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3 years ago

A car accelerates for 10 seconds. During this time, the angular

Physics

1 answer:

bagirrra123 [75]3 years ago

4 0

Answer:

the angular acceleration of the car is 1.5 rad/s²

Explanation:

Given;

initial angular velocity, $\omega_i$ = 10 rad/s

final angular velocity, $\omega_f$ = 25 rad/s

time of motion, t = 10 s

The angular acceleration of the car is calculated as follows;

$a_r = \frac{\omega_f - \omega_i }{t} \\\\a_r = \frac{25-10}{10} = 1.5 \ rad/s^2$

Therefore, the angular acceleration of the car is 1.5 rad/s²

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An archer shoots an arrow with a velocity of 45.0m/s at an angle of 50.0degrees with the horizontal.An assistant standing on the

garri49 [273]

Answer:

a) u = 30.29 m/s

b) t = 2.09 s

Explanation:

given,

velocity = 45 m/s

angle (θ) = 50°

horizontal velocity = 45 cos 50°

time taken to reach 150 m.

times = $\dfrac{150}{45 cos 50^0}$

t = 5.19 s

a) height of arrow

$s = u t +\dfrac{1}{2}gt^2$

$s = v sin \theta \times t+\dfrac{1}{2}gt^2$

$s = 45 sin 50^0 \times 5.19 -\dfrac{1}{2}\times 9.81\times 5.19^2$

s = 46.78 m

v² - u² = 2 g s

u² = 2 × 9.81 × 46.78

u = 30.29 m/s

b) time taken by the apple = $\dfrac{u}{g}=\dfrac{30.29}{9.81}$

= 3.09 s

time after which it has to be thrown = 5.19-3.09 = 2.1 s

5 0

3 years ago

Calculate the current flowing if a charge of 36 kilocoulombs flows in 1 hour.

o-na [289]

Answer:

Current = 10 Amperes.

Explanation:

Given the following dat;

Quantity of charge, Q = 36 kilocoulombs (KC) = 36 * 1000 = 36000C

Time = 1 hour to seconds = 60*60 = 3600 seconds

To find the current;

Quantity of charge = current * time

Substituting in the equation

36000 = current * 3600

Current = 36000/3600

Current = 10 Amperes.

6 0

2 years ago

Describe how the student should measure the time taken for the toy parachute to

PolarNik [594]

The time must be measured with respect to gravity. As it falls, it has free fall that is the force acting on it will be the gravity.With the distance in account, d = 1/2 gt²

t = √(2d/g)

4 0

3 years ago

Someone help this is a test and i need to finish this quick i will give brainliest of it lets me

tekilochka [14]

Sorry, I don't know but I think the correct answer is the first option.

7 0

2 years ago

What percentage of the takeoff velocity did the plane gain when it reached the midpoint of the runway? a plane accelerates from

ElenaW [278]

When is at the end of the runway the velocity of the plane is given by the equation $vf^{2}=0+2*a*s$ where s=1800 m is the runway length. Thus
$vf^{2}=2*5*1800=18000 (m/s)^{2}$
$vf =134.164 (m/s)$

At half runway the velocity of the plane is
$v^{2}=2*5* \frac{1800}{2}=9000 ( \frac{m}{s} )^{2}
$
$v= \sqrt{9000}=94.87 ( \frac{m}{s})$

Therefore at midpoint of runway the percentage of takeoff velocity is
‰ $P= \frac{v}{vf}= \frac{94.87}{134.164}=0.707$

6 0

3 years ago