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bagirrra123 [75]
3 years ago
10

Dry cells become soft even when not in use. Explain.

Physics
1 answer:
mamaluj [8]3 years ago
4 0
This is possible due to self-discharge. Very small internal currents inevitably occur in these cells over time and they will eventually exhaust the chemistry.
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are the properties of a specific element the same as the properties of a molecule made up of that element ? why or why not
konstantin123 [22]
Because it’s science
8 0
3 years ago
A car drives 23m/s east for 35 seconds. What is the displacement?
Lina20 [59]

Answer:

805m

Explanation:

Speed = displacement/time

Speed = 23m/s

Time = 35s

Displacement = speed × time

= 23 × 35

= 805m

5 0
3 years ago
A swimming pool is 50 ft wide and 100 ft long and its bottom is an inclined plane, the shallow end having a depth of 4 ft and th
Nina [5.8K]

Explanation:

We define force as the product of mass and acceleration.

F = ma

It means that the object has zero net force when it is in rest state or it when it has no acceleration. However in the case of liquids. just like the above mentioned case, the water is at rest but it is still exerting a pressure on the walls of the swimming pool. That pressure exerted by the liquids in their rest state is known as hydro static force.

Given Data:

Width of the pool = w = 50 ft

length of the pool = l= 100 ft

Depth of the shallow end = h(s) = 4 ft

Depth of the deep end = h(d) = 10 ft.

weight density = ρg = 62.5 lb/ft

Solution:

a) Force on a shallow end:

F = \frac{pgwh}{2} (2x_{1}+h)

F = \frac{(62.5)(50)(4)}{2}(2(0)+4)

F = 25000 lb

b) Force on deep end:

F = \frac{pgwh}{2} (2x_{1}+h)

F = \frac{(62.5)(50)(10)}{2} (2(0)+10)

F = 187500 lb

c) Force on one of the sides:

As it is mentioned in the question that the bottom of the swimming pool is an inclined plane so sum of the forces on the rectangular part and triangular part will give us the force on one of the sides of the pool.

1) Force on the Rectangular part:

F = \frac{pg(l.h)}{2}(2(x_{1} )+ h)

x_{1} = 0\\h_{s} = 4ft

F = \frac{(62.5)(100)(2)}{2}(2(0)+4)

F =25000lb

2) Force on the triangular part:

F = \frac{pg(l.h)}{6} (3x_{1} +2h)

here

h = h(d) - h(s)

h = 10-4

h = 6ft

x_{1} = 4ft\\

F = \frac{62.5 (100)(6)}{6} (3(4)+2(6))

F = 150000 lb

now add both of these forces,

F = 25000lb + 150000lb

F = 175000lb

d) Force on the bottom:

F = \frac{pgw\sqrt{l^{2} + ((h_{d}) - h(s)) } (h_{d}+h_{s})   }{2}

F = \frac{62.5(50)\sqrt{100^{2}(10-4) } (10+4) }{2}

F = 2187937.5 lb

7 0
3 years ago
One particular descent goes from 2100m to 1600m. Assuming work done against friction is 90% of the potential energy change of th
alisha [4.7K]

Answer:

1/2 M V^2 = .1 M g H       where 10% of PE goes into KE

V^2 = .2 g H = .2 * 9.8 * (2100 - 1600) = 980 m^2 / s^2

V = 31.1 m/s       increase in speed during descent

1 km / hr = 1000 m / 3600 sec = .278 m/s

V = 31.1 m/s / (.278 m/s / km /hr)= 112 km/hr

7 0
2 years ago
Soccer ball is heading east positive direction with momentum of 2kgm/s. It is kicked backwards and a fraction of a second later
attashe74 [19]

Answer:

-4.3 kgm/s

Explanation:

Impulse which the soccer ball experiences is equal to the change in momentum of a body.

It is given mathematically as:

Impulse = m*Δv = mv - mu

Initial momentum of soccer ball = 2 kgm/s

Final momentum of the soccer ball = -2.3 kgm/s

Therefore, Impulse will be:

Impulse = -2.3 - 2

Impulse = -4.3 kgm/s

The impulse put on the soccer ball by the kicker is -4.3 kgm/s.

8 0
4 years ago
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