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Vsevolod [243]
3 years ago
10

Find the resultant of an easterly force of 100 N and a southeast force of 80 N acting at 65 degrees to the 100 N force

Physics
1 answer:
saveliy_v [14]3 years ago
7 0

Answer:

Resultant is 152 N at 28.5 degrees south to the 100 N force

Explanation:

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A 68 kg object, starting from rest, travels from point A to point B at a rate of 30 m/s in 2 hours. What is the applied force on
Natasha2012 [34]

Answer:

\huge\boxed{\sf F = 0.28\ N}

Explanation:

<h3>Given Data:</h3>

Mass = m = 68 kg

Velocity = v = 30 m/s

Time = 2 hours = 2 × 60 × 60 = 7200 s

<h3>Required:</h3>

Force = F = ?

<h3>Formula to be used:</h3>

\displaystyle F = \frac{mv}{t}

<h3>Solution:</h3>

\displaystyle F = \frac{(68)(30)}{7200} \\\\F = \frac{2040}{7200} \\\\F = 0.28 N\\\\\rule[225]{225}{2}

7 0
1 year ago
The emf induced in a coil that is rotating in a magnetic field will be at a maximum at which moment?
adelina 88 [10]
TLDR: It will reach a maximum when the angle between the area vector and the magnetic field vector are perpendicular to one another.

This is an example that requires you to investigate the properties that occur in electric generators; for example, hydroelectric dams produce electricity by forcing a coil to rotate in the presence of a magnetic field, generating a current.

To solve this, we need to understand the principles of electromotive forces and Lenz’ Law; changing the magnetic field conditions around anything with this potential causes an induced current in the wire that resists this change. This principle is known as Lenz’ Law, and can be described using equations that are specific to certain situations. For this, we need the two that are useful here:

e = -N•dI/dt; dI = ABcos(theta)

where “e” describes the electromotive force, “N” describes the number of loops in the coil, “dI” describes the change in magnetic flux, “dt” describes the change in time, “A” describes the area vector of the coil (this points perpendicular to the loops, intersecting it in open space), “B” describes the magnetic field vector, and theta describes the angle between the area and mag vectors.

Because the number of loops remains constant and the speed of the coils rotation isn’t up for us to decide, the only thing that can increase or decrease the emf is the change in magnetic flux, represented by ABcos(theta). The magnetic field and the size of the loop are also constant, so all we can control is the angle between the two. To generate the largest emf, we need cos(theta) to be as large as possible. To do this, we can search a graph of cos(theta) for the highest point. This occurs when theta equals 90 degrees, or a right angle. Therefore, the electromotive potential will reach a maximum when the angle between the area vector and the magnetic field vector are perpendicular to one another.

Hope this helps!
6 0
3 years ago
Here on Earth you hang a mass from a vertical spring and start it oscillating with amplitude 1.9 cm. You observe that it takes 3
Vlada [557]

Answer:

T = 3.23 s

Explanation:

In the simple harmonic movement of a spring with a mass the angular velocity is given by

               w = √ K / m

With the initial data let's look for the ratio k / m

The angular velocity is related to the frequency and period

           w = 2π f = 2π / T

            2π / T = √ k / m

            k₀ / m₀ = (2π / T)²

            k₀ / m₀ = (2π / 3.0)²

            k₀ / m₀ = 4.3865

The period on the new planet is

          2π / T = √ k / m

           T = 2π √ m / k

In this case the amounts are

           m = 6 m₀

           k = 10 k₀

We replace

            T = 2π√6m₀ / 10k₀

            T = 2π √6/10 √m₀ / k₀

            T = 2π √ 0.6  √1 / 4.3865

            T = 3.23 s

5 0
3 years ago
If a topographic map included a 6,000 ft. mountain next to an area of low hills, which would best describe the contour lines on
VMariaS [17]
The answer to the given question above would be option B. If a topographic map included a 6,000 ft. mountain next to an area of low hills, the statement that best describe the contour lines on the map is this: <span>The contour lines around the mountain would be very close together. Hope this helps.</span>
8 0
3 years ago
Read 2 more answers
A uniformly charged conducting sphere of 1.1 m diameter has a surface charge density of 6.2 µC/m2. (a) Find the net charge on th
ira [324]

Answer:

(a) q = 2.357 x 10⁻⁵ C

(b) Φ = 2.66 x 10⁶ N.m²/C

Explanation:

Given;

diameter of the sphere, d = 1.1 m

radius of the sphere, r = 1.1 / 2 = 0.55 m

surface charge density, σ = 6.2 µC/m²

(a)  Net charge on the sphere

q = 4πr²σ

where;

4πr² is surface area of the sphere

q is the net charge on the sphere

σ is the surface charge density

q = 4π(0.55)²(6.2 x 10⁻⁶)

q = 2.357 x 10⁻⁵ C

(b) the total electric flux leaving the surface of the sphere

Φ = q / ε

where;

Φ is the total electric flux leaving the surface of the sphere

ε is the permittivity of free space

Φ = (2.357 x 10⁻⁵) / (8.85 x 10⁻¹²)

Φ = 2.66 x 10⁶ N.m²/C

8 0
3 years ago
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