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faust18 [17]
3 years ago
9

Projectile motion in two dimensions cannot be determined by breaking the problem into two connected one-dimensional problems.

Physics
1 answer:
andriy [413]3 years ago
8 0
The answer I believe is False
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Please help with this question ASAP!!!
Paladinen [302]
The answer is the second choice

7 0
4 years ago
What is newton's third law second law first law ?​
sergeinik [125]

Answer:

first law: an object remains in uniform motion except an external force has acted upon it eg a ball in stable motion doesn't move until one moves or kicks it

second law:the body acted upon by an external force gains a momentum which is directly proportional to the applied force and acts in the direction of the force

third law: to every action there is an equal and opposite reaction eg if u push someone the person moves backward away from you and not towards you

8 0
3 years ago
Read 2 more answers
How long does it take the earth to rotate?
zubka84 [21]
It takes Earth 23 hours and 56 minutes to rotate. To be accurate.
Or you can just say 24 hours.
HOPE THIS HELPS YOU! ^_^
3 0
4 years ago
Read 2 more answers
What is An invention created for fun called ?
Fantom [35]

Answer & Explanation:

A provisional patent application, also called a PPA, shows your invention as being in the process of getting a patent. This means that you will be safe from others copying your idea while your patent application is still being processed.

4 0
4 years ago
A hockey puck slides off the edge of a horizontal platform with an initial velocity of 28.0 m/shorizontally in a city where the
kozerog [31]

Answer:

θ = 12.60°

Explanation:

In order to calculate the angle below the horizontal for the velocity of the hockey puck, you need to calculate both x and y component of the velocity of the puck, and also you need to use the following formula:

\theta=tan^{-1}(\frac{v_y}{v_x})       (1)

θ: angle below he horizontal

vy: y component of the velocity just after the puck hits the ground

vx: x component of the velocity

The x component of the velocity is constant in the complete trajectory and is calculated by using the following formula:

v_x=v_o

vo: initial velocity = 28.0 m/s

The y component is calculated with the following equation:

v_y^2=v_{oy}^2+2gy         (2)

voy: vertical component of the initial velocity = 0m/s

g: gravitational acceleration = 9.8 m/s^2

y: height

You solve the equation (2) for vy and replace the values of the parameters:

v_y=\sqrt{2gy}=\sqrt{2(9.8m/s^2)(2.00m)}=6.26\frac{m}{s}

Finally, you use the equation (1) to find the angle:

\theta=tan^{-1}(\frac{6.26m/s}{28.0m/s})=12.60\°

The angle below the horizontal is 12.60°

7 0
3 years ago
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