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Nookie1986 [14]
2 years ago
14

The total length of the wire of potentiometer is 10m. A potential gradient of 0.0015 V/cm is obtained when a steady current is p

assed through this wire.
Calculate,

i) the distance of null point on connecting a cell of emf 1.018 v

ii) the unknown P.d. if null point is obtained at 940 cm.​
Physics
1 answer:
andreyandreev [35.5K]2 years ago
3 0

Answer:

this picture might help you

Explanation:

if you think correct pls mark brainliest

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A compressed spring has elastic potential energy. Please select the best answer from the choices provided T F
riadik2000 [5.3K]

Answer: True

Explanation:

As the spring is compressed, it acumulates energy, and the spring "wants to release that energy". This acumulated energy, (potential energy) is called "elastic potential energy" because of the elastical nature of the spring, that when compressed it wants to return to the original shape. So the sentence is true

6 0
3 years ago
A rubber ball and a lump of clay have equal mass. They are thrown with equal speed against a wall. The ball bounces back with ne
gregori [183]

Answer:

The ball experiences the greater momentum change

Explanation:

The momentum change of each object is given by:

\Delta p = m \Delta v= m (v-u)

where

m is the mass of the object

v is the final velocity

u is the initial velocity

Both objects have same mass m and same initial velocity u. So we have:

- For the ball, the final velocity is

v=-u

Since it bounces back (so, opposite direction --> negative sign) with same speed (so, the magnitude of the final velocity is still u). So the change in momentum is

\Delta p=m(v-u)=m((-u)-u)=-2mu

- For the clay, the final velocity is

v=0

since it sticks to the wall. So, the change in momentum is

\Delta p = m(v-u)=m(0-u)=-mu

So we see that the greater momentum change (in magnitude) is experienced by the ball.

3 0
3 years ago
Objects with masses of 235 kg and a 535 kg are separated by 0.330 m. (a) find the net gravitational force exerted by these objec
saul85 [17]
The 61.0 kg object<span> ... F = (300kg)(6.673×10−11 </span>N m<span>^2 </span>kg<span>^−2)(61kg)/(.225m)^2. F = 2.412e-5 </span>N<span> towards the 495 </span>kg<span> block. </span>b. [195kg] ===.45m ... (b<span>) You cannot achieve this </span>position<span>. For the </span>net force<span> to become zero, one or both of the </span>masses<span> must ...</span>
7 0
3 years ago
A standard 1 kilogram weight is a cylinder 54.0 mm in height and 55.0 mm in diameter. what is the density of the material
denis-greek [22]

The radius of the cylinder is equal to half the diameter:

r=\frac{d}{2}=\frac{55.0 mm}{2}=27.5 mm

The volume of the cylinder is given by:

V=\pi r^2 h=\pi (27.5 mm)^2 (54.0 mm)=1.28 \cdot 10^5 mm^3

where h is the heigth of the cylinder. Converting into meters,

V=1.28 \cdot 10^{-4} m^3

And the density of the material will be given by the ratio between the mass and the volume:

d=\frac{m}{V}=\frac{1 kg}{1.28 \cdot 10^{-4} m^3}=7812.5 kg/m^3

5 0
3 years ago
What happens to the coefficient of friction when the weight is increased? Why is this?
Crazy boy [7]

Answer:

Usually the coefficient of friction remains unchanged

Explanation:

The coefficient of friction should in the majority of cases, remain constant no matter what your normal force is. When you apply a greater normal force, the frictional force increases, and your coefficient of friction stays the same. Here's another way to think about it: because the force of friction is equal to the normal force times the coefficient of friction, friction is increased when normal force is increased.

Plus, the coefficient of friction is a property of the materials being "rubbed", and this property usually does not depend on the normal force.

6 0
3 years ago
Read 2 more answers
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