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3 years ago

Amy throws a softball through the air. What are the different forces acting on the ball while it’s in the air?

Physics

2 answers:

Tema [17]3 years ago

5 0

Answer:

An applied force, drag force and gravitational force.

Explanation:

When Amy throws a softball through the air, there is some force that acts on it. The forces that are experienced by the ball in the air as follows :

The softball experiences an applied force as a result of Amy’s throw. As the ball moves, it experiences a drag from the air it passes through. It also experiences a downward pull because of gravity.

The applied force is defined as the force applied by the person who is throwing the ball.

Drag is equivalent to the frictional force that is applied in the opposite direction.

The gravitational force always acts in downward direction.

aniked [119]3 years ago

4 0

An applied force is a force that is applied to an object by a person or another object.
An attractive force is a force of an attraction (where object are attracted by each other). Gravitation is an example of attractive force.
Normal force is the component, perpendicular to the surface (surface being a plane) of contact.
The softball experiences an applied force as a result of Amy’s throw. As the ball moves, it experiences attractive force from the air it passes through. It also experiences a downward pull because of the normal force.
Solution A.

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Using Newton's Second Law, can you explain why one of the major advancements in spaceflight was the development of strong cerami

katovenus [111]

Answer:

high density can withstand high acceleration and applied forces

Heavy metals are toxic to humans,

the clay is quite abundant and in general it is not toxic

Explanation:

The selection of materials for the construction of rockets takes into account many aspects, the technical resistance to the demands of space travel, but also the abundance of the material. Heavy metals have two very serious problems. The first one, some of them are a little scarce in nature, but the most serious problem is that almost all of them are toxic to humans, for example: lead and mercury.

On the other hand, the clay is quite abundant and in general it is not toxic to living beings.

If we use Newton's second law

F = m a

let's use the concept of density

rho = m / V

m = rho V

let's substitute

F = rho V a

From this expression we see that a material with high density can withstand high acceleration and applied forces, such as those existing in spacecraft clearance and re-entry to Earth.

Unfortunately with this law there is no criterion to select a material unless its density is high, in addition to this criterion low toxicity criteria for human beings are used,

8 0

3 years ago

Calculate the linear acceleration (in m/s2) of a car, the 0.310 m radius tires of which have an angular acceleration of 15.0 rad

love history [14]

Answer:

a) The linear acceleration of the car is $4.65\,\frac{m}{s^{2}}$ , b) The tires did 7.46 revolutions in 2.50 seconds from rest.

Explanation:

a) A tire experiments a general plane motion, which is the sum of rotation and translation. The linear acceleration experimented by the car corresponds to the linear acceleration at the center of the tire with respect to the point of contact between tire and ground, whose magnitude is described by the following formula measured in meters per square second:

$\| \vec a \| = \sqrt{a_{r}^{2} + a_{t}^{2}}$

Where:

$a_{r}$ - Magnitude of the radial acceleration, measured in meters per square second.

$a_{t}$ - Magnitude of the tangent acceleration, measured in meters per square second.

Let suppose that tire is moving on a horizontal ground, since radius of curvature is too big, then radial acceleration tends to be zero. So that:

$\| \vec a \| = a_{t}$

$\| \vec a \| = r \cdot \alpha$

Where:

$\alpha$ - Angular acceleration, measured in radians per square second.

$r$ - Radius of rotation (Radius of a tire), measured in meters.

Given that $\alpha = 15\,\frac{rad}{s^{2}}$ and $r = 0.31\,m$ . The linear acceleration experimented by the car is:

$\| \vec a \| = (0.31\,m)\cdot \left(15\,\frac{rad}{s^{2}} \right)$

$\| \vec a \| = 4.65\,\frac{m}{s^{2}}$

The linear acceleration of the car is $4.65\,\frac{m}{s^{2}}$ .

b) Assuming that angular acceleration is constant, the following kinematic equation is used:

$\theta = \theta_{o} + \omega_{o}\cdot t + \frac{1}{2}\cdot \alpha \cdot t^{2}$

Where:

$\theta$ - Final angular position, measured in radians.

$\theta_{o}$ - Initial angular position, measured in radians.

$\omega_{o}$ - Initial angular speed, measured in radians per second.

$\alpha$ - Angular acceleration, measured in radians per square second.

$t$ - Time, measured in seconds.

If $\theta_{o} = 0\,rad$ , $\omega_{o} = 0\,\frac{rad}{s}$ , $\alpha = 15\,\frac{rad}{s^{2}}$ , the final angular position is:

$\theta = 0\,rad + \left(0\,\frac{rad}{s}\right)\cdot (2.50\,s) + \frac{1}{2}\cdot \left(15\,\frac{rad}{s^{2}}\right)\cdot (2.50\,s)^{2}$

$\theta = 46.875\,rad$

Let convert this outcome into revolutions: (1 revolution is equal to 2π radians)

$\theta = 7.46\,rev$

The tires did 7.46 revolutions in 2.50 seconds from rest.

3 0

3 years ago

A box with a mass of 12.5kg sits on the floor how high would you need to lift it has a GPE of 568j

Degger [83]

GPE=mgh
m= 12.5kg
g= 9.81 always
h=?

568=12.5*9.81*h
Solve for h
You will get 4.63m

4 0

3 years ago

Which element can form straight chains, branched chains, and rings

Mice21 [21]

Carbon atoms can form straight, and branched chains, and rings

6 0

3 years ago

Read 2 more answers

How does inertia affect someone who isn't wearing a seatbelt

Paul [167]

Basically, when someone is resting in an accelerated vehicle without restraint from a seatbelt, the force of stopping the vehicle will be when inertia occurs, and that force of the vehicle coming to a stop will affect the passenger (without a seatbelt/restraint from another force or object) greatly by throwing them.
For example;
If I were to be riding in a vehicle (without a seatbelt) that's accelerating at 40 m/s^2 and it suddenly gets slammed on the breaks, I will be thrown forward from inside the vehicle.

I hope this helps!

7 0

3 years ago