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nexus9112 [7]
3 years ago
8

Please answer soon

Physics
1 answer:
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

now,

weight of the mass attached = Kx

             m g = k x

             0.35 x 9.8 = k x 0.13

                k = 26.38 N/m

now, using conservation of energy

 \dfrac{1}{2}mv^2 = \dfrac{1}{2}kx'^2

 v = \sqrt{\dfrac{kx'^2}{m}}

 v = \sqrt{\dfrac{26.38 \times 0.15^2}{0.35}}

 v = \sqrt{1.6958}

    v = 1.30 m/s

6 0
3 years ago
Please help<br> types of stimuli the sensory organs receive.
Alborosie

Answer: Photoreceptor cell, Chemoreceptor, Thermoreceptor, Mechanoreceptor

Hope this helps(:

Explanation:

3 0
3 years ago
An artillery shell is launched on a flat, horizontal field at an angle of α = 40.8° with respect to the horizontal and with an i
Lina20 [59]

Answer:

1317.4 m

Explanation:

We are given that

Angle=\alpha=40.8^{\circ}

Initial speed =v_0=346m/s

We have to find the horizontal distance covered  by the shell after 5.03 s.

Horizontal component of initial speed=v_{ox}=v_0cos\theta=346cos40.8=261.9m/s

Vertical component of initial speed=v_{oy}=346sin40.8=226.1m/s

Time=t=5.03 s

Horizontal distance =Horizontal\;velocity\times time

Using the  formula

Horizontal distance=261.9\times 5.03

Horizontal distance=1317.4 m

Hence, the horizontal distance covered by the shell=1317.4 m

8 0
3 years ago
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