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kap26 [50]
3 years ago
10

A spring with constant k = 78 N/m is at the base of a frictionless, 30.0°-inclined plane. A 0.50-kg block is pressed against the

spring, compressing it 0.20 m from its equilibrium position. The block is then released. If the block is not attached to the spring, how far up the incline will it travel before it stops?
Physics
1 answer:
olasank [31]3 years ago
4 0

Explanation:

The given data is as follows.

   Spring constant (k) = 78 N/m,     \theta = 30^{o}

 Mass of block (m) = 0.50 kg

According to the formula of energy conservation,

                mgh sin \theta = \frac{1}{2}kx^{2}

       h = \frac{1}{2} \times \frac{kx^{2}}{mg Sin \theta}

          = \frac{78 \times 0.04}{2 \times 0.5 \times 9.8 \times 0.5}

          = 0.64 m

Thus, we can conclude that the distance traveled by the block is 0.64 m.

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7 0
3 years ago
A 50-kg cart is on an incline of 30 degrees above the horizontal. The cart is at rest. The static coefficient of the slope is 0.
sesenic [268]

The cart is at rest, so it is in equilibrium and there is no net force acting on it. The only forces acting on the cart are its weight (magnitude <em>w</em>), the normal force (mag. <em>n</em>), and the friction force (maximum mag. <em>f</em> ).

In the horizontal direction, we have

<em>n</em> cos(120º) + <em>f</em> cos(30º) = 0

-1/2 <em>n</em> + √3/2 <em>f</em> = 0

<em>n</em> = √3 <em>f</em>

and in the vertical,

<em>n</em> sin(120º) + <em>f</em> sin(30º) + (-<em>w</em>) = 0

<em>n</em> sin(120º) + <em>f</em> sin(30º) = (50 kg) (9.80 m/s²)

√3/2 <em>n</em> + 1/2 <em>f</em> = 490 N

Substitute <em>n</em> = √3 <em>f</em> and solve for <em>f</em> :

√3/2 (√3 <em>f </em>) + 1/2 <em>f</em> = 490 N

2 <em>f</em> = 490 N

<em>f</em> = 245 N

(pointed up the incline)

4 0
3 years ago
Can someone plz help me with this I’m not understanding it and show work plz plz
Triss [41]

There are two forces acting on the teacher:

Force due to weight/gravity (Fg)

Force due to drag (Fd), which is a resistance opposite to the direction of motion. Think of an airplane flying through the sky: there will be air that tries to oppose the plane's direction of motion AKA air-resistance.

The force of gravity is always downward (the direction of gravity).

Like we said before, the force of drag is always opposite to the direction of motion. Since the teacher is falling down, the force of drag is exerted upward.

Look at the attached diagram. The teacher is the circle in the middle. The two arrows indicate the two forces and their directions.

Now let's look at numbers:

Fg = mg = 65kg * 9.81 m/s^2 = ??N

Fd = 320N

To find the "Net Force" we must add up all of the forces exerted on the teacher, BUT we have to take into account the direction of forces.

Let's define downward as our "positive" direction. Since downward is positive, that means our force due to gravity is positive = +Fg

But since our force due to drag is UPWARD that means our force is NEGATIVE = -Fd.

So our total net force is

\Sigma F = F_g - F_d

3 0
4 years ago
One problem caused by technology is that of pollution true or false
HACTEHA [7]

Answer:

True, because tech is what creates thing like cars and other veicals (i know i spelled it wrong)

hope this helps

Explanation:

5 0
3 years ago
Read 2 more answers
f 720-nm and 620-nm light passes through two slits 0.68 mm apart, how far apart are the second-order fringes for these two wavel
valkas [14]

Answer:

0.0003 m = 0.3 mm

Explanation:

For constructive interference in the Young's experiment.

The position of the mth fringe from the central fringe is given by

y = L(mλ/d)

λ = wavelength = 720 nm = 720 × 10⁻⁹ m

L = distance between slits and screen respectively = 1.0 m

d = separation of slits = 0.68 mm = 0.68 × 10⁻³ m

m = 2

y = 1(2 × 720 × 10⁻⁹/(0.68 × 10⁻³) = 0.00212 m = 2.12 mm

For the 620 nm light,

y = 1(2 × 620 × 10⁻⁹/(0.68 × 10⁻³) = 0.00182 m = 1.82 mm

Distance apart = 2.12 - 1.82 = 0.3 mm = 0.0003 m

8 0
3 years ago
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