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Free_Kalibri [48]
3 years ago
5

If you wanted to know how big a tiger is, which measurements would you use?

Physics
1 answer:
valina [46]3 years ago
5 0

Answer:

meter, kilogram

Explanation:

Here we want to know how big the tiger is. This means that we want to measure its size and possibly its mass.

The size is actually a measure of the length of the tiger, and length is measured in meters.

The mass of an object, instead, is a measure of the "amount of matter" in the substance, and it is measured in kilograms.

The other options are wrong because:

- The second is the unit of time

- The candela is the unit of the luminous intensity

- The mole is the unit of the amount of substance, and it is used for gases

- The ampere is the unit of the current

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True or false if an object has energy it must be moving
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Answer:

Explanation:

If an object is moving, it is said to have kinetic energy (KE). Potential energy (PE) is energy that is "stored" because of the position and/or arrangement of the object. The classic example of potential energy is to pick up a brick. When it's on the ground, the brick had a certain amount of energy.

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3 years ago
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3 years ago
In an experiment, students measure the position x of a cart as a function of time t for a cart that starts at rest and moves wit
Minchanka [31]

Given :

Initial velocity , u = 0 m/s² .

To Find :

The acceleration of the cart.

Solution :

Since, acceleration is constant.

Using equation of motion :

x = ut + \dfrac{at^2}{2}\\\\x = \dfrac{at^2}{2}

Putting, t = 1 s  and x = 4 m in above equation, we get :

4 = \dfrac{a(1)^2}{2}\\\\a = 8 \  m/s^2

Therefore, the acceleration of the cart is 8 m/s².

5 0
3 years ago
Plz, help the rubber ball be dropped from the top of a ladder. It bounces on the same spot on the ground several times to a less
Vladimir [108]

Answer:

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6 0
2 years ago
In an RLC series circuit that includes a source of alternating current operating at fixed frequency and voltage, the resistance
maw [93]

Answer:

Capacitive Reactance is 4 times of resistance

Solution:

As per the question:

R = X_{L} = j\omega L = 2\pi fL

where

R = resistance

X_{L} = Inductive Reactance

f = fixed frequency

Now,

For a parallel plate capacitor, capacitance, C:

C = \frac{\epsilon_{o}A}{x}

where

x = separation between the parallel plates

Thus

C ∝ \frac{1}{x}

Now, if the distance reduces to one-third:

Capacitance becomes 3 times of the initial capacitace, i.e., x' = 3x, then C' = 3C and hence Current, I becomes 3I.

Also,

Z = \sqrt{R^{2} + (X_{L} - X_{C})^{2}}

Also,

Z ∝ I

Therefore,

\frac{Z}{I} = \frac{Z'}{I'}

\frac{\sqrt{R^{2} + (R - X_{C})^{2}}}{3I} = \frac{\sqrt{R^{2} + (R - \frac{X_{C}}{3})^{2}}}{I}

{R^{2} + (R - X_{C})^{2}} = 9({R^{2} + (R - \frac{X_{C}}{3})^{2}})

{R^{2} + R^{2} + X_{C}^{2} - 2RX_{C} = 9({R^{2} + R^{2} + \frac{X_{C}^{2}}{9} - 2RX_{C})

Solving the above eqn:

X_{C} = 4R

6 0
3 years ago
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