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3 years ago

What is the relationship between the horizontal and vertical components of velocity for a projectile launched

1 answer:

3 0

Answer: The horizontal velocity of a projectile is constant (a never changing in value. The vertical velocity of a projectile changes by 9.8 m/s each second.

Explanation: I hope that helped!

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A man and his dog are “walking” on flat Street. He is pulling on his stubborn dog with a force of 70 N Directed at a 30° angle f

Delvig [45]

Answer:

X component of force is 60.62 N.

Y component of force is 35 N.

Force of gravity on the dog is 245 N.

Magnitude of normal force is 210 N.

Explanation:

Given:

Force of pull is, $F=70\ N$

Mass of the dog is, $m=25\ kg$

Angle of inclination is, $\theta =30$ °

Acceleration due to gravity is, $g=9.8\ m/s^2$

The free body diagram of the dog is shown below.

The X and Y components of force of pull is given as:

$F_X=F\cos \theta=(70)\cos (30)=60.62\ N\\F_Y=F\sin \theta=(70)\sin(30)=35\ N$

Therefore, the X and Y components of the force are 60.62 N and 35 N respectively.

Force of gravity on the dog is the product of its mass and acceleration due to gravity. Thus,

$F_g=mg=25\times 9.8=245\ N$

Therefore, the force of gravity on the dog is 245 N.

Now, consider the motion in the vertical direction of the dog. As there is no motion in the vertical direction, the net force along the Y direction is 0. In other words, the total Upward force is equal to the total downward force.

From the free body diagram,

$N+F_Y=mg\\N=mg-F_Y\\N=245-35=210\ N$

Therefore, the normal force acting on the dog is 210 N.

6 0

4 years ago

Why does the frequency of a siren get higher as an ambulance using that siren gets closer?

Archy [21]

The answer is D. As the ambulance gets closer, the sound waves are compressed relative to the person; so the frequency increases.

7 0

3 years ago

Read 2 more answers

Eric has a mass of 70 . He is standing on a scale in an elevator that is accelerating downward at 1.7 . What is the approximate

pychu [463]

Answer:

Explanation:

The value the scale shows is the reaction force to the normal force (they are equal by Newton's 3rd Law) that the scale exerts on Eric.

The forces on Eric are his weight (downward) and this normal force (upward), so we can write the net force over him as (also using Newton's 2nd Law):

$F=W-N=ma$

which means

$N=W-ma=mg-ma=m(g-a)$

and for our values this is:

$N=mg-ma=(70kg)(9.8m/s^2-1.7m/s^2)=567N$

5 0

3 years ago

I need help plz help me

kirza4 [7]

Answer:ghhg

Explanation:

8 0

3 years ago

A golf ball strikes a hard, smooth floor at an angle of 28.2 ° and, as the drawing shows, rebounds at the same angle. The mass o

WITCHER [35]

Answer:

I = 1.06886 N s

Explanation:

The expression for momentum is

I = F t = Δp

therefore the momentum is a vector quantity, for which we define a reference system parallel to the floor

Let's find the components of the initial velocity

sin 28.2 = v_y / v

cos 28.2= vₓ / v

v_y = v sin 282

vₓ = v cos 28.2

v_y = 42.8 sin 28.2 = 20.225 m / s

vₓ = 42.8 cos 28.2 = 37.72 m / s

since the ball is heading to the ground, the vertical velocity is negative and the horizontal velocity is positive, it can also be calculated by making

θ = -28.2

v_y = -20.55 m / s

v_x = 37.72 m / s

X axis

Iₓ = Δpₓ = $p_{fx} - p_{ox}$

since the ball moves in the x-axis without changing the velocity, the change in moment must be zero

Δpₓ = m $v_{fx}$ - m v₀ₓ = 0

v_{fx} = v₀ₓ

therefore

Iₓ = 0

Y axis

I_y = Δp_y = p_{fy} -p_{oy}

when the ball reaches the floor its vertical speed is downwards and when it leaves the floor its speed has the same modulus but the direction is upwards

v_{fy} = - v_{oy}

Δp_y = 2 m v_{oy}

Δp_y = 2 0.0260 (20.55)

$\Delta p_{y}$ = 1.0686 N s

the total impulse is

I = Iₓ i ^ + I_y j ^

I = 1.06886 j^ N s

7 0

3 years ago