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3 years ago

A projectile is launched

1 answer:

4 0

The vertical component of the velocity after the given time is -9.8 m/s while the horizontal component of the velocity is 16 m/s.

The given parameters;

initial horizontal velocity, vₓ = 16 m/s
initial vertical velocity, $v_y =0$
time interval 1 seconds

The components of the velocity can be horizontal or vertical velocity.

The vertical component of the velocity is affected by acceleration due to gravity while the horizontal component of the velocity is not affected by gravity.

The vertical component of the velocity is calculated as;

$v_y = v_0_y -gt\\\\v_y = 0 - (1\times 9.8)\\\\v_y = -9.8 \ m/s$

The horizontal component of the velocity is constant since it is not affected by gravity.

The horizontal component of the velocity = 16 m/s

Thus, the vertical component of the velocity after the given time is -9.8 m/s while the horizontal component of the velocity is 16 m/s.

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Your dog is running around the grass in your backyard. He undergoes successive displacements, 3.60 m south, 8.46 m northeast, an

Eduardwww [97]

First, we resolve the northeast displacement into its north and east components. The angle from the positive x-axis of a northeast displacement is 45 degrees. Thus:
North = 8.46sin(45) = 5.98 m
East = 8.46cos(45) = 5.98 m

North displacement = 5.98 - 3.6 = 2.38 m
West displacement = 15.6 - 5.98 = 9.62

Magnitude = √(2.38² + 9.62²)
Magnitude = 9.91 m

Direction:
tan∅ = 2.38 / 9.62
∅ = 13.9° north from east

3 0

3 years ago

Read 2 more answers

The focal length of a concave mirror has a magnitude of 20 cm. what is its radius of curvature?

zepelin [54]

I think the answer is 40cm. If I am wrong please tell me.

7 0

3 years ago

What is the net force on a 4000 kg car that doubles its speed from 15 m/s to 30 m/s over 10 seconds

kozerog [31]

The net force of the car <u>is 6000 N.</u>

Why?

To calculate the net force of the car during the given time, we first need to calculate its acceleration, and then, use Newton's Second Law equation to calculate the net force.

We can calculate the acceleration of the car during the given tieme using the following equation:

$a=\frac{v-v_{o}}{time}$

Substituting the given information, we have:

$a=\frac{30\frac{m}{s} -15\frac{m}{s} }{10s}=\frac{15\frac{m}{s} }{10s}=1.5\frac{m}{s^{2} }$

Now that we have calculated the acceleration, we can calculate the net force:

$Force=mass*acceleration\\\\Force=4000kg*1.5\frac{m}{s^{2}}=6000\frac{kg.m}{s^{2}}=6000N$

Hence, we have that the net force is equal to 6000N.

Have a nice day!

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3 years ago

Two people, Micah and Lyra, with different near points are equally close to an object. Both inspect the object through the same

Gekata [30.6K]

Meh kock is really really hard

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3 years ago

A person exerts a 19-N force on a cart attached to a spring and holds the cart steady. The cart is displaced 0.060 m from its eq

Mariana [72]

To solve this problem, apply the concepts related to Hooke's law. From there we will find the spring constant. Subsequently, applying Energy balance, which includes gravitational potential energy, elastic potential energy and kinetic energy, we will bury the system's energy. Finally, using the displacement expression for the simple harmonic movement, we will find the expression that describes the system.

PART A) The expression for the spring force is

$F=kx$

Here,

k = Spring constant

x = Displacement

Rearranging to find the spring constant we have that

$k = \frac{F}{x}$

$k = \frac{19}{0.06}$

$k = 316.66N/m \approx 320N/m$

PART B ) The gravitational potential energy acts on the spring holds the cart is zero. Since cart is placed in the equilibrium position. The kinetic energy of the cart is zero. Therefore the expression for the total energy is,

$E = (PE)_g+(PE)_{spring}+KE$