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A car goes from rest to a speed of 90 km/h in 10 seconds . What is the car's acceleration in m/s²

zloy xaker [14]

First you must convert Km/hr to m/s. 90 km/hr equals 25m/s (this can be done through a conversion table by plugging in the conversion values). Then you need to see what was given:
vi (initial velocity)= 0m/s
vf (final velocity= 25m/s (90km/hr)
t (time)= 10s

Next you should find an equation that requires only the values you know and gives you the value you're looking for. Sometimes that requires two equations to be used, but in this case you only need one. The best equation for this would be a=(vf-vi)/t. Finally, plug in your values (a=(25-0)/10) to get your answer which would be 2.5m/s^2. Hope this helped!

7 0

3 years ago

An athlete runs at a constant velocity of 5.2 m/s. What is the velocity of the athlete relative to the ground?

dimulka [17.4K]

The relative velocity of the athlete relative to the ground is 5.2 m/s

The given parameters;

constant velocity of the athlete, V = 5.2 m/s

let the velocity of the ground = Vg = 0

The relative velocity concept helps us to determine the velocity of a moving object relative to a stationary observer.

The athlete is the moving object in this question while the ground is stationary.

The relative velocity of the athlete relative to the ground is calculated as follows;

$V/V_g = V - V_g = 5.2 - 0 = 5.2 \ m/s$

Thus, the relative velocity of the athlete relative to the ground is 5.2 m/s

Learn more here: brainly.com/question/24430414

5 0

3 years ago

A 0.10-kg ball, traveling horizontally at 25 m/s, strikes a wall and rebounds at 19 m/s. What is the 7) magnitude of the change

IRISSAK [1]

Answer:

Change in momentum will be -4.4 kgm/sec

So option (A) is correct option

Explanation:

Mass of the ball is given m = 0.10 kg

Initial velocity of ball $v_1=25m/sec$

And velocity after rebound $v_2=-19m/sec$

We have to find the change in momentum

So change in momentum is equal to $=m(v_2-v_1)=0.1\times (-19-25)=-4.4kgm/sec$ ( here negative sign shows only direction )

So option (A) will be correct answer

5 0

3 years ago

A projectile is shot horizontally at 23.4 m/s from the roof of a building 55.0 m tall. (a) Determine the time necessary for the

jeka94

(a) 3.35 s

The time needed for the projectile to reach the ground depends only on the vertical motion of the projectile, which is a uniformly accelerated motion with constant acceleration

a = g = -9.8 m/s^2

towards the ground.

The initial height of the projectile is

h = 55.0 m

The vertical position of the projectile at time t is

$y = h + \frac{1}{2}at^2$

By requiring y = 0, we find the time t at which the projectile reaches the position y=0, which corresponds to the ground:

$0 = h + \frac{1}{2}at^2\\t=\sqrt{-\frac{2h}{a}}=\sqrt{-\frac{2(55.0 m)}{(-9.8 m/s^2)}}=3.35 s$

(b) 78.4 m

The distance travelled by the projectile from the base of the building to the point it lands depends only on the horizontal motion.

The horizontal motion is a uniform motion with constant velocity -

The horizontal velocity of the projectile is

$v_x = 23.4 m/s$

the time it takes the projectile to reach the ground is

t = 3.35 s

So, the horizontal distance covered by the projectile is

$d=v_x t = (23.4 m/s)(3.35 s)=78.4 m$

(c) 23.4 m/s, -32.8 m/s

The motion of the projectile consists of two independent motions:

- Along the horizontal direction, it is a uniform motion, so the horizontal velocity is always constant and it is equal to

$v_x = 23.4 m/s$

so this value is also the value of the horizontal velocity just before the projectile reaches the ground.

- Along the vertical direction, the motion is acceleration, so the vertical velocity is given by

$v_y = u_y +at$

where

$u_y = 0$ is the initial vertical velocity

Using

a = g = -9.8 m/s^2

and

t = 3.35 s

We find the vertical velocity of the projectile just before reaching the ground

$v_y = 0 + (-9.8 m/s^2)(3.35 s)=-32.8 m/s$

and the negative sign means it points downward.

3 0

3 years ago

Two balls with equal masses, m, and equal speed, v, engage in a head on elastic collision. what is the final velocity of each ba

Allushta [10]

The collision is elastic. This means that both momentum and kinetic energy are conserved after the collision.

- Let's start with conservation of momentum. The initial momentum of the total system is the sum of the momenta of the two balls, but we should put a negative sign in front of the velocity of the second ball, because it travels in the opposite direction of ball 1. So ball 1 has mass m and speed v, while ball 2 has mass m and speed -v:
$p_i = p_1-p_2 = mv-mv =0$
So, the final momentum must be zero as well:
$p_f = 0$
Calling v1 and v2 the velocities of the two balls after the collision, the final momentum can be written as
$p_f = mv_1 + mv_2 = 0$
From which
$v_1 = -v_2$

- So now let's apply conservation of kinetic energy. The kinetic energy of each ball is $\frac{1}{2} mv^2$ . Therefore, the total kinetic energy before the collision is
$K_i = \frac{1}{2} mv^2 + \frac{1}{2} mv^2 = mv^2$
the kinetic energy after the collision must be conserved, and therefore must be equal to this value:
$K_f = K_i = mv^2$ (1)
But the final kinetic energy, Kf, is also
$K_f = \frac{1}{2} mv_1^2 + \frac{1}{2}mv_2^2$
Substituting $v_1 = -v_2$ as we found in the conservation of momentum, this becomes
$K_f = mv_2 ^2$
we also said that Kf must be equal to the initial kinetic energy (1), therefore we can write
$mv_2^2 = mv^2$

Therefore, the two final speeds of the balls are
$v_2 = v$
$v_1 = -v_2 = -v$

This means that after the collision, the two balls have same velocity v, but they go in the opposite direction with respect to their original direction.

8 0

3 years ago