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2 years ago

An object moves in a constant velocity, what is the acceleration of the object? *

Physics

1 answer:

MrMuchimi2 years ago

8 0

Answer:

Zero

Explanation:

If an object has a constant velocity then its not increases its velocity therefore acceleration is zero

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A baseball pitcher throws a baseball horizontally at a linear speed of 49.4 m/s. Before being caught, the baseball travels a hor

RideAnS [48]

Answer:

$v = 3.61 m/s$

Explanation:

As we know that ball travels horizontal distance of 24.7 m with uniform speed 49.4 m/s

so we will have

$t = \frac{x}{v_x}$

$t = \frac{24.7}{49.4}$

$t = 0.5 s$

now in the same time ball is turned by angle

$\theta = 52.7 rad$

now we know that

$\theta = \omega t$

$52.7 = \omega (0.5)$

$\omega = 105.4 rad/s$

now the tangential speed of a point at equator is given as

$v = r\omega$

$v = 0.0343(105.4)$

$v = 3.61 m/s$

4 0

3 years ago

You stand on a frictional platform that is rotating at 1.8 rev/s. Your arms are outstretched, and you hold a heavy weight in eac

dusya [7]

Answer:

20.62361 rad/s

489.81804 J

Explanation:

$I_i$ = Initial moment of inertia = 9.3 kgm²

$I_f$ = Final moment of inertia = 5.1 kgm²

$\omega_i$ = Initial angular speed = 1.8 rev/s

$\omega_f$ = Final angular speed

As the angular momentum of the system is conserved

$I_i\omega_i=I_f\omega_f\\\Rightarrow \omega_f=\dfrac{I_i\omega_i}{I_f}\\\Rightarrow \omega_f=\dfrac{9.3\times 1.8}{5.1}\\\Rightarrow \omega_f=3.28235\ rev/s=3.28235\times 2\pi=20.62361\ rad/s$

The resulting angular speed of the platform is 20.62361 rad/s

Change in kinetic energy is given by

$\Delta K=\dfrac{1}{2}(I_f\omega_f^2-I_i\omega_i^2)\\\Rightarrow \Delta K=\dfrac{1}{2}(5.1\times (20.62361)^2-9.3\times (1.8\times 2\pi)^2)\\\Rightarrow \Delta K=489.81804\ J$

The change in kinetic energy of the system is 489.81804 J

As the work was done to move the weight in there was an increase in kinetic energy

6 0

3 years ago

A car covers 400 km in an hour towards west .calculate the velocity

crimeas [40]

Answer:

-400km/hr

Explanation:

Velocity=displacement/time

=400/1

=400Km/hr

=-400km/hr (because west direction)

7 0

2 years ago

8. two +1 C charges are separated by 3000m. What is the magnitude of the electric force between them?

Sidana [21]

Answer:

1000 N

Explanation:

The magnitude of the electrostatic force between two charged object is given by

$F=k\frac{q_1 q_2}{r^2}$

where

k is the Coulomb constant

q1, q2 is the magnitude of the two charges

r is the distance between the two objects

Moreover, the force is:

- Attractive if the two forces have opposite sign

- Repulsive if the two forces have same sign

In this problem:

$q_1=q_2=+1C$ are the two charges

r = 3000 m is their separation

Therefore, the electric force between the charges is:

$F=(9\cdot 10^9)\frac{(1)(1)}{3000^2}=1000 N$

8 0

3 years ago

a ball is shot from the ground straight up into the air with initial velocity of 50 50 ft/sec. assuming that the air resistance

Anna [14]

The height attained by the ball is 11.86m

a ball is shot from the ground straight up into the air its initial and final velocity is

initial velocity, u = 50 ft/s = 50×0.305 = 15.25m/s

final velocity ,v = 0 m/s

gravity =-9.8 m/s²

( negative sign shows acceleration in opposite direction)

height =?

using the newton motion of equation

v² = u² + 2as

where

a= acceleration due to gravity(g)

s = height

v² = u² + 2gs

(0)² = (15.25)² + 2×(-9.8)×s

0 = (15.25)² - 19.6 × s

s= - (15.25)²/ 19.6

s = 11.86m

after ignoring the air resistance the maximum height of the ball is 11.86m

To learn more about motion under gravity -

brainly.com/question/27962354

#SPJ4

3 0

1 year ago