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snow_lady
1 month ago
9

Snowboarder Jump—Energy and Momentum

Physics
1 answer:
serg [3.5K]1 month ago
4 0

Answer:

DEFINED TERMS ARE AS FOLLOWS:

Explanation:

MOMENTUM- REFERS TO THE CAPACITY TO INCREASE OR GROW A CONSTANT FORCE.

KINETIC ENERGY: - IT REPRESENTS THE ENERGY POSSESSED BY A PARTICLE WHEN IN ACTIVE MOTION.

POTENTIAL ENERGY- THIS IS THE ENERGY A PARTICLE HOLDS WHEN IT IS IN A STATE OF REST.

IN THIS SCENARIO, THE SNOWBOARDER IS IN A MOVING STATE, HENCE HE HAS KINETIC ENERGY, AND MOMENTUM IS ESSENTIAL FOR SUSTAINING THAT KINETIC ENERGY OVER TIME.

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Violet light (λ = 400 nm) passing through a diffraction grating for which the slit spacing is 6.0 μm forms a pattern on a screen
Maru [3345]

Answer:

The third-order dark fringe

Explanation:

y = Distance from central bright fringe = 204 mm

λ = Wavelength = 400 nm

L = Distance from screen to source = 1 m

d = Slit spacing = 6 μm

tan\theta =\frac{y}{L}\\\Rightarrow tan\theta =\frac{204}{1000}=0.2012^{\circ}

dsin\theta=m\lambda\\\Rightarrow m=\frac{dsin\theta}{\lambda}\\\Rightarrow m=\frac{6\times 10^{-6}sin0.2012}{400\times 10^{-9}}=2.9982\approx 3

The order of the fringe is 3

Thus, it’s identified as a dark fringe.

3 0
1 month ago
An aluminum "12 gauge" wire has a diameter d of 0.205 centimeters. The resistivity ρ of aluminum is 2.75×10−8 ohm-meters. The el
Sav [3153]

Complete Question

An aluminum "12 gauge" wire measures a diameter of 0.205 centimeters. The resistivity ρ of aluminum is 2.75×10−8 ohm-meters. The electric field E in the wire varies over time as E(t)=0.0004t2−0.0001t+0.0004 newtons per coulomb, where time is recorded in seconds.

At time 5 seconds, I = 1.2 A.

We need to find the charge Q traveling through a cross-section of the conductor from time 0 to time 5 seconds.

Answer:

The charge is  Q =2.094 C

Explanation:

The question indicates that

    The wire’s diameter is  d = 0.205cm = 0.00205 \ m

     The radius of the wire is  r = \frac{0.00205}{2} = 0.001025 \ m

     Aluminum's resistivity is 2.75*10^{-8} \ ohm-meters.

       The electric field variation is described as

         E (t) = 0.0004t^2 - 0.0001 +0.0004

     

The charge is effectively given by the equation

       Q = \int\limits^{t}_{0} {\frac{A}{\rho} E(t) } \, dt

Where A is the area expressed as

       A = \pi r^2 = (3.142 * (0.001025^2)) = 3.30*10^{-6} \ m^2

 Thus,

       \frac{A}{\rho} = \frac{3.3 *10^{-6}}{2.75 *10^{-8}} = 120.03 \ m / \Omega

Therefore

      Q = 120 \int\limits^{t}_{0} { E(t) } \, dt

By substituting values

      Q = 120 \int\limits^{t}_{0} { [ 0.0004t^2 - 0.0001t +0.0004] } \, dt

     Q = 120 [ \frac{0.0004t^3 }{3} - \frac{0.0001 t^2}{2} +0.0004t] } \left | t} \atop {0}} \right.

The question states that t =  5 seconds

           Q = 120 [ \frac{0.0004t^3 }{3} - \frac{0.0001 t^2}{2} +0.0004t] } \left | 5} \atop {0}} \right.

          Q = 120 [ \frac{0.0004(5)^3 }{3} - \frac{0.0001 (5)^2}{2} +0.0004(5)] }

         Q =2.094 C

     

5 0
1 month ago
A car has a mass of 1600 kg. It is stuck in the snow and is being pulled out by a cable that applies a force of 7560 N due north
Ostrovityanka [3204]

Response:

The car's acceleration will be a=0.1375m/sec^2

Reasoning:

We are provided with a mass for the ball, m = 1600 kg

The force directed northward is F= 7560 N

The resistance force that counters the car's motion F_R=7340N

Thus, the overall force acting on the car F_{net}=F-F_R=7560-7340=220N

According to Newton's second law, we understand that F=ma

Therefore 220=1600\times a

a=0.1375m/sec^2

7 0
1 month ago
a professional baseball player can pitch a baseball with a velocity of 44.7m/s towards home plate. If a baseball weighs 1.4 N, h
Sav [3153]
To find the mass using a weight of 1.4 N:
1.4/9.8 = 0.1428 kg
The momentum is calculated as 0.1428 multiplied by 44.7, which is 6.38 kgm/s.
3 0
1 month ago
Read 2 more answers
When you skid to a stop on your bike, you can significantly heat the small patch of tire that rubs against the road surface. Sup
Sav [3153]

Answer:

W_f = 148.17J

Explanation:

The friction created between the tire and the ground generates thermal energy as force is applied during skidding.

The mentioned force relates to half the impact on the rear tire, resulting in a calculated normal force of,

N=\frac{mg}{2} = \frac{90*9.8}{2} = 441N

The work executed is determined by the frictional force and the distance covered,

W_f = fd = \mu_k Nd

Where \mu_k [/ tex] is the coefficient of kinetic frictionN is the normal force previously found d is the distance traveled,Replacing,[tex]W_f = (0.80)(441)(0.42)

The thermal energy produced from the work done is,

W_f = 148.17J

3 0
1 month ago
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