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Neko [114]
3 years ago
14

An unusually high tide is called a _______ tide.

Physics
1 answer:
Monica [59]3 years ago
8 0
 Proxigean Spring <span>Tide</span>
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What's an easy way to create an interference pattern of waves?
igor_vitrenko [27]

Answer:

B

Explanation:

5 0
3 years ago
Two thermometers are calibrated, one in de
zlopas [31]
-40 c = -40 f but k would be 233.15
6 0
3 years ago
Sean climbs a tower that is 71.3 m high to make a jump with a parachute. The mass of Sean plus the parachute is 81.4 kg. If U =
myrzilka [38]

Answer:

U = 56877.4 J

Explanation:

The potential energy of a body is that which it possesses because it is located at a certain height above the surface of the earth and can be calculated using the following formula:

U = mgh Formula (1)

Where:

U is the potential energy in Joules (J)

m is the mass of the body in kilograms (kg)

g is the acceleration due to gravity (m/s²)

h is the height at which the body is found from the surface of the earth in meters (m)

Data

m= 81.4 kg

g= 9.8 m/s²

h = 71.3 m

Potential energy of Sean and the parachute at the top of the tower

We replace data in the formula (1)

U = m*g*h

U = (81.4 kg)*(9.8 m/s²)*(71.3 m)

U = 56877.4 N*m

U = 56877.4 J

3 0
4 years ago
A 15-kg block at rest on a horizontal frictionless surface is attached to a very light ideal spring of force constant 450 N/m. T
den301095 [7]

Answer:

0.266 m

Explanation:

Assuming the lump of patty is 3 Kg then applying the principal of conservation of linear momentum,

P= mv where p is momentum, m is mass and v is the speed of an object. In this case

m_pv_p=v_c(m_p+m_b) where sunscripts p and b represent putty and block respectively, c is common velocity.

Substituting the given values then

3*8=v(15+3)

V=24/18=1.33 m/s

The resultant kinetic energy is transferred to spring hence we apply the law of conservation of energy

0.5(m_p+m_b)v_c^{2}=0.5kx^{2} where k is spring constant and x is the compression of spring. Substituting the given values then

(3+15)*1.33^{2}=450*x^{2}\\x\approx 0.266 m

7 0
3 years ago
A charged cloud system produces an electric field in the air near earth's surface. a particle of charge -2.0 × 10-9 c is acted o
defon

Part a)

Magnitude of electric field is given by force per unit charge

E = \frac{F}{q}

E = \frac{4.3 * 10^{-6}}{2 * 10^{-9}}

E = 2150 N/C

Part b)

Electrostatic force on the proton is given as

F = qE

F = 1.6 * 10^{-19} * 2150

F = 3.44 * 10^{-16} N

PART C)

Gravitational force is given by

F_g = mg

F_g = 1.6 * 10^{-27}*9.8

F_g = 1.57 * 10^{-26} N

PART d)

Ratio of electric force to weight

\frac{F_e}{F_g} = \frac{3.44 * 10^{-16}}{1.57*10^{-26}}

\frac{F_e}{F_g} = 2.2 * 10^{10}

7 0
4 years ago
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