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iragen [17]
2 years ago
9

You drop your keys in a high-speed elevator going up at a constant speed. Part APart complete Do the keys accelerate faster towa

rd the elevator floor than they would if the elevator were not moving? Do the keys accelerate faster toward the elevator floor than they would if the elevator were not moving? Yes No Previous Answers Correct Part B Do the keys accelerate faster toward the elevator floor than they would if the elevator were accelerating downward? Do the keys accelerate faster toward the elevator floor than they would if the elevator were accelerating downward? No Yes
Physics
1 answer:
anzhelika [568]2 years ago
5 0

Answer:

Explained

Explanation:

a) No, the keys were initially moving upward in the elevator only effects the initial velocity of the key and not the rate of change of velocity that is acceleration. So, the keys accelerate with the same acceleration as before.

b)Yes, keys will accelerate towards the floor faster if it is a constant speed than it is moving downward because if the elevator is accelerating downward, the downward change in velocity of the keys is at least partially matched by a downward change in the velocity of the of the elevator.

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With your hand parallel to the floor and your palm upright, you lower a 3-kg book downward. If the force exerted on the book by
Stells [14]
For Newton's second law, the resultant of the forces acting on the book is equal to the product between the mass of the book and its acceleration:
\sum F = ma (1)

There are only two forces acting on the book:
- its weight, directed downward: mg
- the force exerted by the hand on the book, of 20 N, directed upward

so, equation (1) becomes
mg - F = ma
from which we can calculate the book's acceleration, a:
a= g -  \frac{F}{m}= 9.81 m/s^2 - \frac{20 N}{3 kg}=3.14 m/s^2
7 0
2 years ago
Read 2 more answers
A 3.00-kg model airplane has velocity components of 5.00 m/s due east and 8.00 m/s due north. What is the plane’s kinetic energy
GalinKa [24]

Answer:

Kinetic energy, E = 133.38 Joules

Explanation:

It is given that,

Mass of the model airplane, m = 3 kg

Velocity component, v₁ = 5 m/s (due east)

Velocity component, v₂ = 8 m/s (due north)

Let v is the resultant of velocity. It is given by :

v=\sqrt{v_1^2+v_2^2}

v=\sqrt{5^2+8^2}=9.43\ m/s

Let E is the kinetic energy of the plane. It is given by :

E=\dfrac{1}{2}mv^2

E=\dfrac{1}{2}\times 3\ kg\times (9.43\ m/s)^2

E = 133.38 Joules

So, the kinetic energy of the plane is 133.38 Joules. Hence, this is the required solution.

5 0
2 years ago
Read 2 more answers
The note created by a flute will increase the speed of sound increases. When a marching band goes outside on a cold day, what wo
alexgriva [62]

A).

It would decrease because the speed of sound and temperature are proportional.

4 0
2 years ago
Read 2 more answers
A 50-n crate sits on a horizontal floor where the coefficient of static friction between the crate and the floor is 0.50 . A 20-
andreyandreev [35.5K]

The resultant static friction force is equal to 20 N to the left.

Why?

I'm assuming that you forgot to write the question of the exercise, so,  I will try to complete it:

"A 50-n crate sits on a horizontal floor where the coefficient of static friction between the crate and the floor is 0.50 . A 20-n force is applied to the crate acting to the right. What is the resulting static friction force acting on the crate?"

So, if we are going to calculate the resulting static friction force, it means that there is no movement, we must remember that the friction coefficient will give us the maximum force before the crate starts to move.

We can calculate the static friction force by using the following formula:

Fr=F(appliedforce)

Since the crate is not moving (static), the static friction force acting on the crate will be equal to the applied force.

Calculating we have:

Fr=F(appliedforce)

Fr=20N

Hence, the static friction force is equal to 20 N to the left (since the applied force is acting to the right)

So,

FrictionForce=AppliedForce

Since the static friction force is equal to the applied force, the crate does not start to move.

Have a nice day!

8 0
2 years ago
You are exploring a distant planet. When your spaceship is in a circular orbit at a distance of 630 km above the planet's surfac
NemiM [27]

Answer:

The horizontal range of the projectile = 26.63 meters

Explanation:

Step 1: Data given

Distance above the planet's surface = 630 km = 630000

The ship's orbal speed = 4900 m/s

Radius of the planet = 4.48 *10^6 m

Initial speed of the projectile = 13.6 m/s

Angle = 30.8 °

Step 2: Calculate g

g= GM /R² = (v²*(R+h)) /(R²)

⇒ with v= the ship's orbal speed = 4900 m/S

⇒ with R = the radius of the planet = 4.48 *10^6 m

⇒ with h = the distance above the planet's surface = 630000 meter

g = (4900² * ( 4.48*10^6+ 630000)) / ((4.48*10^6)²)

g = 6.11 m/s²

<u>Step 3:</u> Describe the position of the projectile

Horizontal component: x(t) = v0*t *cos∅

Vertical component: y(t) = v0*t *sin∅ -1/2 gt² ( will be reduced to 0 in time )

⇒ with ∅ = 30.8 °

⇒ with v0 = 13.6 m/s

⇒ with t= v(sin∅)/g = 1.14 s

Horizontal range d = v0²/g *2sin∅cos∅  = v0²/g * sin2∅

Horizontal range d =(13.6²)/6.11 * sin(2*30.8)

Horizontal range d =26.63 m

The horizontal range of the projectile = 26.63 meters

6 0
2 years ago
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