Groups of stars that are smaller than galaxies are called<br> (two words)<br> Enter the answer

A stone is thrown horizontally at 15 m/s from the top of a cliff 40 meters high. How far from the base does the stone hit the gr

Reil [10]

Answer:

She stone hit ground 42.86 m far from base of cliff.

Explanation:

Initial height from ground = 40 meter.

We have equation of motion , $s= ut+\frac{1}{2} at^2$ , s is the displacement, u is the initial velocity, a is the acceleration and t is the time.

In this the velocity of body in vertical direction = 0 m/s, acceleration = 9.8 $m/s^2$ , we need to calculate time when s = 40 meter.

Substituting

$40=0*t+\frac{1}{2} *9.8*t^2\\ \\ t = 2.86 seconds$

So it will take 2.86 seconds to reach ground.

So, stone travels horizontally at 15 m/s for 2.86 seconds.

Distance travelled = 15 x 2.86 = 42.86 m

So, the stone hit ground 42.86 m far from base of cliff.

5 0

3 years ago

When are Waves produced

serg [7]

Answer:

Waves are caused by wind, or are created by the friction between wind and surface water

Explanation:

7 0

2 years ago

A toy of mass 0.155 kg is undergoing simple harmonic motion (SHM) on the end of a horizontal spring with force constant 305 N/m

zzz [600]

Answer:

$310.38\times 10^{-4}j$

Explanation:

We have given mass m=0.155 kg

Force constant K = 305 N/m

Distance $X=1.25\times 10^{-2}m$

Velocity $v=.305 m/sec$

The total energy at any position of the motion is give by $E=\frac{1}{2}mv^2+KX^2$ here $\frac{1}{2}mv^2$ is energy due to motion and $KX^2$ is energy due to spring elongation

So total energy $E=\frac{1}{2}\times 0.155\times 0.305^2+305\times 0.0125^2=310.38\times 10^{-4}j$

4 0

2 years ago

A small rock is thrown vertically upward with a speed of 17.0m/s from the edge of the roof of a 26.0m tall building. The rock do

Pachacha [2.7K]

Answer:

A) v = 28.3 m/s

B) t = 4.64 s

Explanation:

A)

Assuming no other forces acting on the rock, since the accelerarion due to gravity close to the surface to the Earth can be taken as constant, we can use one of the kinematic equations in order to get first the maximum height (over the roof level) that the ball reaches:

$v_{f}^{2} - v_{o}^{2} = 2* g* \Delta h (1)$

Taking into account that at this point, the speed of the rock is just zero, this means vf=0 in (1), so replacing by the givens and solving for Δh, we get:

$\Delta h = \frac{-v_{o} ^{2}}{2*g} = \frac{-(17.0m/s)^{2} }{2*(-9.8m/s2)} = 14.8 m (2)$

So, we can use now the same equation, taking into account that the initial speed is zero (when it starts falling from the maximum height) and that the total vertical displacement is the distance between the roof level and the ground (26.0 m) plus the maximum height that we have just found in (2) , 14.8m:
Δh = 26.0 m + 14. 8 m = 40.8 m (3)
Replacing now in (1), we can solve for vf, as follows:

$v_{f} =\sqrt{2*g*\Delta h} = \sqrt{2*9.8m/s2*40.8m} = 28.3 m/s (4)$

B)

In order to find the total elapsed from when the rock is thrown until it hits the street, we can divide this time in two parts:
1) Time elapsed from the the rock is thrown, until it reaches to its maximum height, when vf =0
2) Time elapsed from this point until it hits the street, with vo=0.
For the first part, we can simply use the definition of acceleration (g in this case), making vf =0, as follows:

$v_{f} = v_{o} + a*\Delta t = v_{o} - g*\Delta t = 0 (5)$

Replacing by the givens in (5) and solving for Δt, we get:

$\Delta t = \frac{v_{o}}{g} = \frac{17.0m/s}{9.8m/s2} = 1.74 s (6)$

For the second part, since we know the total vertical displacement from (3), and that vo = 0 since it starts to fall, we can use the kinematic equation for displacement, as follows:

$\Delta h = \frac{1}{2} * g * t^{2} (7)$

Replacing by the givens and solving for t in (7), we get:

$t_{fall} =\sqrt{\frac{2*\Delta h}{g}} = \sqrt{\frac{2*40.8m}{9.8m/s2} } = 2.9 s (8)$

So, total time is just the sum of (6) and (8):
t = 2.9 s + 1.74 s = 4.64 s

5 0

2 years ago

the radius of the tires on a particular vehicle 0.62m if the tires are rotating 5 times per second, what is the velocity of the

tankabanditka [31]

Circumference of the tire = (2 pi) x (radius)

   = (2 pi) x (0.62 meter)

   = 3.9 meters

If the tire never slips or skids, then the speed of the vehicle is

   speed = (distance) / (time to cover the distance)

   = (5 x 3.9 meters) / 1 second

   = 19.48 meters/second .

(about 43.6 miles per hour) .

We can't say anything about the vehicle's velocity, because we have
no information about the direction in which it's heading.

8 0

3 years ago

Groups of stars that are smaller than galaxies are called (two words) Enter the answer