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3 years ago

MULTIPLE CHOICE

Physics

2 answers:

Lady bird [3.3K]3 years ago

6 0

Answer:

abcdefghijklmnopqrstuvwxyz

boyakko [2]3 years ago

3 0

Soft skills are personal attributes that enable someone to interact effectively and harmoniously with other people. so therefore A and F

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The path a projectile takes is known as the Question 1 options: vertical component trajectory horizontal component parabola Ques

Firdavs [7]

1. Trajectory

The path a projectile is called a trajectory in physics. It has a vertical component and even makes a parabola, but if we are talking about physics, it is trajectory.

2. A person sitting in a chair

Projectiles can be defined as an object that is in flight. So it has to be in the air. Since a person sitting in a chair is not in flight, then it is NOT a projectile. (Unless you throw the person in the air while he is in the chair)

3. 490 meters

We have the formula and our given:

d = 1/2gt²

Just plug in the values to get your answer:

d = 1/2(-9.8m/s²)(10s)²

d = (-4.9m/s²)(100s²)

d = -490m

So since height is a scalar value, just take out the negative sign.

4. 65 m/s

Again we have our formula and given:

$v=\dfrac{d}{t}$

So we just plug in our values:

$v=\dfrac{650m}{10s}$

$v=650m/s$

5. True

A projectile, if you will notice its trajectory moves both horizontally and vertically. The horizontal motion is what we call the x-component and the vertical is called the y-component. This is what gives it its' curved path.

6. False

An ellipse is an oval-shaped path. A projectile does not move in a circular/oval path. It travels a curved path. It can be parabolic. It is curved but it does not follow a circular path.

7. The vertical component always equals the horizontal component

This is false. Vertical component is different from the horizontal component. The horizontal component is not influenced by gravity only the vertical component. So in short, the horizontal component is the same throughout, but the vertical component changes over time.

8. Constant

Like mentioned above, the horizontal movement is constant or it does not change. This is because a projectile is defined also as an object that is influenced solely by gravity.

9. Vertical velocity decreases

This is because the movement is against the pull of gravity. It will continue to decrease until it will eventually come to a stop and start to descend. As it descends it increases.

10. Vertical velocity increases

I guess explained it above. As the object descends, the vertical movement increases. This is why you can actually die at certain heights. It increases as the time in flight increases. So the longer in flight, the faster it will get.

3 0

3 years ago

Read 2 more answers

9.5 g bullet has a speed of 1.5 km/s what is the kinetic energy of the bullet

Amiraneli [1.4K]

The answer is 0.000824653J

You need to use the formula Mass * Velocity^2 over 2

3 0

3 years ago

Read 2 more answers

What name is given to the wall of water that makes landfall just ahead of a hurricane?

andreev551 [17]

Pretty sure its a. storm surge

8 0

4 years ago

A projectile is launched horizontally from a 20-m tall edifice with a vox of 25 m/s. How long will it take for the projectile to

NISA [10]

Answer:

a) First let's analyze the vertical problem:

When the projectile is on the air, the only vertical force acting on it is the gravitational force, then the acceleration of the projectile is the gravitational acceleration, and we can write this as:

a(t) = -9.8m/s^2

To get the vertical velocity we need to integrate over time to get:

v(t) = (-9.8m/s^2)*t + v0

where v0 is the initial vertical velocity because the object is thrown horizontally, we do not have any initial vertical velocity, then v0 = 0m/s

v(t) = (-9.8m/s^2)*t

To get the vertical position equation we need to integrate over time again, to get:

p(t) = (1/2)*(-9.8m/s^2)*t^2 + p0

where p0 is the initial position, in this case is the height of the edifice, 20m

then:

p(t) = (-4.9m/s^2)*t^2+ 20m

The projectile will hit the ground when p(t) = 0m, then we need to solve:

(-4.9m/s^2)*t^2+ 20m = 0m

20m = (4.9m/s^2)*t^2

√(20m/ (4.9m/s^2)) = t = 2.02 seconds

The correct option is a.

b) The range will be the total horizontal distance traveled by the projectile, as we do not have any horizontal force, we know that the horizontal velocity is 25 m/s constant.

Now we can use the relationship:

distance = speed*time

We know that the projectile travels for 2.02 seconds, then the total distance that it travels is:

distance = 2.02s*25m/s = 50.5m

Here the correct option is a.

c) Again, the horizontal velocity never changes, is 25m/s constantly, then here the correct option is option b. 25m/s

d) Here we need to evaluate the velocity equation in t = 2.02 seconds, this is the velocity of the projectile when it hits the ground.

v(2.02s) = (-9.8m/s^2)*2.02s = -19.796 m/s

The velocity is negative because it goes down, and it matches with option d, so I suppose that the correct option here is option d (because the sign depends on how you think the problem)

4 0

3 years ago

A thin spherical shell has a radius of 0.70 m. An applied torque of 860 N m gives the shell an angular acceleration of 4.70 rad/

Artyom0805 [142]

Answer:

$I=182.97\ kg-m^2$

Explanation:

Given that,

Radius of a spherical shell, r = 0.7 m

Torque acting on the shell, $\tau=860\ N$

Angular acceleration of the shell, $\alpha =4.7\ m/s^2$

We need to find the rotational inertia of the shell about the axis of rotation. The relation between the torque and the angular acceleration is given by :

$\tau=I\alpha$

I is the rotational inertia of the shell

$I=\dfrac{\tau}{\alpha }\\\\I=\dfrac{860}{4.7}\\\\I=182.97\ kg-m^2$

So, the rotational inertia of the shell is $182.97\ kg-m^2$ .

7 0

3 years ago