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3 years ago

PLEASE USE KINEMATIC EQUATIONS!!!

Physics

1 answer:

liberstina [14]3 years ago

5 0

The total time for car to travel the given distance is 2 s.

<h3>Time of motion of the car</h3>

The time of motion of the car is determined by applying second kinematic equation as shown below;

$s = vt + \frac{1}{2}at^2\\\\$

where;

v is the constant speed = 6m/s
t is the time of motion = ?
a is the acceleration = 1.5 m/s²
s is the distance traveled by the car = 15 m

Thus, the total time for car to travel the given distance is 2 s.

Learn more about time of motion here: brainly.com/question/2364404

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A 6.0-kg ball is sliding to the right at 25.0m/s and strikes a second ball (15-kg) that is initially at rest head-on. After the

katrin [286]

Answer:

15 m/s to the right

Explanation:

Let's say right is positive and left is negative.

Momentum is conserved:

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

(6.0 kg) (25.0 m/s) + (15 kg) (0 m/s) = (6.0 kg) (-12.5 m/s) + (15 kg) v₂

v₂ = 15 m/s

5 0

3 years ago

Which of these help create radio waves?

inysia [295]

The one that help create radio waves is :
Changing electric and magnetic fields applied at right angles

Radio waves are transverse wave, which means that the oscillations occurring perpendicular to the direction of energy transfer

hope this helps

3 0

3 years ago

A wheel is rotating about an axis that is in the z direction The angular velocity ωz is 6.00 rad s at t 0 increases linearly wit

Amanda [17]

A) $+1.67 rad/s^2$

The angular acceleration of the wheel is given by

$\alpha = \frac{\omega_f - \omega_i}{\Delta t}$

where

$\omega_i = -6.00 rad/s$ is the initial angular velocity of the wheel (initially clockwise, so with a negative sign)

$\omega_f = 4.00 rad/s$ is the final angular velocity (anticlockwise, so with a positive sign)

$\Delta t= 6.00 s - 0=6.00 s$ is the time interval

Substituting into the equation, we find the angular acceleration:

$\alpha = \frac{4.00 rad/s - (-6.00 rad/s)}{6.00 s}=+1.67 rad/s^2$

And the acceleration is positive since the angular velocity increases steadily from a negative value to a positive value.

B) 3.6 s

The time interval during which the angular velocity is increasing is the time interval between the instant $t_1$ where the angular velocity becomes positive (so, $\omega_i=0$ ) and the time corresponding to the final instant $t_2 = 6.0 s$ , where $\omega_f = +6.00 rad/s$ . We can find this time interval by using

$\alpha = \frac{\omega_f - \omega_i}{\Delta t}$

And solving for $\Delta t$ we find

$\Delta t = \frac{\omega_f - \omega_i}{\alpha}=\frac{+6.00 rad/s-0}{+1.67 rad/s^2}=3.6 s$

C) 2.4 s

The time interval during which the angular velocity is idecreasing is the time interval between the initial instant $t_1=0$ when $\omega_i=-4.00 rad/s$ ) and the time corresponding to the instant in which the velovity becomes positive $t_2$ , when $\omega_f = 0 rad/s$ . We can find this time interval by using

$\alpha = \frac{\omega_f - \omega_i}{\Delta t}$

And solving for $\Delta t$ we find

$\Delta t = \frac{\omega_f - \omega_i}{\alpha}=\frac{0-(-4.00 rad/s)}{+1.67 rad/s^2}=2.4 s$

D) 5.6 rad

The angular displacement of the wheel is given by the equation

$\omega_f^2 - \omega_i^2 = 2 \alpha \theta$

where we have

$\omega_i = -6.00 rad/s$ is the initial angular velocity of the wheel

$\omega_f = 4.00 rad/s$ is the final angular velocity

$\alpha=+1.67 rad/s^2$ is the angular acceleration

Solving for $\theta$ ,

$\theta=\frac{\omega_f^2-\omega_i^2}{2\alpha}=\frac{((+6.00 rad/s)^2-(-4.00 rad/s)^2}{2(+1.67 rad/s^2)}=5.6 rad$

3 0

3 years ago

Can any one help me on these two please

Svetach [21]

I THINK 5 is A and 6 is C

4 0

3 years ago

What happens when the force of friction is more than the force of a moving object?

ycow [4]

Inertia is a force which brings all objects to a rest position. Fast-moving objects have more inertia than slow-moving objects. False - The speed of an object has no impact upon the amount of inertia that it has. Mostly True - Two objects of the same mass can weigh differently if they are located in different locations.

I don't know if this will help

7 0

3 years ago

Read 2 more answers