All categories

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

You might be interested in

The barrel of a rifle has a length of 0.855 m. A bullet leaves the muzzle of a rifle with a speed of 553 m/s. What is the accele

Novay_Z [31]

Answer:

Acceleration of the bullet will be 1778835.6 $m/sec^2$

Explanation:

We have given length of the barrel refile s= 0.855 m

When the bullet leaves the muzzle its velocity is 553 m/sec

So final velocity v = 553 m/sec

Initial velocity will be 0 that is u = 0 m/sec

According to third equation of motion $v^2=u^2+2as$

$553^2=0^2+2\times a\times 0.855$

$a=178835.6m/sec^2$

5 0

4 years ago

A 50.0-kg projectile is fired at an angle of 30 degrees above thehorizontal with an initial speed of 1.20 x 102 m/s fromthe top

Advocard [28]

Answer:

a) Em₀ = 42.96 104 J , b) $W_{fr}$ = -2.49 105 J , c) vf = 3.75 m / s

Explanation:

The mechanical energy of a body is the sum of its kinetic energy plus the potential energies it has

Em = K + U

a) Let's look for the initial mechanical energy

Em₀ = K + U

Em₀ = ½ m v2 + mg and

Em₀ = ½ 50.0 (1.20 102) 2 + 50 9.8 142

Em₀ = 36 104 + 6.96 104

Em₀ = 42.96 104 J

b) The work of the friction force is equal to the change in the mechanical energy of the body

$W_{fr}$ = Em₂ -Em₀

Em₂ = K + U

Em₂ = ½ m v₂² + m g y₂

Em₂ = ½ 50 85 2 + 50 9.8 427

Em₂ = 180.625 + 2.09 105

Em₂ = 1,806 105 J

$W_{fr}$ = Em₂ -Em₀

$W_{fr}$ = 1,806 105 - 4,296 105

$W_{fr}$ = -2.49 105 J

The negative sign indicates that the work that force and displacement have opposite directions

c) In this case the work of the friction going up is already calculated in part b and the work of the friction going down would be 1.5 that job

We have that the work of friction is equal to the change of mechanical energy

$W_{fr}$ = ΔEm

$W_{fr}$ = Emf - Emo

-1.5 2.49 10⁵ = ½ m vf² - 42.96 10⁴

½ m vf² = -1.5 2.49 10⁵ + 4.296 10⁵

½ 50.0 vf² = 0.561

vf = √ 0.561 25

vf = 3.75 m / s

6 0

3 years ago

PLEASE HELP :)

Tom [10]

1) in the opposite direction always
2) drag force is the force that acts in the opposite direction of the applied force (air resistance if a drag force for example)
3) frictions forces between the sole and the ground as the owner walks
4) chalk would glide very easily, so a creaking sound could never be produced, but also one could not write anything on the board!
5) I apologize for my lack of familiarity with the kind of sport but the probable answer is to raise the friction of their hands in order to decrease slipping and get a tighter, more firm grip

6 0

3 years ago

A block of mass 11.0 kg slides from rest down a frictionless 38.0° incline and is stopped by a strong spring with

ivanzaharov [21]

Answer:

11.72 mm

Explanation:

The gravitational potential energy equals the potential energy of the spring hence

$PE_{gravitational}=PE_{spring}$

$mgh=0.5kx^{2}$ where m is the mass of object, g is the acceleration due to gravity, h is the height, k is the spring constant and x is the extension of the spring