Which of these planets is the odd one out based on their physical features?

A toy car runs off the edge of a table that is 1.807 m high. The car lands 0.3012 m from the base of the table. How long does it

LekaFEV [45]

Answer:

a) $t=0.60 s$

b) $v_{ox} =0.5m/s$

Explanation:

From the exercise we have initial height and final X position

$y_{o}=1.807m$

$X=0.3012m$

a) From the concept of free falling objects we have that

$y=y_{o}+v_{oy}t+\frac{1}{2}gt^{2}$

Since the car runs off the edge of the table, that means the car is moving in x direction with $v_{oy}=0m/s$ and at the end of the motion $y=0m$

$0=1.807m-\frac{1}{2}(9.8)t^{2}$

Solving for t

t=± 0.6072 s

Since the time can not be negative, the answer is t=0.6072 s

b) To find the horizontal velocity of the car, we need to use the time that we just calculate

$X=v_{ox}t$

$v_{ox}=\frac{X}{t}=\frac{0.3012m}{0.6072s} =0.5m/s$

6 0

3 years ago

What is simple harmonic oscillation? express your answer verbally, mathematically, and visually.

Morgarella [4.7K]

Simple harmonic motion is the motion in objects where the restoring for d is directly proportional to the body displacement.

<h3>What is simple harmonic motion?</h3>

Simple harmonic motion a type of periodic motion where the force that restores moving object to it's right position is directly proportional to the magnitude of body's displacement and this normally acts towards the object's equilibrium position.

The formulate for simple harmonic motion are

,F = −kx, where F is the force, x is the displacement, and k is a constanst. An example of a simple harmonic oscillator is the vibration of a mass attached to a vertical spring.

Simple harmonic oscillator can be represented by the equation x ( t ) = A cos ⁡ ( 2 π f t ) x(t) = A\cos(2\pi f t) x(t)=Acos(2πft)x,

where t is period.

x is displacement

A is amplitude.

Learn more about simple harmonic motion here.

brainly.com/question/17315536

3 0

2 years ago

Auroras occur at _____ of a planet. the equator the north magnetic pole only the south magnetic pole only both magnetic poles

goldenfox [79]

The answer is - both magnetic poles

In the earth's southern magnetic pole, auroras are known as aurora austalis or southern lights while in the northern magnetic pole they are known as aurora borealis or northern lights.

They are caused by charged particles borne in solar winds that are escaping from the sun. As they approach the earth, the winds distort the earth's magnetic field which allows some particles to enter the earth's atmosphere, where they excite the gases and make them glow with spectacular display of colors.

3 0

4 years ago

A scholar climbs up 8 m of stairs while using 500 J of work in 10 seconds. What is their power?

ehidna [41]

Answer:

according to the new geometry lesson alcometry which will after 11 years in geometry

7 0

3 years ago

A 30.0-g object connected to a spring with a force constant of 40.0 N/m oscillates with an amplitude of 6.00 cm on a frictionles

Naddika [18.5K]

Answer:

a) 72mJ

b) 0.33m/s

c) 54mJ

d) 18mJ

Explanation:

(a) The total energy of the system is given by:

$E_T=\frac{1}{2}kA^2$

where k is the force constant and A is the amplitude. By replacing we get:

$E_T=\frac{1}{2}(40N/m)(0.06m)^2=0.072J$ = 72mJ

(b) we can get the speed by the conservation of energy (the kinetic energy and potential energy must equal the total energy in any place):

$E_T=E_k+E_p\\\\\frac{1}{2}kA^2=\frac{1}{2}mv^2+\frac{1}{2}kx^2\\\\v=\sqrt{\frac{kA^2-kx^2}{m}}=\sqrt{\frac{(40N/m)((0.06m)^2-(0.0115m)^2)}{0.03kg}}=0.33m/s$

where we have used that x=1.15cm=0.0115m; A=6.00cm=0.06m

(c) Again, by the conservation of energy:

$E_k=\frac{1}{2}k(A^2-x^2)=\frac{1}{2}(40N/m)((0.06m)^2-(0.03m)^2)=0.054J=54mJ$

(d) $E_p=\frac{1}{2}kx^2=\frac{1}{2}(40.0N/m)(0.03m)^2=0.018J=18mJ$

hope this helps!!

6 0

3 years ago