a 1250 kg car accelerates from rest to 6.13m/s over a distance of 8.58m calculate the average force of traction

A ball with an initial velocity of 2 m/s rolls for a period of 3 seconds. If the ball is uniformly accelerating at a rate of 3 m

ikadub [295]

Answer: 11 m/s

vinitial=2 m/s

time=3 s

acceleration = 3 m/s^2

vfinal = ?

The key here is that it is a constant acceleration, so we can use the constant acceleration equations. The easiest one to use would be:

vfinal=vinitial + a*t

We need vfinal, so algebraically we are ready to put in numbers into the equation:

vfinal=vinitial + a*t = 2 m/s + (3 m/s^2)*(3 s ) = 11 m/s is the final velocity

7 0

2 years ago

Two identical stunt professionals A and B stand on the roof of building 1. Person A steps off of the roof and falls vertically o

Vladimir [108]

Answer:

Both of the stunt professionals will sustain injuries of the same seriousness

Explanation:

We are being told that both stunt professionals are standing from the same height, therefore they will attain the same equivalent speed at the bottom if we are to look at it from the principle of conservation of energy.

Now; According to principle of momentum; the momentum at which the first stunt professional A hits the ground be equal as the momentum with which stunt professional B will hit the wall.

Thus; both of the stunt professionals will sustain injuries of the same seriousness

6 0

2 years ago

When a car is 100 meters from its starting position traveling at 60.0 m/s., it starts braking and comes to a stop 350 meters fro

NISA [10]

Remember your kinematic equations for constant acceleration. One of the equations is $x_{f} = x_{i} + v_{i}(t) + \frac{1}{2} at^{2}$ , where $x_{f}$ = final position, $x_{i}$ = initial position, $v_{i}$ = initial velocity, t = time, and a = acceleration.

Your initial position is where you initially were before you braked. That means $x_{i}$ = 100m. You final position is where you ended up after t seconds passed, so $x_{f}$ = 350m. The time it took you to go from 100m to 350m was t = 8.3s. You initial velocity at the initial position before you braked was $v_{i}$ = 60.0 m/s. Knowing these values, plug them into the equation and solve for a, your acceleration:
$350\:m = 100\:m + (60.0\:m/s)(8.3\:s) + \frac{1}{2} a(8.3\:s)^{2}\\
250\:m = (60.0\:m/s)(8.3\:s) + \frac{1}{2} a(8.3\:s)^{2}\\
250\:m = 498\:m +34.445\:s^{2}(a)\\
-248\:m = 34.445\:s^{2}(a)\\
a \approx -7.2 \: m/s^{2}$

Your acceleration is approximately $-7.2 \: m/s^{2}$ .

4 0

2 years ago

1)After catching the ball, Sarah throws it back to Julie. However, Sarah throws it too hard so it is over Julie's head when it r

DENIUS [597]

Answer:

1)

$v_{oy}=11.29\ m/s$

2)

$y=7.39\ m$

Explanation:

Projectile Motion

When an object is launched near the Earth's surface forming an angle $\theta$ with the horizontal plane, it describes a well-known path called a parabola. The only force acting (neglecting the effects of the wind) is the gravity, which acts on the vertical axis.

The heigh of an object can be computed as

$\displaystyle y=y_o+V_{oy}t-\frac{gt^2}{2}$

Where $y_o$ is the initial height above the ground level, $v_{oy}$ is the vertical component of the initial velocity and t is the time

The y-component of the speed is

$v_y=v_{oy}-gt$

1) We'll find the vertical component of the initial speed since we have not enough data to compute the magnitude of $v_o$

The object will reach the maximum height when $v_y=0$ . It allows us to compute the time to reach that point

$v_{oy}-gt_m=0$

Solving for $t_m$

$\displaystyle t_m=\frac{v_{oy}}{g}$

Thus, the maximum heigh is

$\displaystyle y_m=y_o+\frac{v_{oy}^2}{2g}$

We know this value is 8 meters

$\displaystyle y_o+\frac{v_{oy}^2}{2g}=8$

Solving for $v_{oy}$

$\displaystyle v_{oy}=\sqrt{2g(8-y_o)}$

Replacing the known values

$\displaystyle v_{oy}=\sqrt{2(9.8)(8-1.5)}$

$\displaystyle v_{oy}=11.29\ m/s$

2) We know at t=1.505 sec the ball is above Julie's head, we can compute

$\displaystyle y=y_o+V_{oy}t-\frac{gt^2}{2}$

$\displaystyle y=1.5+(11.29)(1.505)-\frac{9.8(1.505)^2}{2}$

$\displaystyle y=1.5\ m+16,991\ m-11.098\ m$

$y=7.39\ m$

5 0

2 years ago

A student measures the pH of a solution to be 6.8. Which should the student add if she wants to decrease the pH of the solution?

zloy xaker [14]

The neutral pH is 7. Less than 7 indicates an acid and more than 7 indicates a base (up to 14).
 NaCl - it's a salt (we can't measure the pH)
H2O - it can be an acid but also a base (the pH it is almost neutral,meaning close to 7 )
HF - it is a strong acid
 KOH - it is a strong base (pH=14)

↓

He needs to use HF (Hydrogen fluoride) to decrease the pH.

7 0

2 years ago

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