = m_{Al} \ v_{Al_f} + m_{Jo} \ v_{Jo_f} = 0

We know that the have an combined mass of 195 kg:

m_{total} = m_{Al} + m_{Jo} = 195 \ kg.

so:

m_{Jo} = 195 \ kg - m_{Al}.

m_{Al} \ v_{Al_f} + (195 \ kg - m_{Al}) \ v_{Jo_f} = 0

m_{Al} \  v_{Al_f} - m_{Al} \  v_{Jo_f}= - 195 \ kg \  v_{Jo_f}

m_{Al} \ (v_{Al_f} - v_{Jo_f})= - 195 \ kg \ v_{Jo_f}

m_{Al} = \frac{ - 195 \ kg \ v_{Jo_f} } {  v_{Al_f} - v_{Jo_f} }

m_{Al} = \frac{195 \ kg  \ v_{Jo_f} } {    v_{Jo_f} - v_{Al_f} }

Now, we can use the values:

v_{Al_f}= 10.2 \frac{m}{s}

v_{Jo_f}= - 11.4 \frac{m}{s}

where the minus sign appears as they are moving at opposite directions

m_{Al} = \frac{195 \ kg  ( - 11.4 \frac{m}{s} ) } {   (- 11.4 \frac{m}{s}) - 10.2 \frac{m}{s} }

m_{Al} = 102.92 \ kg

and this is the Al's mass.

5 0
3 years ago
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sergeinik [125]

Answer:

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Explanation:

5 0
3 years ago
Read 2 more answers
ken, 0.75 kg, moves toward a wall (his path normal to the wall) at 52 m/s. 13.0 ms after he touches the wall he pushes himself o
shtirl [24]

Q: ken, 0.75 kg, moves toward a wall (his path normal to the wall) at 52 m/s. 13.0 ms after he touches the wall he pushes himself off in the opposite direction at 60 m/s. What is the magnitude of the average force the wall exerts on Ken during this rapid maneuver

Answer:

-6461.54 N

Explanation:

From Newton's Fundamental equation,

F = m(v-u)/t.................... Equation 1

Where F = Force exerted in sonic, m = mass of ken, v = final velocity, u = initial velocity, t = time.

Given: m = 0.75 kg, v = - 60 m/s (opposite direction), u = 52 m/s, t = 13 ms = 0.013 s

Substitute into equation 1

F = 0.75(-60-52)/0.013

F = 0.75(-112)/0.013

F = -84/0.013

F = -6461.54 N

Note: The negative sign tells that the force act in opposite direction to the initial motion of ken.

Hence the magnitude of the average force of the wall = -6461.54 N

4 0
3 years ago
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