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4 years ago

How did the theory of relativity change the law of conservation of energy?

1 answer:

3 0

I believe that the first one is correct because if you have a ball that Weighs 5 pounds and you push it down a ramp it will hit the end with more force than a ball that weighs 2 pounds

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Car À moves at a speed of 8m/s for 43 seconds. Car B moves at a speed of 7 m/s for 50 seconds. Which car traveled a longer dista

nadezda [96]

Distance = (speed) x (time)

Car A: Distance = (8 m/s) x (43 s) = 344 meters

Car B: Distance = (7 m/s) x (50 s) = 350 meters

350 meters is a longer distance than 344 meters.

Car-B traveled a longer distance than Car-A did.

5 0

3 years ago

Read 2 more answers

Fitness results come from hard work, not an instant fix Question 1 options: True False

zhannawk [14.2K]

This is a true statement. Fitness is typically a long-term solution, not a short-term repair. When looking for fitness results, it will usually take weeks to months to see any tangible results, even though there could be some short-term benefits such as increased energy. This is why fitness is sometimes seen as a lifetime endeavor and not one that occurs overnight.

5 0

3 years ago

Un niño empuja una carreta con su perro dentro. La masa del perro y la carreta juntos es de 45 kg. La carreta

Zielflug [23.3K]

Answer:

The force exerted by the child is 38.25 Newton

Explanation:

We use the formula F=mxa (m=mass and a= aceleration):

F= 45kg x 0,85 m/s2=38, 25 kgxm/s2= 38, 25 N

8 0

3 years ago

Write equations for both the electric and magnetic fields for an electromagnetic wave in the red part of the visible spectrum th

NeTakaya

Answer:

Explanation:

General equation of the electromagnetic wave:

$E(x, t)= E_0sin[\frac{2\pi}{\lambda}(x-ct)+\phi ]$

where

$\phi =$ Phase angle, 0

c = speed of the electromagnetic wave, 3 × 10⁸

$\lambda =$ wavelength of electromagnetic wave, 698 × 10⁻⁹m

E₀ = 3.5V/m

Electric field equation

$E(x, t)= 3.5sin[\frac{2\pi}{6.98\times10^{-7}}(x-3\times 10^8t)]\\\\E(x, t)= 3.5sin[{9 \times 10^6}(x-3\times 10^8t)]\\\\E(x, t)= 3.5sin[{9 \times 10^6x-2.7\times 10^{15}t)]$

Magnetic field Equation

$B(x, t)= B_0sin[\frac{2\pi}{\lambda}(x-ct)+\phi ]$

Where B₀= E₀/c

$B_0 = \frac{E_0}{c} = \frac{3.5}{3\times10^8}=1.2 \times 10^{-8}T$

$B(x, t)= 1.2\times10^{-8}sin[\frac{2\pi}{6.98\times10^{-7}}(x-3\times 10^8t)]\\\\B(x, t)= 1.2\times10^{-8}sin[{9 \times 10^6}(x-3\times 10^8t)]\\\\B(x, t)= 1.2\times10^{-8}sin[{9 \times 10^6x-2.7\times 10^{15}t)]$

6 0

3 years ago

The decrease of PE for a freely falling object equals its gain in KE, in accord with the conservation of energy. By simple algeb

Delvig [45]

Answer:

$v = \sqrt{20h}$

Explanation:

The potential energy (PE) we are looking here is gravitational potential energy (GPE).

GPE= mgh,

where m is the mass of an object,

g is the gravitational field strength

h is the height of the object

KE= ½mv²,

where m is the mass and v is the velocity

loss in GPE= gain in KE

mgh= ½mv²

gh= ½v² (divide by m throughout)

Assuming that the object is on earth, then g= 10N/Kg

½v²= 10h (substitute g=10)

v²= 20h (×2 on both sides)

v= √20h (square root both sides)

4 0

3 years ago