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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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A uniform disk a uniform hoop and a uniform sphere are released at the same time at the top of an inclined ramp. They all roll w
lara [203]

Answer:

D. The sphere the disk and the hoop

Explanation:

This is because the sphere has inertial of

2/5mR²

Disk 1/2mR²

Hope mR²

So these are moment of inertial which is resistance or opposition to rotation so since the sphere has a smaller moment to inertial it will move faster and reach the ground first then the disk then the hoop in that order

6 0
3 years ago
B. Study the given figure and calculate the area of the leaf
yanalaym [24]

Answer:

19 cm sq

Explanation:

you have to count one full and half area occupied by leaf in each box, then you have to multiply with 1 cm sq.

4 0
3 years ago
Which of the following is not a physical property of mattera. melting pointb.heat of combustionc. viscosityd. boiling point
NNADVOKAT [17]

Answer:

The Correct Answer is Heat of Combustion  

Explanation:

6 0
3 years ago
Which tool would you use to measure the height of a cube of metal
SCORPION-xisa [38]
It depends on the size of the cube, but you could use a ruler for almost any size, or if it is small enough, you could use a caliper to get a very precise measurement.
3 0
3 years ago
Read 2 more answers
Two planets P1 and P2 orbit around a star S in circular orbits with speeds v1 = 40.2 km/s, and v2 = 56.0 km/s respectively. If t
Readme [11.4K]

Answer: 3.66(10)^{33}kg

Explanation:

We are told both planets describe a circular orbit around the star S. So, let's approach this problem begining with the angular velocity \omega of the planet P1 with a period T=750years=2.36(10)^{10}s:

\omega=\frac{2\pi}{T}=\frac{V_{1}}{R} (1)

Where:

V_{1}=40.2km/s=40200m/s is the velocity of planet P1

R is the radius of the orbit of planet P1

Finding R:

R=\frac{V_{1}}{2\pi}T (2)

R=\frac{40200m/s}{2\pi}2.36(10)^{10}s (3)

R=1.5132(10)^{14}m (4)

On the other hand, we know the gravitational force F between the star S with mass M and the planet P1 with mass m is:

F=G\frac{Mm}{R^{2}} (5)

Where G is the Gravitational Constant and its value is 6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}

In addition, the centripetal force F_{c} exerted on the planet is:

F_{c}=\frac{m{V_{1}}^{2}}{R^{2}} (6)

Assuming this system is in equilibrium:

F=F_{c} (7)

Substituting (5) and (6) in (7):

G\frac{Mm}{R^{2}}=\frac{m{V_{1}}^{2}}{R^{2}} (8)

Finding M:

M=\frac{V^{2}R}{G} (9)

M=\frac{(40200m/s)^{2}(1.5132(10)^{14}m)}{6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}} (10)

Finally:

M=3.66(10)^{33}kg (11) This is the mass of the star S

4 0
4 years ago
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