a. How high did the car climb up the hill after bouncing? Did it go higher lower or the same height as the drop position?

Infer the direction of the net force acting on a car as it slows down and turns right.

Kobotan [32]

Net force would be towards the right and back (opposite direction of motion) since it's slowing down (decelerating) and turning right.

6 0

3 years ago

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If you can throw a stone straight up to height h, what’s the maximum horizontal distance you could throw it over level ground?

Mariana [72]

To answer this question, first we take note that the maximum height that can be reached by an object thrown straight up at a certain speed is calculated through the equation,
Hmax = v²sin²θ/2g
where v is the velocity, θ is the angle (in this case, 90°) and g is the gravitational constant. Since all are known except for v, we can then solve for v whichi s the initial velocity of the projectile.

Once we have the value of v, we multiply this by the total time traveled by the projectile to solve for the value of the range (that is the total horizontal distance).

8 0

3 years ago

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Calcular la aceleración que produce una fuerza de 40 N sobre un cuerpo con 88 Kg de masa. Expresar el resultado en m⁄s^2 *

tigry1 [53]

Answer:

a = 0.45 m/s²

Explanation:

The given question is ''Calculate the acceleration that produces a force of 40 N on a body with 88 kg of mass".

Given that,

Force, F = 40 N

Mass of the body, m = 88 kg

The net force acting on the body is given by :

F = ma

Where

a is the acceleration of the body

$a=\dfrac{F}{m}\\\\a=\dfrac{40\ N}{88\ kg}\\\\a=0.45\ m/s^2$

So, the required acceleration is 0.45 m/s².

4 0

3 years ago

A dentist causes the bit of a high-speed drill to accelerate from an angular speed of 1.10 104 rad/s to an angular speed of 3.14

Anestetic [448]

Answer:

3.63 s

Explanation:

We can solve the problem by using the equivalent SUVAT equations for the angular motion.

To find the angular acceleration, we can use the following equation:

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

where

$\omega_f = 3.14\cdot 10^4 rad/s$ is the final angular speed

$\omega_i = 1.10 \cdot 10^4 rad/s$ is the initial angular speed

$\theta= 2.00 \cdot 10^4 rad$ is the angular distance covered

$\alpha$ is the angular acceleration

Re-arranging the formula, we can find $\alpha$ :

$\alpha=\frac{\omega_f^2-\omega_i^2}{2\theta}=\frac{(3.14\cdot 10^4 rad/s)^2-(1.10\cdot 10^4 rad/s)^2}{2(2.00\cdot 10^4 rad)}=2.16\cdot 10^4 rad/s^2$

Now we want to know the time the bit takes starting from rest to reach a speed of $\omega_f=7.85\cdot 10^4 rad/s$ . So, we can use the following equation: