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Alik [6]
3 years ago
8

What is the connection between the x- and y-motions of a projectile?

Physics
2 answers:
Vikentia [17]3 years ago
7 0

Answer:

D. height

Explanation:

Sunny_sXe [5.5K]3 years ago
5 0
Answer c, velocity would be the answer.
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Unlike the majority of psychologists, psychiatrists __________. A. can diagnose disorders B. use psychotherapy C. treat people w
ludmilkaskok [199]
D prescribe medication

3 0
3 years ago
Read 2 more answers
If an object is propelled upward from a height of 128 feet at an initial velocity of 112 feet per​ second, then its height h aft
Marina CMI [18]

Explanation:

The equation of motion of an object is given by :

h(t)=-16t^2+112t+128

Where

t is the time in seconds

We need to find the time when the object hits the ground. When the object hits the ground, h(t) = 0

So,

-16t^2+112t+128=0

-t^2+7t+8=0

On solving above equation using online calculator, t = 8 seconds. So, the object hit the ground after 8 seconds. Hence, this is the required solution.

8 0
3 years ago
In the year 2055, a rocket was launched from a research laboratory on Mars. Mars has essentially no atmosphere. The test rocket
algol [13]

Answer:

h = 618.64 m

Explanation:

First we need to calculate the height gained by rocket while the fuel is burning. We use 2nd equation of motion for that purpose:

h₁ = Vit + (1/2)at²

where,

h₁ = height gained during the burning of fuel

Vi = Initial Velocity = 0 m/s

t = time = 7 s

a = acceleration = 8 m/s²

Therefore,

h₁ = (0 m/s)(7 s) + (1/2)(8 m/s²)(7 s)²

h₁ = 196 m

Now we use 1st equation of motion to find final speed Vf:

Vf = Vi + at

Vf = 0 m/s + (8 m/s²)(7 s)

Vf =  56 m/s

Now, we calculate height covered in free fall motion. Using 3rd equation of motion:

2ah₂ = Vf² - Vi²

where,

a = - 3.71 m/s²

h₂ = height gained during free fall motion = ?

Vf = Final Velocity = 0 m/s (since, rocket will stop at highest point)

Vi = 56 m/s

Therefore,

(2)(-3.71 m/s²)h₂ = (0 m/s)² - (56 m/s)²

h₂ = 422.64 m

So the total height gained will be:

h = h₁ + h₂

h = 196 m + 422.64 m

<u>h = 618.64 m</u>

4 0
2 years ago
g You want to make simultaneous measurements of the position and momentum of an electron and a proton that are moving along a st
Maslowich

Answer:

1832

Explanation:

From;

Δp Δx = h/4π

Δp = uncertainty in momentum

Δx = uncertainty in position

h= Plank's constant

But p =mv hence, Δp= Δmv

m= mass, v= velocity

mass of electron = 9.11 * 10^-31 Kg

Mass of proton = 1.67 * 10^-27 Kg

since m is a constant,

Δv = h/Δxm4π

For proton;

Δv = 6.6 * 10^-34/4 * 3.14 * 1.67 * 10^-27 * 1 * 10^-10

Δv = 315 ms-1

For electron;

Δv = 6.6 * 10^-34/4 * 3.14 * 9.11 * 10^-31 * 1 * 10^-10

Δv = 577000 ms-1

Ratio of uncertainty of electron to that of proton = 577000 ms-1/315 ms-1= 1832

7 0
2 years ago
Un auto si muove lungo una strada rettilinea​
Valentin [98]

Answer:

Un'auto si muove lungo un percorso rettilineo con velocità variabile come mostrato in figura. Quando l'auto è in possesso di A, la sua velocità è 10 ms-1 e quando è in posizione B, la sua velocità è 20 ms-1. Se l'auto impiega 5 secondi per spostarsi da A a B, trova l'accelerazione dell'auto.

Explanation:

8 0
3 years ago
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