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Rom4ik [11]
3 years ago
13

A toy car (0.50 kg) runs on a frictionless track and has an initial kinetic energy of 2.2 J, as the drawing shows. The numbers b

eneath each hill give the heights of the hills. Over which of the hills will the car coast?
Physics
1 answer:
steposvetlana [31]3 years ago
6 0

Answer:

Height will be equal to 0.4489 m

Explanation:

We have given mass of the toy m = 0.50 kg

Initial kinetic energy K = 2.2 J

We have to fond the height of the hill over which car roast

When car will roast the hill its kinetic energy will be converted into potential energy and at maximum height all kinetic energy will be converted into potential energy

So at maximum height mgh=K

Acceleration due to gravity g=9.8m/sec^2

So 0.50×9.8×h = 2.2

h = 0.4489 m

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Determine the kinetic energy of 1000-kg roller coaster car that is moving with speed of 20.0m/s
nevsk [136]
B, i got the same question
7 0
3 years ago
A current of 1.41 A in a long, straight wire produces a magnetic field of 5.61 uT at a certain distance from the wire. Find
pshichka [43]

Answer:

0.050 m

Explanation:

The strength of the magnetic field produced by a current-carrying wire is given by

B=\frac{\mu_0 I}{2\pi r}

where

\mu_0=4\pi \cdot 10^{-7} H/m is the vacuum permeability

I is the current in the wire

r is the distance from the wire

And the magnetic field around the wire forms concentric circles, and it is tangential to the circles.

In this problem, we have:

I=1.41 A (current in the wire)

B=5.61\mu T=5.61\cdot 10^{-6} T (strength of magnetic field)

Solving  for r, we find the distance  from the wire:

r=\frac{\mu_0 I}{2\pi B}=\frac{(4\pi \cdot 10^{-7})(1.41)}{2\pi (5.61\cdot 10^{-6})}=0.050 m

4 0
3 years ago
A car speeding down the highway honks its horn, which has a frequency 392 Hz, but a resting bystander hears the frequency 440 Hz
Natali [406]

Answer:

37.42 m/s

Explanation:

We know that apparent frequency, \bar f is given by

\bar f=f\frac {V}{V-V_s} where f is the given frequency in this case 392, V is the speed of sound in air which is given as 343 and V_s is the speed of car which is unknown, \bar f is given as 440 Hz

440=392\times \frac {343}{343-V_s}\\343-V_s=392\times \frac {343}{440}=305.5818182\\V_s=343-305.5818182=37.41818182\approx 37.42 m/s

8 0
3 years ago
Can someone help me?​
Gwar [14]
24- series
25- parallel
26- no, because they’re connected in series
27- yes, because they’re connected in parallel
7 0
3 years ago
A 70.0 kg ice hockey goalie, originally at rest, has a 0.110 kg hockey puck slapped at him at a velocity of 31.5 m/s. Suppose th
NISA [10]

Answer

given,

mass of the goalie(m₁) = 70 kg

mass of the puck (m₂)= 0.11 kg

velocity of the puck = 31.5 m/s

elastic collision

v_1=\dfrac{m_2-m_1}{m_1+m_2}v_1+\dfrac{2m_2}{m_1+m_2}v_2

v_{pf}=\dfrac{0.11-70}{0.11+70}31.5+\dfrac{2m_2}{m_1+m_2}\times (0)

v_{pf}=-31.4\ m/s

v'_2 = \dfrac{2m_1v_1}{m_1+m_2}-\dfrac{(m_2-m_1)v_2}{m_2+m_1}

v_{gf} = \dfrac{2\times 0.11\times 31.5}{0.11+70}-\dfrac{(0.11-70)\times 0}{m_1+m_2}

v_{gf} = \dfrac{2\times 0.11\times 31.5}{0.11+70}

v_{gf} = 0.0988\ m/s

4 0
3 years ago
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