At what time were the racers at the same position?<br> O 15<br> O 3s<br> O 5s<br> 0 4s

ozzi

Where is the picture cant6 see it

7 0

3 years ago

A rock is thrown off a 50.0 m high cliff. How fast must the rock leave the cliff top to land on level ground below, 90 m from th

blagie [28]

Answer:

The rock must leave the cliff at a velocity of 28.2 m/s

Explanation:

The position vector of the rock at a time t can be calculated using the following equation:

r = (x0 + v0x · t, y0 + 1/2 · g · t²)

Where:

r = position vector at time t.

x0 = initial horizontal position.

v0x = initial horizontal velocity.

t = time.

g = acceleration due to gravity (-9.81 m/s² considering the upward direction as positive).

Please, see the attached figure for a graphical description of the problem. Notice that the origin of the frame of reference is located at the edge of the cliff so that x0 and y0 = 0.

When the rock reaches the ground, the position vector will be (see r1 in the figure):

r1 = (90 m, -50 m)

Then, using the equation of the vector position written above:

90 m = x0 + v0x · t

-50 m = y0 + 1/2 · g · t²

Since x0 and y0 = 0:

90 m = v0x · t

-50 m = 1/2 · g · t²

Let´s use the equation of the y-component of the vector r1 to find the time it takes the rock to reach the ground and with that time we can calculate v0x:

-50 m = 1/2 · g · t²

-50 m = -1/2 · 9.81 m/s² · t²

-50 m / -1/2 · 9.81 m/s² = t²

t = 3.19 s

Now, using the equation of the x-component of r1:

90 m = v0x · t

90 m = v0x · 3.19 s

v0x = 90 m / 3.19 s

v0x = 28.2 m/s

8 0

3 years ago

How does friction affect the distance of an object?

stiv31 [10]

It slows the object down so it cannot move well and evetually the object cannot be pushed and farther

4 0

3 years ago

In the nuclear fission process mass is converted into energy. Determine the total mass that must be converted to energy in one y

brilliants [131]

Answer:

1) Mass that needs to be converted at 100% efficiency is 0.3504 kg

2) Mass that needs to be converted at 30% efficiency is 1.168 kg

Explanation:

By the principle of mass energy equivalence we have

$E=mc^{2}$

where,

'E' is the energy produced

'm' is the mass consumed

'c' is the velocity of light in free space

Now the energy produced by the reactor in 1 year equals

$Energy=Power\times time\\\\\therefore Energy=1\times 10^{9}\times 365\times 24\times 3600\\\\Energy=31.536\times 10^{15}Joules$

Thus the mass that is covertred at 100% efficiency is

$mass=\frac{Energy}{c^{2}}\\\\mass=\frac{31.536\times 10^{15}}{(3\times 10^{8})^{2}}\\\\mass=\frac{31.536\times 10^{15}}{9\times 10^{16}}\\\\\therefore mass=0.3504kg$