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Leokris [45]
3 years ago
5

Starting from rest, a coin and a ring roll down a ramp without slipping. Which of the following are true:

Physics
1 answer:
svetlana [45]3 years ago
4 0

Answer:

The answer is D

Explanation:

The object with the heavy mass will reach first because the cause it is heaver so it will go faster

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In an experiment, a shearwater (a seabird) was taken from its nest, flown a distance 5220 km away, and released. It found its wa
Korolek [52]

Answer:

4.6834625323 m/s

0 m/s

Explanation:

s = Displacement

t = Time

Velocity is given by

v=\dfrac{s}{t}\\\Rightarrow v=\dfrac{5220000}{12.9\times 24\times 60\times 60}\\\Rightarrow v=4.6834625323\ m/s

The bird's average velocity for the return flight is 4.6834625323 m/s

In the whole episode the bird went 5220 km away from its nest and came back. This means the displacement is zero.

Hence, the average velocity for the whole episode is 0 m/s

7 0
3 years ago
Read 2 more answers
The electric field in a region of space increases from 0 to 2150 N/C in 5.00 s. What is the magnitude of the induced magnetic fi
Feliz [49]

To solve this problem we will use the Ampere-Maxwell law, which   describes the magnetic fields that result from a transmitter wire or loop in electromagnetic surveys. According to Ampere-Maxwell law:

\oint \vec{B}\vec{dl} = \mu_0 \epsilon_0 \frac{d\Phi_E}{dt}

Where,

B= Magnetic Field

l = length

\mu_0 = Vacuum permeability

\epsilon_0 = Vacuum permittivity

Since the change in length (dl) by which the magnetic field moves is equivalent to the perimeter of the circumference and that the electric flow is the rate of change of the electric field by the area, we have to

B(2\pi r) = \mu_0 \epsilon_0 \frac{d(EA)}{dt}

Recall that the speed of light is equivalent to

c^2 = \frac{1}{\mu_0 \epsilon_0}

Then replacing,

B(2\pi r) = \frac{1}{C^2} (\pi r^2) \frac{d(E)}{dt}

B = \frac{r}{2C^2} \frac{dE}{dt}

Our values are given as

dE = 2150N/C

dt = 5s

C = 3*10^8m/s

D = 0.440m \rightarrow r = 0.220m

Replacing we have,

B = \frac{r}{2C^2} \frac{dE}{dt}

B = \frac{0.220}{2(3*10^8)^2} \frac{2150}{5}

B =5.25*10^{-16}T

Therefore the magnetic field around this circular area is B =5.25*10^{-16}T

3 0
3 years ago
Llammmmaaaa or chickennnss?
Semenov [28]

Answer:

chickennnss

Explanation:chickennnss

8 0
3 years ago
Read 2 more answers
What is numbers 2 and 4? Please help, and many thanks to those who do chose to help.
Elan Coil [88]

Answer:

2. 6 km/h

4. 150 miles/h

5 0
3 years ago
A devout halloweener not only dressed as an astronaut, but travelled to the moon for the full experience. The astronaut jumps on
nikdorinn [45]

Answer:

2.78 m

Explanation:

At the peak, the velocity is 0.

Given:

a = -1.6 m/s²

v₀ = 2.98 m/s

v = 0 m/s

x₀ = 0 m

Find:

x

v² = v₀² + 2a(x - x₀)

(0 m/s)² = (2.98 m/s)² + 2(-1.6 m/s²) (x - 0 m)

x = 2.775 m

Rounded to 3 sig-figs, the astronaut halloweener reaches a maximum height of 2.78 meters.

5 0
3 years ago
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