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3 years ago

1. A drag racer accelerates from rest at 18ft/sec^2. How long does it take to acquire a speed of 60mph? What is required?

Physics

1 answer:

nignag [31]3 years ago

6 0

(1) The time taken for the drag racer to accelerate is 4.89 s

(2) The speed of the contestant (a) in feet per second is 30.89 ft/s (b) in miles per hr is 21.12 miles per hr.

(1) To calculate the time required to accelerate 18 ft/sec² from rest to a velocity of 60 mph, we use the formula below.

Formula:

t = (v-u)/a........... Equation 1

Where:

t = time
v = Final velocity
u = initial velocity
a = acceleration.

From the question,

Given:

a = 18 ft/sec² = (18×0.3048) = 5.4864 m/s²
v = 60 mph = (60×0.44704) = 26.82 m/s
u = 0 m/s ( from rest)

Substitute these values into equation 1

t = (26.82-0)/5.4864
t = 4.89 seconds

(2) To calculate the speed of the contestant, we use the formula below

Formula:

s = d/t............ Equation 1

Where:

s = speed of the contestant
d = distance
t = time.

From the question,

Given:

d = 100 m
t = 10.6 s

Substitute these values into equation 1

s = 100/10.6
s = 9.43 m/s

(a) In feet = (9.43/0.3048) = 30.94 ft/s

(b) in miles per hr = (9.43×2.24) = 21.12 miles per hr

Hence, (1) The time taken for the drag racer to accelerate is 4.89 s (2) The speed of the contestant (a) in feet per second is 30.89 ft/s (b) in miles per hr is 21.12 miles per hr.

Learn more about speed here: brainly.com/question/4931057

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nasty-shy [4]

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3 years ago

Given vectors D (3.00 m, 315 degrees wrt x-axis) and E (4.50 m, 53.0 degrees wrt x-axis), find the resultant R= D + E. (a) Write

Eva8 [605]

Answer:

R = ( 4.831 m , 1.469 m )

Magnitude of R = 5.049 m

Direction of R relative to the x axis= 16°54'33'

Explanation:

Knowing the magnitude and directions relative to the x axis, we can find the Cartesian representation of the vectors using the formula

$\vec{A}= | \vec{A} | \ ( \ cos(\theta) \ , \ sin (\theta) \ )$

where $| \vec{A} |$ its the magnitude and θ.

So, for our vectors, we will have:

$\vec{D}= 3.00 m \ ( \ cos(315) \ , \ sin (315) \ )$

$\vec{D}= ( 2.121 m , -2.121 m )$

and

$\vec{E}= 4.50 m \ ( \ cos(53.0) \ , \ sin (53.0) \ )$

$\vec{E}= ( 2.71 m , 3.59 m )$

Now, we can take the sum of the vectors

$\vec{R} = \vec{D} + \vec{E}$

$\vec{R} = ( 2.121 \ m , -2.121 \ m ) + ( 2.71 \ m , 3.59 \ m )$

$\vec{R} = ( 2.121 \ m + 2.71 \ m , -2.121 \ m + 3.59 \ m )$

$\vec{R} = ( 4.831 \ m , 1.469 \ m )$

This is R in Cartesian representation, now, to find the magnitude we can use the Pythagorean theorem

$|\vec{R}| = \sqrt{R_x^2 + R_y^2}$

$|\vec{R}| = \sqrt{(4.831 m)^2 + (1.469 m)^2}$

$|\vec{R}| = \sqrt{23.338 m^2 + 2.158 m^2}$

$|\vec{R}| = \sqrt{25.496 m^2}$

$|\vec{R}| = 5.049 m$

To find the direction, we can use

$\theta = arctan(\frac{R_y}{R_x})$

$\theta = arctan(\frac{1.469 \ m}{4.831 \ m})$

$\theta = arctan(0.304)$

$\theta = 16\°54'33''$

As we are in the first quadrant, this is relative to the x axis.

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marishachu [46]

Answer:

Explanation:

Given

mass of automobile $m=1000\ kg$

velocity of automobile $v=10\ m/s$

If the automobile crashes into the brick wall then it's kinetic energy is converted into thermal and demolishing energy to wall.

Car completely stops as it crashes into the wall so Energy converted into demolition and thermal energy

$E=\frac{1}{2}mv^2$

$E=\frac{1}{2}\times 1000\times (10)^2$

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