True or False: Waves with shorter wavelengths have more energy.<br><br> False<br><br> True

What body parts were scientists wanting to image that prompted the development of the CT scanner

zalisa [80]

Answer:

The head

Explanation:

4 0

3 years ago

How long will it take to travel 200 m traveling at 10 m/s? Follow example below.

polet [3.4K]

Answer:

$variables - d = 200m \: v = 10 m {s}^{ - 1} \\ equation \: - v = \frac{d}{t} \\ 10 = \frac{200}{t} \\ cross \: multiply \\ 10t = 200 \\ \frac{10t}{10} = \frac{200}{10} \\ t = 20s$

It will take 10 seconds to travel 200m at a speed of 10m/s

Explanation:

HOPE THAT THIS IS HELPFUL.

HAVE A GREAT DAY.

4 0

3 years ago

Read 2 more answers

As a result of nuclear fusion, the sun produces what 2 types of energy ??

ELEN [110]

Answer:

heat energy and nuclear energy

Explanation:

as a resukt of nuclear fusion, the sun produces two types of energy.

1. heat energy

2. nuclear energy

3 0

3 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

4 years ago

2) Two ice skaters have masses m1 and m2 and are initially stationary. Their skates are identical. They push against one another

worty [1.4K]

Answer:

m_1 / m_2 = sqrt (1 / 2)

Explanation:

Given:

- Initial velocity of both skaters V_i = 0

- Velocity of skater 1 after push = V_1

- Velocity of skater after push = V_2

- Distance traveled by skater 1 = s_1

- Distance traveled by skater 2 = s_2

- s_1 = 2*s_2

- Accelerations of both skaters to halt is equal

Find:

What is the ratio m1/m2 of their masses

Solution:

- Apply conservation of momentum for two skaters just before and after the push as follows:

P_i = P_f

0 = m_1*V_1 - m_2*V_2

- Evaluate: m_1 / m_2 = ( V_2 / V_1 )

- Apply Conservation of Energy on both skaters as follows:

- Skater 1:

0.5*m_1*V_1^2 = u_k*m_1*g*s_1

-Simplify: 0.5*V_1^2 = u_k*g*(2*s_2)

- Skater 2:

0.5*m_2*V_2^2 = u_k*m_2*g*s_2

-Simplify: 0.5*V_2^2 = u_k*g*s_2

- Divide the two energy equations for skaters:

(V_1 / V_2)^2 = 2

(V_2 / V_1)^2 = 1 / 2

- simplify: (V_2 / V_1) = sqrt (1 / 2)

-Hence from earlier momentum conservation results:

m_1 / m_2 = ( V_2 / V_1 ) = sqrt (1 / 2)

6 0

3 years ago

True or False: Waves with shorter wavelengths have more energy. False True