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2 years ago

11

Suppose the coefficient of static friction between the road and the tires on a car is 0.638 and the car has no negative lift. Wh

at speed will put the car on the verge of sliding as it rounds a level curve of 25.5 m radius?

1 answer:

larisa [96]2 years ago

4 0

Answer:

12.6332454263 m/s

Explanation:

m = Mass of car

v = Velocity of the car

$\mu$ = Coefficient of static friction = 0.638

g = Acceleration due to gravity = 9.81 m/s²

r = Radius of turn = 25.5 m

When the car is on the verge of sliding we have the force equation

$\dfrac{mv^2}{r}=\mu mg\\\Rightarrow v=\sqrt{\mu gr}\\\Rightarrow v=\sqrt{0.638\times 9.81\times 25.5}\\\Rightarrow v=12.6332454263\ m/s$

The speed of the car that will put it on the verge of sliding is 12.6332454263 m/s

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Object A has mass mA = 9 kg and initial momentum vector pA,i = < 20, -6, 0 > kg · m/s, just before it strikes object B, wh

love history [14]

Answer:

$p= kg m/s$

Explanation:

Momentum is a vector quantity that represents the "amount of motion" of an object.

Mathematically, the momentum of an object is given by

$p=mv$

where

m is the mass of the object

v is the velocity

Since momentum is a vector, it also has a direction, which is the same as the velocity.

Therefore, if we have two objects, the total momentum of the two objects will be obtained from the vector sum of the individual momenta of the two objects.

In this problem we have:

$p_A= kg m/s$ is the momentum of object A

$p_B = kg m/s$ is the momentum of object B

Therefore, the total momentum of objects A and B can be obtained by adding each components of A to the corresponding component of B, so:

$p_x = 20 +6 = 26 kg m/s\\p_y = -6 +6 = 0 kg m/s\\p_z = 0 + 0 = 0 kg m/s$

So the total initial momentum is

$p= kg m/s$

6 0

2 years ago

A positive point charge q1 = +5.00 × 10−4C is held at a fixed position. A small object with mass 4.00×10−3kg and charge q2 = −3.

Lelechka [254]

Answer:

Therefore the speed of q₂ is 1961.19 m/s when it is 0.200 m from from q₁.

Explanation:

Energy conservation law: In isolated system the amount of total energy remains constant.

The types of energy are

Kinetic energy.
Potential energy.

Kinetic energy $=\frac{1}{2} mv^2$

Potential energy = $\frac{Kq_1q_2}{d}$

Here, q₁= +5.00×10⁻⁴C

q₂=-3.00×10⁻⁴C

d= distance = 4.00 m

V = velocity = 800 m/s

Total energy(E) =Kinetic energy+Potential energy

$=\frac{1}{2} mv^2$ + $\frac{Kq_1q_2}{d}$

$=\frac{1}{2} \times 4.00\times 10^{-3}\times(800)^2 +\frac{9\times10^9\times 5\times10^{-4}\times(-3\times10^{-4})}{4}$

=(1280-337.5)J

=942.5 J

Total energy of a system remains constant.

Therefore,

E $=\frac{1}{2} mv^2$ + $\frac{Kq_1q_2}{d}$

$\Rightarrow 942.5 = \frac{1}{2} \times 4 \times10^{-3} \times V^2 +\frac{9\times10^{9}\times5\times 10^{-4}\times(-3\times 10^{-4})}{0.2}$

$\Rightarrow 942.5 = 2\times10^{-3}v^2 -6750$

$\Rightarrow 2 \times10^{-3}\times v^2= 942.5+6750$

$\Rightarrow v^2 = \frac{7692.5}{2\times 10^{-3}}$

$\Rightarrow v= 1961.19$ m/s

Therefore the speed of q₂ is 1961.19 m/s when it is 0.200 m from from q₁.

5 0

2 years ago

what is the kinetic energy of a 1 kilogram ball is thrown into the air with an initial velocity of 3 m/sec

s344n2d4d5 [400]

Data:
m (mass) = 1 Kg
s (speed) = 3 m/s
Kinetic energy = ? (Joule)

Formula (Kinetic energy)
$E_{k} = \frac{m*s^2}{2}$

Solving:
$E_{k} = \frac{m*s^2}{2}$
$E_{k} = \frac{1*3^2}{2}$
$E_{k} = \frac{1*9}{2}$
$E_{k} = \frac{9}{2}$
$\boxed{\boxed{E_{k} = 4.5\:J}}\end{array}}\qquad\quad\checkmark$

3 0

3 years ago

Describe how work is done by a skater pulling in her arms during a spin. In particular, identify the force she exerts on each ar

liberstina [14]

Answer:

∑ τ =0, L₀ = $L_{f}$

Explanation:

In a circular turning movement, when the arms are extended and then contracted in two possibilities:

- They are lowered the force of gravity is what pulls them, the tension of the muscle becomes zero to allow this movement.

In this movement the force is vertical(gravity) and the movement of the center of mass of each arm is vertical, so that the work is the weight value of the arm by the distance traveled by the center of mass.

- Another possibility is that the arms have stuck to the body, in this case the person's muscles perform the force, this force is horizontal and the displacement is the horizontal of the center of mass of the arms from the extended position to the contracted

In these movements the torque of the external force is equal for each arm, but in the opposite direction, so they are canceled where a net torque of zero, this causes the angular momentum to be preserved, which changes is the moment of inertia of the system and therefore you must also change the angular velocity to keep your product constant

∑ τ =0

L₀ = $L_{f}$

I₀ w₀ = I w

4 0

3 years ago

In terms of saturation (unsaturated, saturated, super-saturated). How would you classify the following?

miskamm [114]

<u>Answer:</u>

<em>1. A NaCl solution with a concentration of 50g/100mL of water at 40°C:</em> The NaCl solution with a given concentration is saturated at this temperature .As the temperature increases the solution will more dissolves.

<em>2. A sugar solution with a concentration of 200g/100mL of water at 40°C: </em>The sugar solution with a given concentration is saturated at this temperature. As the temperature increases the solution will more dissolves.

<em>3. A sugar solution with a concentration of 240g/100mL of water at 40°C:</em> The sugar solution with a given concentration is saturated at given temperature.

7 0

2 years ago