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Nastasia [14]
2 years ago
6

You are driving your car around a roundabout when you get a flat tire and you decelerate at a constant rate to a stop. The diame

ter of the roundabout is 100m. It takes you 20 sec to come to a complete stop. While slowing down, you continue to drive in a circle and you stop halfway around the loop. What must your speed have been before the pop?
Physics
1 answer:
ycow [4]2 years ago
3 0

Answer:

2.5 meters per second

Explanation:

stops half way which is 50m and if its at a constant speed of 2.5 meters multiply that by the seconds and you get 50m

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Please Help!
irga5000 [103]

The maximum force that the tires can exert on the road before slipping is 16200 N.

From the information in the question;

The coefficient of static friction =  0.9

The mass of the car = 1800 kg

Using the formula;

μ = F/R

μ  = coefficient of static friction

F = force on the tires

R = the reaction force

But recall that the reaction is equal in magnitude to the weight of the car.

W=R

Hence; R = 1800 kg × 10 ms-2 = 18000 N

Making F the subject of the formula;

F = μR

Substituting values;

F =  18000 N × 0.9

F = 16200 N

Hence, the maximum force that the tires can exert on the road before slipping is 16200 N.

Learn more: brainly.com/question/18754989

6 0
3 years ago
A box is dropped from a spacecraft moving horizontally at 27.0 m/s at a distance of 155 m above the surface of a moon. The rate
gavmur [86]

Answer:

a) 10.54 sec

b) 284.58 m

c) 29.406 m/s

d) 39.92 m/s

Explanation:

Given data:

velocity of spacecraft = 27.0 m/s

rate of free fall acceleration is 2.79 m/s^2

distance of moving aircraft from mooon surface is 155 m

a. from kinematic eqaution of motion we have

y = Vi\times t + (\frac{1}{2}) a\times t^2

where y = 155 m

           Vi = 0  as this relation  is for vertical motion, so the 27.0 m/s is not included

and a = 2.79 m/s^2.

Solving for t we get

t = 10.54 sec

b.

we know that V = \frac{d}{t}

d = v\times t

   = 27 \times 10.54 = 284.58 m

c.  from the kinematic formula

v = u + at

v = 0 + 2.79\times 10.57

v = 29.4066 m/a

d. v = \sqrt { 27^2 + 29.406^2}

     v = 39.92 m/s

4 0
3 years ago
Read 2 more answers
What occurs when air moves from an area of high pressure to an area of lower pressure
vazorg [7]
Air flowing from areas of high pressure to low pressure creates wind.
5 0
3 years ago
The highest recorded climatic temperature in Europe is 321 K in Athens, Greece.
stealth61 [152]

For the temperature 321 K, the temperature is 118.13 degree Fahrenheit.

<u>Explanation:</u>

                     degree Fahrenheit = (K - 273.15) * (9/5) + 32

Fahrenheit is a thermodynamic temperature scale, where the freezing point of water is 32 degree Fahrenheit. Based upon the definitions of the Centigrade scale and the experimental evidence, the absolute zero is -273.15 degree Celsius.

                    degree Fahrenheit of 321 K = (321 - 273.15) * (1.8) + 32

                                                                   = 118.13 degree Fahrenheit.

7 0
3 years ago
The energy required to remove an electron from a gaseous atom is called
IRISSAK [1]

Answer: The energy required to remove an electron from a gaseous atom is called ionization energy.

Explanation:

Ionization energy is defined as the energy required to remove the most loosely bound electron from a neutral gaseous atom.

When we move across a period from left to right then there occurs a decrease in atomic size of the atoms. Therefore, ionization energy increases along a period but decreases along a group.

Smaller is the size of an atom more will be the force of attraction between its protons and electrons. Hence, more amount of energy is required to remove an electron.

Thus, we can conclude that the energy required to remove an electron from a gaseous atom is called ionization energy.

6 0
3 years ago
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