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Debora [2.8K]
3 years ago
12

a car travels between the 100 meter and the 250 meter highway marker in 10 seconds calculate the velocity of the car during this

interval
Physics
1 answer:
kozerog [31]3 years ago
4 0

Given:\\\Delta s= 250m-100m=150m\\t=10s\\\\Find:\\v=?\\\\Roz.\\\\v=\frac{\Delta s}{t} \\\\v=\frac{150m}{10s} =15\frac{m}{s}

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How does water flowing over a waterfall involve both kinetic energy and potential energy?
Vinvika [58]

Answer:

While the water falls v increases and h decreases, so the kinetic energy increases and the gravitational potential energy decreases, and this happens in a way that the total energy is always the same. (If there is no friction)

Explanation:

7 0
3 years ago
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Fiora starts riding her bike at 20 mi/h. after a while, she slows down to 12 mi/h, and maintains that speed for the rest of the
hammer [34]
<span>d = r*t

t = hours at 20 mi/hr


20t + 12*(4.5 - t) = 70
8t = 16
t = 2 hours

d at 20 mi/hr = 20*2 = 40 miles

40/20 + 30/12 = 4.5 hours

Fiora travels a total distance of 4.5 hours</span>
3 0
3 years ago
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A 0.155 kg arrow is shot upward
solniwko [45]

Answer:

2.43J

Explanation:

Given parameters:

Mass of the arrow = 0.155kg

Velocity = 31.4m /s

Unknown:

Kinetic energy when it leaves the bow = ?

Solution:

The kinetic energy of a body is the energy in motion of the body;

 it can be derived using the expression below:

 

   K.E  = \frac{1}{2}  m v²

m is the mass

v is the velocity

 Solve for K.E;

    K.E  =  \frac{1}{2}  x 0.155 x 31.4 = 2.43J

3 0
3 years ago
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Which elements will bond ionically with barium such that the formula would be written as BaX2? A) Nitrogen, chlorine, and sodium
erma4kov [3.2K]

The answer would be:

B. Chlorine, iodine and Fluorine

Barium has 2 valence electrons. To satisfy the BaX₂ , this would mean that Barium will need to give one of each of its electrons. The elements that need 1 electron would be those that have 7 valence electrons to complete the octet. These elements would fall in group 7 or halogens. Chlorine, iodine and fluorine are all in Group 7, so this would be the best choice.

7 0
3 years ago
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Two Earth satellites, A and B, each of mass m, are to be launched into circular orbits about Earth's center. Satellite A is to o
Pachacha [2.7K]

(a) 0.448

The gravitational potential energy of a satellite in orbit is given by:

U=-\frac{GMm}{r}

where

G is the gravitational constant

M is the Earth's mass

m is the satellite's mass

r is the distance of the satellite from the Earth's centre, which is sum of the Earth's radius (R) and the altitude of the satellite (h):

r = R + h

We can therefore write the ratio between the potentially energy of satellite B to that of satellite A as

\frac{U_B}{U_A}=\frac{-\frac{GMm}{R+h_B}}{-\frac{GMm}{R+h_A}}=\frac{R+h_A}{R+h_B}

and so, substituting:

R=6370 km\\h_A = 5970 km\\h_B = 21200 km

We find

\frac{U_B}{U_A}=\frac{6370 km+5970 km}{6370 km+21200 km}=0.448

(b) 0.448

The kinetic energy of a satellite in orbit around the Earth is given by

K=\frac{1}{2}\frac{GMm}{r}

So, the ratio between the two kinetic energies is

\frac{K_B}{K_A}=\frac{\frac{1}{2}\frac{GMm}{R+h_B}}{\frac{1}{2}\frac{GMm}{R+h_A}}=\frac{R+h_A}{R+h_B}

Which is exactly identical to the ratio of the potential energies. Therefore, this ratio is also equal to 0.448.

(c) B

The total energy of a satellite is given by the sum of the potential energy and the kinetic energy:

E=U+K=-\frac{GMm}{R+h}+\frac{1}{2}\frac{GMm}{R+h}=-\frac{1}{2}\frac{GMm}{R+h}

For satellite A, we have

E_A=-\frac{1}{2}\frac{GMm}{R+h_A}=-\frac{1}{2}\frac{(6.67\cdot 10^{-11})(5.98\cdot 10^{24}kg)(28.8 kg)}{6.37\cdot 10^6 m+5.97\cdot 10^6 m}=-4.65\cdot 10^8 J

For satellite B, we have

E_B=-\frac{1}{2}\frac{GMm}{R+h_B}=-\frac{1}{2}\frac{(6.67\cdot 10^{-11})(5.98\cdot 10^{24}kg)(28.8 kg)}{6.37\cdot 10^6 m+21.2\cdot 10^6 m}=-2.08\cdot 10^8 J

So, satellite B has the greater total energy (since the energy is negative).

(d) -2.57\cdot 10^8 J

The difference between the energy of the two satellites is:

E_B-E_A=-2.08\cdot 10^8 J-(-4.65\cdot 10^8 J)=-2.57\cdot 10^8 J

4 0
3 years ago
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