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Sliva [168]
3 years ago
11

Formula One racers speed up much more quickly than normal passenger vehicles, and they also can stop in a much shorter distance.

A Formula One racer traveling at 90m/s can stop in a distance of 110m. What is the magnitude of the car's acceleration as it slows during braking?
Physics
1 answer:
klasskru [66]3 years ago
4 0

Answer:

The magnitude of the car's acceleration as it slows during braking is 36.81 m/s²

Explanation:

From the question, the given values are as follows:

Initial velocity, u = 90 m/s

final velocity, v = 0 m/s

distance, s = 110 m

acceleration, a = ?

Using the equation of motion, v² = u² + 2as

(90)² + 2 * 110 * a = 0

8100 + 220a = 0

220a = -8100

a = -8100/220

a = -36.81 m/s²

The value for acceleration is negative showing that car is decelerating to a stop. The magnitude of the car's acceleration as it slows during braking is therefore 36.81 m/s²

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When the distance between two masses is doubled, the gravitational attraction between
mrs_skeptik [129]

The new gravitational attraction will be 1/4 as much

Explanation:

The magnitude of the gravitational force between two objects is given by

F=G\frac{m_1 m_2}{r^2}

where

G=6.67\cdot 10^{-11} m^3 kg^{-1}s^{-2} is the gravitational constant

m1, m2 are the masses of the two objects

r is the separation between them

In this problem, the original force between the two objects is F, when they are separated by a distance r.

Later, the distance between the two objects is doubled, so the new distance is

r'=2r

Therefore, the new force will be

F'=G\frac{m_1 m_2}{(2r)^2}=\frac{1}{4}(\frac{Gm_1 m_2}{r^2})=\frac{F}{4}

Therefore, the new force will be one-fourth as much.

Learn more about gravitational force:

brainly.com/question/1724648

brainly.com/question/12785992

#LearnwithBrainly

5 0
3 years ago
A plane accelerates from rest at a constant rate of 5.00 m/s2 along a runway that is 1800 m long. Assume that the plane reaches
tiny-mole [99]

Answer:

26.8 seconds

Explanation:

To solve this problem we have to use 2 kinematics equations: *I can't use subscripts for some reason on here so I am going to use these variables:

v = final velocity

z = initial velocity

x = distance

t = time

a = acceleration

{v}^{2}  =  {z}^{2}  + 2ax

v = z + at

First let's find the final velocity the plane will have at the end of the runway using the first equation:

{v}^{2}  =  {0}^{2}  + 2(5)(1800)

v = 60 \sqrt{5}

Now we can plug this into the second equation to find t:

60 \sqrt{5}  = 0 + 5t

t = 12 \sqrt{5}

Then using 3 significant figures we round to 26.8 seconds

3 0
2 years ago
ow long must a simple pendulum be if it is to make exactly ten swings per second? (That is, one complete vibration takes exactly
Igoryamba
The period T of a pendulum is given by:
T=2 \pi  \sqrt{ \frac{L}{g} }
where L is the length of the pendulum while g=9.81 m/s^2 is the gravitational acceleration.

In the pendulum of the problem, one complete vibration takes exactly 0.200 s, this means its period is T=0.200 s. Using this data, we can solve the previous formula to find L:
L=g ( \frac{T}{2\pi} )^2=(9.81 m/s^2)( \frac{0.2 s}{2 \pi} )^2=1 \cdot 10^{-3} m=1 mm
4 0
3 years ago
Carbon dioxide in a piston-‐‐cylinder is expanded in a polytropic manner. The initialtemperature and pressure are 400 K and 550
Arisa [49]

Answer:

 q_poly = 14.55 KJ/kg

Explanation:

Given:

Initial State:

P_i = 550 KPa

T_i = 400 K

Final State:

T_f = 350 K

Constants:

R = 0.189 KJ/kgK

k = 1.289 = c_p / c_v

n = 1.2   (poly-tropic index)

Find:

Determine the heat transfer per kg in the process.

Solution:

-The heat transfer per kg of poly-tropic process is given by the expression:

                            q_poly = w_poly*(k - n)/(k-1)

- Evaluate w_poly:

                            w_poly = R*(T_f - T_i)/(1-n)

                            w_poly = 0.189*(350 - 400)/(1-1.2)

                            w_poly = 47.25 KJ/kg

-Hence,

                           q_poly = 47.25*(1.289 - 1.2)/(1.289-1)

                           q_poly = 14.55 KJ/kg

4 0
3 years ago
A projectile has an initial horizontal velocity of 15 meters per second and an initial vertical velocity of 25 meters per second
Artyom0805 [142]

Answer:

75 m

Explanation:

The horizontal motion of the projectile is a uniform motion with constant speed, since there are no forces acting along the horizontal direction (if we neglect air resistance), so the horizontal acceleration is zero.

The horizontal component of the velocity of the projectile is

v_x = 15 m/s

and it is constant during the motion;

the total time of flight is

t = 5 s

Therefore, we can apply the formula of the uniform motion to find the horizontal displacement of the projectile:

d= v_x t =(15 m/s)(5 s)=75 m

5 0
3 years ago
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