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kozerog [31]
3 years ago
12

What is used to measure the amount of sunshine ​

Physics
2 answers:
saul85 [17]3 years ago
8 0

Answer:

modern sunshine sensorssunshine recordersCampbell-Stokes

lozanna [386]3 years ago
6 0

Answer:

Campbell-Stokes sunshine recorders or modern sunshine sensors.

Explanation:

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PLZZZZ HELP QUICK . 30 points. Emma weighs 560 N, she has decided to stay in shape so she is working out every morning, This mor
Phantasy [73]

Force equals mass time acceleration. Weight is a force and it can replace force in the equation. The acceleration would be gravity, which is an acceleration.

1.)

Fw (weight) = m (mass) · g (gravity, 9.8 m/s²)

Fw = m * 9.81 m/s²

560N = m · 9.81 m/s²

m ≈ 57.08 kg

2.)

d = 350 meters

t = 65 seconds

velocity = d/t

velocity = 350 meters / 65 seconds

velocity ≈ 5.38 meters/sec

3.)

Force = 35N

Distance = 2 meters

Work = Force · Distance

Work = 35N · 2 meters

Work = 70 J

3 0
3 years ago
A positive charge is moved from point a to point b along an equipotential surface. How much work is performed or required in mov
snow_lady [41]

Answer:

The work done is zero

Explanation:

No work is performed or required in moving the positive charge from point A to point B

3 0
4 years ago
A force of 7.50 N is applied to a spring whose spring constant is .298 N/cm. Find it’s length
gizmo_the_mogwai [7]

Answer:

to find the length just divide 7.50 and . 298

5 0
3 years ago
SERIOUSLY HELP NOWWW LIKE NOWW I REALLY NEED THIS ANSWERED
Orlov [11]

Answer:

the second one Test the rate of decay of specific elements in rock samples.

4 0
3 years ago
Consult Multiple-Concept Example 5 for insight into solving this problem. A skier slides horizontally along the snow for a dista
Gennadij [26K]

Answer:

The speed with which the skier was going is approximately 2.9906 m/s

Explanation:

The given parameters are;

The distance the skier slides before coming to rest, s = 12.4 m

The coefficient of friction between the skier and the snow, \mu _k = 0.0368

Therefore, for conservation of energy, we have;

Initial kinetic energy = Work done by the kinetic friction force

Initial kinetic energy = 1/2·m·v²

The work done by the kinetic friction force = \mu _k×m×g×s

Where;

m = The mass of the skier

v = The speed with which the skier was going

g = The acceleration due to gravity = 9.8 m/s²

s = The distance the skier slides before coming to rest = 12.4 m

\mu _k = The kinematic friction = 0.0368

Therefore, for conservation of energy, we have;

1/2·m·v² = \mu _k×m×g×s

1/2·v² = \mu _k×g×s

v² = 2×\mu _k×g×s  = 2 × 0.0368 × 9.8 × 12.4 = 8.943872

v = √(8.943872) ≈ 2.99063070271 ≈ 2.9906 m/s

The speed with which the skier was going = v ≈ 2.9906 m/s.

3 0
3 years ago
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