Argon, neon, and xenon are examples of

What would a force diagram for something WHILE it is being thrown DOWNWARDS look like? <br><br> Ty

Dovator [93]

Answer:

it look the same just to tell you

5 0

3 years ago

Which law states that for any closed loop, the sum of the length segments times the magnetic field in the direction of the segme

dedylja [7]

That's Ampere's Law. ( C ).
The magnetic permeability is the proportionality constant.

3 0

3 years ago

Read 2 more answers

Which statement is true about the force of gravity?

kumpel [21]

Answer:

The moon's gravity pulls the Earth to make tides.

Explanation:

The Moons Gravity Pulls On The Earth With Different Strenght Making High Tide And Low Tide.

Hope This Helps!

4 0

2 years ago

When a golf ball is hit, it travels at 41 meters per second. The mass of a golf ball is 0.045kg. Calculate the kinetic energy of

Fittoniya [83]

Answer:

75.645 J

Explanation:

The kinetic energy is related to the mass and velocity by the formula ...

KE = 1/2mv²

For the given mass of 0.045 kg, and velocity of 41 m/s, the kinetic energy is ...

KE = 1/2(0.045 kg)(41 m/s)² = 75.645 J

__

The unit of energy, joule, is a derived unit equal to 1 kg·m²/s².

4 0

2 years ago

A tugboat tows a ship at a constant velocity. The tow harness consists of a single tow cable attached to the tugboat at point A

Y_Kistochka [10]

Answer:

The tensions in $T_{BC}$ is approximately 4,934.2 lb and the tension in $T_{BD}$ is approximately 6,035.7 lb

Explanation:

The given information are;

The angle formed by the two rope segments are;

The angle, Φ, formed by rope segment BC with the line AB extended to the center (midpoint) of the ship = 26.0°

The angle, θ, formed by rope segment BD with the line AB extended to the center (midpoint) of the ship = 21.0°

Therefore, we have;

The tension in rope segment BC = $T_{BC}$

The tension in rope segment BD = $T_{BD}$

The tension in rope segment AB = $T_{AB}$ = Pulling force of tugboat = 1200 lb

By resolution of forces acting along the line A_F gives;

$T_{BC}$ × cos(26.0°) + $T_{BD}$ × cos(21.0°) = $T_{AB}$ = 1200 lb

$T_{BC}$ × cos(26.0°) + $T_{BD}$ × cos(21.0°) = 1200 lb............(1)

Similarly, we have for equilibrium, the sum of the forces acting perpendicular to tow cable = 0, therefore, we have;

$T_{BC}$ × sin(26.0°) + $T_{BD}$ × sin(21.0°) = 0...........................(2)

Which gives;

$T_{BC}$ × sin(26.0°) = - $T_{BD}$ × sin(21.0°)

$T_{BC}$ = - $T_{BD}$ × sin(21.0°)/(sin(26.0°)) ≈ - $T_{BD}$ × 0.8175

Substituting the value of, $T_{BC}$ , in equation (1), gives;

- $T_{BD}$ × 0.8175 × cos(26.0°) + $T_{BD}$ × cos(21.0°) = 1200 lb

- $T_{BD}$ × 0.7348 + $T_{BD}$ ×0.9336 = 1200 lb

$T_{BD}$ ×0.1988 = 1200 lb

$T_{BD}$ ≈ 1200 lb/0.1988 = 6,035.6938 lb

$T_{BD}$ ≈ 6,035.6938 lb

$T_{BC}$ ≈ - $T_{BD}$ × 0.8175 = 6,035.6938 × 0.8175 = -4934.1733 lb

$T_{BC}$ ≈ -4934.1733 lb

From which we have;

The tensions in $T_{BC}$ ≈ -4934.2 lb and $T_{BD}$ ≈ 6,035.7 lb.

8 0

3 years ago