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spayn [35]
2 years ago
14

A 2600-m-high mountain is located on the equator. how much faster does a climber on top of the mountain move than a surfer at a

nearby beach? the earth's radius is 6400 km.
Physics
1 answer:
puteri [66]2 years ago
5 0

The climber move 0.19 m/s  faster than surfer on the nearby beach.

Since both the person are on the earth, and moves with the constant angular velocity of earth, however there linear velocity is different.

Number of seconds in a day, t=24*60*60=86400 sec

The linear speed on the beach is calculated as

V1=\frac{2πr}{t}

Here, t is the time

Plugging the values in the above equation

V1=\frac{2π*6.4*10^6}{86400}=465.421 m/s

Velocity on the mountain is calculated as

V2=\frac{2π(r+h)}{t}

Plugging the values in the above equation

V2=\frac{2π(6.4*10^6+2600}{86400}=465.61 m/s

Therefore person on the mountain moves faster than the person on the beach by 465.61-465.421=0.19 m/s

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If one replaces the conducting cube with one that has positive charge carriers, in what direction does the induced electric fiel
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Answer:

There will be no change in the direction of the electric field .

Explanation:

The direction will remain the same because the sign of the charges has no effect on it.

When one replaces the conducting cube with one that has positive charge carriers there will be no change in the direction of the field as there is no defined relationship between the direction of the electric field and sign of the charge.

3 0
2 years ago
To show that displacement current is necessary to make Ampère's law consistent for a charging capacitor Ampère's law relates the
Sladkaya [172]

Answer: The Ampère -Max-well law

Explanation:

The Ampère -Max-well law relates magnetic flux and electric current. It determines the relationship between current in association with a magnetic field and also magnetic field in association to related current.

7 0
2 years ago
A shot putter releases the shot some distance above the level ground with a velocity of 12.0 m/s, 51.0 ∘above the horizontal. Th
alina1380 [7]

A) Zero

The motion of the shot is a projectile's motion: this means that there is only one force acting on the projectile, which is gravity. However, gravity only acts in the vertical direction: so, there are no forces acting in the horizontal direction. Therefore, the x-component of the acceleration is zero.

B) -9.8 m/s^2

The vertical acceleration is given by the only force acting in the vertical direction, which is gravity:

F=mg

where m is the projectile's mass and g is the gravitational acceleration. Therefore, the y-component of the shot's acceleration is equal to the acceleration due to gravity:

a_y = g = -9.8 m/s^2

where the negative sign means it points downward.

C) 7.6 m/s

The x-component of the shot's velocity is given by:

v_x = v_0 cos \theta

where

v_0 = 12.0 m/s is the initial velocity

\theta=51.0^{\circ} is the angle of the shot

Substituting into the equation, we find

v_x = (12.0 m/s)(cos 51^{\circ})=7.6 m/s

D) 9.3 m/s

The y-component of the shot's velocity is given by:

v_y = v_0 sin \theta

where

v_0 = 12.0 m/s is the initial velocity

\theta=51.0^{\circ} is the angle of the shot

Substituting into the equation, we find

v_y = (12.0 m/s)(sin 51^{\circ})=9.3 m/s

E) 7.6 m/s

We said at point A) that the acceleration along the x-direction is zero: therefore, the velocity along the x-direction does not change, so the x-component of the velocity at the end of the trajectory is equal to the x-velocity at the beginning:

v_x = 7.6 m/s

F) -11.1 m/s

The y-component of the velocity at time t is given by:

v_y(t) = v_y + at

where

v_y = 9.3 m/s is the initial y-velocity

a = g = -9.8 m/s^2 is the vertical acceleration

t is the time

Since the total time of the motion is t=2.08 s, we can substitute this value into the equation, and we find:

v_y(2.08 s)=9.3 m/s + (-9.8 m/s^2)(2.08 s)=-11.1 m/s

where the negative sign means the vertical velocity is now downward.

3 0
2 years ago
A cylinder of radius R and height H is floating upright in
emmainna [20.7K]

Answer:

Pressure difference between Top and Bottom of the cylinder is given as

\Delta P = \frac{gH}{2}(\rho_A + \rho_B)

Explanation:

As we know that the force due to pressure is balanced by the weight of the cylinder

So we will have

F = mg

so we have

\Delta P \pi R^2 = mg

so we have

\Delta P \pi R^2 = \pi R^2(\rho_A(\frac{H}{2}) + \rho_B(\frac{H}{2}))g

so we have

\Delta P = \frac{gH}{2}(\rho_A + \rho_B)

6 0
2 years ago
A 1700kg rhino charges at a speed of 50.0km/h. what is the magnitude of the average force needed to bring the rhino to a stop in
stepladder [879]
Impulse equals Change in Momentum
F = average applied force = to be determined
Δt = time during which the force is applied = 0.50 s
m = mass = 1,700 kg
Δp = change in momentum = to be determined
Δv = change in velocity = to be determined
v1 = initial velocity = 50.0 km/h = 50,000 m/h = 13.9 m/s
v2 = final velocity = 0.00 km/h = 0.00 m/s

F∙Δt = Δp
F∙Δt = m∙Δv
F∙Δt = m∙(v2 - v1)

F = m∙(v2 - v1) / Δt
F = 1,700 kg∙(0.00 m/s - 13.9 m/s) / 0.50 s
<span>F = -47,222 N The negative sign means that the force vector is </span>
<span>applied AGAINST the momentum vector of the rhinoceros.</span>
7 0
2 years ago
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