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3 years ago

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A driver in a car,originally moving 12.2 m/s, applies the brakes until the car comes to a stop. The car moves a distance of 36.5

m while braking. How much time did it take for the car to stop? Assume constant acceleration during braking.

1 answer:

mojhsa [17]3 years ago

8 0

First we have to calculate the acceleration of the car,

$a =\frac{u-v}{t}$

Here, u is initial velocity of the car and its value is given 12.2 m/s and v is final velocity of the car and it comes to stop, so its value zero.

$a=\frac{0-12.2 m/s}{t} =\frac{-12.2 \ m/s}{t}$ .

As during the braking the acceleration is constant, from the kinematic equation,

$s=ut + \frac{1}{2} a t^2$

Here, s is the distance traveled by the car during braking and its value is given 36. 5 m.

Substituting all the values in kinematic equation, we get

$36.5 m =(12.2 m/s) t + \frac{1}{2} (- \frac{12.2 m/s}{t}) t^2 \\\\ 36.5 \ m = (12.2 \ m/s) t -(6.1 \ m/s) t \\\\\ t = \frac{36.5 m}{6.1 m/s} = 5.98 s$

Therefore, car will stop after 5.98 s.

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If the acceleration of the projective is: a = c s m/s 2 Where c is a constant that depends on the initial gas pressure behind th

Sedbober [7]

Answer:

c = 4,444.44

Explanation:

You have the following expression for the acceleration of the projectile:

$a=cs$ (1)

s: distance to the ground of the projectile

To find the value of the constant c you use the following formula:

$v^2=v_o^2+2a \Delta s$ (2)

vo: initial velocity = 0 m/s

v: final speed = 200 m/s

Δs: distance traveled by the projectile = 3m - 1.5m = 1.5m

You replace the expression (1) into the expression (2):

$v^2=2(cs)\Delta s$

You do the constant c in the last equation, then you replace the values of v, s and Δs:

$c=\frac{v^2}{2s\Delta s}=\frac{(200m/s)^2}{2(3m/s^2)(1.5m)}=4444.44$

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3 years ago

A 5.0 kg block hangs from the ceiling by a mass-less rope. A Second block with a mass of 10.0 kg is attached to the first block

gayaneshka [121]

The tension in the first and second rope are; 147 Newton and 98 Newton respectively.

Given the data in the question

Mass of first block; $m_1 = 5.0kg$

Mass of second block, $m_2 =10kg$

Tension on first rope; $T_1 =\ ?$

Tension on second rope; $T_2 =\ ?$

To find the Tension in each of the ropes, we make use of the equation from Newton's Second Laws of Motion:

$F = m\ *\ a$

Where F is the force, m is the mass of the object and a is the acceleration ( In this case the block is under gravity. Hence ''a" becomes acceleration due to gravity $g = 9.8m/s^2$ )

For the First Rope

Total mass hanging on it; $m_T = m_1 + m_2 = 5.0kg + 10.0kg = 15.0kg$

So Tension of the rope;

$F = m\ * \ g\\\\F = 15.0kg \ * 9.8m/s^2\\\\F = 147 kg.m/s^2\\\\F = 147N$

Therefore, the tension in the first rope is 147 Newton

For the Second Rope

Since only the block of mass 10kg is hang from the second, the tension in the second rope will be;

$F = m\ * \ g\\\\F = 10.0kg \ * 9.8m/s^2\\\\F = 98 kg.m/s^2\\\\F = 98N$

Therefore, the tension in the second rope is 98 Newton

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Which equation can be used to solve for acceleration? <br><br>

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The cabinet is mounted on coasters and has a mass of 45 kg. The casters are locked to prevent the tires from rotating. The coeff

stira [4]

Answer:

the force P required for impending motion is 132.3 N

the largest value of "h" allowed if the cabinet is not to tip over is 0.8 m

Explanation:

Given that:

mass of the cabinet m = 45 kg

coefficient of static friction μ = 0.30

A free flow body diagram illustrating what the question represents is attached in the file below;

The given condition from the question let us realize that ; the casters are locked to prevent the tires from rotating.

Thus; considering the forces along the vertical axis ; we have :

$\sum f_y =0$

The upward force and the downward force is :

$N_A+N_B = mg$

where;

$\mathbf { N_A \ and \ N_B}$ are the normal contact force at center point A and B respectively .

$N_A+N_B = 45*9.8$

$N_A+N_B = 441$ ------- equation (1)

Considering the forces on the horizontal axis:

$\sum f_x = 0$

$F_A +F_B = P$

where ;

$\mathbf{ F_A \ and \ F_B }$ are the static friction at center point A and B respectively.

which can be written also as:

$\mu_s N_A + \mu_s N_B = P$

$\mu_s( N_A + N_B) = P$

replacing our value from equation (1)

P = 0.30 ( 441)

P = 132.3 N

Thus; the force P required for impending motion is 132.3 N

b) Since the horizontal distance between the casters A and B is 480 mm; Then half the distance = 480 mm/2 = 240 mm = 0.24 cm

the largest value of "h" allowed for the cabinet is not to tip over is calculated by determining the limiting condition of the unbalanced torque whose effect is canceled by the normal reaction at $N_A$ and it is shifted to $N_B$ :

Then:

$\sum M _B = 0$

$P*h = mg*0.24$

$h =\frac{45*9.8*0.24}{132.3}$

h = 0.8 m

Thus; the largest value of "h" allowed if the cabinet is not to tip over is 0.8 m

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3 years ago