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3 years ago

A 325 N force applied to an object for 5.5 s. What’s the impulse?

1 answer:

4 0

According to Newton second law of motion, the resultant force is directly proportional to the rate of change in momentum while maintaining other factors constant. Therefore, F = (mv-mu)/t where F is the resultant force , m is the mass of the object, v is the final velocity and u is the initial velocity.
Hence, Ft = mv-mu, but impulse is given by force multiplied by time, thus, impulse is equivalent to the change in momentum.
Impulse = Ft
= 325 × 2.2 sec
= 715 Ns

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A heavy object falls with the acceleration as a light object during free fall. why?

alexgriva [62]

A heavy object falls with the acceleration as a light object during free fall because:

Heavy things have a large gravitational force and also have less acceleration. So both the effects exactly cancel and make the falling objects to have the same acceleration irrespective of mass.

Free fall is a unique motion which has only gravitational force that acts on an object. Objects that undergo free fall experience only have the influence of gravity and not any other force.

So when we apply newton's second law of gravity which is:

$\text {acceleration}=\frac{\text {Force}}{\text {mass}}$

where,

F is the force

m is the mass

a is the acceleration

For example: When a 1000 kg elephant and a 1 kg rat fall from the same height,

The acceleration can be calculated as follows:

For elephant: F = 10000 N and m = 1000 kg. So,

$\text {acceleration}=\frac{10000}{1000}=10\ \mathrm{m} / \mathrm{s}^{2}$

For rat : F = 10 N and m = 1 kg

Thus, $\text {acceleration}=\frac{10}{1}=1\ \mathrm{m} / \mathrm{s}^{2}$

Hence, it shows that both the animals have the same acceleration irrespective of their mass.

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3 years ago

A 8.0-g bullet with an initial velocity of 115.0 m/s lodges in a block of wood and comes to rest at a

Gnesinka [82]

A. When your fire a bullet into the air, it typically takes between 20 and 90 second for it to come down,depending on the angle it was fired at,its muzzle velocity and caliber.

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3 years ago

Can someone please help a struggling physics student?

zhenek [66]

What is the maximum height the ball will reach in the air?

Kinematics equation used:

$v_f^2=v_i^2+2ad$ , where $v_f$ is final velocity, $v_i$ is initial velocity, $a$ is acceleration, and $d$ is distance travelled. From SI units, velocity should be in $m/s$ , acceleration should be in $m/s^2$ , and distance should be in $m$

We're given that the initial velocity is 12.0 m/s in the y-direction. At the maximum height, the vertical velocity of the ball will be 0 m/s, otherwise it would not be at maximum height. This is our final velocity.

The only acceleration in the system is acceleration due to gravity, which is approximately $9.8\:\mathrm{ m/s^2}$ . However, the acceleration is acting down, whereas the ball is moving up. To express its direction, acceleration should be plugged in as $-9.8\:\mathrm{m/s^2}$ . We have three variables, and we are solving for the fourth, which is distance travelled. This will be the maximum height of the ball.

Substitute $v_i=12$ , $v_f=0$ , $a=-9.8$ to solve for $d$ :

$0^2=12^2+2(-9.8)(d),\\0=144-19.6d,\\-19.6d=-144,\\d=\frac{-144}{-19.6}=7.34693877551\approx \boxed{7.35\text{ m}}$

What is the velocity of the ball when it hits the ground?

This question tests a physics concept rather than a physics formula. The vertical velocity of the ball when it hits the ground is equal in magnitude but opposite in direction to the ball's initial vertical velocity. This is because the ball spends equal time travelling to its max height as it does travelling from max height to the ground (ball is accelerating from initial velocity to 0 and then from 0 to some velocity over the same distance and time). Since the ball has an initial vertical velocity of +12.0 m/s, its velocity when it hits the ground will be $\boxed{-12.0\text{ m/s}}$ . (The negative sign represents the direction. Because velocity is a vector, it is required.)

**Since my initial answer exceeds the character limit, I've attached the first question to Part B as an image. Please refer to the attached image for the answer and explanation to the first question of Part B. Apologies for the inconvenience.**

What is the direction of the velocity of the ball when it hits the ground? Express your answer in terms of the angle (in degrees ) of the ball's velocity with respect to the horizontal direction (see figure).

This question uses a similar concept as the second question of Part A. The vertical velocity of the ball at launch is equal in magnitude but opposite in direction to the ball's final velocity. The horizontal component is equal in both magnitude and direction throughout the entire launch, since there are no horizontal forces acting on the system. Therefore, the angle below the horizontal of the ball's velocity when it hits the ground is equal to the angle of the ball to the horizontal at launch.

To find this, we need to use basic trigonometry for a right triangle. In any right triangle, the tangent/tan of an angle is equal to its opposite side divided by its adjacent side.

Let the angle to the horizontal at launch be $\theta$ . The angle's opposite side is represented by the vertical velocity at launch (12.0 m/s) and the angle's adjacent side is represented by the horizontal velocity at launch (2.3 m/s). Therefore, we have the following equation:

$\tan \theta=\frac{12.0\text{m/s}}{2.3\text{ m/s}}$

Take the inverse tangent of both sides:

$\arctan (\tan \theta)=\arctan (\frac{12.0}{2.3})$

Simplify using $\arctan(\tan \theta)=\theta \text{ for }\theta \in (-90^{\circ}, 90^{\circ})$ :

$\theta=\arctan(\frac{12.3}{2.3}),\\\theta =79.14989537\approx \boxed{79.15^{\circ}}$

We can express our answer by saying that the direction of the velocity of the ball when it hits the ground is $\boxed{\text{approximately }79.15^{\circ} \text{ below the horizontal}}$ or $\boxed{\text{approximately }-79.15^{\circ} \text{ to the horizontal}}$ .