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Ahat [919]
3 years ago
14

You found the distance that you will cover while braking. You can use the expected amount of time for braking to check your answ

er for distance.
Using the information given in the introduction and the value for distance that you found above, determine the time required for you to stop once you press the brakes. Note that the time for braking is just the time interval t3-t2.
This is the information given in the introduction:_______.
You are driving down the highway late one night at 20 m/swhen a deer steps onto the road 35m in front of you. Your reaction time before stepping on the brakes is 0.50s, and the maximum deceleration of your car is 10m/s2. How much distance is between you and the deer when you come to a stop?
Physics
1 answer:
katrin [286]3 years ago
3 0

Answer:

Explanation:

First we need to determine the distance covered during deceleration. According to the equation of motion.

S = ut+1/2at²

Given:

u = 20m/s

t = 0.50s

a = -10m/s (deceleration is negative acceleration)

S = 20²+1/2(-10)(0.5)²

S = 400-5(0.5)²

S = 400-5(0.25)

S = 400-1.25

S = 398.75m

If the deer steps onto the road 35m in front of you, the distance between you and the deer when you come to a stop will be 398.75-35 = 363.75m

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A uniform ladder of mass m and length L stands on a floor at angle α, leaning against a frictionless wall. The static coefficien
luda_lava [24]

To solve this problem it will be necessary to apply the equilibrium conditions. At the same time, it is necessary to make a free body diagram that allows clarifying the origin of the forces with their respective components used to generate the system of equations that allow us to determine the value of the necessary Angle.

As show in free body diagram we have that

F_w = Force due to wall

F_r= Frictional force

N = Normal Force

For equilibrium the moments about point A must be zero, then

\sum M_A = 0

(F_w sin\theta)(L)-(mgcos\alpha)(\frac{L}{2})=0

F_w = \frac{mg}{2tan\alpha}

And similarly the equilibrium with the force show us that,

\sum F_x = 0

N = mg

\sum F_y = 0

F_w = F_r = \mu_s N

From above equation and replacing we have then

\frac{mg}{2tan\alpha} = \mu_sN

\frac{mg}{2tan\alpha} = \mu_s (mg)

tan\alpha = \frac{1}{2\mu_s}

Therefore for equilibrium

tan\alpha >\frac{1}{2\mu_s}

5 0
3 years ago
What is the amount of heat, in calories, given off from a 5 g piece of aluminum when it cools from 80°C to 20°C? The specific he
vagabundo [1.1K]

Answer:

Q (heat) = S * m * (T2 - T1)  where Q is heat gained (loss), S the specific heat capacity of the substance and T2, T1 are the final and initial temps

Q = .215 cal  / (g deg C) * 5 g * (20 - 80) deg C = -64.5 cal

Since the question specifies the heat emitted, then  64.5 cal is the

heat emitted (loss).

8 0
3 years ago
Astronauts wear special device on the ear why give reason
likoan [24]

Answer:

Valsalva device

Explanation: Valsalva device allows astronauts to equalize the pressure in their ears by performing the Valsalva maneuver inside the suit without using their hands to block their nose.

4 0
2 years ago
Read 2 more answers
a 1811kg car speeds up from rest when a traffic light turns green until it reaches a speed of 23m/s. how much work was done on t
sattari [20]

Answer:

The work done by the frictional force, W = 612565.32 J

Explanation:

Given data,

The mass of the car, m = 1811 kg

The initial velocity of the car, u = 0 m/s

The final velocity of the car, v = 23 m/s

Let the time period of the car be, t = 3 s

The acceleration of the car,

                              a = (v-u) /t

                                 = (23 - 0)/ 3

                                = 7.67 m/s²

The normal force acting on the car,

                         F = mg

                             = 1811 kg x 9.8 m/s²

                              = 17747.8 N

The displacement of the car,

                             s = ut + ½ at²

                                 = 0 + ½ x 7.67 x 3²

                                 = 54.52 m

The work done by the frictional force,

                              W = F · S

                              W = 17747.8 N x 54.52 m

                                   = 612565.32 J

Hence, the work done by the frictional force, W = 612565.32 J

3 0
3 years ago
Six dogs pull a two-person sled with a total mass of 220 kg. The coefficient of kinetic friction between the sled and the snow i
irina [24]

Answer:

a) 2457J

b) 558W

c) 337N

Explanation:

Assuming dogs started from rest.

v=a*t\\t=\frac{a}{v}\\v=12\frac{km}{h}\frac{1000m}{km}*\frac{1h}{3600s}\\\\v=3.3m/s\\\\t=\frac{3.3m/s}{0.75m/s^2}\\t=4.4s

and the displacement is given by:

d=\frac{1}{2}*a*t^2\\d=7.3m

Using the energy conservation formula:

K_i+U_i+W_d+W_f=K_f+U_f

Because the motion started from rest the initial kinetic energy is zero, the motion occurred in-ground level so the gravitational energy is zero too.

the work done by the friction force is given by:

W_f=F_f*d*cos(\theta)\\W_f=\µ*m*g*d*cos(180)\\W_f=0.08*220kg*9.8m/s^2*7.3m*(-1)\\W_f=-1259J

so:

W_d=\frac{1}{2}*220kg*(3.3m/s)^2+1259J\\W_d=2457J

The power is given by:

P=\frac{W}{t}\\\\P=\frac{2457J}{4.4s}\\\\P=558W

and the force exerted by the dogs:

W_d=F_d*d*cos(\theta)\\F_d=\frac{W_d}{d*cos(0)}\\\\F=\frac{2457J}{7.3m*(1)}\\\\F=337N

4 0
2 years ago
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