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ArbitrLikvidat [17]
2 years ago
9

Why is it that when riding in a car, you don't feel like you're moving?

Physics
1 answer:
UkoKoshka [18]2 years ago
6 0

This may shock you:

We NEVER feel speed, velocity, or motion, as long as it's constant.

We only feel CHANGES in speed, velocity, or motion.

That means speeding up, slowing down, or changing direction.

As long as we're moving in a straight line at a constant speed, we don't feel anything.

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More people end up in U.S. emergency rooms because of fall-related injuries than from any other cause. At what speed v would som
miss Akunina [59]

Answer:

v_{f}=6.47m/s

Explanation:

Given data

Distance d=7.00 ft= 7*(1/3.281) =2.1336m

Initial velocity vi=0m/s

To find

Final velocity

Solution

From Kinematic equation we know that:

v_{f}^{2} =v_{i}^{2}+2gd\\v_{f}^{2}=0+2(9.81m/s^{2} ) (2.1336m)\\v_{f}^{2}=41.86\\v_{f}=\sqrt{41.86}\\v_{f}=6.47m/s

6 0
3 years ago
If a reaction is exothermic, then which of the following statements is always true? If a reaction is exothermic, then which of t
sesenic [268]

Explanation:

Reactants ⇒ product + heat (exothermic reaction) ....(1)

We have given that the reaction is exothermic, so the heat is rejected from the reaction. We know that heat is the form of energy.

From equation (1)

from the given equation we can see that reactants have more energy than products.

So the reactants have higher potential energy in comparison to the products.

7 0
3 years ago
A sample of copper with a mass of 1.80 kg, initially at a temperature of 150.0°C, is in a well-insulated container. Water at a t
user100 [1]

Answer:

the mass of water is 0.3 Kg

Explanation:

since the container is well-insulated, the heat released by the copper is absorbed by the water , therefore:

Q water + Q copper = Q surroundings =0 (insulated)

Q water = - Q copper

since Q = m * c * ( T eq - Ti ) , where m = mass, c = specific heat, T eq = equilibrium temperature and Ti = initial temperature

and denoting w as water and co as copper :

m w * c w * (T eq - Tiw) = - m co * c co * (T eq - Ti co) =  m co * c co * (T co - Ti eq)

m w = m co * c co * (T co - Ti eq) / [ c w * (T eq - Tiw) ]

We take the specific heat of water as c= 1 cal/g °C = 4.186 J/g °C . Also the specific heat of copper can be found in tables → at 25°C c co = 0.385 J/g°C

if we assume that both specific heats do not change during the process (or the change is insignificant)

m w = m co * c co * (T eq - Ti co) / [ c w * (T eq - Tiw) ]

m w= 1.80 kg *  0.385 J/g°C ( 150°C - 70°C) /( 4.186 J/g°C ( 70°C- 27°C))

m w= 0.3 kg

7 0
3 years ago
The speaker at the concert has the sound intensity level of 100 dB if we listen from the distance 5 m.How far from the speaker d
Mademuasel [1]

Answer:

28.11m far from the speaker the intensity drops to 85 dB.

Explanation:

In the equation for the Decibel scale

  (1). \: \:\beta =10 log(\dfrac{I}{I_0})

The ratio of the intensities can be written as

$ \frac{I}{I_0} = \dfrac{\frac{P}{A} }{\frac{P}{A_0} } $

\dfrac{I}{I_0} = \dfrac{A_0}{A}.

And since

A = 4\pi r^2

and

A_0 = 4 \pi r_0^2,

\dfrac{A_0}{A} = \dfrac{4 \pi r_0^2}{4 \pi r^2}  = \dfrac{r_0^2}{r^2}

meaning

\dfrac{I}{I_0} = \dfrac{r_0^2}{r^2}.

Putting this into equation (1), we get:

\boxed{ (2).\: \: \beta = 10log(\dfrac{r_0^2}{r^2})}

Now, if the intensity is 100 dB when the distance is 5 meters, we have:

100dB=10 log(\dfrac{r_02}{(5m)^2})

10= log(\dfrac{r_0^2}{25})

by taking both sides to the exponent:  

10^{10}= \dfrac{r_0^2}{25}

r^2 = 25 *10^{10}\\r = 5 *10^5

Now equation (2) becomes

\beta = 10log(\dfrac{25*10^{10}}{r^2})

when the intensity level is 85 dB we have

85 = 10log(\dfrac{25*10^{10}}{r^2})

8.5 = log(\dfrac{25*10^{10}}{r^2})

take both sides to exponents and we get:

10^{8.5} =10^{ log(\dfrac{25*10^{10}}{r^2})}

10^{8.5} =\dfrac{25*10^{10}}{r^2}

r^2 = \dfrac{25*10^{10}}{10^{8.5}}

\boxed{r = 28.11m}

Thus, 28.11m far from the speaker the intensity drops to 85 dB.

7 0
3 years ago
Helpppppp
aleksley [76]

Negative celestial declinations are all positions in the sky that are directly over south surface latitudes.

Sirius is SOUTH of the celestial equator.

During the course of a year, it traces a circular path directly over the parallel of 16° South latitude, taking it over Peru, La Paz in Bolivia, St. Helena island in the Atlantic, Namibia, Angola, Zambia, Mozambique, Malawi, Madagascar, Australia, French Polynesia, American Samoa, and the Pacific Ocean.

7 0
3 years ago
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