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dolphi86 [110]
3 years ago
9

Recall some things you already know about projectile motion. Does a force in the vertical direction affect the horizontal compon

ent of an object’s velocity? In this situation, can you use the horizontal velocity component to find the time required to travel some horizontal distance?
Physics
1 answer:
azamat3 years ago
7 0

Answer:

vertical force cannot change the velocity on the x-axis.   t =x/v₀ₓ

Explanation:

The force is a vector magnitude, so the forces on the x-axis affect the acceleration on this axis. Consequently a vertical force cannot change the velocity on the x-axis.

     F_{y} = m g

    Fₓ = 0

The horizontal velocity in projectile motion is constant, if we neglect the air resistance, so it can be used to find the time of a horizontal displacement

       x = v₀ₓ t

       t =x/v₀ₓ

The only magnitude that is the same for both movements is the time that is a scalar

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svlad2 [7]

Answer:

see the attachment

Explanation:

take coordinate system correctly. use formulas of projectile motion

Download pdf
3 0
2 years ago
A 2-kg box sits on a horizontal table. the force of friction between the box and the table is 10 n. the box is pushed to the rig
dangina [55]
By Newton's 2nd law of motion, F = ma, where F is force, m is mass, and a is acceleration.

Rearranging this equation to find acceleration would give us:
a = F/m

The horizontal force to the right is 10N, because the box is pushed to the right with a force of 20N, and the friction force of 10N opposes that, so:
20N - 10N = 10N

The mass is 2kg.

Putting these values into the equation gives us:
a = F/m
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6 0
3 years ago
slader Question: A Model Rocket Is Launched Straight Upward With An Initial Speed Of 50m/s. Iit Accelerates With A Constant Upwa
xenn [34]

Answer:

Maximum height reached by the rocket is

y_{max} = 308 m

total time of the motion of rocket is given as

T = 16.44 s

Explanation:

Initial speed of the rocket is given as

v_i = 50 m/s

acceleration of the rocket is given as

a = 2 m/s^2

engine stops at height h = 150 m

so the final speed of the rocket at this height is given as

v_f^2 - v_i^2 = 2 a d

v_f^2 - 50^2 = 2(2)(150)

v_f = 55.68 m/s

so maximum height reached by the rocket is given as the height where its final speed becomes zero

so we will have

v_f^2 - v_i^2 = 2 a d

0 - 55.68^2 = 2(-9.81)(y - 150)

y_{max} = 308 m

Now the total time of the motion of rocket is given as

1) time to reach the height of 150 m

v_f - v_i = at

55.68 - 50 = 2 t

t_1 = 2.84 s

2) time to reach ground from this height

\Delta y = v_y t + \frac{1}{2}gt^2

-150 = 55.68 t - \frac{1}{2}(9.81) t^2

t_2 = 13.6 s

so total time of the motion of rocket is given as

T = 13.6 + 2.84 = 16.44 s

3 0
3 years ago
1) Calculate the ideal efficiency of a heat engine that takes in energy at 800 K and expels heat to a
masha68 [24]

Answer:

Could you explain that more better?

Explanation:

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3 years ago
A proton moving with a velocity of 4.0 × 104 m/s enters a magnetic field of 0.20 t. if the angle between the velocity of the pro
svlad2 [7]

Answer:

Magnitude of the force on proton = F = 1.1085 × 10^-15 N

Explanation:

Charge on proton = q = 1.60 × 10^-19 C

Velocity of proton = V = 4.0 × 10^4 m/s

Magnetic field = B = 0.20 T  

Angle between V and B = θ = 60

We know that,  

F = qVBsin θ = (1.60 × 10^-19)( 4.0 × 10^4)( 0.20)sin(60)

F = 1.1085 × 10^-15 N    

8 0
3 years ago
Read 2 more answers
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