1.<br> What is the most direct type of heat transfer?<br> radiation<br> conduction

A person wearing rollerblades starts from rest and accelerates at a constant rate. She reaches a speed of 13 mph (5.18 m/s) exac

kvv77 [185]

Answer: a. ) What is the magnitude of their average acceleration (in m/s2)

5 0

3 years ago

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The thunderbolt bobsled team is training for Olympic Gold. During practice they start a run with a speed of 0.57 m/s, they compl

aleksandr82 [10.1K]

$acceleration=\frac{\Delta\ velocity}{\Delta\ time}\\\\
v_{initial}=0,57m/s\\
distance=1360m\\ \Delta\ time=89,49seconds\\\\
v_{final}-v_{initial}=\frac{distance}{time}\\
v_{final}=\frac{distance}{time}+v_{initial}\\
v_{final}=\frac{1360}{89,49}+0,57\\\\v_{final}=15,77\frac{m}{s}\\\\
acceleration=\frac{15,77-0,57}{89,49}=0,17\frac{m}{s^2}\\\\ \boxed{acceleration=0,17\frac{m}{s^2}}$

6 0

3 years ago

PLEASE HELP 15 POINTS AND BRAINIEST!!!<br> Reporting fake answers

Zolol [24]

Answer:

(3) The period of the satellite is independent of its mass, an increase in the mass of the satellite will not affect its period around the Earth.

(4) he gravitational force between the Sun and Neptune is 6.75 x 10²⁰ N

Explanation:

(3) The period of a satellite is given as;

$T = 2\pi \sqrt{\frac{r^3}{GM} }$

where;

T is the period of the satellite

M is mass of Earth

r is the radius of the orbit

Thus, the period of the satellite is independent of its mass, an increase in the mass of the satellite will not affect its period around the Earth.

(4)

Given;

mass of the ball, m₁ = 1.99 x 10⁴⁰ kg

mass of Neptune, m₂ = 1.03 x 10²⁶ kg

mass of Sun, m₃ = 1.99 x 10³⁰ kg

distance between the Sun and Neptune, r = 4.5 x 10¹² m

The gravitational force between the Sun and Neptune is calculated as;

$F_g = \frac{Gm_2m_3}{r^2} \\\\F_g = \frac{6.67\times 10^{-11} \times 1.03 \times 10^{26}\times 1.99\times 10^{30}}{(4.5\times 10^{12})^2} \\\\F_g = 6.751 \times 10^{20} \ N$

5 0

3 years ago

A 7.3 cm diameter loop of wire is initially oriented so that its plane is perpendicular to a magnetic field of 0.61 T pointing u

kenny6666 [7]

Answer:

induced emf = 28.65 mV

Explanation:

given data

diameter = 7.3 cm

magnetic field = 0.61

time period = 0.13 s

to find out

magnitude of the induced emf

solution

we know radius is diameter / 2

radius = 7.3 / 2

radius = 3.65 m

so induced emf is dπ/dt = Adb/dt

induced emf = A × ΔB / Δt

induced emf = πr² × ΔB / Δt

induced emf = π (0..65)² × ( 0.61 - (-0.28)) / 0.13

induced emf = 0.0286538 V

so induced emf = 28.65 mV

3 0

3 years ago

Most radiation exposure comes from

Mrrafil [7]

Most radiation exposure comes from natural sources. These so-called "natural sources" include Radon, Internal, Terrestrial, and Cosmic. Among that, Radon proves to be the largest source of radiation. Radon is a naturally occurring radioactive gas that comes from the breakdown of uranium.

5 0

4 years ago

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1. What is the most direct type of heat transfer? radiation conduction