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Physics

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kupik [55]3 years ago

3 0

Equal to the force of friction exerted upon the car . ( how this is right )

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A cosmic ray proton moving toward Earth at 5.00 x 107 m/s experiences a magnetic force of 1.7 x 10-16 N. What is the strength of

Blizzard [7]

Answer:

the strength of the magnetic field is 3 x 10⁻⁵ T

Explanation:

Given;

velocity of the cosmic ray, v = 5 x 10⁷ m/s

force experienced by the ray, f = 1.7 x 10⁻¹⁶ N

angle between the ray's velocity and the magnetic field, θ = 45⁰

The strength of the magnetic field is calculated as;

$F = qvB \ sin(\theta)\\\\B = \frac{F}{qv\times sin(\theta)} \\\\where;\\\\B \ is \ the \ strength \ of \ the \ magnetic \ field\\\\q \ is \ the \ charge \ of \ the \ cosmic \ ray \ proton = 1.602 \times 10^{-19} \ C\\\\B = \frac{1.7\times 10^{-16}}{(1.602 \times 10^{-19})\times (5\times 10^7) \times sin \ (45)} \\\\B = 3 \times 10^{-5} \ T$

Therefore, the strength of the magnetic field is 3 x 10⁻⁵ T

7 0

3 years ago

What can be added to an atom to cause a nonvalence electron in the atom to temporarily become a valence electron

Inessa05 [86]

Answer:

providing energy to an atom can allow the electron in its non valence shell to obtain energy and move to a higher energy orbital and act as a valence electron.

Explanation:

8 0

3 years ago

Read 2 more answers

A light spring stretches 0.13 m when a 0.35 kg mass is hung from it. The mass is pulled down from this equilibrium position an a

Alenkasestr [34]

Answer:

v = 1.30 m/s

Explanation:

given,

mass hung = 0.35 Kg

spring stretched when load is hanged (x)= 0.13 m