A chemical reaction is the only way to separate a(n)

Which wave causes the medium to vibrate only in a direction parallel to the wave’s motion?

vivado [14]

None of above/D (but I'm not very sure)

5 0

4 years ago

Read 2 more answers

Plz help ASAP I'll mark as brainliest

gogolik [260]

Hi there!

1.

Hooke's law states that:

F = -kx

k = Spring constant (N/m)

x = DISPLACEMENT from equilibrium (m)

Essentially, the force of a spring is PROPORTIONAL to its spring constant and its displacement from its equilibrium point.

2.

The force of the spring (T) is not proportional to the spring's length (l), but rather its DISPLACEMENT from its equilibrium length. (Δl)

3.

The equilibrium length is where the force of the spring (T) = 0N. Looking at the graph, the line intersects this value at l = 30cm.

4.

We can begin by looking at the given graph.

When the spring force = 4N, the total length of the spring is 35 cm.

Now, the EQUILIBRIUM length is 30 cm, so the total elongation is:

35 - 30 = 5 cm.

5.1.

If the spring elongates by 10 cm, the total length of the spring is:

30 + 10 = 40 cm

According to the graph, a length of 40 cm corresponds to a force of 8N.

5.2.

We can solve for the weight of the ball using the following:

W (weight) = m (mass) · acceleration due to gravity (10N/kg)

Using a summation of forces:

∑F = T - W

The elongation that we are solving for occurs at the equilibrium point (net force = 0 N), so:

0 = T - W

T = W = 8 N

5.3.

0 = T - Mg

T = Mg

Use the prior value of T and gravity to solve:

8 = 10M

m = 0.8 kg

8 0

3 years ago

Causing a collision while speeding puts points on your license

Marizza181 [45]

Yes, I believe it will put points on your license. I think in some cases you will get your license revoked

5 0

3 years ago

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A flat, 181 ‑turn, current‑carrying loop is immersed in a uniform magnetic field. The area of the loop is 4.97 cm2 and the angle

Sidana [21]

Answer:

B = 0.135T

Explanation:

To find the magnitude of the magnetic field you use the following formula, for the torque produced by a magnetic field B in a loop:

$\tau=NIABsin\theta$ (1)

τ: torque = 1.51*10^-5 Nm

I: current = 2.47mA = 2.47*10^-3 A

B: magnitude of the magnetic field

A: area of the loop = 4.97cm^2 = 4.97(10^-2m)^2=4.97*10^-4m^2

N: turns = 181

θ: angle between B and the magnetic dipole (same as the direction of the normal to the plane)

You replace the values of the parameters in (1). Furthermore you do B the subject of the formula:

$B=\frac{\tau}{NIAsin\tetha}=\frac{1.51*10^{-5}Nm}{(181)(2.47*10^{-3}A)(4.97*10^{-4}m^2)(sin30.1\°)}=0.135T$

3 0

3 years ago

A 68 kg person runs off a dock at 2.79 m/s and lands on a stationary 131 kg boat. What is their joint velocity afterward?

kolbaska11 [484]

P=mv
pi =pf
(68x2.79)+(131x0) = 199v
v= 0.95 m/s

4 0

3 years ago