A sports car accelerates at a constant rate from rest to a speed of 90 km/hr in 8 s. What is its acceleration?

1 answer:

3 0

First convert 90km/hr to m/s.

Initiate velocity = 0m/s (car was at rest)

Final velocity is 25m/s (90km/hr converted)

25m/s - 0m/s / 8s = 3.125 m/s^s

Therefore the answer is option A (3.13m/s^2)

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A ball is dropped from a cliff. determine how far the ball fell after 7.5 seconds

grin007 [14]

Answer:

The ball fell 275.625 meters after 7.5 seconds

Explanation:

If an object is left on free air (no friction), it describes an accelerated motion in the vertical direction, powered exclusively by the acceleration of gravity. The formulas needed to compute the different magnitudes involved are

$V_f=gt$

$\displaystyle y=\frac{gt^2}{2}$

Where $V_f$ is the final speed of the object in free fall, assumed positive downwards, t is the time elapsed since the release and y is the vertical distance traveled by the object

The ball was dropped from a cliff. We need to calculate the vertical distance the ball went down in t=7.5 seconds. We'll use the formula

$\displaystyle y=\frac{gt^2}{2}$

$\displaystyle y=\frac{(9.8)(7.5)^2}{2}$

$Y=275.625\ m$

5 0

3 years ago

In si units, the electric field in an electromagnetic wave is described by ey = 104 sin(1.40 107x − ωt). (a) find the amplitude

melamori03 [73]

Answers:
(a) $B_o = 0.3466$ μT
(b) $\lambda = 0.4488$ μm
(c) f = $6.68 * 10^{14}Hz$

Explanation:
Given electric field(in y direction) equation:
$E_y = 104sin(1.40 * 10^7 x -\omega t)$

(a) The amplitude of electric field is $E_o = 104$ . Hence

The amplitude of magnetic field oscillations is $B_o = \frac{E_o}{c}$
Where c = speed of light

Therefore,
$B_o = \frac{104}{3*10^8} = 0.3466$ μT (Where T is in seconds--signifies the oscillations)

(b) To find the wavelength use:
$\frac{2 \pi }{\lambda} = 1.40 * 10^7$
$\lambda = \frac{2 \pi}{1.40} * 10^{-7}$
$\lambda = 0.4488$ μm

(c) Since c = fλ
=> f = c/λ

Now plug-in the values
f = (3*10^8)/(0.4488*10^-6)
f = $6.68 * 10^{14}Hz$

6 0

3 years ago

A triply charged ion with velocity 7.00 × 10^6 m/s moves in a path of radius 36.0 cm in a magnetic field of 0.55 T in a mass spe

saw5 [17]

Answer:

Mass of ion will be $22\times 10^{-13}kg$