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3 years ago

Why do you think the standards changed over time?

Physics

1 answer:

Eddi Din [679]3 years ago

7 0

Answer:

https://www.quora.com/Why-do-you-think-standards-of-beauty-have-changed-over-the-years

Explanation:

your answer is in this link

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A toy car is moving across a table with a velocity of 7.0 m/s and drives off the end. The table is 1.8 m tall. How long will the

KatRina [158]

Answer:

Option D - 0.2 s

Explanation:

We are given;

Initial velocity; u = 7 m/s

Height of table; h = 1.8m

Now,since we want to find the time the car spent in the air, we will simply use one of Newton's equation of motion.

Thus;

h = ut + ½gt²

Plugging in the relevant values, we have;

1.8 = 7t + ½(9.8)t²

4.9t² + 7t - 1.8 = 0

Using quadratic formula to find the roots of the equation gives us;

t = -1.65 or 0.22

We can't have negative t value, thus we will pick the positive one.

So, t = 0.22 s

This is approximately 0.2 s

7 0

3 years ago

Fill in the blanks. The electrostatic force between two objects is proportional to the ____________________ of the distance ____

Shkiper50 [21]

Write an equation to calculate the force between two objects if the product of their charges is 10.0 × 10-4 C. (Note: Use the variable R for the distance between the charges.)

F = 900 ÷_________

6 0

3 years ago

Read 2 more answers

At one instant, a 17.0-kg sled is moving over a horizontal surface of snow at 4.10 m/s. After 6.15 s has elapsed, the sled stops

jasenka [17]

Answer:

force = 11.33 $kg-m/s^{2}$

Explanation:

given data:

sled mass = 17.0 kg

inital velocity (U) = 4.10 m/s

elapsed time (T) 6.15 s

final velocity (V) = 0

final momentum P2 = 0

Initial momentum of sledge is

$P_{1}=mU$

$P_{1}= 17.0 * 4.10 = 69.7 kg- m/s$

from newton second law of motion

$F=\frac{\Delta P}{\Delta t}$

$F = \frac{P_{1}-P_{2}}{T}$

Kgm/s^2

$F = \frac{69.7-0}{6.15}= 11.33[tex]kg-m/s^{2}$ [/tex]

8 0

3 years ago

A battery with an internal resistance ofrand an emf of 10.00 V is connected to a loadresistorR=r. As the battery ages, the inter

yanalaym [24]

Answer:

The current is reduced to half of its original value.

Explanation:

Assuming we can apply Ohm's Law to the circuit, as the internal resistance and the load resistor are in series, we can find the current I₁ as follows:

$I_{1} = \frac{V}{R_{int} +r_{L} }$

where Rint = r and RL = r
Replacing these values in I₁, we have:

$I_{1} = \frac{V}{R_{int} +r_{L} } = \frac{V}{2*r} (1)$

When the battery ages, if the internal resistance triples, the new current can be found using Ohm's Law again:

$I_{2} = \frac{V}{R_{int} +r_{L} } = \frac{V}{(3*r) +r} = \frac{V}{4*r} (2)$

We can find the relationship between I₂, and I₁, dividing both sides, as follows:

$\frac{I_{2} }{I_{1} } = \frac{V}{4*r} *\frac{2*r}{V} = \frac{1}{2}$

The current when the internal resistance triples, is half of the original value, when the internal resistance was r, equal to the resistance of the load.

7 0

3 years ago

If the angular velocity of the pully is -8.4rad/s at a given time, and its anglar acceleration is -2.8rad/s2, what is the angula

snow_lady [41]

Here we know that

$\omega_i = - 8.4 rad/s$