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Andrej [43]
3 years ago
13

Two skiers arrive at the bottom of a ramp travelling at the same speed. Friction may be ignored in this problem. One skier is an

adult and one is a child. The adult has 3 times the mass of the child. Which skier makes it to a greater height on the ramp before coming to a stop?
Physics
1 answer:
inysia [295]3 years ago
6 0

Answer:

Both will reach to same height

Explanation:

Here we can see that friction is to be ignored

so we can say that work done by all the non conservative forces is change in mechanical energy

Since all non conservative forces here is zero

so mechanical energy is conserved here

so here we can say that sum of initial kinetic energy and potential energy = sum of final kinetic energy and potential energy

So we will have

\frac{1}{2}mv^2 = mgH

now maximum height is given as

H = \frac{v^2}{2g}

so here we can say that greatest height will be independent of the mass so they both will reach at same height

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A wire carrying a current of 26.9 A is bent into a circular arc with a radius of 0.6 cm that sweeps out 0.900 radians. What is t
melomori [17]

The magnetic field at the center of the arc is 4 × 10^(-4) T.

To find the answer, we need to know about the magnetic field due to a circular arc.

<h3>What's the mathematical expression of magnetic field at the center of a circular arc?</h3>
  • According to Biot savert's law, magnetic field at the center of a circular arc is
  • B=(μ₀ I/4π)× (arc/radius²)
  • As arc is given as angle × radius, so

        B=( μ₀I/4π)×(angle/radius)

<h3>What will be the magnetic field at the center of a circular arc, if the arc has current 26.9 A, radius 0.6 cm and angle 0.9 radian?</h3>

B=(μ₀ I/4π)× (0.9/0.006)

  = (10^(-7)× 26.9)× (0.9/0.006)

  = 4 × 10^(-4) T

Thus, we can conclude that the magnitude of magnetic field at the center of the circular arc is 4 × 10^(-4) T.

Learn more about the magnetic field of a circular arc here:

brainly.com/question/15259752

#SPJ4

5 0
2 years ago
The tungsten filament of a light bulb has an operating temperature of about 2 100 K. If the emitting area of the filament is 1.0
Murljashka [212]

Answer:

75 W

Explanation:

T = temperature of the filament = 2100 K

A = Emitting area of the filament = 1 cm² = 10⁻⁴ m²

e = Emissivity = 0.68

\sigma = Stefan's constant = 5.67 x 10⁻⁸ Wm⁻²K⁻⁴

Using Stefan's law, Power output of the light bulb is given as

P = \sigma e AT^{4} \\P = (5.67\times10^{-8}) (0.68) (10^{-4}) (2100)^{4}\\P = 75 W

5 0
3 years ago
The recoil of a shotgun can be significant. Suppose a 3.6-kg shotgun is held tightly by an arm and shoulder with a combined mass
devlian [24]

Answer:

The velocity of the recoil is v=1.001 \frac{m}{s}

Explanation:

Kinetic Energy

m_{bullet}*v_{bullet}=m_{gun}*v_{recoil}\\m_{gun}= 15.0kg+3.6kg

The mass of th gun is the both mass the shotgun and the arm shoulder combination

m_{bullet}=0.049kg\\v_{bullet}=380\frac{m}{s} \\m_{bullet}*v_{bullet}=m_{gun}*v_{recoil}\\0.049kg*380\frac{m}{s}=(15.kg+3.6kg)* v_{recoil}\\v_{recoil}=-\frac{18.62 kg \frac{m}{s} }{18.6 kg}\\ v_{recoil}=-1.0010 \frac{m}{s}

The velocity is negative because is in opposite direction of the bullet

6 0
3 years ago
Read 2 more answers
Two balls of equal mass are thrown horizontally with the same initial velocity. They hit identical stationary boxes resting on a
11111nata11111 [884]

Answer:

The box 1 moves faster.

Explanation:

lets

Mass =m  kg

Initial velocity = u m/s

Initial velocity of box = 0 m/s

Let stake mass of block = m

When ball bounces back:

The final speed of the box = v

Final speed of ball = - u

Pi = Pf  ( From linear momentum conservation)

m x u + m x 0 = m ( - u) + m v

mu + mu = m v

v= 2 u

When ball get stuck :

The final speed of ball and box = v

Pi = Pf  ( From linear momentum conservation)

m x u + m x 0 = (m+m) v

v= u /2

So the box 1 moves faster.

6 0
3 years ago
- Which of the following cannot be a unit for work?<br>A. Joule <br>B. N.m<br>C. Kilowatt<br>D. All​
shusha [124]

Answer:

joule

Explanation:

is for work it is also repreasent by SI

4 0
3 years ago
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