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HACTEHA [7]
3 years ago
6

A driver drove for 3 hours at a certain speed. if he drove 12 more miles at the same speed, the distance would be 132 miles. at

what speed did he drive?
Physics
1 answer:
Zepler [3.9K]3 years ago
4 0

Answer:

40mph

Explanation:

1st leg DATA:

time = 3 hrs ; speed = r mph ; distance = 3r miles

------------

2nd leg DATA:

speed = r mph ; distance = 12 miles

 

--------------------------------

3r + 12 = 132

3r = 120

rate = 40 mph

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1 arcsecond is 1/3600 of a degree, 0.00028 degree.  

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3 years ago
A rotating merry go-round makes one complete revolution in 4.2 s. (a) what is the linear speed of a child seated 1.3 m from the
gregori [183]
The radius, r, of the child from the center of the wheel is
r = 1.3 m

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ω = (2π rad)/(4.2 s) = 1.496 rad/s

The linear speed of the child is the tangential velocity, given by
v = rω
   = (1.3 m)*(1.496 rad/s)
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3 years ago
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I need help on 6, 7, 8, and 9
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5 0
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The light coming out of a concave lens:
torisob [31]
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4 0
3 years ago
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If the electron just misses the upper plate as it emerges from the field, find the magnitude of the electric field.
Damm [24]

Answer:

The magnitude of the electric field be 171.76 N/C so that the electron misses the plate.

Explanation:

As data is incomplete here, so by seeing the complete question from the search the data is

vx_0=1.1 x 10^6

ax=0 As acceleration is zero in the horizontal axis so

Equation of motion in horizontal direction is given as

s_x=v_x_0 t

t=\frac{s_x}{v_x}\\t=\frac{2 \times 10^{-2}}{1.1 \times 6}\\t=1.82 \times 10^{-8} s

Now for the vertical distance

vy_o=0

than the equation of motion becomes

s_y=v_y_0 t+\frac{1}{2} at^2\\s_y=\frac{1}{2} at^2\\0.5 \times 10^{-2}=\frac{1}{2} a(1.82 \times 10^{-8})^2\\a=3.02 \times 10^{13} m/s^2

Now using this acceleration the value of electric field is calculated as

E=\frac{F}{q}\\E=\frac{ma}{q}\\E=\frac{m_ea}{q_e}\\

Here a is calculated above, m is the mass of electron while q is the charge of electron, substituting values in the equation

E=\frac{9.1\times 10^{-31} \times 3.02 \times 10^{13} }{1.6 \times 10^{-19}}\\E=171.76 N/C

So the magnitude of the electric field be 171.76 N/C so that the electron misses the plate.

5 0
3 years ago
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