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2 years ago

Length of table is 1.0m,1.00m and 1.000m.Which one is more accurate?

Physics

1 answer:

natita [175]2 years ago

6 0

Answer:

1.00 m is a more accurate measured length.

Explanation:

Denote length of the table by L.

For L=1.0 m, there is one significant digit after the decimal.

Care 1: When one more significant digit after decimal considered, the exact number can be from 0.95 to 1.05.

So, the possible span of error $\Delta E_1= 1.05-0.95= 0.1m$

For L=1.00 m, there is two significant digits after the decimal.

Case 2: When one more significant digit after decimal considered, the exact number can be from 0.095 to 1.005.

So, the possible span of error $\Delta E_2= 1.005-0.095= 0.01m$

Case 3: For L=1.000 m, there is three significant digits after the decimal.

When one more significant digit after decimal considered, the exact number can be from 0.0095 to 1.0005.

So, the possible span of error $\Delta E_3= 1.0005-0.0095= 0.001m$

As $\Delta E_1 >\Delta E_2>\Delta E_3$

So, the least error is in the third case when L=1.00m, hence, L= 1.00m is more accurate.

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A heavy object falls with the acceleration as a light object during free fall. why?

alexgriva [62]

<u>A heavy object falls with the acceleration as a light object during free fall because:</u>

Heavy things have a large gravitational force and also have less acceleration. So both the effects exactly cancel and make the falling objects to have the same acceleration irrespective of mass.

Free fall is a unique motion which has only gravitational force that acts on an object. Objects that undergo free fall experience only have the influence of gravity and not any other force.

So when we apply newton's second law of gravity which is:

$\text {acceleration}=\frac{\text {Force}}{\text {mass}}$

where,

F is the force

m is the mass

a is the acceleration

For example: When a 1000 kg elephant and a 1 kg rat fall from the same height,

The acceleration can be calculated as follows:

For elephant: F = 10000 N and m = 1000 kg. So,

$\text {acceleration}=\frac{10000}{1000}=10\ \mathrm{m} / \mathrm{s}^{2}$

For rat : F = 10 N and m = 1 kg

Thus, $\text {acceleration}=\frac{10}{1}=1\ \mathrm{m} / \mathrm{s}^{2}$

Hence, it shows that both the animals have the same acceleration irrespective of their mass.

6 0

3 years ago

What is the mass of a cart that accelerates at 3.0 m/s if a 0.5 N force is applied?

frozen [14]

The mass of the cart is 0.167 kg

Explanation:

We can solve the problem by using Newton's second law, which states that the net force applied to an object is equal to the product between its mass and its acceleration:

$F=ma$

where

F is the net force acting on an object

m is its mass

a is its acceleration

For the cart in this problem, we have

F = 0.5 N is the net force applied

$a=3.0 m/s^2$ is the acceleration

Substituting, we find the mass:

$m=\frac{F}{a}=\frac{0.5}{3.0}=0.167 kg$

Learn more about Newton's second law:

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6 0

3 years ago

Please help me......

Novosadov [1.4K]

The first one, Climate

6 0

3 years ago

Water is leaking out of an inverted conical tank at a rate of 1.5 cm3 /min at the same time that water is being pumped into the

iris [78.8K]

Answer:

a) Check Explanation

b) Check Explanation

c) The rate at which water is being pumped into the tank = 2.631 cm³/min

Explanation:

Let the rate of flow of water into the tank be k cm³/min

a) The image of the conical tank is presented in the attached image

Note, the radius and height of a cone are related through the similar triangles principle.

As shown in the attached image, it is evident that

r/h = 3/10

r = 3h/10 = 0.3 h

b) The quantities given in the problem.

- Shape of the tank, conical tank, Hence volume of the tank = πr²h/3

- total height of the tank, H = 10 cm

- Radius of the tank at the top, R = D/2 = 6/2 = 3 cm

- rate at which water is leaking from the tank = 1.5 cm³/min

- water is being pumped into the tank at constant rate of k cm³/min

- As at height of water, h = 2 cm, the rate of rise in water level = 1 cm/min

c) volume of the tank at any time = πr²h/3

Rate of change in the volume of water in the tank = (rate of flow into the tank) - (Rate of water flow out of the tank)

dV/dt = k - 1.5

V = πr²h/3 and r = 0.3 h, r² = 0.09 h²

V = 0.03πh³

dV/dt = (dV/dh) × (dh/dt)

dV/dh = 0.09π h²

dV/dt = 0.09π h² (dh/dt)

dV/dt = k - 1.5

0.09π h² (dh/dt) = k - 1.5

But at h = 2 cm, (dh/dt) = 1.0 cm/min

0.09π h² (dh/dt) = k - 1.5

0.09π 2² (1) = k - 1.5

k - 1.5 = 1.131

k = 1.5 + 1.131 = 2.631 cm³/min

5 0

2 years ago

A dog walking to the right at 1.5\,\dfrac{\text m}{\text s}1.5sm1, point, 5, space, start fraction, m, divided by, s, end fract

Tatiana [17]

Answer:

8.6 m/s

Explanation:

We can find the final velocity of the dog by using the following SUVAT equation:

$v^2-u^2=2ad$

where

u is the initial velocity

a is the acceleration

d is the distance covered

For the dog in the problem, we have

u = 1.5 m/s

$a = 12 m/s^2$

And the distance covered is

d = 3.0 m

Therefore, we can re-arrange the equation to find the final velocity, v:

$v=\sqrt{u^2+2ad}=\sqrt{1.5^2+2(12)(3.0)}=8.6 m/s$

6 0

2 years ago

Length of table is 1.0m,1.00m and 1.000m.Which one is more accurate?​

Length of table is 1.0m,1.00m and 1.000m.Which one is more accurate?