Match up the following set of words with their definitions.

A projectile is launched with an initial velocity of 25 m/s at an angle of 30° above the horizontal. The projectile reaches maxi

Alchen [17]

Answer:

x ≈ 56 m

Explanation:

vertical initial velocity = $v_{0y}(t)$ = 25 m/s* sin(30°)= 12.5 m/s

height = h

$h =v_{0y}t+\frac{at^{2}}{2} \\\\65m = 12.5m/s*t + \frac{9.8m/s^{2}*t^{2}}{2} \\\\t=2.584 s$

t- time is found solving quadratic equation.

horizontal velocity = $v_{0x}=25m/s*cos(30^{o})=21.65 m/s$

Horizontal velocity is constant, so distance $x=v_{0x}*t =21.65 m/s *2.584 s=55.9 = 56 m$

6 0

3 years ago

A 750-kg automobile is moving at 16.8 m/s at a height of 5.00 m above the bottom of a hill when it runs out of gasoline. The car

anygoal [31]

Answer:h=19.4 m

Explanation:

Given

mass of automobile $m=750\ kg$

Initial height of automobile $h_o=5\ m$

Velocity at this instant $v=16.8\ m/s$

If the car stops somewhere at a height $h$

Thus conserving total energy we get

$K_i+U_i=K_f+U_f$

$\frac{1}{2}mv^2+mgh_o=\frac{1}{2}m(0)^2+mgh$

$\frac{v^2}{2g}+h_o=h$

$h=5+\frac{16.8^}{2\times 9.8}$

$h=5+14.4$

$h=19.4\ m$

6 0

3 years ago

YALL THIS QUESTION IS DUE IN 2 HOURS AND I HAVE 3 MORE PAGED TO DO PLEASE HELP ME WITH THIS PHYSICS HOMEWORK PROBLEM!!!! I ATTAC

goldfiish [28.3K]

Answer:

I dont know :)

Explanation:

7 0

2 years ago

a brick is suspended above the ground at a height of 6.6 m. it has a mass of 5.3 kg. what is the potential energy of the brick

Svetradugi [14.3K]

The formula for potential energy is
E(p) = mgh

(Mass x gravity x height)

Therefore energy = (5.3)(9.8)(6.6)
= 342.8 J

How did I get 9.8?
9.8 is the constant for gravity

8 0

3 years ago

(a) Calculate the force (in N) needed to bring a 1100 kg car to rest from a speed of 85.0 km/h in a distance of 125 m (a fairly

nasty-shy [4]

(a) -2451 N

We can start by calculating the acceleration of the car. We have:

$u=85.0 km/h = 23.6 m/s$ is the initial velocity

v = 0 is the final velocity of the car

d = 125 m is the stopping distance

So we can use the following equation

$v^2 - u^2 = 2ad$

To find the acceleration of the car, a:

$a=\frac{v^2-u^2}{2d}=\frac{0-(23.6 m/s)^2}{2(125 m)}=-2.23 m/s^2$

Now we can use Newton's second Law:

F = ma

where m = 1100 kg to find the force exerted on the car in order to stop it; we find:

$F=(1100 kg)(-2.23 m/s^2)=-2451 N$

and the negative sign means the force is in the opposite direction to the motion of the car.

(b) $-1.53\cdot 10^5 N$

We can use again the equation

$v^2 - u^2 = 2ad$

To find the acceleration of the car. This time we have

$u=85.0 km/h = 23.6 m/s$ is the initial velocity

v = 0 is the final velocity of the car

d = 2.0 m is the stopping distance

Substituting and solving for a,

$a=\frac{v^2-u^2}{2d}=\frac{0-(23.6 m/s)^2}{2(2 m)}=-139.2 m/s^2$

So now we can find the force exerted on the car by using again Newton's second law:

$F=ma=(1100 kg)(-139.2 m/s^2)=-1.53\cdot 10^5 N$

As we can see, the force is much stronger than the force exerted in part a).

8 0

3 years ago