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charle [14.2K]
2 years ago
6

a 1250 kg car accelerates from rest to 6.13m/s over a distance of 8.58m calculate the average force of traction

Physics
1 answer:
iogann1982 [59]2 years ago
3 0
Use formula, v^2= u^2 + 2as.
The "v" and the "s" of the formula are given.
Since u is 0, just use f=ma.
I hope this helped!
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Which of the following statements about electric field lines associated with electric charges is false? Electric field lines can
Tcecarenko [31]

Answer:

The false statement is 'Electric field lines form closed loops'.

Explanation:

  • Electric field lines originate from positive end and terminates at negative end,i.e., field lines are inward in direction to the negative charges and outward from the positive charges.
  • These lines when close together represents high intensity and when far apart shows low intensity of the field.
  • These lines do not intersect, as the tangent drawn on these lines provides us with the field direction and intersection of these lines means two field directions which is not possible.
  • These lines unlike magnetic field lines do not form closed loops as they do not turn around but originate at positive end and terminates at negative end which ensures no loop formation.
8 0
2 years ago
A 7.25-kg bowling ball is being swung horizontally in a clockwise direction (as viewed from above) at a constant speed in a circ
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Answer:

a_c=9.66\frac{m}{s^2}

Explanation:

The centripetal acceleration is given by:

a_c=\frac{v^2}{r}(1)

Here v is the linear speed and r is the radius of the circular motion. v is defined as the distance traveled to make one revolution (2\pi r) divided into the time takes to make one revolution, that is, the period (T).

v=\frac{2\pi r}{T}(2)

Replacing (2) in (1) and replacing the given values:

a_c=\frac{(\frac{2\pi r}{T})^2}{r}\\a_c=\frac{4\pi^2 r}{T^2}\\a_c=\frac{4\pi^2 (1.85m)}{(2.75s)^2}\\a_c=9.66\frac{m}{s^2}

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2 years ago
The electric field near the earth's surface has magnitude of about 150n/c. what is the acceleration experienced by an electron n
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Felectric = q*E 
<span> Ftranslational = m*a 
</span><span> Felectric = Ftranslational
</span> <span>q*E = m*a 
</span><span> Solve for a 
</span><span> a = q/m*E </span>
<span> Our sign convention is "up is positive" 
</span><span> q = 1.6*10^-19 C 
</span><span> m = 1.67*10^-27 kg 
</span><span> E = -150 N/C (- because it is down and up is positive) 
</span> a =<span> -6,4*10^5</span><span> m/s^2 (downward) 
</span> answer
 a = -6,4*10^5 m/s^2 (downward) 
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Vader's light saber is red, while Obi-Wan's light saber is blue, meaning that Obi-Wan's light saber is emitting _____ compared t
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I assume it woukd be higher energy light waves. when fire is at its hottest state its blue because its burning off so much.
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A bar of silicon is 4 cm long with a circular cross section. If the resistance of the bar is 270 ω at room temperature, what is
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solution:

consider the following data\\&#10;length of slicon bar with circular cross section is 4cm or 0.04m\\&#10;at room temperature resistance of the slicon bar is 270\Omega \\&#10;represent the resistance in mathematical from\\&#10;r=p\frac{1}{A}---1\\&#10;where r is resistance and l is the length \\&#10;A is cross sectional area\\&#10;it is clear that resistivity of the silicon meterial is 6.4\times^2 \Omega.m\\&#10;substitute 6.4\times10^2 for p,270\Omega for R and 0.04m for l i equation (1).\\270=(604\times10^2)\frac{0.04}{A}\\&#10;rewrite the equation\\&#10;a=(6.4\times10^2)\frac{(0.04)}{270}\\&#10;=0.9481m^2\\&#10;write the formula for the circular cross sectional area of silicon bar.\\&#10;A=\pi r^2\\&#10;substitute 0.9481 for A in the above equation\\&#10;\pi r^2=0.9481&#10;r^2=\frac{0.9481}{3.14},since \pi =3.14\\&#10;0.30194\\&#10;further simplified\\&#10;r^2=0.30194\\&#10;\sqrt{0.30194}\\&#10;\cong 0.1509m\\&#10;\cong 150.1mm

7 0
2 years ago
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