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4 years ago

Why is their displacement the same when the distance each traveled was different?

1 answer:

4 0

Displacement is how far something is from its starting position

for example if X moved 10 feet, turned around and moved 30 feet, its distance would be 40, but the displacement would be 20,
and if Y Moved 2 feet, and turned around and moved 22 feet, its distance would be 24, but its displacement would also be 20

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Calculate the acceleration of a 3000 kg truck that had a net force of 500 N applied to it. Express your answer in m/s^2???

WINSTONCH [101]

Newton's 2nd law of motion . . . . . Force = (mass) x (acceleration)

Divide each side by (mass), and you have

Acceleration = (force) / (mass)

We know the force and mass for this truck, so we can fillum in:

Acceleration = (500 N) / (3,000 kg) = 0.167 m/s² .

6 0

3 years ago

Does food have energy?

FinnZ [79.3K]

Answer:

chemical energy that goes for all food actually

Explanation:

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3 years ago

What is the current in the circuit at the instant the capacitor begins to charge (the instant the switch is first closed)?

daser333 [38]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The current is $I = 4.2 \ A$

Explanation:

From the question we are told that

The voltage of battery is $V = 25 V$

The capacitance of capacitor is $C = 15 \mu F$

The resistance of the resistor is $R = 6 \Omega$

Generally the current at the instant the capacitor starts charging is

$I = \frac{V}{R}$

substituting values

$I = \frac{25}{6}$

$I = 4.2 \ A$

8 0

3 years ago

Mars’ axial tilt is 25. 19°. Earth’s axial tilt is 23. 5°. What does this most likely indicate about mars?.

sweet-ann [11.9K]

In comparison to Earth, Mars will have more extreme weather/climatic conditions

<h3>what is Axial tilt?</h3>

Axial tilt is the angular measure between the axis of rotation and the axis of revolution. it affects the extents of varying climatic conditions in a direct proportional relationship. This means that the more the axial tilt the more the difference in extremes of the climatic condition.

Therefore comparing the two axial tilts one can arrive at. Mars having more axial tilt will have more extreme seasonal conditions than Earth.

Read more on axial tilt here:

4 0

2 years ago

A spring (k=15.19kN/m)is is compresses 25cm and held in place on a 36.87° incline. A block (M=10kg) is placed on the spring. Whe

Savatey [412]

Answer:

The maximum vertical displacement is 2.07 meters.

Explanation:

We can solve this problem using energy. Since there is a frictional force acting on the block, we need to consider the work done by this force. So, the initial potential energy stored in the spring is transferred to the block and it starts to move upwards. Let's name the point at which the block leaves the ramp "1" and the highest point of its trajectory in the air "2". Then, we can say that:

$E_0=E_1\\\\U_e_0=K_1+U_g_1+W_f_1$

Where $U_e_0$ is the elastic potential energy stored in the spring, $K_1$ is the kinetic energy of the block at point 1, $U_g_1$ is the gravitational potential energy of the block at point 1, and $W_f_1$ is the work done by friction at point 1.

Now, rearranging the equation we obtain:

$\frac{1}{2}kx^{2}=\frac{1}{2}mv_1^{2}+mgh_1+\mu Ns_1$

Where $k$ is the spring constant, $x$ is the compression of the spring, $m$ is the mass of the block, $v_1$ is the speed at point 1, $g$ is the acceleration due to gravity, $h_1$ is the vertical height of the block at point 1, $\mu$ is the coefficient of kinetic friction, $N$ is the magnitude of the normal force and $s_1$ is the displacement of the block along the ramp to point 1.

Since the force is in an inclined plane, the normal force is equal to:

$N=mg\cos\theta$

Where $\theta$ is the angle of the ramp.

We can find the height $h_1$ using trigonometry:

$h_1=s_1\sin\theta$

Then, our equation becomes:

$\frac{1}{2}kx^{2}=\frac{1}{2}mv_1^{2}+mgs_1\sin\theta+\mu mgs_1\cos\theta\\\\\implies v_1=\sqrt{\frac{2(\frac{1}{2}kx^{2}-mgs_1\sin\theta-\mu mgs_1\cos\theta)}{m}}=\sqrt{\frac{kx^{2}}{m}-2gs_1(\sin\theta+\mu \cos\theta)}$

Plugging in the known values, we get:

$v_1=\sqrt{\frac{(15190N/m)(0.25m)^{2}}{10kg}-2(9.8m/s^{2})(1.12m)(\sin36.87\°+(0.300) \cos36.87\°)}\\\\v_1=8.75m/s$

Now, we can obtain the height from point 1 to point 2 using the kinematics equations. We care about the vertical axis, so first we calculate the vertical component of the velocity at point 1:

$v_1_y=v_1\sin\theta=(8.75m/s)\sin36.87\°=5.25m/s$

Now, we have:

$y=\frac{v_1_y^{2}}{2g}\\\\y=\frac{(5.25m/s)^{2}}{2(9.8m/s^{2})}\\\\y=1.40m$

Finally, the maximum vertical displacement $h_2$ is equal to the height $h_1$ plus the vertical displacement $y$ :

$h_2=h_1+y=s_1\sin\theta +y\\\\h_2=(1.12m)\sin36.87\°+1.40m\\\\h_2=2.07m$

It means that the maximum vertical displacement of the block after it becomes airborne is 2.07 meters.

7 0

3 years ago