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3 years ago

How does surface currents would be affected if eArth didn't rotate

1 answer:

6 0

Surface ocean currents are generally wind-driven. However, the rotation of the Earth affects the way the waters move through currents. Without rotation, currents may not exist.

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. A grindstone increases in angular speed from 4.00 rad/s to 12.00 rad/s in 4.00 s. Through what angle does it turn during that

ZanzabumX [31]

Answer:

$\theta=32rad$

Explanation:

In an uniformly accelerated circular motion, the angle traveled by the object is given by:

$\theta=\frac{\omega_f+\omega_0}{2}t$

Here $\omega_f$ is the final angular speed, $\omega_0$ is the initial angular speed and t is the time of the motion. Replacing the given values:

$\theta=\frac{12\frac{rad}{s}+4\frac{rad}{s}}{2}(4s)\\\theta=32rad$

5 0

4 years ago

When you hit a .27 kg volleyball the contact time is 50 ms and the average force is 125 N. If you serve the volleyball (from res

yulyashka [42]

(A) P(v) = 0.135v

(B) P(h) = 0.234v

Explanation:

Given-

Mass of the ball, m = 0.27kg

Force, F = 125N

angle of projection, θ = 30°

Let v be the velocity of the ball.

A) vertical component of the momentum of the volleyball

We know,

P(vertical) = mvsinθ

P(V) = 0.27 X v X sin 30°

P(V) = 0.27 X v X 0.5

P(V) = 0.135v

B) horizontal component of the momentum of the volleyball

We know,

P(Horizontal) = mvcosθ

P(h) = 0.27 X v X cos 30°

P(h) = 0.27 X v X 0.866

P(h) = 0.234v

8 0

3 years ago

You have a summer job at a company that developed systems to safely lower large loads down ramps. Your team is investigating a m

Fofino [41]

Answer:

Note that the emf induced is

emf = B d v cos (A)

---> v = emf / [B d cos (A)]

where

B = magnetic field

d = distance of two rails

v = constant speed

A = angle of rails with respect to the horizontal

Also, note that

I = emf/R

where R = resistance of the bar

Thus,

I = B d v cos (A) / R

Thus, the bar experiences a magnetic force of

F(B) = B I d = B^2 d^2 v cos (A) / R, horizontally, up the incline.

Thus, the component of this parallel to the incline is

F(B //) = F(B) cos(A) = B I d = B^2 d^2 v cos^2 (A) / R

As this is equal to the component of the weight parallel to the incline,

B^2 d^2 v cos^2 (A) / R = m g sin (A)

where m = the mass of the bar.

Solving for v,

v = [R m g sin (A) / B^2 d^2 cos^2 (A)] [ANSWER, the constant speed, PART A]

******************************

v = [R m g sin (A) / B^2 d^2 cos^2 (A)]

Plugging in the units,

m/s = [ [ohm * kg * m/s^2] / [T^2 m^2] ]

Note that T = kg / (s * C), and ohm = J * s/C^2

Thus,

m/s = [ [J * s/C^2 * kg * m/s^2] / [(kg / (s * C))^2 m^2] ]

= [ [J * s/C^2 * kg * m/s^2] / [(kg^2 m^2) / (s^2 C^2)]

As J = kg*m^2/s^2, cancelling C^2,,

= [ [kg*m^2/s^2 * s * kg * m/s^2] / [(kg^2 m^2) / (s^2)]

Cancelling kg^2,

= [ [m^2/s^2 * s * m/s^2] / [(m^2) / (s^2)]

Cancelling m^2/s^2,

= [s * m/s^2]

Cancelling s,

=m/s [DONE! WE SHOWED THE UNITS ARE CORRECT! ]

8 0

3 years ago

A paintball’s mass is 0.0032kg. A typical paintball strikes a target moving at 85.3 m/s.

vekshin1

Answer:

A) If the paintball stops completely the magnitude of the change in the paintball’s momentum is $p=0.273kg*m/s$

B) If the paintball bounces off its target and afterward moves in the opposite direction with the same speed, the change in the paintball’s momentum is $p=0.546kg*m/s$

C) A paintball bouncing off your skin in the opposite direction with the same speed hurts more than a paintball exploding upon your skin because of the strength exerted is twice than if it explodes.

Explanation:

A) We use the formula of momentum $p=mv$ , so we have $p=0.0032kg*85.3m/s=0.273kg*m/s$

B) We use the same formula above, then due we have a change of direction at the same speed, therefore the change in the momentum is the double so

$p=2*0.0032kg*85.3m/s=0.546kg*m/s$ .

C) The average strength of the force an object exerts during impact is determined by the amount the object’s momentum changes. therefore

$F=\frac{\Delta p}{\Delta t}$ , as we don't have any data about the impact time but we know momentum is twice, time does no matter and strength is twice too.

4 0

3 years ago

A technician wishes to determine the wavelength of the light in a laser beam. To do so, she directs the beam toward a partition

Nana76 [90]

Answer:

λ = 5.656 x 10⁻⁷ m = 565.6 nm

Explanation:

Using the formula of fringe spacing from the Young's Double Slit experiment, which is given as follows:

$\Delta x = \frac{\lambda L}{d}\\\\\lambda = \frac{\Delta x\ d}{L}$

where,

λ = wavelength = ?

Δx = fringe spacing = 1.6 cm = 0.016 m

L = Distance between slits and screen = 4.95 m

d = slit separation = 0.175 mm = 0.000175 m

Therefore,

$\lambda = \frac{(0.016\ m)(0.000175\ m)}{4.95\ m}\\\\$

λ = 5.656 x 10⁻⁷ m = 565.6 nm

6 0

3 years ago