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Naya [18.7K]
3 years ago
6

A car is traveling at 100 km/h when the driver sees an accident 80 m ahead and slams on the brakes. what minimum constant decele

ration is required to stop the car in time to avoid a pileup? (round your answer to two decimal places.)
Physics
1 answer:
Effectus [21]3 years ago
4 0
Assuming the driver starts slamming the brakes immediately, the car moves by uniformly decelerated motion, so we can use the following relationship
2aS = v_f^2 - v_i^2 (1)
where 
a is the deleceration
S is the distance covered after a time t
v_f is the velocity at time t
v_i=100 km/h = 27.8 m/s is the initial speed of the car

The accident is 80 m ahead of the car, so the minimum deceleration required to avoid the accident is the value of a such that S=80 m and v_f=0 (the car should stop exactly at S=80 m to avoid the accident). Using these data, we can solve  the equation (1) to find a:
a=- \frac{v_i^2}{2 S}= -\frac{(27.8 m/s)^2}{2 \cdot 80 m} =-4.83 m/s^2
And the negative sign means it is a deceleration.

You might be interested in
The magnification of a microscope is increased when_________.
azamat

Answer:

Option B

Explanation:

Magnification of Microscope is  

M = M_o \times M_e  

Mo= Magnification of objective lens and  

Me= magnification of the eyepiece.  

Both magnifications( of objective and eyepiece) are inversely proportional to the focal length.  

Magnification,  

M \propto \dfrac{1}{f}

when the focal length is less magnification will be high and when the magnification is the low focal length of the microscope will be more.

Thus. Magnification will increase by decreasing the focal length.

The correct answer is Option B

6 0
3 years ago
For years, the tallest tower in the United States was the Phoenix Shot Tower in Baltimore, Maryland. The shot tower was used fro
serious [3.7K]

Answer:

The answer to your question is:

a) t = 3.81 s

b) vf =  37.4 m/s

Explanation:

Data

height = 71.3 m = 234 feet

t = 0 m/s

vf = ?

vo = 0 m/s

Formula

h = vot + 1/2gt²

vf = vo + gt

Process

a)

               h = vot + 1/2gt²

             71.3 = 0t + 1/2(9.81)t²

             2(71.3) = 9,81t²

              t² = 2(71.3)/9.81

              t² = 14.53

              t = 3.81 s

b)

      vf = 0 + (9.81)(3.81)

      vf = 37.4 m/s

3 0
3 years ago
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
In 2h2so4, the number of hydrogen atoms is 1 2 4 8
pishuonlain [190]
C.) In this, number of Hydrogen atoms is 4
7 0
3 years ago
On earth, two parts of a space probe weigh 14500 N and 4800 N. These parts are separated by a center-to-center distance of 18 m
Nastasia [14]

Answer:

F = 1.489*10^{-7}  N

Explanation: Weight of space probes on earth is given by:W= m*g

W= weight of the object( in N)

m= mass of the object (in kg)

g=acceleration due to gravity(9.81 \frac{m}{s^{2} })

Therefore,

m_{1} = \frac{14500}{9.81}

m_{1} = 1478.08  kg

Similarly,

m_{2} = \frac{4800}{9.81}

m_{2} = 489.29  kg

Now, considering these two parts as uniform spherical objects

Also, according to Superposition principle, gravitational net force experienced by an object is sum of all individual forces on the object.

Force between these two objects is given by:

F =  \frac{Gm_{1} m_{2}}{R^{2} }

G= gravitational constant (6.67 * 10^{-11} m^{3} kg^{-1} s^{-2})

m_{1} , m_{2}= masses of the object

R= distance between their centres (in m)(18 m)

Substituiting all these values into the above formula

F = 1.489*10^{-7}  N

This is the magnitude of force experienced by each part in the direction towards the other part, i.e the gravitational force is attractive in nature.

7 0
3 years ago
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